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It also proves quantitative refinement-defect formulas under explicit norm hypotheses, including exact transfer of a bounded generator defect to every fractional power $0<\\alpha<1$ for bounded nonnegative self-adjoint generators. It does not claim that an arbitrary infinite-dimensional operator generates a semigroup, that every stochastic process admits a finite packet representation, that every posterior family is conjugate, that finite Bayesian sufficiency automatically extends to infinite disintegrations, or that every PDE continuum limit follows from a finite test. Domain, closability, sectoriality, measurability, tightness, and compactness hypotheses remain necessary in their respective infinite-dimensional settings. Classical sources are retained in the separate end-of-series reference ledger rather than reproduced in this paper.\n\\end{minipage}}\n\\end{…","Let $X$ be a finite-dimensional real or complex vector space, let $A\\in\\operatorname{End}(X)$, and let $s\\in X$ be a primitive seed. The basic continuous transfer is\n\\[\nT(t)=e^{tA},\\qquad t\\ge 0,\n\\]\nwhile a positive operator $L$ is more naturally written\n\\[\nH(t)=e^{-tL}.\n\\]\nBoth satisfy the semigroup law. The sign convention changes with the domain: Markov generators and evolution equations are usually written with $e^{tA}$; positive diffusion and elliptic operators are usually written with $e^{-tL}$.","dimensional real or complex vector space, let $A\\in\\operatorname{End}(X)$, and let $s\\in X$ be a primitive seed. The basic continuous transfer is\n\\[\nT(t)=e^{tA},\\qquad t\\ge 0,\n\\]\nwhile a positive operator $L$ is more naturally written\n\\[\nH(t)=e^{-tL}.\n\\]\nBoth satisfy the semigroup law. The sign convention changes with the domain: Markov generators and evolution equations are usually written with $e^{tA}$; positive diffusion and elliptic operators are usually written with $e^{-tL}$.","d let $s\\in X$ be a primitive seed. The basic continuous transfer is\n\\[\nT(t)=e^{tA},\\qquad t\\ge 0,\n\\]\nwhile a positive operator $L$ is more naturally written\n\\[\nH(t)=e^{-tL}.\n\\]\nBoth satisfy the semigroup law. The sign convention changes with the domain: Markov generators and evolution equations are usually written with $e^{tA}$; positive diffusion and elliptic operators are usually written with $e^{-tL}$.","The common operator chain is\n\\[\n\\boxed{\n\\text{seed}\\longrightarrow\\text{transfer}\\longrightarrow\\text{project}\n\\longrightarrow\\text{condition}\\longrightarrow\\text{refine}\n\\longrightarrow\\text{complete}\\longrightarrow\\text{reconstruct}.\n}\n\\]","\\begin{definition}[Transfer-semigroup seed system]\\label{core:record:001}\nA finite transfer-semigroup seed system is a tuple\n\\[\n\\mathfrak S=(X,s,A,\\{P_\\alpha\\}_{\\alpha\\in I},\\tau),\n\\]\nwhere $X$ is the state space, $s$ is the seed, $A$ is a transfer generator, the $P_\\alpha$ are pairwise orthogonal idempotents summing to the identity, and $\\tau$ specifies the topology in which a completion or limit is interpreted.\n\\end{definition}","\\begin{theorem}[Finite semigroup law]\\label{core:record:002}\nFor every matrix $A$ and all $s,t\\in\\bbR$,\n\\[\n\\boxed{e^{(s+t)A}=e^{sA}e^{tA}.}\n\\]\nMoreover, $T(0)=I$ and $T'(t)=AT(t)=T(t)A$.\n\\end{theorem}","\\begin{proof}\nBoth exponentials are absolutely convergent matrix power series. Since every power of $A$ commutes with every other power, the Cauchy product gives\n\\[\ne^{sA}e^{tA}\n=\\sum_{m,n\\ge0}\\frac{s^mt^n}{m!n!}A^{m+n}\n=\\sum_{r\\ge0}\\frac{(s+t)^r}{r!}A^r.\n\\]\nTermwise differentiation is valid in finite dimension and gives the derivative identities.\n\\end{proof}","\\begin{boundary}[Infinite-dimensional generators]\\label{core:record:003}\nFor an unbounded operator, writing $e^{tA}$ is not a definition unless a semigroup-generation theorem applies. The finite matrix identities in this paper remain algebraically correct models, but a continuum realization must state the operator domain, closure, growth bound, and topology of convergence.\n\\end{boundary}","Let\n\\[\nP_\\alpha P_\\beta=\\delta_{\\alpha\\beta}P_\\alpha,\n\\qquad\n\\sum_{\\alpha\\in I}P_\\alpha=I.\n\\]\nFor any operator $B$, define\n\\[\nB_{\\beta\\alpha}=P_\\beta BP_\\alpha.\n\\]\nThen\n\\[\nB=\\sum_{\\alpha,\\beta}B_{\\beta\\alpha}.\n\\]\nFor the semigroup, write\n\\[\nT_{\\beta\\alpha}(t)=P_\\beta e^{tA}P_\\alpha.\n\\]","\\begin{theorem}[Packet semigroup convolution]\\label{core:record:004}\nFor every $s,t$ and every pair of packets $\\alpha,\\beta$,\n\\[\n\\boxed{\nT_{\\beta\\alpha}(s+t)\n=\\sum_{\\gamma\\in I}T_{\\beta\\gamma}(s)T_{\\gamma\\alpha}(t).\n}\n\\]\nThus the semigroup law becomes ordinary matrix multiplication in the packet indices.\n\\end{theorem}","\\begin{proof}\nInsert the resolution of the identity between the two semigroup factors:\n\\[\n\\begin{aligned}\nP_\\beta e^{(s+t)A}P_\\alpha\n&=P_\\beta e^{sA}\\left(\\sum_\\gamma P_\\gamma\\right)e^{tA}P_\\alpha\\\\\n&=\\sum_\\gamma\n(P_\\beta e^{sA}P_\\gamma)(P_\\gamma e^{tA}P_\\alpha).\n\\end{aligned}\n\\]\n\\end{proof}","\\begin{corollary}[Reducing packets]\\label{core:record:005}\nIf $P_\\beta AP_\\alpha=0$ for every $\\beta\\ne\\alpha$, then\n\\[\nP_\\beta e^{tA}P_\\alpha=0\n\\quad(\\beta\\ne\\alpha)\n\\]\nfor every $t$. Each packet evolves independently under $P_\\alpha AP_\\alpha$.\n\\end{corollary}","\\begin{proof}\nThe hypothesis makes $A$ block diagonal. Every power $A^m$ is block diagonal with respect to the same projectors, and therefore so is the exponential series.\n\\end{proof}","\\begin{figure}[htbp]\n\\centering\n\\includegraphics[width=0.66\\textwidth]{02_packet_semigroup_blocks.pdf}\n\\caption{Norms of the packet blocks $P_\\beta e^{tA}P_\\alpha$. Off-diagonal entries quantify transfer between packets.}\n\\label{core:fig:packet-blocks}\n\\end{figure}","\\begin{remark}[Path expansion]\\label{core:record:006}\nExpanding the exponential and inserting $I=\\sum_\\alpha P_\\alpha$ between every pair of generators gives\n\\[\nP_{\\alpha_m}A^mP_{\\alpha_0}\n=\n\\sum_{\\alpha_1,\\ldots,\\alpha_{m-1}}\nP_{\\alpha_m}AP_{\\alpha_{m-1}}\\cdots P_{\\alpha_1}AP_{\\alpha_0}.\n\\]\nEvery coefficient is therefore a sum over packet-transfer paths. The numerical library verifies that truncating this path series converges to the ordinary matrix exponential.\n\\end{remark}","Let $X_f$ and $X_c$ be fine and coarse spaces, with generators $A_f$ and $A_c$, and let\n\\[\nC:X_f\\to X_c\n\\]\nbe a coarsening map. Define the generator-level refinement defect\n\\[\n\\boxed{\\Delta=CA_f-A_cC.}\n\\]\nThe corresponding semigroup defect is\n\\[\n\\mathfrak D(t)=Ce^{tA_f}-e^{tA_c}C.\n\\]","\\begin{theorem}[Exact Duhamel refinement identity]\\label{core:record:007}\nFor every $t\\ge0$,\n\\[\n\\boxed{\nCe^{tA_f}-e^{tA_c}C\n=\n\\int_0^t e^{(t-r)A_c}\\Delta e^{rA_f}\\,dr.\n}\n\\]\nIn particular, $\\Delta=0$ implies exact semigroup intertwining at every time.\n\\end{theorem}","\\begin{proof}\nSet\n\\[\nF(r)=e^{(t-r)A_c}Ce^{rA_f}.\n\\]\nThen\n\\[\nF'(r)=e^{(t-r)A_c}(CA_f-A_cC)e^{rA_f}.\n\\]\nIntegrating from $0$ to $t$ gives $F(t)-F(0)=Ce^{tA_f}-e^{tA_c}C$ and proves the formula.\n\\end{proof}","\\begin{corollary}[Norm bound]\\label{core:record:008}\nSuppose\n\\[\n\\norm{e^{tA_f}}\\le M_f e^{\\omega_f t},\n\\qquad\n\\norm{e^{tA_c}}\\le M_c e^{\\omega_c t}.\n\\]\nThen\n\\[\n\\norm{\\mathfrak D(t)}\n\\le\nM_fM_c\\norm{\\Delta}\n\\int_0^t e^{\\omega_c(t-r)}e^{\\omega_fr}\\,dr.\n\\]\nFor contraction semigroups,\n\\[\n\\boxed{\\norm{\\mathfrak D(t)}\\le t\\norm{\\Delta}.}\n\\]\n\\end{corollary}","For $z$ in the common resolvent set, let\n\\[\nR_f(z)=(zI-A_f)^{-1},\n\\qquad\nR_c(z)=(zI-A_c)^{-1}.\n\\]","\\begin{theorem}[Resolvent refinement identity]\\label{core:record:009}\n\\[\n\\boxed{\nCR_f(z)-R_c(z)C\n=R_c(z)\\Delta R_f(z).\n}\n\\]\n\\end{theorem}","\\begin{proof}\nMultiply the left side by $zI-A_c$ on the left and $zI-A_f$ on the right. The result is\n\\[\n(zI-A_c)C-C(zI-A_f)=CA_f-A_cC=\\Delta.\n\\]\nMultiplication by the two resolvents recovers the stated identity.\n\\end{proof}","Use column probability vectors. A finite continuous-time Markov generator $Q$ has nonnegative off-diagonal entries and column sums zero. The stochastic semigroup is\n\\[\nK(t)=e^{tQ}.\n\\]","\\begin{theorem}[Finite Markov semigroup]\\label{core:record:010}\nIf $Q$ has nonnegative off-diagonal entries and column sums zero, then $e^{tQ}$ has nonnegative entries and column sums one for every $t\\ge0$.\n\\end{theorem}","\\begin{proof}\nChoose $\\lambda\\ge\\max_i(-Q_{ii})$. Then $B=Q+\\lambda I$ has nonnegative entries, so $e^{tB}$ is entrywise nonnegative by its power series. Hence\n\\[\ne^{tQ}=e^{-\\lambda t}e^{tB}\n\\]\nis nonnegative. Since $\\bm 1^TQ=0$, differentiation gives\n\\[\n\\frac{d}{dt}\\bm 1^Te^{tQ}=\\bm 1^TQe^{tQ}=0,\n\\]\nso $\\bm 1^Te^{tQ}=\\bm 1^T$.\n\\end{proof}","\\begin{figure}[htbp]\n\\centering\n\\includegraphics[width=0.78\\textwidth]{03_random_walk_transfer.pdf}\n\\caption{The point mass $\\delta_0$ propagated by repeated symmetric random-walk transfer.}\n\\label{core:fig:random-walk}\n\\end{figure}","\\begin{example}[Symmetric random walk]\\label{core:record:011}\nStarting from $\\delta_0$ and applying the transfer\n\\[\nT=\\frac12(S+S^{-1})\n\\]\nfor $n$ steps gives\n\\[\n\\bbP(X_n=k)=\n2^{-n}\\binom{n}{(n+k)/2}\n\\]\nwhen $k\\equiv n\\pmod2$, and zero otherwise. Hence\n\\[\n\\Exp[X_n]=0,\n\\qquad\n\\Var(X_n)=n.\n\\]\nThe parity packet is exact: all mass remains in the one residue class permitted by $n$.\n\\end{example}","Let $K$ be a column-stochastic fine transition matrix, let $C$ aggregate fine states into coarse packets, and let $\\bar K$ be a coarse transition matrix. Define\n\\[\n\\Delta=CK-\\bar KC.\n\\]","\\begin{theorem}[Exact lumpability]\\label{core:record:012}\nIf $\\Delta=0$, then\n\\[\n\\boxed{CK^n=\\bar K^nC}\n\\]\nfor every $n\\ge0$. Thus fine evolution followed by coarse reconstruction equals coarse evolution from the beginning.\n\\end{theorem}","\\begin{proof}\nThe result is immediate for $n=0$. If $CK^n=\\bar K^nC$, then\n\\[\nCK^{n+1}=\\bar K^nCK=\\bar K^{n+1}C.\n\\]\n\\end{proof}","\\begin{theorem}[Finite-horizon defect formula]\\label{core:record:013}\nFor arbitrary $\\Delta=CK-\\bar KC$,\n\\[\n\\boxed{\nCK^n-\\bar K^nC\n=\n\\sum_{j=0}^{n-1}\\bar K^{n-1-j}\\Delta K^j.\n}\n\\]\nIf $K$ and $\\bar K$ are contractions in the chosen norm, then\n\\[\n\\norm{CK^n-\\bar K^nC}\\le n\\norm{\\Delta}.\n\\]\n\\end{theorem}","\\begin{proof}\nThe identity follows by telescoping:\n\\[\n\\bar K^{n-1-j}CK^{j+1}-\\bar K^{n-j}CK^j\n=\\bar K^{n-1-j}(CK-\\bar KC)K^j.\n\\]\nSumming from $j=0$ to $n-1$ leaves $CK^n-\\bar K^nC$. The norm bound follows termwise.\n\\end{proof}","Let $\\Omega=\\{1,\\ldots,n\\}$ carry strictly positive probabilities $\\pi_i$, and let $\\mathcal Q=\\{B_1,\\ldots,B_m\\}$ be a partition. Define\n\\[\n(P_{\\mathcal Q}f)_i\n=\n\\frac{\\sum_{j\\in B(i)}\\pi_jf_j}{\\sum_{j\\in B(i)}\\pi_j},\n\\]\nwhere $B(i)$ is the block containing $i$.","\\begin{theorem}[Finite conditional-expectation projector]\\label{core:record:014}\nWith the weighted inner product\n\\[\n\\langle f,g\\rangle_\\pi=\\sum_i\\pi_if_i\\overline{g_i},\n\\]\nthe operator $P_{\\mathcal Q}$ satisfies\n\\[\nP_{\\mathcal Q}^2=P_{\\mathcal Q},\n\\qquad\n\\langle P_{\\mathcal Q}f,g\\rangle_\\pi\n=\n\\langle f,P_{\\mathcal Q}g\\rangle_\\pi.\n\\]\nIts range is the space of functions constant on each partition block.\n\\end{theorem}","\\begin{proof}\nAfter one application, the result is constant on every block, so applying the same block average again changes nothing. Weighted self-adjointness follows by expanding both inner products block by block; both equal\n\\[\n\\sum_{B\\in\\mathcal Q}\n\\frac{1}{\\pi(B)}\n\\left(\\sum_{i\\in B}\\pi_i f_i\\right)\n\\overline{\\left(\\sum_{j\\in B}\\pi_j g_j\\right)}.\n\\]\n\\end{proof}","\\begin{theorem}[Tower law and martingale packets]\\label{core:record:015}\nIf $\\mathcal Q_c$ is coarser than $\\mathcal Q_f$, then\n\\[\n\\boxed{\nP_cP_f=P_fP_c=P_c.\n}\n\\]\nFor a nested tower $P_0\\le P_1\\le\\cdots\\le P_r$, define\n\\[\nD_0=P_0,\n\\qquad\nD_j=P_j-P_{j-1}.\n\\]\nThen\n\\[\nD_iD_j=\\delta_{ij}D_j,\n\\qquad\nP_r=\\sum_{j=0}^rD_j.\n\\]\n\\end{theorem}","\\begin{proof}\nA coarse-block average of a fine-block average is the coarse-block average, and a fine-block average of a coarse-constant function leaves it unchanged. The difference identities follow from $P_iP_j=P_{\\min(i,j)}$.\n\\end{proof}","\\begin{classical}[Gaussian chaos realization]\\label{core:record:016}\nFor the standard Gaussian measure, the normalized probabilists' Hermite polynomials form orthonormal chaos packets. The Ornstein--Uhlenbeck semigroup acts diagonally:\n\\[\nT_tH_n=e^{-nt}H_n.\n\\]\nThus conditional expectation, martingale differences, and Gaussian chaos are all projector realizations of stochastic refinement.\n\\end{classical}","Let $p=(p_i)$ be a probability vector and let $\\lambda=(\\lambda_i)$ be a nonnegative likelihood with positive evidence\n\\[\nZ_\\lambda(p)=\\sum_i\\lambda_ip_i.\n\\]\nDefine\n\\[\n\\boxed{\n\\cB_\\lambda(p)\n=\n\\frac{\\lambda\\odot p}{Z_\\lambda(p)}.\n}\n\\]\nThis map is nonlinear solely because of its scalar reconstruction factor.","\\begin{theorem}[Independent-likelihood composition]\\label{core:record:017}\nFor nonnegative likelihoods $\\lambda$ and $\\mu$, assume $Z_\\lambda(p)>0$ and $Z_{\\lambda\\odot\\mu}(p)>0$. Equivalently, both successive updates have positive evidence. Then\n\\[\n\\boxed{\n\\cB_\\mu(\\cB_\\lambda(p))\n=\n\\cB_{\\lambda\\odot\\mu}(p).\n}\n\\]\n\\end{theorem}","\\begin{proof}\nThe first normalization appears in both numerator and denominator of the second update and cancels:\n\\[\n\\frac{\\mu_i(\\lambda_ip_i/Z_\\lambda)}{\\sum_j\\mu_j\\lambda_jp_j/Z_\\lambda}\n=\n\\frac{\\mu_i\\lambda_ip_i}{\\sum_j\\mu_j\\lambda_jp_j}.\n\\]\n\\end{proof}","Let the state indices be partitioned into packets $I_\\alpha$. Define\n\\[\nZ_\\alpha=\\sum_{i\\in I_\\alpha}\\lambda_ip_i,\n\\qquad\nm_\\alpha=\\frac{Z_\\alpha}{\\sum_\\beta Z_\\beta},\n\\qquad\nq_{\\alpha,i}=\\begin{cases}\n\\lambda_ip_i/Z_\\alpha,&Z_\\alpha>0,\\ i\\in I_\\alpha,","Let the state indices be partitioned into packets $I_\\alpha$. Define\n\\[\nZ_\\alpha=\\sum_{i\\in I_\\alpha}\\lambda_ip_i,\n\\qquad\nm_\\alpha=\\frac{Z_\\alpha}{\\sum_\\beta Z_\\beta},\n\\qquad\nq_{\\alpha,i}=\\begin{cases}\n\\lambda_ip_i/Z_\\alpha,&Z_\\alpha>0,\\ i\\in I_\\alpha,\\\\\n0,&\\text{otherwise}.\n\\end{cases}\n\\]","\\begin{theorem}[Bayesian packet reconstruction]\\label{core:record:018}\n\\[\n\\boxed{\n\\cB_\\lambda(p)\n=\n\\sum_\\alpha m_\\alpha q_\\alpha.\n}\n\\]\nOnly packets with positive evidence carry a conditional probability vector. A zero-evidence packet has zero posterior mass and contributes the zero vector. The nonlinear coupling between packets occurs only through the total evidence used to normalize the packet evidences.\n\\end{theorem}","\\begin{proof}\nIf $Z_\\alpha=0$, nonnegativity implies $\\lambda_i p_i=0$ for every $i\\in I_\\alpha$, so the reconstruction identity holds with both sides zero. For $Z_\\alpha>0$ and $i\\in I_\\alpha$,\n\\[\nm_\\alpha q_{\\alpha,i}\n=\n\\frac{Z_\\alpha}{Z}\\frac{\\lambda_ip_i}{Z_\\alpha}\n=\n\\frac{\\lambda_ip_i}{Z}.\n\\]\nThe packet supports are disjoint, so summing reconstructs the full posterior.\n\\end{proof}","\\begin{theorem}[Likelihood perturbation bound]\\label{core:record:019}\nLet $p$ be fixed and let $\\lambda,\\mu\\ge0$ have positive evidences $Z_\\lambda,Z_\\mu$. Then\n\\[\n\\boxed{\n\\norm{\\cB_\\lambda(p)-\\cB_\\mu(p)}_1\n\\le\n\\frac{2\\sum_i|\\lambda_i-\\mu_i|p_i}{\\min(Z_\\lambda,Z_\\mu)}.\n}\n\\]\n\\end{theorem}","\\begin{proof}\nAdd and subtract $\\mu\\odot p/Z_\\lambda$ and use the triangle inequality:\n\\[\n\\norm{\\frac{\\lambda\\odot p}{Z_\\lambda}-\\frac{\\mu\\odot p}{Z_\\mu}}_1\n\\le\n\\frac{\\norm{(\\lambda-\\mu)\\odot p}_1}{Z_\\lambda}\n+\n\\norm{\\mu\\odot p}_1\\left|\\frac1{Z_\\lambda}-\\frac1{Z_\\mu}\\right|.\n\\]\nThe second term equals $|Z_\\lambda-Z_\\mu|/Z_\\lambda$ and is bounded by the same weighted $L^1$ difference divided by $Z_\\lambda$. Interchanging $\\lambda$ and $\\mu$ gives the minimum-evidence form.\n\\end{proof}","btract $\\mu\\odot p/Z_\\lambda$ and use the triangle inequality:\n\\[\n\\norm{\\frac{\\lambda\\odot p}{Z_\\lambda}-\\frac{\\mu\\odot p}{Z_\\mu}}_1\n\\le\n\\frac{\\norm{(\\lambda-\\mu)\\odot p}_1}{Z_\\lambda}\n+\n\\norm{\\mu\\odot p}_1\\left|\\frac1{Z_\\lambda}-\\frac1{Z_\\mu}\\right|.\n\\]\nThe second term equals $|Z_\\lambda-Z_\\mu|/Z_\\lambda$ and is bounded by the same weighted $L^1$ difference divided by $Z_\\lambda$. Interchanging $\\lambda$ and $\\mu$ gives the minimum-evidence form.\n\\end{proof}","ot p/Z_\\lambda$ and use the triangle inequality:\n\\[\n\\norm{\\frac{\\lambda\\odot p}{Z_\\lambda}-\\frac{\\mu\\odot p}{Z_\\mu}}_1\n\\le\n\\frac{\\norm{(\\lambda-\\mu)\\odot p}_1}{Z_\\lambda}\n+\n\\norm{\\mu\\odot p}_1\\left|\\frac1{Z_\\lambda}-\\frac1{Z_\\mu}\\right|.\n\\]\nThe second term equals $|Z_\\lambda-Z_\\mu|/Z_\\lambda$ and is bounded by the same weighted $L^1$ difference divided by $Z_\\lambda$. Interchanging $\\lambda$ and $\\mu$ gives the minimum-evidence form.\n\\end{proof}","Let the fine state set $I$ be partitioned into nonempty coarse packets $I_a$, $a\\in A$.  The coarsening map on probability vectors is\n\\[\n(C_\\pi p)_a=P_a=\\sum_{i\\in I_a}p_i.\n\\]\nFor a fine likelihood $\\lambda=(\\lambda_i)_{i\\in I}$ and a prior $p$, define the effective coarse likelihood\n\\[\nr_a(p)\n=\n\\frac{\\sum_{i\\in I_a}\\lambda_i p_i}{P_a}\n\\qquad(P_a>0).\n\\]\nIts value on a zero-mass packet is irrelevant and may be set to zero.","\\begin{theorem}[Exact effective-likelihood factorization]\\label{core:record:020}\nWhenever the fine evidence is positive,\n\\[\n\\boxed{\nC_\\pi\\cB_\\lambda(p)\n=\n\\cB_{r(p)}(C_\\pi p).}\n\\]\nThus the only fine information needed for the coarse posterior, beyond the coarse prior masses, is the vector of conditional packet evidences $r(p)$.\n\\end{theorem}","\\begin{proof}\nFor $P=C_\\pi p$ and $Z=\\sum_i\\lambda_ip_i=\\sum_aP_ar_a(p)$,\n\\[\n\\bigl(C_\\pi\\cB_\\lambda(p)\\bigr)_a\n=\n\\frac{\\sum_{i\\in I_a}\\lambda_ip_i}{Z}\n=\n\\frac{P_ar_a(p)}{\\sum_bP_br_b(p)}\n=\n\\bigl(\\cB_{r(p)}(P)\\bigr)_a.\n\\]\n\\end{proof}","\\begin{theorem}[Bayesian refinement criterion]\\label{core:record:021}\nAssume that the partition contains at least two coarse packets and that the fine and coarse likelihoods are strictly positive.  The identity\n\\[\n\\boxed{\nC_\\pi\\cB_\\lambda(p)\n=\n\\cB_{\\bar\\lambda}(C_\\pi p)}\n\\]\nholds for every fine prior $p$ if and only if there exists one scalar $c>0$ such that\n\\[\n\\boxed{\n\\lambda_i=c\\,\\bar\\lambda_a\n\\qquad(i\\in I_a).}\n\\]\nEquivalently, the likelihood is constant on each coarse packet, with the coarse values determined up to the globally invisible Bayesian scaling $c$.\n\\end{theorem}","lihoods are strictly positive.  The identity\n\\[\n\\boxed{\nC_\\pi\\cB_\\lambda(p)\n=\n\\cB_{\\bar\\lambda}(C_\\pi p)}\n\\]\nholds for every fine prior $p$ if and only if there exists one scalar $c>0$ such that\n\\[\n\\boxed{\n\\lambda_i=c\\,\\bar\\lambda_a\n\\qquad(i\\in I_a).}\n\\]\nEquivalently, the likelihood is constant on each coarse packet, with the coarse values determined up to the globally invisible Bayesian scaling $c$.\n\\end{theorem}","\\begin{proof}\nIf $\\lambda_i=c\\bar\\lambda_a$ on $I_a$, then $P_a r_a(p)=c\\bar\\lambda_aP_a$ for every prior, including zero-mass packets, and the preceding theorem gives the desired identity because Bayesian normalization removes the common scalar $c$.","Conversely, fix two fine states $i,i'\\in I_a$ and one state $j\\in I_b$ in another coarse packet.  Vary the distribution of a fixed coarse mass inside $I_a$ between $i$ and $i'$ while leaving every coarse packet mass unchanged.  The right side depends only on those coarse masses and is therefore unchanged.  The left-side posterior mass of packet $a$ depends on the conditional average of $\\lambda_i$ and $\\lambda_{i'}$.  It can be independent of that variation only if $\\lambda_i=\\lambda_{i'}$.  Hence $\\lambda$ is constant on every packet; write the value on $I_a$ as $\\lambda_a$.","sely, fix two fine states $i,i'\\in I_a$ and one state $j\\in I_b$ in another coarse packet.  Vary the distribution of a fixed coarse mass inside $I_a$ between $i$ and $i'$ while leaving every coarse packet mass unchanged.  The right side depends only on those coarse masses and is therefore unchanged.  The left-side posterior mass of packet $a$ depends on the conditional average of $\\lambda_i$ and $\\lambda_{i'}$.  It can be independent of that variation only if $\\lambda_i=\\lambda_{i'}$.  Hence $\\lambda$ is constant on every packet; write the value on $I_a$ as $\\lambda_a$.","in another coarse packet.  Vary the distribution of a fixed coarse mass inside $I_a$ between $i$ and $i'$ while leaving every coarse packet mass unchanged.  The right side depends only on those coarse masses and is therefore unchanged.  The left-side posterior mass of packet $a$ depends on the conditional average of $\\lambda_i$ and $\\lambda_{i'}$.  It can be independent of that variation only if $\\lambda_i=\\lambda_{i'}$.  Hence $\\lambda$ is constant on every packet; write the value on $I_a$ as $\\lambda_a$.","he distribution of a fixed coarse mass inside $I_a$ between $i$ and $i'$ while leaving every coarse packet mass unchanged.  The right side depends only on those coarse masses and is therefore unchanged.  The left-side posterior mass of packet $a$ depends on the conditional average of $\\lambda_i$ and $\\lambda_{i'}$.  It can be independent of that variation only if $\\lambda_i=\\lambda_{i'}$.  Hence $\\lambda$ is constant on every packet; write the value on $I_a$ as $\\lambda_a$.","n $i$ and $i'$ while leaving every coarse packet mass unchanged.  The right side depends only on those coarse masses and is therefore unchanged.  The left-side posterior mass of packet $a$ depends on the conditional average of $\\lambda_i$ and $\\lambda_{i'}$.  It can be independent of that variation only if $\\lambda_i=\\lambda_{i'}$.  Hence $\\lambda$ is constant on every packet; write the value on $I_a$ as $\\lambda_a$.","$i'$ while leaving every coarse packet mass unchanged.  The right side depends only on those coarse masses and is therefore unchanged.  The left-side posterior mass of packet $a$ depends on the conditional average of $\\lambda_i$ and $\\lambda_{i'}$.  It can be independent of that variation only if $\\lambda_i=\\lambda_{i'}$.  Hence $\\lambda$ is constant on every packet; write the value on $I_a$ as $\\lambda_a$.","Now use priors supported on one state of packet $a$ and one state of packet $b$.  Equality of the two coarse posterior odds gives\n\\[\n\\frac{\\lambda_a}{\\lambda_b}\n=\n\\frac{\\bar\\lambda_a}{\\bar\\lambda_b}.\n\\]\nThis holds for every pair of coarse packets, so $\\lambda_a=c\\bar\\lambda_a$ for one common $c>0$.\n\\end{proof}","\\begin{remark}[The one-packet boundary]\\label{core:record:022}\nIf the coarse partition has only one packet, every posterior coarsens to the scalar $1$, so commutation is automatic and no within-packet likelihood information can be identified.  The nontrivial criterion begins with at least two coarse packets.\n\\end{remark}","When sufficiency fails, the defect still has an exact finite factorization.  Fix a declared coarse likelihood $\\bar\\lambda$, put\n\\[\nP=C_\\pi p,\n\\qquad\n\\delta(p)=r(p)-\\bar\\lambda,\n\\qquad\n\\bar b=\\cB_{\\bar\\lambda}(P),\n\\qquad\nZ_r=\\langle r(p),P\\rangle.\n\\]","\\begin{theorem}[Bayesian information-loss packet]\\label{core:record:023}\nAssume nonnegative fine and declared coarse likelihoods, with both $Z_r>0$ and $Z_{\\bar\\lambda}>0$. The coarsening defect is\n\\[\n\\boxed{\nC_\\pi\\cB_\\lambda(p)\n-\n\\cB_{\\bar\\lambda}(P)\n=\n\\frac{1}{Z_r}\n\\left(\\diag(P)-\\bar bP^{\\mathsf T}\\right)\n\\delta(p).}\n\\]\nThe factor matrix annihilates $\\bar\\lambda$.  This records the common-scale null direction at the declared coarse likelihood; the full exact defect remains homogeneous under simultaneous likelihood rescaling because the numerator and evidence denominator scale together.  Moreover,\n\\[\n\\boxed{\n\\left\\|C_\\pi\\cB_\\lambda(p)-\\cB_{\\bar\\lambda}(P)\\right\\|_1\n\\le\n\\frac{2\\sum_aP_a|\\delta_a(p)|}\n{\\min\\{Z_r,Z_{\\bar\\lambda}\\}}}\n\\]\nand\n\\[\n\\sum_aP_a|\\delta_a(p)|\n\\le\n\\sum_i p_i\\left|\\lambda_i-\\bar\\lambda_{\\pi(i)}\\right|.\n\\]\n\\end{theo","\\begin{theorem}[Bayesian information-loss packet]\\label{core:record:023}\nAssume nonnegative fine and declared coarse likelihoods, with both $Z_r>0$ and $Z_{\\bar\\lambda}>0$. The coarsening defect is\n\\[\n\\boxed{\nC_\\pi\\cB_\\lambda(p)\n-\n\\cB_{\\bar\\lambda}(P)\n=\n\\frac{1}{Z_r}\n\\left(\\diag(P)-\\bar bP^{\\mathsf T}\\right)\n\\delta(p).}\n\\]\nThe factor matrix annihilates $\\bar\\lambda$.  This records the common-scale null direction at the declared coarse likelihood; the full exact defect remains homogeneous under simultaneous likelihood rescaling because the numerator and evidence denominator scale together.  Moreover,\n\\[\n\\boxed{\n\\left\\|C_\\pi\\cB_\\lambda(p)-\\cB_{\\bar\\lambda}(P)\\right\\|_1\n\\le\n\\frac{2\\sum_aP_a|\\delta_a(p)|}\n{\\min\\{Z_r,Z_{\\bar\\lambda}\\}}}\n\\]\nand\n\\[\n\\sum_aP_a|\\delta_a(p)|\n\\le\n\\sum_i p_i\\left|\\lambda_i-\\bar\\lambda_{\\pi(i)}\\right|.\n\\]\n\\end{theorem}","C_\\pi\\cB_\\lambda(p)\n-\n\\cB_{\\bar\\lambda}(P)\n=\n\\frac{1}{Z_r}\n\\left(\\diag(P)-\\bar bP^{\\mathsf T}\\right)\n\\delta(p).}\n\\]\nThe factor matrix annihilates $\\bar\\lambda$.  This records the common-scale null direction at the declared coarse likelihood; the full exact defect remains homogeneous under simultaneous likelihood rescaling because the numerator and evidence denominator scale together.  Moreover,\n\\[\n\\boxed{\n\\left\\|C_\\pi\\cB_\\lambda(p)-\\cB_{\\bar\\lambda}(P)\\right\\|_1\n\\le\n\\frac{2\\sum_aP_a|\\delta_a(p)|}\n{\\min\\{Z_r,Z_{\\bar\\lambda}\\}}}\n\\]\nand\n\\[\n\\sum_aP_a|\\delta_a(p)|\n\\le\n\\sum_i p_i\\left|\\lambda_i-\\bar\\lambda_{\\pi(i)}\\right|.\n\\]\n\\end{theorem}","tes $\\bar\\lambda$.  This records the common-scale null direction at the declared coarse likelihood; the full exact defect remains homogeneous under simultaneous likelihood rescaling because the numerator and evidence denominator scale together.  Moreover,\n\\[\n\\boxed{\n\\left\\|C_\\pi\\cB_\\lambda(p)-\\cB_{\\bar\\lambda}(P)\\right\\|_1\n\\le\n\\frac{2\\sum_aP_a|\\delta_a(p)|}\n{\\min\\{Z_r,Z_{\\bar\\lambda}\\}}}\n\\]\nand\n\\[\n\\sum_aP_a|\\delta_a(p)|\n\\le\n\\sum_i p_i\\left|\\lambda_i-\\bar\\lambda_{\\pi(i)}\\right|.\n\\]\n\\end{theorem}","\\begin{proof}\nThe effective-likelihood theorem reduces the defect to\n\\[\n\\cB_{\\bar\\lambda+\\delta}(P)-\\cB_{\\bar\\lambda}(P).\n\\]\nSince $Z_r=Z_{\\bar\\lambda}+P^{\\mathsf T}\\delta$, direct subtraction gives\n\\[\n\\frac{1}{Z_r}\n\\left(\\diag(P)-\\bar bP^{\\mathsf T}\\right)\\delta.\n\\]\nAlso,\n\\[\n\\left(\\diag(P)-\\bar bP^{\\mathsf T}\\right)\\bar\\lambda\n=P\\odot\\bar\\lambda-\\bar bZ_{\\bar\\lambda}=0.\n\\]\nThe first norm bound is the likelihood perturbation theorem applied on the coarse simplex. For a zero-mass packet, both sides of the following estimate vanish. For $P_a>0$,\n\\[\nP_a|\\delta_a|\n=\nP_a\\left|\\sum_{i\\in I_a}\\frac{p_i}{P_a}\n(\\lambda_i-\\bar\\lambda_a)\\right|\n\\le\n\\sum_{i\\in I_a}p_i|\\lambda_i-\\bar\\lambda_a|,\n\\]\nand summing over $a$ proves the second inequality.\n\\end{proof}","r\\lambda}+P^{\\mathsf T}\\delta$, direct subtraction gives\n\\[\n\\frac{1}{Z_r}\n\\left(\\diag(P)-\\bar bP^{\\mathsf T}\\right)\\delta.\n\\]\nAlso,\n\\[\n\\left(\\diag(P)-\\bar bP^{\\mathsf T}\\right)\\bar\\lambda\n=P\\odot\\bar\\lambda-\\bar bZ_{\\bar\\lambda}=0.\n\\]\nThe first norm bound is the likelihood perturbation theorem applied on the coarse simplex. For a zero-mass packet, both sides of the following estimate vanish. For $P_a>0$,\n\\[\nP_a|\\delta_a|\n=\nP_a\\left|\\sum_{i\\in I_a}\\frac{p_i}{P_a}\n(\\lambda_i-\\bar\\lambda_a)\\right|\n\\le\n\\sum_{i\\in I_a}p_i|\\lambda_i-\\bar\\lambda_a|,\n\\]\nand summing over $a$ proves the second inequality.\n\\end{proof}","P^{\\mathsf T}\\delta$, direct subtraction gives\n\\[\n\\frac{1}{Z_r}\n\\left(\\diag(P)-\\bar bP^{\\mathsf T}\\right)\\delta.\n\\]\nAlso,\n\\[\n\\left(\\diag(P)-\\bar bP^{\\mathsf T}\\right)\\bar\\lambda\n=P\\odot\\bar\\lambda-\\bar bZ_{\\bar\\lambda}=0.\n\\]\nThe first norm bound is the likelihood perturbation theorem applied on the coarse simplex. For a zero-mass packet, both sides of the following estimate vanish. For $P_a>0$,\n\\[\nP_a|\\delta_a|\n=\nP_a\\left|\\sum_{i\\in I_a}\\frac{p_i}{P_a}\n(\\lambda_i-\\bar\\lambda_a)\\right|\n\\le\n\\sum_{i\\in I_a}p_i|\\lambda_i-\\bar\\lambda_a|,\n\\]\nand summing over $a$ proves the second inequality.\n\\end{proof}","right)\\bar\\lambda\n=P\\odot\\bar\\lambda-\\bar bZ_{\\bar\\lambda}=0.\n\\]\nThe first norm bound is the likelihood perturbation theorem applied on the coarse simplex. For a zero-mass packet, both sides of the following estimate vanish. For $P_a>0$,\n\\[\nP_a|\\delta_a|\n=\nP_a\\left|\\sum_{i\\in I_a}\\frac{p_i}{P_a}\n(\\lambda_i-\\bar\\lambda_a)\\right|\n\\le\n\\sum_{i\\in I_a}p_i|\\lambda_i-\\bar\\lambda_a|,\n\\]\nand summing over $a$ proves the second inequality.\n\\end{proof}","\\begin{corollary}[Canonical coarsest likelihood-resolving partition]\\label{core:record:025}\nFor a finite family of likelihoods $\\{\\lambda^{(y)}\\}_{y\\in Y}$, define\n\\[\ni\\sim j\n\\quad\\Longleftrightarrow\\quad\n\\lambda_i^{(y)}=\\lambda_j^{(y)}\n\\text{ for every }y\\in Y.\n\\]\nThe equivalence classes form the unique coarsest likelihood-resolving partition: every likelihood is constant on each packet. Every update therefore closes on this partition whenever its evidence is positive. If all likelihoods are strictly positive and there are at least two profile classes, this is also the unique coarsest nontrivial partition on which every update closes for every prior. The one-packet partition always closes, independently of the profiles, and is excluded from this last assertion.\n\\end{corollary}","\\begin{certificate}[Bayesian refinement]\\label{core:record:026}\nFor the finite test partition $\\{0,1\\}\\sqcup\\{2,3\\}$, a blockwise-constant likelihood gave commutation residual $1.11\\times10^{-16}$.  A nonconstant likelihood produced an exact defect-factorization residual $7.29\\times10^{-17}$.  Its measured $L^1$ defect was $4.8787\\times10^{-2}$, below the certified bound $1.1016\\times10^{-1}$.  Two priors with identical coarse masses but different hidden within-packet shapes produced coarse posteriors differing by $1.2400\\times10^{-1}$, which is a direct necessity witness.\n\\end{certificate}","lockwise-constant likelihood gave commutation residual $1.11\\times10^{-16}$.  A nonconstant likelihood produced an exact defect-factorization residual $7.29\\times10^{-17}$.  Its measured $L^1$ defect was $4.8787\\times10^{-2}$, below the certified bound $1.1016\\times10^{-1}$.  Two priors with identical coarse masses but different hidden within-packet shapes produced coarse posteriors differing by $1.2400\\times10^{-1}$, which is a direct necessity witness.\n\\end{certificate}","Fix a score vector $g$. The likelihood flow is\n\\[\n\\lambda_i(t)=e^{tg_i},\n\\]\nand the Bayesian flow is\n\\[\n\\Phi_t(p)_i\n=\n\\frac{e^{tg_i}p_i}{\\sum_je^{tg_j}p_j}.\n\\]","\\begin{theorem}[Bayesian nonlinear semigroup]\\label{core:record:028}\n\\[\n\\boxed{\n\\Phi_{s+t}=\\Phi_s\\circ\\Phi_t.\n}\n\\]\nIts generator is the centered multiplicative field\n\\[\n\\boxed{\n\\frac{d}{dt}p_i(t)\n=\n\\bigl(g_i-\\langle g,p(t)\\rangle\\bigr)p_i(t).\n}\n\\]\n\\end{theorem}","\\begin{proof}\nThe semigroup law is the likelihood-composition theorem with $e^{sg}e^{tg}=e^{(s+t)g}$. Differentiating the normalized exponential formula gives\n\\[\n\\dot p_i=g_ip_i-p_i\\frac{\\sum_jg_je^{tg_j}p_j(0)}{\\sum_je^{tg_j}p_j(0)}.\n\\]\nThe quotient is $\\langle g,p(t)\\rangle$.\n\\end{proof}","\\begin{example}[Beta--Binomial conjugate packet]\\label{core:record:029}\nA $\\operatorname{Beta}(a,b)$ prior and $s$ successes in $n$ trials produce\n\\[\n\\operatorname{Beta}(a+s,b+n-s).\n\\]\nThe prior exponents are additive packets: the likelihood increments them by $(s,n-s)$. In the numerical example,\n\\[\n(a,b)=(2,3),\\quad (s,n)=(7,10),\n\\]\nso the posterior is $\\operatorname{Beta}(9,6)$ with mean $0.6$.\n\\end{example}","Let $K$ be column stochastic and let $\\lambda^{(r)}$ be the likelihood associated with the $r$th observation. The filter recursion is\n\\[\n\\widehat p_r=Kp_{r-1},\n\\qquad\np_r=\\cB_{\\lambda^{(r)}}(\\widehat p_r).\n\\]","\\begin{proposition}[Prediction-update reconstruction]\\label{core:record:030}\nEvery hidden Markov filtering step is the composition of a linear Markov transfer and a normalized nonlinear likelihood transfer. Assume a probability initial vector, a column-stochastic transition matrix, and nonnegative likelihoods with positive evidence at every step. Then every $p_r$ is a probability vector.\n\\end{proposition}","\\begin{proof}\nColumn stochasticity gives $\\widehat p_r\\ge0$ and $\\bm1^T\\widehat p_r=1$. Bayesian normalization then gives $p_r\\ge0$ and $\\bm1^Tp_r=1$.\n\\end{proof}","Let $L=L^*\\ge0$ on a finite-dimensional Hilbert space. Its spectral resolution is\n\\[\nL=\\sum_j\\lambda_jP_j.\n\\]\nFor $\\alpha>0$, define\n\\[\n\\boxed{L^\\alpha=\\sum_j\\lambda_j^\\alpha P_j.}\n\\]\nThe associated fractional semigroup is\n\\[\nT_\\alpha(t)=e^{-tL^\\alpha}\n=\\sum_je^{-t\\lambda_j^\\alpha}P_j.\n\\]","\\begin{theorem}[Fractional packet calculus]\\label{core:record:031}\nFor $\\alpha,\\beta>0$,\n\\[\nL^\\alpha L^\\beta=L^{\\alpha+\\beta},\n\\]\nand for $s,t\\ge0$,\n\\[\n\\boxed{T_\\alpha(s)T_\\alpha(t)=T_\\alpha(s+t).}\n\\]\nEvery spectral packet $P_j$ reduces $L^\\alpha$ and $T_\\alpha(t)$.\n\\end{theorem}","\\begin{proof}\nAll statements hold coefficientwise on the mutually orthogonal spectral projectors. For example,\n\\[\nL^\\alpha L^\\beta\n=\\sum_{j,k}\\lambda_j^\\alpha\\lambda_k^\\beta P_jP_k\n=\\sum_j\\lambda_j^{\\alpha+\\beta}P_j.\n\\]\n\\end{proof}","\\begin{theorem}[Poisson subordination]\\label{core:record:032}\nFor $t>0$ and $L\\ge0$,\n\\[\n\\boxed{\ne^{-t\\sqrt L}\n=\n\\frac{t}{2\\sqrt\\pi}\n\\int_0^\\infty\ns^{-3/2}e^{-t^2/(4s)}e^{-sL}\\,ds.\n}\n\\]\nThe integral converges in operator norm in finite dimension.\n\\end{theorem}","\\begin{proof}\nBy the spectral theorem it suffices to prove the scalar identity\n\\[\ne^{-t\\sqrt\\lambda}\n=\n\\frac{t}{2\\sqrt\\pi}\n\\int_0^\\infty\ns^{-3/2}e^{-t^2/(4s)-\\lambda s}\\,ds\n\\]\nfor $\\lambda\\ge0$. This is the classical Gaussian integral after the substitution $u=t/(2\\sqrt s)$. Applying the identity to every eigenvalue and recombining the spectral projectors proves the operator formula.\n\\end{proof}","Let $A\\ge0$ act on a fine Hilbert space, let $B\\ge0$ act on a coarse Hilbert space, and let $C$ be a bounded coarsening map.  Set\n\\[\nD=CA-BC,\n\\qquad\nD_\\alpha=CA^\\alpha-B^\\alpha C,\n\\qquad 0<\\alpha<1.\n\\]","\\begin{theorem}[Fractional intertwining theorem]\\label{core:record:033}\\label{core:thm:fractional-naturality}\nFor bounded nonnegative self-adjoint $A$ and $B$,\n\\[\n\\boxed{\nD_\\alpha\n=\n\\frac{\\sin(\\pi\\alpha)}{\\pi}\n\\int_0^\\infty\n t^\\alpha(tI+B)^{-1}D(tI+A)^{-1}\\,dt.\n}\n\\]\nThe improper integral converges in operator norm.  Moreover,\n\\[\n\\boxed{\n\\norm{D_\\alpha}\n\\le\nK_\\alpha\\norm{C}^{1-\\alpha}\\norm{D}^{\\alpha},\n\\qquad\nK_\\alpha=\n\\frac{2^{1-\\alpha}\\sin(\\pi\\alpha)}\n{\\pi\\alpha(1-\\alpha)}.\n}\n\\]\nIf $A,B\\ge mI$ for some $m>0$, then the stronger linear estimate holds:\n\\[\n\\boxed{\n\\norm{D_\\alpha}\n\\le\n\\alpha m^{\\alpha-1}\\norm{D}.\n}\n\\]\nIn particular, exact generator intertwining $CA=BC$ implies exact fractional intertwining $CA^\\alpha=B^\\alpha C$.\n\\end{theorem}","actional-naturality}\nFor bounded nonnegative self-adjoint $A$ and $B$,\n\\[\n\\boxed{\nD_\\alpha\n=\n\\frac{\\sin(\\pi\\alpha)}{\\pi}\n\\int_0^\\infty\n t^\\alpha(tI+B)^{-1}D(tI+A)^{-1}\\,dt.\n}\n\\]\nThe improper integral converges in operator norm.  Moreover,\n\\[\n\\boxed{\n\\norm{D_\\alpha}\n\\le\nK_\\alpha\\norm{C}^{1-\\alpha}\\norm{D}^{\\alpha},\n\\qquad\nK_\\alpha=\n\\frac{2^{1-\\alpha}\\sin(\\pi\\alpha)}\n{\\pi\\alpha(1-\\alpha)}.\n}\n\\]\nIf $A,B\\ge mI$ for some $m>0$, then the stronger linear estimate holds:\n\\[\n\\boxed{\n\\norm{D_\\alpha}\n\\le\n\\alpha m^{\\alpha-1}\\norm{D}.\n}\n\\]\nIn particular, exact generator intertwining $CA=BC$ implies exact fractional intertwining $CA^\\alpha=B^\\alpha C$.\n\\end{theorem}","For bounded nonnegative self-adjoint $A$ and $B$,\n\\[\n\\boxed{\nD_\\alpha\n=\n\\frac{\\sin(\\pi\\alpha)}{\\pi}\n\\int_0^\\infty\n t^\\alpha(tI+B)^{-1}D(tI+A)^{-1}\\,dt.\n}\n\\]\nThe improper integral converges in operator norm.  Moreover,\n\\[\n\\boxed{\n\\norm{D_\\alpha}\n\\le\nK_\\alpha\\norm{C}^{1-\\alpha}\\norm{D}^{\\alpha},\n\\qquad\nK_\\alpha=\n\\frac{2^{1-\\alpha}\\sin(\\pi\\alpha)}\n{\\pi\\alpha(1-\\alpha)}.\n}\n\\]\nIf $A,B\\ge mI$ for some $m>0$, then the stronger linear estimate holds:\n\\[\n\\boxed{\n\\norm{D_\\alpha}\n\\le\n\\alpha m^{\\alpha-1}\\norm{D}.\n}\n\\]\nIn particular, exact generator intertwining $CA=BC$ implies exact fractional intertwining $CA^\\alpha=B^\\alpha C$.\n\\end{theorem}","\\[\n\\boxed{\nD_\\alpha\n=\n\\frac{\\sin(\\pi\\alpha)}{\\pi}\n\\int_0^\\infty\n t^\\alpha(tI+B)^{-1}D(tI+A)^{-1}\\,dt.\n}\n\\]\nThe improper integral converges in operator norm.  Moreover,\n\\[\n\\boxed{\n\\norm{D_\\alpha}\n\\le\nK_\\alpha\\norm{C}^{1-\\alpha}\\norm{D}^{\\alpha},\n\\qquad\nK_\\alpha=\n\\frac{2^{1-\\alpha}\\sin(\\pi\\alpha)}\n{\\pi\\alpha(1-\\alpha)}.\n}\n\\]\nIf $A,B\\ge mI$ for some $m>0$, then the stronger linear estimate holds:\n\\[\n\\boxed{\n\\norm{D_\\alpha}\n\\le\n\\alpha m^{\\alpha-1}\\norm{D}.\n}\n\\]\nIn particular, exact generator intertwining $CA=BC$ implies exact fractional intertwining $CA^\\alpha=B^\\alpha C$.\n\\end{theorem}","The improper integral converges in operator norm.  Moreover,\n\\[\n\\boxed{\n\\norm{D_\\alpha}\n\\le\nK_\\alpha\\norm{C}^{1-\\alpha}\\norm{D}^{\\alpha},\n\\qquad\nK_\\alpha=\n\\frac{2^{1-\\alpha}\\sin(\\pi\\alpha)}\n{\\pi\\alpha(1-\\alpha)}.\n}\n\\]\nIf $A,B\\ge mI$ for some $m>0$, then the stronger linear estimate holds:\n\\[\n\\boxed{\n\\norm{D_\\alpha}\n\\le\n\\alpha m^{\\alpha-1}\\norm{D}.\n}\n\\]\nIn particular, exact generator intertwining $CA=BC$ implies exact fractional intertwining $CA^\\alpha=B^\\alpha C$.\n\\end{theorem}","rm.  Moreover,\n\\[\n\\boxed{\n\\norm{D_\\alpha}\n\\le\nK_\\alpha\\norm{C}^{1-\\alpha}\\norm{D}^{\\alpha},\n\\qquad\nK_\\alpha=\n\\frac{2^{1-\\alpha}\\sin(\\pi\\alpha)}\n{\\pi\\alpha(1-\\alpha)}.\n}\n\\]\nIf $A,B\\ge mI$ for some $m>0$, then the stronger linear estimate holds:\n\\[\n\\boxed{\n\\norm{D_\\alpha}\n\\le\n\\alpha m^{\\alpha-1}\\norm{D}.\n}\n\\]\nIn particular, exact generator intertwining $CA=BC$ implies exact fractional intertwining $CA^\\alpha=B^\\alpha C$.\n\\end{theorem}","\\begin{proof}\nFor $0<\\alpha<1$, the Balakrishnan formula gives\n\\[\nA^\\alpha\n=\n\\frac{\\sin(\\pi\\alpha)}{\\pi}\n\\int_0^\\infty t^{\\alpha-1}A(tI+A)^{-1}\\,dt,\n\\]\nand likewise for $B$.  Using\n\\[\nA(tI+A)^{-1}=I-t(tI+A)^{-1}\n\\]\nand the rectangular resolvent identity\n\\[\n(tI+B)^{-1}C-C(tI+A)^{-1}\n=\n(tI+B)^{-1}(CA-BC)(tI+A)^{-1},\n\\]\nthe difference of the two Balakrishnan integrals becomes the displayed resolvent-defect formula.","For small $t$,\n\\[\n\\norm{CA(tI+A)^{-1}-B(tI+B)^{-1}C}\n\\le 2\\norm C,\n\\]\nbecause the two positive resolvent factors are contractions.  For large $t$, the resolvent identity gives the bound $\\norm D/t$.  Splitting the integral at\n\\[\nr=\\frac{\\norm D}{2\\norm C}\n\\]\nwhen $C$ and $D$ are nonzero yields\n\\[\n\\begin{aligned}\n\\norm{D_\\alpha}\n&\\le\n\\frac{\\sin(\\pi\\alpha)}{\\pi}\n\\left(\n2\\norm C\\int_0^r t^{\\alpha-1}\\,dt\n+\n\\norm D\\int_r^\\infty t^{\\alpha-2}\\,dt\n\\right)","For small $t$,\n\\[\n\\norm{CA(tI+A)^{-1}-B(tI+B)^{-1}C}\n\\le 2\\norm C,\n\\]\nbecause the two positive resolvent factors are contractions.  For large $t$, the resolvent identity gives the bound $\\norm D/t$.  Splitting the integral at\n\\[\nr=\\frac{\\norm D}{2\\norm C}\n\\]\nwhen $C$ and $D$ are nonzero yields\n\\[\n\\begin{aligned}\n\\norm{D_\\alpha}\n&\\le\n\\frac{\\sin(\\pi\\alpha)}{\\pi}\n\\left(\n2\\norm C\\int_0^r t^{\\alpha-1}\\,dt\n+\n\\norm D\\int_r^\\infty t^{\\alpha-2}\\,dt\n\\right)\\\\\n&=\nK_\\alpha\\norm C^{1-\\alpha}\\norm D^\\alpha.\n\\end{aligned}\n\\]\nThe zero cases are immediate.","If $A,B\\ge mI$, then both resolvents have norm at most $(t+m)^{-1}$, so\n\\[\n\\norm{D_\\alpha}\n\\le\n\\frac{\\sin(\\pi\\alpha)}{\\pi}\\norm D\n\\int_0^\\infty\\frac{t^\\alpha}{(t+m)^2}\\,dt.\n\\]\nAfter $t=mu$, the beta integral equals\n\\[\nm^{\\alpha-1}\\frac{\\alpha\\pi}{\\sin(\\pi\\alpha)},\n\\]\nwhich gives the linear estimate.\n\\end{proof}","\\begin{corollary}[Fractional semigroup refinement]\\label{core:record:034}\\label{core:cor:fractional-semigroup-refinement}\nFor $s\\ge0$,\n\\[\n\\boxed{\nCe^{-sA^\\alpha}-e^{-sB^\\alpha}C\n=\n-\\int_0^s\n e^{-(s-r)B^\\alpha}D_\\alpha e^{-rA^\\alpha}\\,dr.\n}\n\\]\nHence\n\\[\n\\norm{Ce^{-sA^\\alpha}-e^{-sB^\\alpha}C}\n\\le s\\norm{D_\\alpha}.\n\\]\nUnder the common gap $A,B\\ge mI$,\n\\[\n\\boxed{\n\\norm{Ce^{-sA^\\alpha}-e^{-sB^\\alpha}C}\n\\le\ns e^{-sm^\\alpha}\\alpha m^{\\alpha-1}\\norm D.\n}\n\\]\n\\end{corollary}","\\begin{proof}\nDifferentiate\n\\[\nr\\longmapsto e^{-(s-r)B^\\alpha}Ce^{-rA^\\alpha}\n\\]\nand integrate from $0$ to $s$.  Positive fractional semigroups are contractions; under the gap, the product of their two norm bounds is $e^{-sm^\\alpha}$.\n\\end{proof}","\\begin{corollary}[Fractional refinement towers]\\label{core:record:035}\nLet $D_\\ell=C_\\ell A_{\\ell+1}-A_\\ell C_\\ell$ in a bounded nonnegative self-adjoint tower, with $\\sup_\\ell\\norm{C_\\ell}\\le M$.  If\n\\[\n\\sum_\\ell\\norm{D_\\ell}^{\\alpha}<\\infty,\n\\]\nthen the fractional defects\n\\[\nC_\\ell A_{\\ell+1}^{\\alpha}-A_\\ell^{\\alpha}C_\\ell\n\\]\nhave summable operator norms. If the tower has a uniform gap $m>0$, it is enough that $\\sum_\\ell\\norm{D_\\ell}<\\infty$. On every bounded time interval, the series of corresponding fractional-semigroup defect norms converges uniformly. The maps may have different domain and codomain spaces; no operator sum across unrelated spaces is asserted.\n\\end{corollary}","[Fractional refinement towers]\\label{core:record:035}\nLet $D_\\ell=C_\\ell A_{\\ell+1}-A_\\ell C_\\ell$ in a bounded nonnegative self-adjoint tower, with $\\sup_\\ell\\norm{C_\\ell}\\le M$.  If\n\\[\n\\sum_\\ell\\norm{D_\\ell}^{\\alpha}<\\infty,\n\\]\nthen the fractional defects\n\\[\nC_\\ell A_{\\ell+1}^{\\alpha}-A_\\ell^{\\alpha}C_\\ell\n\\]\nhave summable operator norms. If the tower has a uniform gap $m>0$, it is enough that $\\sum_\\ell\\norm{D_\\ell}<\\infty$. On every bounded time interval, the series of corresponding fractional-semigroup defect norms converges uniformly. The maps may have different domain and codomain spaces; no operator sum across unrelated spaces is asserted.\n\\end{corollary}","\\begin{boundary}[Unbounded and nonnormal generators]\\label{core:record:036}\nThe theorem resolves fractional naturality for bounded nonnegative self-adjoint refinement systems and gives the exact estimate needed for finite discretization towers.  For unbounded positive generators, the same resolvent argument requires a declared common-domain condition and a bounded extension of $CA-BC$.  Sectorial or nonnormal generators require their own contour calculus and resolvent bounds.  Those domain questions are not inferred from the finite certificates.\n\\end{boundary}","On the interval $(0,1)$ with Dirichlet boundary conditions, the Green kernel for $-u''=f$ is\n\\[\n\\boxed{G(x,y)=\\min(x,y)-xy.}\n\\]\nThus the point source $\\delta_y$ generates the solution kernel, and a general source is reconstructed by\n\\[\nu(x)=\\int_0^1G(x,y)f(y)\\,dy.\n\\]","For $n$ interior points with spacing $h=1/(n+1)$, define\n\\[\nL_h=\\frac1{h^2}\n\\begin{pmatrix}\n2&-1&&","For $n$ interior points with spacing $h=1/(n+1)$, define\n\\[\nL_h=\\frac1{h^2}\n\\begin{pmatrix}\n2&-1&&\\\\\n-1&2&\\ddots&\\\\\n&\\ddots&\\ddots&-1\\\\\n&&-1&2\n\\end{pmatrix}.\n\\]","\\begin{theorem}[Exact discrete Green reconstruction at grid nodes]\\label{core:record:038}\nLet $x_i=ih$. Then\n\\[\n\\boxed{\n\\frac1h(L_h^{-1})_{ij}\n=G(x_i,x_j).\n}\n\\]\nThus the scaled discrete inverse reproduces the continuum Dirichlet Green kernel exactly at the grid nodes.\n\\end{theorem}","\\begin{proof}\nThe inverse of the unscaled tridiagonal matrix has entries\n\\[\n\\frac{\\min(i,j)(n+1-\\max(i,j))}{n+1}.\n\\]\nMultiplying by $h^2$ and then dividing by $h$ gives\n\\[\nh\\min(i,j)\\left(1-\\frac{\\max(i,j)}{n+1}\\right)\n=\\min(x_i,x_j)-x_ix_j.\n\\]\n\\end{proof}","The $k$th discrete eigenvalue is\n\\[\n\\lambda_{k,h}=\\frac4{h^2}\\sin^2\\left(\\frac{k\\pi h}{2}\\right),\n\\]\nwhereas the continuum eigenvalue is $(k\\pi)^2$.","\\begin{theorem}[Second-order low-mode convergence]\\label{core:record:039}\nFor fixed $k$,\n\\[\n\\lambda_{k,h}=(k\\pi)^2+O(h^2).\n\\]\nMore precisely, for sufficiently small $h$,\n\\[\n0\\le (k\\pi)^2-\\lambda_{k,h}\n\\le \\frac{(k\\pi)^4}{12}h^2.\n\\]\n\\end{theorem}","\\begin{proof}\nFor $x\\ge0$, $0\\le x^2-\\sin^2x\\le x^4/3$. Set $x=k\\pi h/2$ and multiply by $4/h^2$.\n\\end{proof}","Let $K$ be a column-stochastic matrix, let $V$ be a potential on the states, and let $f$ be terminal data. Define\n\\[\nM_V=\\diag(e^{-V})K^T.\n\\]\nThe backward Feynman--Kac value after $n$ steps is\n\\[\nu_n=M_V^nf.\n\\]","\\begin{theorem}[Discrete Feynman--Kac path formula]\\label{core:record:040}\nFor an initial state $i$,\n\\[\n\\boxed{\n(u_n)_i\n=\n\\Exp_i\\left[\n\\exp\\left(-\\sum_{r=0}^{n-1}V(X_r)\\right)f(X_n)\n\\right].\n}\n\\]\n\\end{theorem}","\\begin{proof}\nFor $n=1$,\n\\[\n(M_Vf)_i=e^{-V(i)}\\sum_jK_{j i}f_j,\n\\]\nwhich is the stated expectation. Iterating and expanding the matrix product sums over every path $(i=i_0,i_1,\\ldots,i_n)$ with probability $\\prod_rK_{i_{r+1},i_r}$ and weight $\\prod_re^{-V(i_r)}$.\n\\end{proof}","Let $X=P X\\oplus Q X$ and write\n\\[\nA=\n\\begin{pmatrix}\nA_{PP}&A_{PQ}","Let $X=P X\\oplus Q X$ and write\n\\[\nA=\n\\begin{pmatrix}\nA_{PP}&A_{PQ}\\\\\nA_{QP}&A_{QQ}\n\\end{pmatrix}.\n\\]","\\begin{theorem}[Schur--Feshbach projected resolvent]\\label{core:record:041}\nAssume both $zI-A$ and $zI-A_{QQ}$ are invertible. Define\n\\[\n\\boxed{\nF_P(z)\n=zI-A_{PP}-A_{PQ}(zI-A_{QQ})^{-1}A_{QP}.\n}\n\\]\nThen\n\\[\n\\boxed{\n\\left.P(zI-A)^{-1}\\right|_{\\Ran P}=F_P(z)^{-1}.\n}\n\\]\n\\end{theorem}","\\begin{proof}\nSolve the hidden block equation\n\\[\n(zI-A_{QQ})x_Q=A_{QP}x_P+y_Q\n\\]\nfor $x_Q$ and substitute it into the retained equation. The resulting retained coefficient is $F_P(z)$. Invertibility of the full operator and of the hidden block implies invertibility of this Schur complement. Setting $y_Q=0$ gives the stated identity as an operator on $\\Ran P$. Equivalently, this is the block inverse formula.\n\\end{proof}","The self-energy\n\\[\n\\Sigma_P(z)=A_{PQ}(zI-A_{QQ})^{-1}A_{QP}\n\\]\ncontains the exact effect of the eliminated packets. A bare truncation $A_{PP}$ discards this term and is generally not exact.","For the homogeneous system $\\dot x=Ax$, variation of constants gives\n\\[\nx_Q(t)=e^{tA_{QQ}}x_Q(0)\n+\\int_0^te^{(t-r)A_{QQ}}A_{QP}x_P(r)\\,dr.\n\\]\nSubstitution yields\n\\[\n\\dot x_P(t)\n=A_{PP}x_P(t)+A_{PQ}e^{tA_{QQ}}x_Q(0)\n+\\int_0^t\\cK(t-r)x_P(r)\\,dr,\n\\]\nwhere\n\\[\n\\boxed{\n\\cK(t)=A_{PQ}e^{tA_{QQ}}A_{QP}.\n}\n\\]","\\begin{theorem}[Memory/self-energy correspondence]\\label{core:record:042}\nFor finite complex block matrices, assume $\\operatorname{Re}z>\\operatorname{Re}\\lambda$ for every $\\lambda\\in\\sigma(A_{QQ})$. Then the following improper integral is absolutely convergent and\n\\[\n\\boxed{\n\\int_0^\\infty e^{-zt}\\cK(t)\\,dt\n=A_{PQ}(zI-A_{QQ})^{-1}A_{QP}\n=\\Sigma_P(z).\n}\n\\]\n\\end{theorem}","\\begin{proof}\nPut $B=A_{QQ}$ and $T_z(t)=e^{-zt}e^{tB}=e^{t(B-zI)}$.\nThe generalized-eigenspace decomposition in finite dimensions writes $T_z$ as a\nfinite sum of constant matrix coefficients times $t^k e^{(\\lambda-z)t}$.\nThis follows by expanding the nilpotent part on each generalized eigenspace;\nequivalently, apply the decomposition to left multiplication by $B$ on the matrix\nalgebra and evaluate the exponential at $I$.\nThe strict spectral inequality makes every such term absolutely integrable on\n$(0,\\infty)$ and forces it to tend to zero. Hence $T_z$ is Bochner integrable\nand $T_z(t)\\to0$, including when $B$ is nonnormal.","_{QQ}$ and $T_z(t)=e^{-zt}e^{tB}=e^{t(B-zI)}$.\nThe generalized-eigenspace decomposition in finite dimensions writes $T_z$ as a\nfinite sum of constant matrix coefficients times $t^k e^{(\\lambda-z)t}$.\nThis follows by expanding the nilpotent part on each generalized eigenspace;\nequivalently, apply the decomposition to left multiplication by $B$ on the matrix\nalgebra and evaluate the exponential at $I$.\nThe strict spectral inequality makes every such term absolutely integrable on\n$(0,\\infty)$ and forces it to tend to zero. Hence $T_z$ is Bochner integrable\nand $T_z(t)\\to0$, including when $B$ is nonnormal.","tion in finite dimensions writes $T_z$ as a\nfinite sum of constant matrix coefficients times $t^k e^{(\\lambda-z)t}$.\nThis follows by expanding the nilpotent part on each generalized eigenspace;\nequivalently, apply the decomposition to left multiplication by $B$ on the matrix\nalgebra and evaluate the exponential at $I$.\nThe strict spectral inequality makes every such term absolutely integrable on\n$(0,\\infty)$ and forces it to tend to zero. Hence $T_z$ is Bochner integrable\nand $T_z(t)\\to0$, including when $B$ is nonnormal.","sum of constant matrix coefficients times $t^k e^{(\\lambda-z)t}$.\nThis follows by expanding the nilpotent part on each generalized eigenspace;\nequivalently, apply the decomposition to left multiplication by $B$ on the matrix\nalgebra and evaluate the exponential at $I$.\nThe strict spectral inequality makes every such term absolutely integrable on\n$(0,\\infty)$ and forces it to tend to zero. Hence $T_z$ is Bochner integrable\nand $T_z(t)\\to0$, including when $B$ is nonnormal.","ents times $t^k e^{(\\lambda-z)t}$.\nThis follows by expanding the nilpotent part on each generalized eigenspace;\nequivalently, apply the decomposition to left multiplication by $B$ on the matrix\nalgebra and evaluate the exponential at $I$.\nThe strict spectral inequality makes every such term absolutely integrable on\n$(0,\\infty)$ and forces it to tend to zero. Hence $T_z$ is Bochner integrable\nand $T_z(t)\\to0$, including when $B$ is nonnormal.",")t}$.\nThis follows by expanding the nilpotent part on each generalized eigenspace;\nequivalently, apply the decomposition to left multiplication by $B$ on the matrix\nalgebra and evaluate the exponential at $I$.\nThe strict spectral inequality makes every such term absolutely integrable on\n$(0,\\infty)$ and forces it to tend to zero. Hence $T_z$ is Bochner integrable\nand $T_z(t)\\to0$, including when $B$ is nonnormal.","Let $J=\\int_0^\\infty T_z(t)\\,dt$. Integrating both derivative identities\n$T_z'=(B-zI)T_z=T_z(B-zI)$ and using $T_z(0)=I$ gives\n\\[\n(zI-B)J=J(zI-B)=I.\n\\]\nThus $J=(zI-B)^{-1}$. Continuous left and right multiplication carry the\nBochner integral through the two rectangular transfer blocks, yielding the claim.\n\\end{proof}","Let $A$ be a finite subgenerator: its off-diagonal entries are nonnegative and its column sums are nonpositive. These sign conditions alone allow nonabsorbing states; eventual absorption is a separate hypothesis. Let $p_0$ be an initial transient-state distribution. Define\n\\[\np(t)=e^{tA}p_0,\n\\qquad\nS(t)=\\bm1^Tp(t).\n\\]\nThe exit-rate row vector is\n\\[\na^T=-\\bm1^TA\\ge0.\n\\]","\\begin{theorem}[Semigroup survival identities]\\label{core:record:043}\nFor $t\\ge0$, the finite-time exit density and, wherever $S(t)>0$, the hazard are\n\\[\n\\boxed{\nf(t)=a^Te^{tA}p_0=-S'(t),\n\\qquad\nh(t)=\\frac{f(t)}{S(t)}.\n}\n\\]\nConsequently, $S$ is nonincreasing and\n\\[\nS(t)=\\exp\\left(-\\int_0^th(r)\\,dr\\right)\n\\]\nwhenever $S(t)>0$. The density may be defective: its total mass is $1-\\lim_{t\\to\\infty}S(t)$. It is a probability density when absorption is eventual, so that this limit is zero.\n\\end{theorem}","\\begin{proof}\nDifferentiate:\n\\[\nS'(t)=\\bm1^TAe^{tA}p_0=-a^Te^{tA}p_0.\n\\]\nThe hazard identity is the definition of conditional instantaneous exit rate. Integrating $(\\log S)'=-h$ gives the final formula.\n\\end{proof}","\\midrule\nMarkov & point mass $\\delta_i$ & $K^n$ or $e^{tQ}$ & state lumping and $CK^n=\\bar K^nC$","Conditional & random variable & $P_{\\mathcal Q}$ & nested partitions and tower law","Bayesian & prior $p$ & $\\cB_\\lambda$ & packet evidences and normalized synthesis","Fractional & spectral source & $e^{-tL^\\alpha}$ & spectral projectors $P_j$","PDE & source $f$ or $\\delta_y$ & resolvent/Green operator & eigenmode or spatial packet synthesis","Feynman--Kac & terminal data $f$ & $(\\diag(e^{-V})K^T)^n$ & weighted path reconstruction","Survival & initial transient law & $e^{tA}$ & sum over transient packets","For clarity, the finite-dimensional holomorphic calculus used below has the\nfollowing canonical definition. For $A$ acting on $\\mathbb C^d$, its generalized\neigenspaces $E_\\mu=\\ker(A-\\mu I)^d$ form a direct-sum decomposition over the\neigenvalues. Define the unique linear operator $F(A)$ by\n\\[\n\\left.F(A)\\right|_{E_\\mu}\n=\\left.\\sum_{k=0}^{d-1}\\frac{F^{(k)}(\\mu)}{k!}(A-\\mu I)^k\\right|_{E_\\mu}.\n\\]\nThe formula is independent of a basis and of increasing the truncation index:\nthe extra nilpotent powers vanish. It depends only on the finite Taylor jets of\n$F$ at the spectrum. If $F$ is holomorphic on an open neighborhood $U$ of the\nspectrum, each eigenvalue admits a radius $r_\\mu>0$ with\n$\\overline{D(\\mu,r_\\mu)}\\subset U$, and Cauchy's formula gives\n\\[\n\\frac{F^{(k)}(\\mu)}{k!}\n=\\frac{1}{2\\pi i}\\oint_{|z-\\mu|=r_","For clarity, the finite-dimensional holomorphic calculus used below has the\nfollowing canonical definition. For $A$ acting on $\\mathbb C^d$, its generalized\neigenspaces $E_\\mu=\\ker(A-\\mu I)^d$ form a direct-sum decomposition over the\neigenvalues. Define the unique linear operator $F(A)$ by\n\\[\n\\left.F(A)\\right|_{E_\\mu}\n=\\left.\\sum_{k=0}^{d-1}\\frac{F^{(k)}(\\mu)}{k!}(A-\\mu I)^k\\right|_{E_\\mu}.\n\\]\nThe formula is independent of a basis and of increasing the truncation index:\nthe extra nilpotent powers vanish. It depends only on the finite Taylor jets of\n$F$ at the spectrum. If $F$ is holomorphic on an open neighborhood $U$ of the\nspectrum, each eigenvalue admits a radius $r_\\mu>0$ with\n$\\overline{D(\\mu,r_\\mu)}\\subset U$, and Cauchy's formula gives\n\\[\n\\frac{F^{(k)}(\\mu)}{k!}\n=\\frac{1}{2\\pi i}\\oint_{|z-\\mu|=r_\\mu}\n  \\frac{F(z)}{(z-\\m","For clarity, the finite-dimensional holomorphic calculus used below has the\nfollowing canonical definition. For $A$ acting on $\\mathbb C^d$, its generalized\neigenspaces $E_\\mu=\\ker(A-\\mu I)^d$ form a direct-sum decomposition over the\neigenvalues. Define the unique linear operator $F(A)$ by\n\\[\n\\left.F(A)\\right|_{E_\\mu}\n=\\left.\\sum_{k=0}^{d-1}\\frac{F^{(k)}(\\mu)}{k!}(A-\\mu I)^k\\right|_{E_\\mu}.\n\\]\nThe formula is independent of a basis and of increasing the truncation index:\nthe extra nilpotent powers vanish. It depends only on the finite Taylor jets of\n$F$ at the spectrum. If $F$ is holomorphic on an open neighborhood $U$ of the\nspectrum, each eigenvalue admits a radius $r_\\mu>0$ with\n$\\overline{D(\\mu,r_\\mu)}\\subset U$, and Cauchy's formula gives\n\\[\n\\frac{F^{(k)}(\\mu)}{k!}\n=\\frac{1}{2\\pi i}\\oint_{|z-\\mu|=r_\\mu}\n  \\frac{F(z)}{(z-\\mu)^{k+1}}\\,dz.\n\\]\nConsequently the same canonical ope","For clarity, the finite-dimensional holomorphic calculus used below has the\nfollowing canonical definition. For $A$ acting on $\\mathbb C^d$, its generalized\neigenspaces $E_\\mu=\\ker(A-\\mu I)^d$ form a direct-sum decomposition over the\neigenvalues. Define the unique linear operator $F(A)$ by\n\\[\n\\left.F(A)\\right|_{E_\\mu}\n=\\left.\\sum_{k=0}^{d-1}\\frac{F^{(k)}(\\mu)}{k!}(A-\\mu I)^k\\right|_{E_\\mu}.\n\\]\nThe formula is independent of a basis and of increasing the truncation index:\nthe extra nilpotent powers vanish. It depends only on the finite Taylor jets of\n$F$ at the spectrum. If $F$ is holomorphic on an open neighborhood $U$ of the\nspectrum, each eigenvalue admits a radius $r_\\mu>0$ with\n$\\overline{D(\\mu,r_\\mu)}\\subset U$, and Cauchy's formula gives\n\\[\n\\frac{F^{(k)}(\\mu)}{k!}\n=\\frac{1}{2\\pi i}\\oint_{|z-\\mu|=r_\\mu}\n  \\frac{F(z)}{(z-\\mu)^{k+1}}\\,dz.\n\\]\nConsequently the same canonical operator is obtained from any such local\nCauchy circles, even when $U$ is disconnected. This is th","For clarity, the finite-dimensional holomorphic calculus used below has the\nfollowing canonical definition. For $A$ acting on $\\mathbb C^d$, its generalized\neigenspaces $E_\\mu=\\ker(A-\\mu I)^d$ form a direct-sum decomposition over the\neigenvalues. Define the unique linear operator $F(A)$ by\n\\[\n\\left.F(A)\\right|_{E_\\mu}\n=\\left.\\sum_{k=0}^{d-1}\\frac{F^{(k)}(\\mu)}{k!}(A-\\mu I)^k\\right|_{E_\\mu}.\n\\]\nThe formula is independent of a basis and of increasing the truncation index:\nthe extra nilpotent powers vanish. It depends only on the finite Taylor jets of\n$F$ at the spectrum. If $F$ is holomorphic on an open neighborhood $U$ of the\nspectrum, each eigenvalue admits a radius $r_\\mu>0$ with\n$\\overline{D(\\mu,r_\\mu)}\\subset U$, and Cauchy's formula gives\n\\[\n\\frac{F^{(k)}(\\mu)}{k!}\n=\\frac{1}{2\\pi i}\\oint_{|z-\\mu|=r_\\mu}\n  \\frac{F(z)}{(z-\\mu)^{k+1}}\\,dz.\n\\]\nConsequently the same canonical operator is obtained from any such local\nCauchy circles, even when $U$ is disconnected. This is the standard finite\nTaylor-jet definition of the holomorphic matrix function.","eigenspaces $E_\\mu=\\ker(A-\\mu I)^d$ form a direct-sum decomposition over the\neigenvalues. Define the unique linear operator $F(A)$ by\n\\[\n\\left.F(A)\\right|_{E_\\mu}\n=\\left.\\sum_{k=0}^{d-1}\\frac{F^{(k)}(\\mu)}{k!}(A-\\mu I)^k\\right|_{E_\\mu}.\n\\]\nThe formula is independent of a basis and of increasing the truncation index:\nthe extra nilpotent powers vanish. It depends only on the finite Taylor jets of\n$F$ at the spectrum. If $F$ is holomorphic on an open neighborhood $U$ of the\nspectrum, each eigenvalue admits a radius $r_\\mu>0$ with\n$\\overline{D(\\mu,r_\\mu)}\\subset U$, and Cauchy's formula gives\n\\[\n\\frac{F^{(k)}(\\mu)}{k!}\n=\\frac{1}{2\\pi i}\\oint_{|z-\\mu|=r_\\mu}\n  \\frac{F(z)}{(z-\\mu)^{k+1}}\\,dz.\n\\]\nConsequently the same canonical operator is obtained from any such local\nCauchy circles, even when $U$ is disconnected. This is the standard finite\nTaylor-jet definition of the holomorphic matrix function.","(A-\\mu I)^d$ form a direct-sum decomposition over the\neigenvalues. Define the unique linear operator $F(A)$ by\n\\[\n\\left.F(A)\\right|_{E_\\mu}\n=\\left.\\sum_{k=0}^{d-1}\\frac{F^{(k)}(\\mu)}{k!}(A-\\mu I)^k\\right|_{E_\\mu}.\n\\]\nThe formula is independent of a basis and of increasing the truncation index:\nthe extra nilpotent powers vanish. It depends only on the finite Taylor jets of\n$F$ at the spectrum. If $F$ is holomorphic on an open neighborhood $U$ of the\nspectrum, each eigenvalue admits a radius $r_\\mu>0$ with\n$\\overline{D(\\mu,r_\\mu)}\\subset U$, and Cauchy's formula gives\n\\[\n\\frac{F^{(k)}(\\mu)}{k!}\n=\\frac{1}{2\\pi i}\\oint_{|z-\\mu|=r_\\mu}\n  \\frac{F(z)}{(z-\\mu)^{k+1}}\\,dz.\n\\]\nConsequently the same canonical operator is obtained from any such local\nCauchy circles, even when $U$ is disconnected. This is the standard finite\nTaylor-jet definition of the holomorphic matrix function.","n over the\neigenvalues. Define the unique linear operator $F(A)$ by\n\\[\n\\left.F(A)\\right|_{E_\\mu}\n=\\left.\\sum_{k=0}^{d-1}\\frac{F^{(k)}(\\mu)}{k!}(A-\\mu I)^k\\right|_{E_\\mu}.\n\\]\nThe formula is independent of a basis and of increasing the truncation index:\nthe extra nilpotent powers vanish. It depends only on the finite Taylor jets of\n$F$ at the spectrum. If $F$ is holomorphic on an open neighborhood $U$ of the\nspectrum, each eigenvalue admits a radius $r_\\mu>0$ with\n$\\overline{D(\\mu,r_\\mu)}\\subset U$, and Cauchy's formula gives\n\\[\n\\frac{F^{(k)}(\\mu)}{k!}\n=\\frac{1}{2\\pi i}\\oint_{|z-\\mu|=r_\\mu}\n  \\frac{F(z)}{(z-\\mu)^{k+1}}\\,dz.\n\\]\nConsequently the same canonical operator is obtained from any such local\nCauchy circles, even when $U$ is disconnected. This is the standard finite\nTaylor-jet definition of the holomorphic matrix function.","ator $F(A)$ by\n\\[\n\\left.F(A)\\right|_{E_\\mu}\n=\\left.\\sum_{k=0}^{d-1}\\frac{F^{(k)}(\\mu)}{k!}(A-\\mu I)^k\\right|_{E_\\mu}.\n\\]\nThe formula is independent of a basis and of increasing the truncation index:\nthe extra nilpotent powers vanish. It depends only on the finite Taylor jets of\n$F$ at the spectrum. If $F$ is holomorphic on an open neighborhood $U$ of the\nspectrum, each eigenvalue admits a radius $r_\\mu>0$ with\n$\\overline{D(\\mu,r_\\mu)}\\subset U$, and Cauchy's formula gives\n\\[\n\\frac{F^{(k)}(\\mu)}{k!}\n=\\frac{1}{2\\pi i}\\oint_{|z-\\mu|=r_\\mu}\n  \\frac{F(z)}{(z-\\mu)^{k+1}}\\,dz.\n\\]\nConsequently the same canonical operator is obtained from any such local\nCauchy circles, even when $U$ is disconnected. This is the standard finite\nTaylor-jet definition of the holomorphic matrix function.","\\[\n\\left.F(A)\\right|_{E_\\mu}\n=\\left.\\sum_{k=0}^{d-1}\\frac{F^{(k)}(\\mu)}{k!}(A-\\mu I)^k\\right|_{E_\\mu}.\n\\]\nThe formula is independent of a basis and of increasing the truncation index:\nthe extra nilpotent powers vanish. It depends only on the finite Taylor jets of\n$F$ at the spectrum. If $F$ is holomorphic on an open neighborhood $U$ of the\nspectrum, each eigenvalue admits a radius $r_\\mu>0$ with\n$\\overline{D(\\mu,r_\\mu)}\\subset U$, and Cauchy's formula gives\n\\[\n\\frac{F^{(k)}(\\mu)}{k!}\n=\\frac{1}{2\\pi i}\\oint_{|z-\\mu|=r_\\mu}\n  \\frac{F(z)}{(z-\\mu)^{k+1}}\\,dz.\n\\]\nConsequently the same canonical operator is obtained from any such local\nCauchy circles, even when $U$ is disconnected. This is the standard finite\nTaylor-jet definition of the holomorphic matrix function.","}(\\mu)}{k!}(A-\\mu I)^k\\right|_{E_\\mu}.\n\\]\nThe formula is independent of a basis and of increasing the truncation index:\nthe extra nilpotent powers vanish. It depends only on the finite Taylor jets of\n$F$ at the spectrum. If $F$ is holomorphic on an open neighborhood $U$ of the\nspectrum, each eigenvalue admits a radius $r_\\mu>0$ with\n$\\overline{D(\\mu,r_\\mu)}\\subset U$, and Cauchy's formula gives\n\\[\n\\frac{F^{(k)}(\\mu)}{k!}\n=\\frac{1}{2\\pi i}\\oint_{|z-\\mu|=r_\\mu}\n  \\frac{F(z)}{(z-\\mu)^{k+1}}\\,dz.\n\\]\nConsequently the same canonical operator is obtained from any such local\nCauchy circles, even when $U$ is disconnected. This is the standard finite\nTaylor-jet definition of the holomorphic matrix function.","F$ at the spectrum. If $F$ is holomorphic on an open neighborhood $U$ of the\nspectrum, each eigenvalue admits a radius $r_\\mu>0$ with\n$\\overline{D(\\mu,r_\\mu)}\\subset U$, and Cauchy's formula gives\n\\[\n\\frac{F^{(k)}(\\mu)}{k!}\n=\\frac{1}{2\\pi i}\\oint_{|z-\\mu|=r_\\mu}\n  \\frac{F(z)}{(z-\\mu)^{k+1}}\\,dz.\n\\]\nConsequently the same canonical operator is obtained from any such local\nCauchy circles, even when $U$ is disconnected. This is the standard finite\nTaylor-jet definition of the holomorphic matrix function.","\\begin{theorem}[Naturality under exact intertwining]\\label{core:record:044}\nFor finite complex matrices, let $F$ be holomorphic on a neighborhood of the union of the spectra of $A_f$ and $A_c$. Real matrices are treated by complexification. If\n\\[\nCA_f=A_cC,\n\\]\nthen\n\\[\n\\boxed{CF(A_f)=F(A_c)C.}\n\\]\nIn particular, exact generator refinement propagates simultaneously to semigroups, resolvents, integer powers, and analytic transfer functions.\n\\end{theorem}","\\begin{proof}\nThe intertwining identity implies\n$C(A_f-\\mu I)^k=(A_c-\\mu I)^kC$ for every $k\\ge0$.\nIt therefore sends the generalized $\\mu$-eigenspace of $A_f$ into that of $A_c$.\nUse a common truncation index at least as large as both dimensions in the\ncanonical formula above. Multiplying its finite sum by $C$ now gives the desired\nidentity on every generalized eigenspace. The direct-sum decomposition gives\nit on the full space. The argument covers nilpotent Jordan parts and uses only\nlocal spectral Taylor jets; it does not require one Taylor series to converge\non both spectra. The zero-dimensional case is immediate.\n\\end{proof}","\\midrule\nPacket-block semigroup composition residual & $2.554\\times10^{-15}$","Operator reconstruction residual & $0$","Random-walk mass error & $0$","Exact Markov generator-lumping defect & $5.551\\times10^{-17}$","Exact finite-horizon lumping error & $2.220\\times10^{-16}$","Perturbed ten-step lumping error & $6.523\\times10^{-3}$","Linear defect bound for that example & $2.000\\times10^{-1}$","Bayesian composition residual & $5.551\\times10^{-17}$","Bayesian packet reconstruction residual & $0$","Bayesian refinement commutation residual & $1.110\\times10^{-16}$","Bayesian defect-factorization residual & $7.286\\times10^{-17}$","Bayesian flow generator residual & $1.959\\times10^{-11}$","Conditional-expectation tower residual & $0$","Hermite Gram residual & $9.992\\times10^{-15}$","Fractional semigroup residual & $2.082\\times10^{-17}$","Maximum square-root subordination error & $1.665\\times10^{-16}$","Exact fractional-naturality residual & $0$","Fractional-power defect / gap certificate & $0.51059 / 0.95079$","Fractional-semigroup defect / certificate & $0.12969 / 0.44480$","Observed low-mode PDE convergence order & $1.999854$","Feynman--Kac enumeration residual & $1.665\\times10^{-16}$","Feshbach projected-resolvent residual & $5.551\\times10^{-17}$","Memory Laplace-transform residual & $6.939\\times10^{-18}$","Duhamel refinement identity residual & $2.108\\times10^{-16}$","Survival monotonicity violation & $0$","Hidden-Markov normalization error & $1.110\\times10^{-16}$","\\begin{certificate}[Test suite]\\label{core:record:046}\nThe September 6 release records the following automated-test result:\n\\[\n\\boxed{54/54\\text{ automated tests}.}\n\\]\nThe tests cover projector reconstruction, packet-block semigroup composition, exact and approximate Markov lumping, Bayesian composition and flow, the exact Bayesian refinement criterion, information-loss factorization, likelihood-profile partitions, conditional-expectation towers, Hermite orthogonality, fractional semigroups, square-root subordination, fractional intertwining and semigroup-defect bounds, discrete Green kernels, PDE convergence, Feynman--Kac enumeration, Schur--Feshbach elimination, memory kernels, refinement defects, hidden Markov filtering, and survival curves.\n\\end{certificate}","\\begin{conjecture}[Uniform refinement control]\\label{core:record:048}\nFor a nested family of physically or analytically meaningful discretizations, there exist norms and constants independent of resolution such that\n\\[\n\\norm{C_\\ell e^{tA_{\\ell+1}}-e^{tA_\\ell}C_\\ell}\n\\le\nM\\,t\\,\\norm{C_\\ell A_{\\ell+1}-A_\\ell C_\\ell}\n\\]\non every bounded time interval. Summable generator defects would then produce a coherent inverse-limit evolution.\n\\end{conjecture}","The accompanying Python package implements the formulas and finite models used in the paper.\n\\begin{center}\n\\texttt{projector\\_transfer\\_semigroups}\n\\end{center}\nThe computational organization mirrors the paper:\n\\[\n\\begin{array}{c|l}\n\\texttt{packets.py} & \\text{projectors, blocks, reconstruction, block composition}\\\\\n\\texttt{semigroup.py} & \\text{matrix semigroups, defects, Feshbach operators, memory}\\\\\n\\texttt{markov.py} & \\text{random walks, exact lumping, survival}\\\\\n\\texttt{bayes.py} & \\text{normalized updates, packet evidence, conjugacy, flows, HMM}\\\\\n\\texttt{conditional.py} & \\text{conditional expectation and Gaussian chaos}\\\\\n\\texttt{fractional.py} & \\text{spectral powers, subordination, and fractional refinement certificates}\\\\\n\\texttt{pde.py} & \\text{Dirichlet Laplacians, Green kernels, eigenvalue error}\\\\\n\\texttt{feynman\\_kac.py} & \\text{weighted transfer and exhaustive path enumeration}.\n\\end{array}\n\\]","The central structural identity is\n\\[\n\\boxed{\n\\text{generator defect}\n\\longrightarrow\n\\text{Duhamel semigroup defect}\n\\longrightarrow\n\\text{controlled refinement error}.\n}\n\\]\nTogether with exact packet reconstruction, this makes the semigroup paper the natural bridge from the foundational calculus to the quantum, computational, and arithmetic realizations that follow.","\\paragraph{General nonnormal memory (13.2).}\nThe actual damped exponential is decomposed into finitely many terms of the form\n$t^k e^{(\\mu-z)t}$. Integrability and convergence to zero follow from\n$\\operatorname{Re}\\mu<\\operatorname{Re}z$. The formal proof uses generalized\neigenspaces of left multiplication on the matrix algebra, so nilpotent Jordan\nparts are included. Both inverse identities follow by integrating the two\nexponential derivative formulas. A continuous linear sandwich map gives the\nliteral rectangular matrix memory integral and the actual inverse. No normality\nor externally supplied growth bound is required.","The first part of the article treats bounded semigroups on Banach spaces. A finite packet system resolves every operator into input and output blocks. The semigroup law becomes block convolution. Splitting the generator into its packet diagonal and transfer part gives a convergent Dyson expansion over directed packet paths. For a coarse map \\(C\\), the generator defect\n\\[\n\\Delta=CA_f-A_cC\n\\]\nproduces the entire semigroup defect through a Duhamel integral. The same defect controls resolvents and holomorphic functions of the generators.","For a positive operator \\(L\\), spectral packets define \\(L^\\alpha\\) and \\(\\e^{-tL^\\alpha}\\). Exact intertwining passes to every fractional power. When both fine and coarse operators have a common positive lower bound \\(m\\), the Balakrishnan formula gives the quantitative estimate\n\\[\n\\norm{CL_f^\\alpha-L_c^\\alpha C}\n\\leq\n\\alpha m^{\\alpha-1}\n\\norm{CL_f-L_cC},\n\\qquad 0<\\alpha<1.\n\\]\nA Duhamel argument then controls the fractional semigroup defect. This closes a finite version of fractional naturality under explicit spectral hypotheses.","oth Poisson solutions. Finite state Feynman Kac transfer is proved in discrete and continuous time. Schur Feshbach elimination yields the exact projected resolvent, the retained memory equation, and the Laplace correspondence between memory and self energy. Every stable finite hidden system gives a polynomial exponential memory expansion. Primary blocks can be ranked optimally for the certified \\(L^1\\) bound, and a Volterra estimate carries kernel error to the retained solution. Survival, event density, moments, and competing risks are then read directly from a transient semigroup.","\\midrule\nOperator evolution & $\\e^{tA}$ & generator intertwining","Markov chain & $K^n$ or $\\e^{tQ}$ & equality of aggregated transition masses","Conditional expectation & $P_{\\mathcal Q}$ & nesting of the underlying partitions","Bayesian update & $\\Bcal_\\lambda$ & likelihood constant inside each coarse packet","Fractional diffusion & $\\e^{-tL^\\alpha}$ & spectral or resolvent intertwining","Let \\(X\\) be a complex Banach space. A finite packet system is a family of bounded idempotents\n\\[\n\\Pcal=(P_\\alpha)_{\\alpha\\in I}\n\\]\nsuch that\n\\[\nP_\\alpha P_\\beta=\\delta_{\\alpha\\beta}P_\\alpha,\n\\qquad\n\\sum_{\\alpha\\in I}P_\\alpha=I_X.\n\\]\nNo orthogonality is needed in this section.","\\begin{definition}[Packet resolved transfer system]\\label{extended:record:001}\nA packet resolved transfer system is a triple\n\\[\n(X,A,\\Pcal)\n\\]\nconsisting of a Banach state space, a bounded generator, and a finite packet system. A marked transfer system also includes an initial state or source $s\\in X$. The topology and all limiting operations are supplied by the ambient Banach space structure.\n\\end{definition}","\\begin{definition}[Packet blocks]\\label{extended:record:002}\nFor \\(B\\in\\mathcal B(X)\\), define\n\\[\nB_{\\beta\\alpha}=P_\\beta BP_\\alpha.\n\\]\nIf \\(A\\in\\mathcal B(X)\\) and \\(T(t)=\\e^{tA}\\), write\n\\[\nT_{\\beta\\alpha}(t)=P_\\beta T(t)P_\\alpha.\n\\]\n\\end{definition}","\\begin{theorem}[Semigroup and packet convolution]\\label{extended:record:003}\\label{extended:thm:packet_convolution}\nFor every bounded operator \\(A\\), the family \\(T(t)=\\e^{tA}\\) is a uniformly continuous group and\n\\[\nT(s+t)=T(s)T(t).\n\\]\nFor every \\(\\alpha,\\beta\\in I\\),\n\\[\nT_{\\beta\\alpha}(s+t)\n=\n\\sum_{\\gamma\\in I}\nT_{\\beta\\gamma}(s)T_{\\gamma\\alpha}(t).\n\\]\n\\end{theorem}","\\begin{proof}\nThe exponential series converges in operator norm. Since every power of \\(A\\) commutes with every other power, the Cauchy product gives the group law. Inserting \\(I_X=\\sum_\\gamma P_\\gamma\\) between the two factors gives the block formula.\n\\end{proof}","\\begin{proposition}[Block differential system]\\label{extended:record:004}\\label{extended:prop:block_differential}\nThe packet blocks are differentiable in operator norm and satisfy\n\\[\n\\frac{\\dd}{\\dd t}T_{\\beta\\alpha}(t)\n=\n\\sum_{\\gamma\\in I}\nA_{\\beta\\gamma}T_{\\gamma\\alpha}(t)\n=\n\\sum_{\\gamma\\in I}\nT_{\\beta\\gamma}(t)A_{\\gamma\\alpha}.\n\\]\nTheir initial values are\n\\[\nT_{\\beta\\alpha}(0)=\\delta_{\\beta\\alpha}P_\\alpha.\n\\]\n\\end{proposition}","\\begin{proof}\nDifferentiate \\(T(t)\\) in operator norm and insert the packet resolution between \\(A\\) and \\(T(t)\\), then between \\(T(t)\\) and \\(A\\).\n\\end{proof}","\\begin{theorem}[Laplace resolvent and packet transform]\\label{extended:record:005}\\label{extended:thm:laplace_packet_resolvent}\nLet \\(A\\in\\mathcal B(X)\\), let \\(T(t)=\\e^{tA}\\), and let\n\\[\n\\operatorname{Re}z>\\norm A.\n\\]\nThen the Bochner integral converges in operator norm and\n\\[\n(zI-A)^{-1}\n=\n\\int_0^\\infty \\e^{-zt}T(t)\\dd t.\n\\]\nFor every pair of packets,\n\\[\nP_\\beta(zI-A)^{-1}P_\\alpha\n=\n\\int_0^\\infty\n\\e^{-zt}T_{\\beta\\alpha}(t)\\dd t.\n\\]\n\\end{theorem}","\\begin{proof}\nThe estimate \\(\\norm{T(t)}\\leq\\e^{t\\norm A}\\) gives absolute convergence of the operator valued integral. Let\n\\[\nR(z)=\\int_0^\\infty\\e^{-zt}T(t)\\dd t.\n\\]\nSince \\(T'(t)=AT(t)=T(t)A\\), integration by parts gives\n\\[\n(zI-A)R(z)\n=\n-\\int_0^\\infty\n\\frac{\\dd}{\\dd t}\n\\bigl(\\e^{-zt}T(t)\\bigr)\n\\dd t\n=I.\n\\]\nThe same calculation on the other side gives\n\\(\nR(z)(zI-A)=I\n\\).\nMultiplication by \\(P_\\beta\\) and \\(P_\\alpha\\) may be passed under the norm convergent integral, which gives the packet formula.\n\\end{proof}","f}\nThe estimate \\(\\norm{T(t)}\\leq\\e^{t\\norm A}\\) gives absolute convergence of the operator valued integral. Let\n\\[\nR(z)=\\int_0^\\infty\\e^{-zt}T(t)\\dd t.\n\\]\nSince \\(T'(t)=AT(t)=T(t)A\\), integration by parts gives\n\\[\n(zI-A)R(z)\n=\n-\\int_0^\\infty\n\\frac{\\dd}{\\dd t}\n\\bigl(\\e^{-zt}T(t)\\bigr)\n\\dd t\n=I.\n\\]\nThe same calculation on the other side gives\n\\(\nR(z)(zI-A)=I\n\\).\nMultiplication by \\(P_\\beta\\) and \\(P_\\alpha\\) may be passed under the norm convergent integral, which gives the packet formula.\n\\end{proof}","\\begin{corollary}[Resolvent expansion over packet paths]\\label{extended:record:006}\\label{extended:cor:resolvent_packet_paths}\nWith \\(A=D+N\\) as in the directed path decomposition, let\n\\[\nR_D(z)=(zI-D)^{-1}.\n\\]\nIf\n\\[\n\\norm{R_D(z)N}<1,\n\\]\nthen\n\\[\n(zI-A)^{-1}\n=\n\\sum_{m=0}^\\infty\n\\bigl(R_D(z)N\\bigr)^mR_D(z).\n\\]\nAfter multiplication by \\(P_\\beta\\) and \\(P_\\alpha\\), the term of order \\(m\\) is a sum over directed packet paths of length \\(m\\) from \\(\\alpha\\) to \\(\\beta\\).\n\\end{corollary}","n over packet paths]\\label{extended:record:006}\\label{extended:cor:resolvent_packet_paths}\nWith \\(A=D+N\\) as in the directed path decomposition, let\n\\[\nR_D(z)=(zI-D)^{-1}.\n\\]\nIf\n\\[\n\\norm{R_D(z)N}<1,\n\\]\nthen\n\\[\n(zI-A)^{-1}\n=\n\\sum_{m=0}^\\infty\n\\bigl(R_D(z)N\\bigr)^mR_D(z).\n\\]\nAfter multiplication by \\(P_\\beta\\) and \\(P_\\alpha\\), the term of order \\(m\\) is a sum over directed packet paths of length \\(m\\) from \\(\\alpha\\) to \\(\\beta\\).\n\\end{corollary}","et paths]\\label{extended:record:006}\\label{extended:cor:resolvent_packet_paths}\nWith \\(A=D+N\\) as in the directed path decomposition, let\n\\[\nR_D(z)=(zI-D)^{-1}.\n\\]\nIf\n\\[\n\\norm{R_D(z)N}<1,\n\\]\nthen\n\\[\n(zI-A)^{-1}\n=\n\\sum_{m=0}^\\infty\n\\bigl(R_D(z)N\\bigr)^mR_D(z).\n\\]\nAfter multiplication by \\(P_\\beta\\) and \\(P_\\alpha\\), the term of order \\(m\\) is a sum over directed packet paths of length \\(m\\) from \\(\\alpha\\) to \\(\\beta\\).\n\\end{corollary}","l{extended:record:006}\\label{extended:cor:resolvent_packet_paths}\nWith \\(A=D+N\\) as in the directed path decomposition, let\n\\[\nR_D(z)=(zI-D)^{-1}.\n\\]\nIf\n\\[\n\\norm{R_D(z)N}<1,\n\\]\nthen\n\\[\n(zI-A)^{-1}\n=\n\\sum_{m=0}^\\infty\n\\bigl(R_D(z)N\\bigr)^mR_D(z).\n\\]\nAfter multiplication by \\(P_\\beta\\) and \\(P_\\alpha\\), the term of order \\(m\\) is a sum over directed packet paths of length \\(m\\) from \\(\\alpha\\) to \\(\\beta\\).\n\\end{corollary}","\\begin{proof}\nFactor\n\\[\nzI-A=(zI-D)(I-R_D(z)N)\n\\]\nand expand the second inverse by its norm convergent Neumann series. The diagonal resolvent preserves each packet and every factor \\(N\\) changes the packet along an active transfer edge.\n\\end{proof}","\\begin{theorem}[Reducing packet criterion]\\label{extended:record:007}\\label{extended:thm:reducing_packets}\nThe following conditions are equivalent.\n\\begin{enumerate}\n\\item \\(AP_\\alpha=P_\\alpha A\\) for every \\(\\alpha\\).\n\\item \\(A_{\\beta\\alpha}=0\\) whenever \\(\\beta\\neq\\alpha\\).\n\\item \\(T_{\\beta\\alpha}(t)=0\\) whenever \\(\\beta\\neq\\alpha\\) and \\(t\\in\\RR\\).\n\\item Every packet space \\(P_\\alpha X\\) is invariant under \\(T(t)\\) for every \\(t\\).\n\\end{enumerate}\nWhen these conditions hold,\n\\[\nT(t)|_{P_\\alpha X}\n=\n\\e^{tA_{\\alpha\\alpha}}.\n\\]\n\\end{theorem}","[Reducing packet criterion]\\label{extended:record:007}\\label{extended:thm:reducing_packets}\nThe following conditions are equivalent.\n\\begin{enumerate}\n\\item \\(AP_\\alpha=P_\\alpha A\\) for every \\(\\alpha\\).\n\\item \\(A_{\\beta\\alpha}=0\\) whenever \\(\\beta\\neq\\alpha\\).\n\\item \\(T_{\\beta\\alpha}(t)=0\\) whenever \\(\\beta\\neq\\alpha\\) and \\(t\\in\\RR\\).\n\\item Every packet space \\(P_\\alpha X\\) is invariant under \\(T(t)\\) for every \\(t\\).\n\\end{enumerate}\nWhen these conditions hold,\n\\[\nT(t)|_{P_\\alpha X}\n=\n\\e^{tA_{\\alpha\\alpha}}.\n\\]\n\\end{theorem}","iterion]\\label{extended:record:007}\\label{extended:thm:reducing_packets}\nThe following conditions are equivalent.\n\\begin{enumerate}\n\\item \\(AP_\\alpha=P_\\alpha A\\) for every \\(\\alpha\\).\n\\item \\(A_{\\beta\\alpha}=0\\) whenever \\(\\beta\\neq\\alpha\\).\n\\item \\(T_{\\beta\\alpha}(t)=0\\) whenever \\(\\beta\\neq\\alpha\\) and \\(t\\in\\RR\\).\n\\item Every packet space \\(P_\\alpha X\\) is invariant under \\(T(t)\\) for every \\(t\\).\n\\end{enumerate}\nWhen these conditions hold,\n\\[\nT(t)|_{P_\\alpha X}\n=\n\\e^{tA_{\\alpha\\alpha}}.\n\\]\n\\end{theorem}","thm:reducing_packets}\nThe following conditions are equivalent.\n\\begin{enumerate}\n\\item \\(AP_\\alpha=P_\\alpha A\\) for every \\(\\alpha\\).\n\\item \\(A_{\\beta\\alpha}=0\\) whenever \\(\\beta\\neq\\alpha\\).\n\\item \\(T_{\\beta\\alpha}(t)=0\\) whenever \\(\\beta\\neq\\alpha\\) and \\(t\\in\\RR\\).\n\\item Every packet space \\(P_\\alpha X\\) is invariant under \\(T(t)\\) for every \\(t\\).\n\\end{enumerate}\nWhen these conditions hold,\n\\[\nT(t)|_{P_\\alpha X}\n=\n\\e^{tA_{\\alpha\\alpha}}.\n\\]\n\\end{theorem}","\\begin{proof}\nThe first two conditions are equivalent because the packet resolution gives\n\\[\nAP_\\alpha=\\sum_\\beta A_{\\beta\\alpha},\n\\qquad\nP_\\alpha A=\\sum_\\gamma A_{\\alpha\\gamma}.\n\\]\nA block diagonal generator has block diagonal powers, so the second condition implies the third. Differentiating the off diagonal blocks at zero proves the converse. The equivalence with invariance follows from the definition of the blocks. The final identity follows by restricting the exponential series.\n\\end{proof}","Set\n\\[\nD=\\sum_{\\alpha\\in I}P_\\alpha AP_\\alpha,\n\\qquad\nN=A-D.\n\\]\nThe operator \\(D\\) preserves every packet and \\(N\\) contains only transfers between distinct packets.","\\begin{theorem}[Dyson expansion over packet paths]\\label{extended:record:008}\\label{extended:thm:dyson_paths}\nFor \\(t\\geq0\\),\n\\[\n\\begin{aligned}\n\\e^{tA}\n={}&\\e^{tD}\n+\n\\sum_{m=1}^{\\infty}\n\\int_{0<s_1<\\cdots<s_m<t}\n\\e^{(t-s_m)D}N\\e^{(s_m-s_{m-1})D}\n\\cdots\nN\\e^{s_1D}\n\\dd s_1\\cdots\\dd s_m.\n\\end{aligned}\n\\]\nThe series converges absolutely in operator norm. After multiplication by \\(P_\\beta\\) and \\(P_\\alpha\\), the term of order \\(m\\) is a sum over directed packet paths of length \\(m\\) from \\(\\alpha\\) to \\(\\beta\\).\n\\end{theorem}","ion over packet paths]\\label{extended:record:008}\\label{extended:thm:dyson_paths}\nFor \\(t\\geq0\\),\n\\[\n\\begin{aligned}\n\\e^{tA}\n={}&\\e^{tD}\n+\n\\sum_{m=1}^{\\infty}\n\\int_{0<s_1<\\cdots<s_m<t}\n\\e^{(t-s_m)D}N\\e^{(s_m-s_{m-1})D}\n\\cdots\nN\\e^{s_1D}\n\\dd s_1\\cdots\\dd s_m.\n\\end{aligned}\n\\]\nThe series converges absolutely in operator norm. After multiplication by \\(P_\\beta\\) and \\(P_\\alpha\\), the term of order \\(m\\) is a sum over directed packet paths of length \\(m\\) from \\(\\alpha\\) to \\(\\beta\\).\n\\end{theorem}","bel{extended:thm:dyson_paths}\nFor \\(t\\geq0\\),\n\\[\n\\begin{aligned}\n\\e^{tA}\n={}&\\e^{tD}\n+\n\\sum_{m=1}^{\\infty}\n\\int_{0<s_1<\\cdots<s_m<t}\n\\e^{(t-s_m)D}N\\e^{(s_m-s_{m-1})D}\n\\cdots\nN\\e^{s_1D}\n\\dd s_1\\cdots\\dd s_m.\n\\end{aligned}\n\\]\nThe series converges absolutely in operator norm. After multiplication by \\(P_\\beta\\) and \\(P_\\alpha\\), the term of order \\(m\\) is a sum over directed packet paths of length \\(m\\) from \\(\\alpha\\) to \\(\\beta\\).\n\\end{theorem}","d:thm:dyson_paths}\nFor \\(t\\geq0\\),\n\\[\n\\begin{aligned}\n\\e^{tA}\n={}&\\e^{tD}\n+\n\\sum_{m=1}^{\\infty}\n\\int_{0<s_1<\\cdots<s_m<t}\n\\e^{(t-s_m)D}N\\e^{(s_m-s_{m-1})D}\n\\cdots\nN\\e^{s_1D}\n\\dd s_1\\cdots\\dd s_m.\n\\end{aligned}\n\\]\nThe series converges absolutely in operator norm. After multiplication by \\(P_\\beta\\) and \\(P_\\alpha\\), the term of order \\(m\\) is a sum over directed packet paths of length \\(m\\) from \\(\\alpha\\) to \\(\\beta\\).\n\\end{theorem}","ths}\nFor \\(t\\geq0\\),\n\\[\n\\begin{aligned}\n\\e^{tA}\n={}&\\e^{tD}\n+\n\\sum_{m=1}^{\\infty}\n\\int_{0<s_1<\\cdots<s_m<t}\n\\e^{(t-s_m)D}N\\e^{(s_m-s_{m-1})D}\n\\cdots\nN\\e^{s_1D}\n\\dd s_1\\cdots\\dd s_m.\n\\end{aligned}\n\\]\nThe series converges absolutely in operator norm. After multiplication by \\(P_\\beta\\) and \\(P_\\alpha\\), the term of order \\(m\\) is a sum over directed packet paths of length \\(m\\) from \\(\\alpha\\) to \\(\\beta\\).\n\\end{theorem}","\\begin{proof}\nWrite the variation of constants identity\n\\[\n\\e^{tA}\n=\n\\e^{tD}\n+\n\\int_0^t\\e^{(t-s)D}N\\e^{sA}\\dd s.\n\\]\nIterating the identity gives the displayed series. The norm of the term of order \\(m\\) is at most\n\\[\n\\e^{t\\norm D}\\frac{(t\\norm N)^m}{m!}.\n\\]\nThe majorant is summable. Inserting the packet resolution around every factor \\(N\\) leaves only products whose consecutive packet labels are joined by nonzero transfer blocks.\n\\end{proof}","\\begin{corollary}[Termination on an acyclic transfer graph]\\label{extended:record:009}\\label{extended:cor:acyclic_termination}\nSuppose the directed graph with an edge \\(\\alpha\\to\\beta\\) whenever \\(P_\\beta NP_\\alpha\\neq0\\) is acyclic. If its longest directed path has length \\(q\\), then every Dyson term of order greater than \\(q\\) vanishes.\n\\end{corollary}","\\begin{proof}\nEvery nonzero term of order \\(m\\) contains a directed transfer path of length \\(m\\). An acyclic graph has no path longer than \\(q\\).\n\\end{proof}","Let \\(X_f\\) and \\(X_c\\) be complex Banach spaces. Let \\(A_f\\in\\mathcal B(X_f)\\), \\(A_c\\in\\mathcal B(X_c)\\), and \\(C\\in\\mathcal B(X_f,X_c)\\). Define\n\\[\n\\Delta=CA_f-A_cC\n\\]\nand\n\\[\n\\mathfrak D(t)=C\\e^{tA_f}-\\e^{tA_c}C.\n\\]","\\begin{theorem}[Duhamel defect identity]\\label{extended:record:010}\\label{extended:thm:duhamel_defect}\nFor every \\(t\\geq0\\),\n\\[\n\\mathfrak D(t)\n=\n\\int_0^t\n\\e^{(t-s)A_c}\\Delta\\e^{sA_f}\n\\dd s.\n\\]\nIf\n\\[\n\\norm{\\e^{tA_f}}\\leq M_f\\e^{\\omega_f t},\n\\qquad\n\\norm{\\e^{tA_c}}\\leq M_c\\e^{\\omega_c t},\n\\]\nthen\n\\[\n\\norm{\\mathfrak D(t)}\n\\leq\nM_fM_c\\norm\\Delta\n\\int_0^t\n\\e^{\\omega_c(t-s)+\\omega_fs}\n\\dd s.\n\\]\nFor contraction semigroups,\n\\[\n\\norm{\\mathfrak D(t)}\\leq t\\norm\\Delta.\n\\]\n\\end{theorem}","l defect identity]\\label{extended:record:010}\\label{extended:thm:duhamel_defect}\nFor every \\(t\\geq0\\),\n\\[\n\\mathfrak D(t)\n=\n\\int_0^t\n\\e^{(t-s)A_c}\\Delta\\e^{sA_f}\n\\dd s.\n\\]\nIf\n\\[\n\\norm{\\e^{tA_f}}\\leq M_f\\e^{\\omega_f t},\n\\qquad\n\\norm{\\e^{tA_c}}\\leq M_c\\e^{\\omega_c t},\n\\]\nthen\n\\[\n\\norm{\\mathfrak D(t)}\n\\leq\nM_fM_c\\norm\\Delta\n\\int_0^t\n\\e^{\\omega_c(t-s)+\\omega_fs}\n\\dd s.\n\\]\nFor contraction semigroups,\n\\[\n\\norm{\\mathfrak D(t)}\\leq t\\norm\\Delta.\n\\]\n\\end{theorem}","\\begin{proof}\nFor fixed \\(t\\), set\n\\[\nF(s)=\\e^{(t-s)A_c}C\\e^{sA_f}.\n\\]\nThen\n\\[\nF'(s)=\\e^{(t-s)A_c}\\Delta\\e^{sA_f}.\n\\]\nIntegrating from zero to \\(t\\) gives the identity. The estimates follow by taking norms under the integral.\n\\end{proof}","\\begin{proposition}[Infinitesimal recovery of the defect]\\label{extended:record:011}\\label{extended:prop:infinitesimal_defect}\nThe generator defect is the first order semigroup defect:\n\\[\n\\lim_{t\\to0}\n\\frac{\\mathfrak D(t)}{t}\n=\n\\Delta\n\\]\nin operator norm. Consequently,\n\\[\nC\\e^{tA_f}=\\e^{tA_c}C\n\\quad\\hbox{for every }t\n\\]\nif and only if \\(CA_f=A_cC\\).\n\\end{proposition}","\\begin{proof}\nThe operator norm expansions\n\\[\n\\e^{tA_f}=I+tA_f+o(t),\n\\qquad\n\\e^{tA_c}=I+tA_c+o(t)\n\\]\ngive the limit. The equivalence follows from this limit and \\cref{extended:thm:duhamel_defect}.\n\\end{proof}","\\begin{theorem}[Resolvent defect identity]\\label{extended:record:012}\\label{extended:thm:resolvent_defect}\nLet \\(z\\) belong to the resolvent sets of \\(A_f\\) and \\(A_c\\). With\n\\[\nR_f(z)=(zI-A_f)^{-1},\n\\qquad\nR_c(z)=(zI-A_c)^{-1},\n\\]\none has\n\\[\nCR_f(z)-R_c(z)C\n=\nR_c(z)\\Delta R_f(z).\n\\]\n\\end{theorem}","\\begin{proof}\nMultiply the proposed identity on the left by \\(zI-A_c\\) and on the right by \\(zI-A_f\\). Both sides reduce to\n\\[\n(zI-A_c)C-C(zI-A_f)=\\Delta.\n\\]\n\\end{proof}","\\begin{theorem}[Holomorphic defect formula]\\label{extended:record:013}\\label{extended:thm:holomorphic_defect}\nLet \\(\\Gamma\\) be the positively oriented rectifiable boundary of a finite union of pairwise disjoint bounded Jordan domains. Assume that both spectra lie inside this union and that \\(\\Gamma\\) lies in the common resolvent set. Thus its winding number is one about every spectral point. If \\(g\\) is holomorphic on a neighborhood of the closure of the union, then\n\\[\nCg(A_f)-g(A_c)C\n=\n\\frac{1}{2\\pi\\ii}\n\\int_\\Gamma\n g(z)R_c(z)\\Delta R_f(z)\n\\dd z.\n\\]\nIn particular,\n\\[\n\\begin{aligned}\n\\norm{Cg(A_f)-g(A_c)C}\n\\leq{}&\n\\frac{\\operatorname{length}(\\Gamma)}{2\\pi}\n\\sup_{z\\in\\Gamma}\\abs{g(z)}\n\\sup_{z\\in\\Gamma}\\norm{R_c(z)}\n\\sup_{z\\in\\Gamma}\\norm{R_f(z)}\n\\norm\\Delta.\n\\end{aligned}\n\\]\n\\end{theorem}","begin{theorem}[Holomorphic defect formula]\\label{extended:record:013}\\label{extended:thm:holomorphic_defect}\nLet \\(\\Gamma\\) be the positively oriented rectifiable boundary of a finite union of pairwise disjoint bounded Jordan domains. Assume that both spectra lie inside this union and that \\(\\Gamma\\) lies in the common resolvent set. Thus its winding number is one about every spectral point. If \\(g\\) is holomorphic on a neighborhood of the closure of the union, then\n\\[\nCg(A_f)-g(A_c)C\n=\n\\frac{1}{2\\pi\\ii}\n\\int_\\Gamma\n g(z)R_c(z)\\Delta R_f(z)\n\\dd z.\n\\]\nIn particular,\n\\[\n\\begin{aligned}\n\\norm{Cg(A_f)-g(A_c)C}\n\\leq{}&\n\\frac{\\operatorname{length}(\\Gamma)}{2\\pi}\n\\sup_{z\\in\\Gamma}\\abs{g(z)}\n\\sup_{z\\in\\Gamma}\\norm{R_c(z)}\n\\sup_{z\\in\\Gamma}\\norm{R_f(z)}\n\\norm\\Delta.\n\\end{aligned}\n\\]\n\\end{theorem}","el{extended:thm:holomorphic_defect}\nLet \\(\\Gamma\\) be the positively oriented rectifiable boundary of a finite union of pairwise disjoint bounded Jordan domains. Assume that both spectra lie inside this union and that \\(\\Gamma\\) lies in the common resolvent set. Thus its winding number is one about every spectral point. If \\(g\\) is holomorphic on a neighborhood of the closure of the union, then\n\\[\nCg(A_f)-g(A_c)C\n=\n\\frac{1}{2\\pi\\ii}\n\\int_\\Gamma\n g(z)R_c(z)\\Delta R_f(z)\n\\dd z.\n\\]\nIn particular,\n\\[\n\\begin{aligned}\n\\norm{Cg(A_f)-g(A_c)C}\n\\leq{}&\n\\frac{\\operatorname{length}(\\Gamma)}{2\\pi}\n\\sup_{z\\in\\Gamma}\\abs{g(z)}\n\\sup_{z\\in\\Gamma}\\norm{R_c(z)}\n\\sup_{z\\in\\Gamma}\\norm{R_f(z)}\n\\norm\\Delta.\n\\end{aligned}\n\\]\n\\end{theorem}","a finite union of pairwise disjoint bounded Jordan domains. Assume that both spectra lie inside this union and that \\(\\Gamma\\) lies in the common resolvent set. Thus its winding number is one about every spectral point. If \\(g\\) is holomorphic on a neighborhood of the closure of the union, then\n\\[\nCg(A_f)-g(A_c)C\n=\n\\frac{1}{2\\pi\\ii}\n\\int_\\Gamma\n g(z)R_c(z)\\Delta R_f(z)\n\\dd z.\n\\]\nIn particular,\n\\[\n\\begin{aligned}\n\\norm{Cg(A_f)-g(A_c)C}\n\\leq{}&\n\\frac{\\operatorname{length}(\\Gamma)}{2\\pi}\n\\sup_{z\\in\\Gamma}\\abs{g(z)}\n\\sup_{z\\in\\Gamma}\\norm{R_c(z)}\n\\sup_{z\\in\\Gamma}\\norm{R_f(z)}\n\\norm\\Delta.\n\\end{aligned}\n\\]\n\\end{theorem}","\\begin{corollary}[Exact naturality]\\label{extended:record:014}\\label{extended:cor:exact_naturality}\nIf \\(CA_f=A_cC\\), then\n\\[\nCg(A_f)=g(A_c)C\n\\]\nfor every holomorphic function covered by \\cref{extended:thm:holomorphic_defect}. The conclusion includes exponentials, resolvents, and polynomials.\n\\end{corollary}","\\begin{proof}\nSet \\(\\Delta=0\\) in \\cref{extended:thm:holomorphic_defect}.\n\\end{proof}","Let\n\\[\nX_0\\xleftarrow{C_0}X_1\\xleftarrow{C_1}X_2\\xleftarrow{}\\cdots\n\\]\nbe Banach spaces with bounded connecting maps. Define the bounded inverse limit\n\\[\nX_\\infty\n=\n\\left\\{\n(x_\\ell)_{\\ell\\geq0}:\nC_\\ell x_{\\ell+1}=x_\\ell,\n\\quad\n\\sup_\\ell\\norm{x_\\ell}<\\infty\n\\right\\}\n\\]\nwith the supremum norm.","\\begin{lemma}[Completeness of the bounded inverse limit]\\label{extended:record:015}\\label{extended:lem:inverse_limit_banach}\nThe space \\(X_\\infty\\) is Banach.\n\\end{lemma}","\\begin{proof}\nThe product of the spaces \\(X_\\ell\\), equipped with the bounded supremum norm, is Banach. The equations \\(C_\\ell x_{\\ell+1}=x_\\ell\\) define a closed linear subspace.\n\\end{proof}","\\begin{theorem}[Exact projective limit semigroup]\\label{extended:record:016}\\label{extended:thm:exact_projective_limit}\nFor each \\(\\ell\\), let \\(T_\\ell(t)\\) be a strongly continuous semigroup on \\(X_\\ell\\). Assume\n\\[\nC_\\ell T_{\\ell+1}(t)=T_\\ell(t)C_\\ell\n\\]\nfor every \\(\\ell\\) and \\(t\\geq0\\), and assume the uniform bound\n\\[\n\\sup_\\ell\\norm{T_\\ell(t)}\\leq M\\e^{\\omega t}.\n\\]\nDefine\n\\[\nT_\\infty(t)(x_\\ell)_\\ell\n=\n(T_\\ell(t)x_\\ell)_\\ell.\n\\]\nThen \\(T_\\infty(t)\\) is a bounded semigroup on \\(X_\\infty\\). It is strongly continuous if\n\\[\n\\lim_{t\\downarrow0}\n\\sup_\\ell\n\\norm{T_\\ell(t)x_\\ell-x_\\ell}=0\n\\]\nfor every \\((x_\\ell)_\\ell\\in X_\\infty\\).","roup]\\label{extended:record:016}\\label{extended:thm:exact_projective_limit}\nFor each \\(\\ell\\), let \\(T_\\ell(t)\\) be a strongly continuous semigroup on \\(X_\\ell\\). Assume\n\\[\nC_\\ell T_{\\ell+1}(t)=T_\\ell(t)C_\\ell\n\\]\nfor every \\(\\ell\\) and \\(t\\geq0\\), and assume the uniform bound\n\\[\n\\sup_\\ell\\norm{T_\\ell(t)}\\leq M\\e^{\\omega t}.\n\\]\nDefine\n\\[\nT_\\infty(t)(x_\\ell)_\\ell\n=\n(T_\\ell(t)x_\\ell)_\\ell.\n\\]\nThen \\(T_\\infty(t)\\) is a bounded semigroup on \\(X_\\infty\\). It is strongly continuous if\n\\[\n\\lim_{t\\downarrow0}\n\\sup_\\ell\n\\norm{T_\\ell(t)x_\\ell-x_\\ell}=0\n\\]\nfor every \\((x_\\ell)_\\ell\\in X_\\infty\\).","ended:thm:exact_projective_limit}\nFor each \\(\\ell\\), let \\(T_\\ell(t)\\) be a strongly continuous semigroup on \\(X_\\ell\\). Assume\n\\[\nC_\\ell T_{\\ell+1}(t)=T_\\ell(t)C_\\ell\n\\]\nfor every \\(\\ell\\) and \\(t\\geq0\\), and assume the uniform bound\n\\[\n\\sup_\\ell\\norm{T_\\ell(t)}\\leq M\\e^{\\omega t}.\n\\]\nDefine\n\\[\nT_\\infty(t)(x_\\ell)_\\ell\n=\n(T_\\ell(t)x_\\ell)_\\ell.\n\\]\nThen \\(T_\\infty(t)\\) is a bounded semigroup on \\(X_\\infty\\). It is strongly continuous if\n\\[\n\\lim_{t\\downarrow0}\n\\sup_\\ell\n\\norm{T_\\ell(t)x_\\ell-x_\\ell}=0\n\\]\nfor every \\((x_\\ell)_\\ell\\in X_\\infty\\).","\\(\\ell\\), let \\(T_\\ell(t)\\) be a strongly continuous semigroup on \\(X_\\ell\\). Assume\n\\[\nC_\\ell T_{\\ell+1}(t)=T_\\ell(t)C_\\ell\n\\]\nfor every \\(\\ell\\) and \\(t\\geq0\\), and assume the uniform bound\n\\[\n\\sup_\\ell\\norm{T_\\ell(t)}\\leq M\\e^{\\omega t}.\n\\]\nDefine\n\\[\nT_\\infty(t)(x_\\ell)_\\ell\n=\n(T_\\ell(t)x_\\ell)_\\ell.\n\\]\nThen \\(T_\\infty(t)\\) is a bounded semigroup on \\(X_\\infty\\). It is strongly continuous if\n\\[\n\\lim_{t\\downarrow0}\n\\sup_\\ell\n\\norm{T_\\ell(t)x_\\ell-x_\\ell}=0\n\\]\nfor every \\((x_\\ell)_\\ell\\in X_\\infty\\).","Assume\n\\[\nC_\\ell T_{\\ell+1}(t)=T_\\ell(t)C_\\ell\n\\]\nfor every \\(\\ell\\) and \\(t\\geq0\\), and assume the uniform bound\n\\[\n\\sup_\\ell\\norm{T_\\ell(t)}\\leq M\\e^{\\omega t}.\n\\]\nDefine\n\\[\nT_\\infty(t)(x_\\ell)_\\ell\n=\n(T_\\ell(t)x_\\ell)_\\ell.\n\\]\nThen \\(T_\\infty(t)\\) is a bounded semigroup on \\(X_\\infty\\). It is strongly continuous if\n\\[\n\\lim_{t\\downarrow0}\n\\sup_\\ell\n\\norm{T_\\ell(t)x_\\ell-x_\\ell}=0\n\\]\nfor every \\((x_\\ell)_\\ell\\in X_\\infty\\).","If \\(T_\\ell(t)=\\e^{tA_\\ell}\\), the generators are bounded, and\n\\[\n\\sup_\\ell\\norm{A_\\ell}<\\infty,\n\\qquad\nC_\\ell A_{\\ell+1}=A_\\ell C_\\ell,\n\\]\nthen the continuity condition holds and\n\\[\nA_\\infty(x_\\ell)_\\ell=(A_\\ell x_\\ell)_\\ell\n\\]\nis the bounded generator of \\(T_\\infty\\).\n\\end{theorem}","\\begin{proof}\nExact intertwining shows that \\((T_\\ell(t)x_\\ell)_\\ell\\) remains compatible. The uniform growth bound makes the coordinatewise operator bounded on the supremum norm. The semigroup law holds in every coordinate. The stated continuity condition is exactly strong continuity in the supremum norm.","For bounded generators with common bound \\(M_A\\),\n\\[\n\\sup_\\ell\n\\norm{(\\e^{tA_\\ell}-I)x_\\ell}\n\\leq\n(\\e^{tM_A}-1)\\sup_\\ell\\norm{x_\\ell}.\n\\]\nThis tends to zero with \\(t\\). Exact generator intertwining makes \\((A_\\ell x_\\ell)_\\ell\\) compatible, and coordinatewise differentiation identifies the generator.\n\\end{proof}","Assume now that each \\(C_\\ell:X_{\\ell+1}\\to X_\\ell\\) has a bounded right inverse\n\\[\nR_\\ell:X_\\ell\\to X_{\\ell+1},\n\\qquad\nC_\\ell R_\\ell=I_{X_\\ell}.\n\\]\nFor \\(m<n\\), set\n\\[\nC_{m,n}=C_mC_{m+1}\\cdots C_{n-1},\n\\qquad\nR_{n,m}=R_{n-1}\\cdots R_m.\n\\]\nThe empty products are\n\\[\nC_{m,m}=I_{X_m},\n\\qquad\nR_{m,m}=I_{X_m}.\n\\]","\\begin{definition}[Reconstructed fine evolution]\\label{extended:record:018}\nFor a semigroup \\(T_n(t)\\) on \\(X_n\\), define the evolution seen from level \\(m\\) by\n\\[\nU_{m,n}(t)=C_{m,n}T_n(t)R_{n,m}.\n\\]\n\\end{definition}","\\begin{theorem}[Weighted summable defect completion]\\label{extended:record:019}\\label{extended:thm:summable_defect_completion}\nDefine the adjacent semigroup defect\n\\[\nE_n(t)=C_nT_{n+1}(t)-T_n(t)C_n.\n\\]\nFix a coarse level \\(m\\) and a time horizon \\(T>0\\). For \\(n\\geq m\\), set\n\\[\nw_{m,n}\n=\n\\norm{C_{m,n}}\\,\n\\norm{R_nR_{n,m}}.\n\\]\nIf\n\\[\n\\sum_{n=m}^{\\infty}\nw_{m,n}\n\\sup_{0\\leq t\\leq T}\\norm{E_n(t)}<\\infty,\n\\]\nthen \\(U_{m,n}(t)\\) converges in operator norm uniformly for \\(0\\leq t\\leq T\\). Its limit is denoted by \\(U_m(t)\\), and\n\\[\n\\sup_{0\\leq t\\leq T}\n\\norm{U_m(t)-U_{m,n}(t)}\n\\leq\n\\sum_{j=n}^{\\infty}w_{m,j}\n\\sup_{0\\leq t\\leq T}\\norm{E_j(t)}.\n\\]\n\\end{theorem}","}[Weighted summable defect completion]\\label{extended:record:019}\\label{extended:thm:summable_defect_completion}\nDefine the adjacent semigroup defect\n\\[\nE_n(t)=C_nT_{n+1}(t)-T_n(t)C_n.\n\\]\nFix a coarse level \\(m\\) and a time horizon \\(T>0\\). For \\(n\\geq m\\), set\n\\[\nw_{m,n}\n=\n\\norm{C_{m,n}}\\,\n\\norm{R_nR_{n,m}}.\n\\]\nIf\n\\[\n\\sum_{n=m}^{\\infty}\nw_{m,n}\n\\sup_{0\\leq t\\leq T}\\norm{E_n(t)}<\\infty,\n\\]\nthen \\(U_{m,n}(t)\\) converges in operator norm uniformly for \\(0\\leq t\\leq T\\). Its limit is denoted by \\(U_m(t)\\), and\n\\[\n\\sup_{0\\leq t\\leq T}\n\\norm{U_m(t)-U_{m,n}(t)}\n\\leq\n\\sum_{j=n}^{\\infty}w_{m,j}\n\\sup_{0\\leq t\\leq T}\\norm{E_j(t)}.\n\\]\n\\end{theorem}","cord:019}\\label{extended:thm:summable_defect_completion}\nDefine the adjacent semigroup defect\n\\[\nE_n(t)=C_nT_{n+1}(t)-T_n(t)C_n.\n\\]\nFix a coarse level \\(m\\) and a time horizon \\(T>0\\). For \\(n\\geq m\\), set\n\\[\nw_{m,n}\n=\n\\norm{C_{m,n}}\\,\n\\norm{R_nR_{n,m}}.\n\\]\nIf\n\\[\n\\sum_{n=m}^{\\infty}\nw_{m,n}\n\\sup_{0\\leq t\\leq T}\\norm{E_n(t)}<\\infty,\n\\]\nthen \\(U_{m,n}(t)\\) converges in operator norm uniformly for \\(0\\leq t\\leq T\\). Its limit is denoted by \\(U_m(t)\\), and\n\\[\n\\sup_{0\\leq t\\leq T}\n\\norm{U_m(t)-U_{m,n}(t)}\n\\leq\n\\sum_{j=n}^{\\infty}w_{m,j}\n\\sup_{0\\leq t\\leq T}\\norm{E_j(t)}.\n\\]\n\\end{theorem}","t_completion}\nDefine the adjacent semigroup defect\n\\[\nE_n(t)=C_nT_{n+1}(t)-T_n(t)C_n.\n\\]\nFix a coarse level \\(m\\) and a time horizon \\(T>0\\). For \\(n\\geq m\\), set\n\\[\nw_{m,n}\n=\n\\norm{C_{m,n}}\\,\n\\norm{R_nR_{n,m}}.\n\\]\nIf\n\\[\n\\sum_{n=m}^{\\infty}\nw_{m,n}\n\\sup_{0\\leq t\\leq T}\\norm{E_n(t)}<\\infty,\n\\]\nthen \\(U_{m,n}(t)\\) converges in operator norm uniformly for \\(0\\leq t\\leq T\\). Its limit is denoted by \\(U_m(t)\\), and\n\\[\n\\sup_{0\\leq t\\leq T}\n\\norm{U_m(t)-U_{m,n}(t)}\n\\leq\n\\sum_{j=n}^{\\infty}w_{m,j}\n\\sup_{0\\leq t\\leq T}\\norm{E_j(t)}.\n\\]\n\\end{theorem}","fine the adjacent semigroup defect\n\\[\nE_n(t)=C_nT_{n+1}(t)-T_n(t)C_n.\n\\]\nFix a coarse level \\(m\\) and a time horizon \\(T>0\\). For \\(n\\geq m\\), set\n\\[\nw_{m,n}\n=\n\\norm{C_{m,n}}\\,\n\\norm{R_nR_{n,m}}.\n\\]\nIf\n\\[\n\\sum_{n=m}^{\\infty}\nw_{m,n}\n\\sup_{0\\leq t\\leq T}\\norm{E_n(t)}<\\infty,\n\\]\nthen \\(U_{m,n}(t)\\) converges in operator norm uniformly for \\(0\\leq t\\leq T\\). Its limit is denoted by \\(U_m(t)\\), and\n\\[\n\\sup_{0\\leq t\\leq T}\n\\norm{U_m(t)-U_{m,n}(t)}\n\\leq\n\\sum_{j=n}^{\\infty}w_{m,j}\n\\sup_{0\\leq t\\leq T}\\norm{E_j(t)}.\n\\]\n\\end{theorem}","\\begin{proof}\nSince \\(C_nR_n=I\\),\n\\[\n\\begin{aligned}\nU_{m,n+1}(t)-U_{m,n}(t)\n={}&\nC_{m,n}\n\\bigl(C_nT_{n+1}(t)-T_n(t)C_n\\bigr)\nR_nR_{n,m}.\n\\end{aligned}\n\\]\nTherefore\n\\[\n\\norm{U_{m,n+1}(t)-U_{m,n}(t)}\n\\leq\nw_{m,n}\\norm{E_n(t)}.\n\\]\nThe weighted summability hypothesis makes the sequence uniformly Cauchy. Summing the tail estimates gives the stated bound.\n\\end{proof}","\\begin{corollary}[Contractive summable defect completion]\\label{extended:record:020}\\label{extended:cor:contractive_defect_completion}\nAssume\n\\[\n\\norm{C_\\ell}\\leq1,\n\\qquad\n\\norm{R_\\ell}\\leq1,\n\\qquad\n\\norm{T_\\ell(t)}\\leq1\n\\]\nfor every \\(\\ell\\) and \\(t\\geq0\\). If\n\\[\n\\sum_{\\ell=0}^{\\infty}\n\\sup_{0\\leq t\\leq T}\\norm{E_\\ell(t)}<\\infty,\n\\]\nthen the conclusion of \\cref{extended:thm:summable_defect_completion} holds, and\n\\[\n\\sup_{0\\leq t\\leq T}\n\\norm{U_m(t)-U_{m,n}(t)}\n\\leq\n\\sum_{\\ell=n}^{\\infty}\n\\sup_{0\\leq t\\leq T}\\norm{E_\\ell(t)}.\n\\]\n\\end{corollary}","ontractive summable defect completion]\\label{extended:record:020}\\label{extended:cor:contractive_defect_completion}\nAssume\n\\[\n\\norm{C_\\ell}\\leq1,\n\\qquad\n\\norm{R_\\ell}\\leq1,\n\\qquad\n\\norm{T_\\ell(t)}\\leq1\n\\]\nfor every \\(\\ell\\) and \\(t\\geq0\\). If\n\\[\n\\sum_{\\ell=0}^{\\infty}\n\\sup_{0\\leq t\\leq T}\\norm{E_\\ell(t)}<\\infty,\n\\]\nthen the conclusion of \\cref{extended:thm:summable_defect_completion} holds, and\n\\[\n\\sup_{0\\leq t\\leq T}\n\\norm{U_m(t)-U_{m,n}(t)}\n\\leq\n\\sum_{\\ell=n}^{\\infty}\n\\sup_{0\\leq t\\leq T}\\norm{E_\\ell(t)}.\n\\]\n\\end{corollary}","\\begin{proof}\nEvery weight \\(w_{m,n}\\) is at most one.\n\\end{proof}","\\begin{corollary}[Generator defect criterion]\\label{extended:record:021}\\label{extended:cor:generator_defect_completion}\nAssume\n\\[\n\\norm{C_\\ell}\\leq1,\n\\qquad\n\\norm{R_\\ell}\\leq1,\n\\]\nand suppose \\(T_\\ell(t)=\\e^{tA_\\ell}\\) are contraction semigroups with bounded generators. Set\n\\[\n\\Delta_\\ell=C_\\ell A_{\\ell+1}-A_\\ell C_\\ell.\n\\]\nIf\n\\[\n\\sum_{\\ell=0}^{\\infty}\\norm{\\Delta_\\ell}<\\infty,\n\\]\nthen the conclusion of \\cref{extended:cor:contractive_defect_completion} holds on every bounded time interval, and\n\\[\n\\sup_{0\\leq t\\leq T}\n\\norm{U_m(t)-U_{m,n}(t)}\n\\leq\nT\\sum_{\\ell=n}^{\\infty}\\norm{\\Delta_\\ell}.\n\\]\n\\end{corollary}","defect_completion}\nAssume\n\\[\n\\norm{C_\\ell}\\leq1,\n\\qquad\n\\norm{R_\\ell}\\leq1,\n\\]\nand suppose \\(T_\\ell(t)=\\e^{tA_\\ell}\\) are contraction semigroups with bounded generators. Set\n\\[\n\\Delta_\\ell=C_\\ell A_{\\ell+1}-A_\\ell C_\\ell.\n\\]\nIf\n\\[\n\\sum_{\\ell=0}^{\\infty}\\norm{\\Delta_\\ell}<\\infty,\n\\]\nthen the conclusion of \\cref{extended:cor:contractive_defect_completion} holds on every bounded time interval, and\n\\[\n\\sup_{0\\leq t\\leq T}\n\\norm{U_m(t)-U_{m,n}(t)}\n\\leq\nT\\sum_{\\ell=n}^{\\infty}\\norm{\\Delta_\\ell}.\n\\]\n\\end{corollary}","\\begin{theorem}[Asymptotic closure criterion]\\label{extended:record:022}\\label{extended:thm:asymptotic_closure}\nAssume the weighted convergence in \\cref{extended:thm:summable_defect_completion} holds for every bounded time interval, and define\n\\[\nP_{m,n}=R_{n,m}C_{m,n}\n\\]\non \\(X_n\\). Suppose that for every \\(T>0\\),\n\\[\n\\rho_{m,n}(T)\n=\n\\sup_{0\\leq s,t\\leq T}\n\\norm{\nC_{m,n}T_n(s)(I-P_{m,n})T_n(t)R_{n,m}\n}\n\\longrightarrow0.\n\\]\nThen\n\\[\nU_m(s+t)=U_m(s)U_m(t)\n\\]\nfor all \\(s,t\\geq0\\). If the convergence in \\cref{extended:thm:summable_defect_completion} is locally uniform and every \\(T_n\\) is norm continuous, then \\(U_m\\) is a norm continuous semigroup.\n\\end{theorem}","criterion]\\label{extended:record:022}\\label{extended:thm:asymptotic_closure}\nAssume the weighted convergence in \\cref{extended:thm:summable_defect_completion} holds for every bounded time interval, and define\n\\[\nP_{m,n}=R_{n,m}C_{m,n}\n\\]\non \\(X_n\\). Suppose that for every \\(T>0\\),\n\\[\n\\rho_{m,n}(T)\n=\n\\sup_{0\\leq s,t\\leq T}\n\\norm{\nC_{m,n}T_n(s)(I-P_{m,n})T_n(t)R_{n,m}\n}\n\\longrightarrow0.\n\\]\nThen\n\\[\nU_m(s+t)=U_m(s)U_m(t)\n\\]\nfor all \\(s,t\\geq0\\). If the convergence in \\cref{extended:thm:summable_defect_completion} is locally uniform and every \\(T_n\\) is norm continuous, then \\(U_m\\) is a norm continuous semigroup.\n\\end{theorem}","2}\\label{extended:thm:asymptotic_closure}\nAssume the weighted convergence in \\cref{extended:thm:summable_defect_completion} holds for every bounded time interval, and define\n\\[\nP_{m,n}=R_{n,m}C_{m,n}\n\\]\non \\(X_n\\). Suppose that for every \\(T>0\\),\n\\[\n\\rho_{m,n}(T)\n=\n\\sup_{0\\leq s,t\\leq T}\n\\norm{\nC_{m,n}T_n(s)(I-P_{m,n})T_n(t)R_{n,m}\n}\n\\longrightarrow0.\n\\]\nThen\n\\[\nU_m(s+t)=U_m(s)U_m(t)\n\\]\nfor all \\(s,t\\geq0\\). If the convergence in \\cref{extended:thm:summable_defect_completion} is locally uniform and every \\(T_n\\) is norm continuous, then \\(U_m\\) is a norm continuous semigroup.\n\\end{theorem}","completion} holds for every bounded time interval, and define\n\\[\nP_{m,n}=R_{n,m}C_{m,n}\n\\]\non \\(X_n\\). Suppose that for every \\(T>0\\),\n\\[\n\\rho_{m,n}(T)\n=\n\\sup_{0\\leq s,t\\leq T}\n\\norm{\nC_{m,n}T_n(s)(I-P_{m,n})T_n(t)R_{n,m}\n}\n\\longrightarrow0.\n\\]\nThen\n\\[\nU_m(s+t)=U_m(s)U_m(t)\n\\]\nfor all \\(s,t\\geq0\\). If the convergence in \\cref{extended:thm:summable_defect_completion} is locally uniform and every \\(T_n\\) is norm continuous, then \\(U_m\\) is a norm continuous semigroup.\n\\end{theorem}","ed time interval, and define\n\\[\nP_{m,n}=R_{n,m}C_{m,n}\n\\]\non \\(X_n\\). Suppose that for every \\(T>0\\),\n\\[\n\\rho_{m,n}(T)\n=\n\\sup_{0\\leq s,t\\leq T}\n\\norm{\nC_{m,n}T_n(s)(I-P_{m,n})T_n(t)R_{n,m}\n}\n\\longrightarrow0.\n\\]\nThen\n\\[\nU_m(s+t)=U_m(s)U_m(t)\n\\]\nfor all \\(s,t\\geq0\\). If the convergence in \\cref{extended:thm:summable_defect_completion} is locally uniform and every \\(T_n\\) is norm continuous, then \\(U_m\\) is a norm continuous semigroup.\n\\end{theorem}","\\begin{proof}\nAt a finite level,\n\\[\n\\begin{aligned}\nU_{m,n}(s)U_{m,n}(t)-U_{m,n}(s+t)\n={}&\nC_{m,n}T_n(s)\n(P_{m,n}-I)\nT_n(t)R_{n,m}.\n\\end{aligned}\n\\]\nIts norm is at most \\(\\rho_{m,n}(T)\\) for \\(s,t\\leq T\\). Pass to the limit using uniform convergence. Norm continuity follows from a uniform limit of norm continuous functions on compact intervals.\n\\end{proof}","\\begin{remark}\\label{extended:record:023}\nSummable generator defects control consistency between adjacent levels. The quantity \\(\\rho_{m,n}(T)\\) controls closure of the reconstructed fine range under the fine evolution. These are different requirements.\n\\end{remark}","\\begin{example}[Summable defects do not force a semigroup limit]\\label{extended:record:024}\\label{extended:ex:summable_not_semigroup}\nLet \\(X_0=\\RR\\) and let \\(X_n=\\RR^2\\) for every \\(n\\geq1\\). Set\n\\[\nC_0(x,y)=x,\n\\qquad\nR_0x=(x,0),\n\\]\nand use identity connecting maps and reconstructions above level one. Let \\(T_0(t)=I\\) and let every higher level carry the rotation group\n\\[\nT_n(t)\n=\n\\begin{pmatrix}\n\\cos t&-\\sin t","\\begin{example}[Summable defects do not force a semigroup limit]\\label{extended:record:024}\\label{extended:ex:summable_not_semigroup}\nLet \\(X_0=\\RR\\) and let \\(X_n=\\RR^2\\) for every \\(n\\geq1\\). Set\n\\[\nC_0(x,y)=x,\n\\qquad\nR_0x=(x,0),\n\\]\nand use identity connecting maps and reconstructions above level one. Let \\(T_0(t)=I\\) and let every higher level carry the rotation group\n\\[\nT_n(t)\n=\n\\begin{pmatrix}\n\\cos t&-\\sin t\\\\\n\\sin t&\\cos t\n\\end{pmatrix}.\n\\]\nOnly the first adjacent defect is nonzero, so every defect sum is finite. Yet for \\(n\\geq1\\),\n\\[\nU_{0,n}(t)=C_0T_n(t)R_0=\\cos t.\n\\]\nThe limiting family is not a semigroup because in general\n\\[\n\\cos(s+t)\\neq\\cos s\\cos t.\n\\]\nThe failure is exactly the excursion from the reconstructed line into its orthogonal complement between two time steps.\n\\end{example}","\\sin t&\\cos t\n\\end{pmatrix}.\n\\]\nOnly the first adjacent defect is nonzero, so every defect sum is finite. Yet for \\(n\\geq1\\),\n\\[\nU_{0,n}(t)=C_0T_n(t)R_0=\\cos t.\n\\]\nThe limiting family is not a semigroup because in general\n\\[\n\\cos(s+t)\\neq\\cos s\\cos t.\n\\]\nThe failure is exactly the excursion from the reconstructed line into its orthogonal complement between two time steps.\n\\end{example}","\\begin{proposition}[Range leakage criterion]\\label{extended:record:025}\\label{extended:prop:range_leakage}\nUnder the contractive assumptions of \\cref{extended:cor:contractive_defect_completion}, define\n\\[\n\\eta_{m,n}(T)\n=\n\\sup_{0\\leq t\\leq T}\n\\norm{(I-P_{m,n})T_n(t)R_{n,m}}.\n\\]\nThen\n\\[\n\\rho_{m,n}(T)\\leq\\eta_{m,n}(T).\n\\]\nConsequently, \\(\\eta_{m,n}(T)\\to0\\) for every \\(T>0\\) is sufficient for the limiting family \\(U_m\\) to satisfy the semigroup law.\n\\end{proposition}","ion}[Range leakage criterion]\\label{extended:record:025}\\label{extended:prop:range_leakage}\nUnder the contractive assumptions of \\cref{extended:cor:contractive_defect_completion}, define\n\\[\n\\eta_{m,n}(T)\n=\n\\sup_{0\\leq t\\leq T}\n\\norm{(I-P_{m,n})T_n(t)R_{n,m}}.\n\\]\nThen\n\\[\n\\rho_{m,n}(T)\\leq\\eta_{m,n}(T).\n\\]\nConsequently, \\(\\eta_{m,n}(T)\\to0\\) for every \\(T>0\\) is sufficient for the limiting family \\(U_m\\) to satisfy the semigroup law.\n\\end{proposition}","\\begin{proof}\nFor \\(s,t\\leq T\\), the contraction assumptions give\n\\[\n\\begin{aligned}\n&\\norm{\nC_{m,n}T_n(s)(I-P_{m,n})T_n(t)R_{n,m}\n}","\\begin{proof}\nFor \\(s,t\\leq T\\), the contraction assumptions give\n\\[\n\\begin{aligned}\n&\\norm{\nC_{m,n}T_n(s)(I-P_{m,n})T_n(t)R_{n,m}\n}\\\\\n&\\qquad\\leq\n\\norm{(I-P_{m,n})T_n(t)R_{n,m}}.\n\\end{aligned}\n\\]\nTake the supremum and apply \\cref{extended:thm:asymptotic_closure}.\n\\end{proof}","\\begin{theorem}[Generator control of range leakage]\\label{extended:record:026}\\label{extended:thm:generator_range_leakage}\nLet \\(T(t)=\\e^{tA}\\) on a Banach space \\(X\\), let \\(P\\) be a bounded projection, and set \\(Q=I-P\\). Assume\n\\[\n\\norm{T(t)}\\leq M\\e^{\\omega t}\n\\]\nand\n\\[\n\\norm{\\e^{t(QAQ)|_{QX}}}\n\\leq M_Q\\e^{\\omega_Qt}.\n\\]\nThen\n\\[\n\\norm{QT(t)P}\n\\leq\nM M_Q\\norm P^2\\norm{QAP}\n\\int_0^t\n\\e^{\\omega_Q(t-s)+\\omega s}\n\\dd s.\n\\]\nIn particular, for orthogonal projections and contraction bounds,\n\\[\n\\sup_{0\\leq t\\leq T}\n\\norm{QT(t)P}\n\\leq\nT\\norm{QAP}.\n\\]\n\\end{theorem}",":thm:generator_range_leakage}\nLet \\(T(t)=\\e^{tA}\\) on a Banach space \\(X\\), let \\(P\\) be a bounded projection, and set \\(Q=I-P\\). Assume\n\\[\n\\norm{T(t)}\\leq M\\e^{\\omega t}\n\\]\nand\n\\[\n\\norm{\\e^{t(QAQ)|_{QX}}}\n\\leq M_Q\\e^{\\omega_Qt}.\n\\]\nThen\n\\[\n\\norm{QT(t)P}\n\\leq\nM M_Q\\norm P^2\\norm{QAP}\n\\int_0^t\n\\e^{\\omega_Q(t-s)+\\omega s}\n\\dd s.\n\\]\nIn particular, for orthogonal projections and contraction bounds,\n\\[\n\\sup_{0\\leq t\\leq T}\n\\norm{QT(t)P}\n\\leq\nT\\norm{QAP}.\n\\]\n\\end{theorem}","\\begin{proof}\nFor \\(x\\in PX\\), write\n\\[\np(t)=PT(t)x,\n\\qquad\nq(t)=QT(t)x.\n\\]\nThe hidden component satisfies\n\\[\nq'(t)=QAQq(t)+QAPp(t),\n\\qquad\nq(0)=0.\n\\]\nVariation of constants gives\n\\[\nq(t)=\n\\int_0^t\n\\e^{(t-s)(QAQ)|_{QX}}\nQAPp(s)\n\\dd s.\n\\]\nSince\n\\(\n\\norm{p(s)}\\leq\\norm P\\norm{T(s)}\\norm x\n\\)\nand \\(x=Px\\), the first estimate follows. The contraction case is immediate.\n\\end{proof}","\\begin{corollary}[A usable closure condition for refinement towers]\\label{extended:record:027}\\label{extended:cor:generator_closure_tower}\nAssume the contractive summability hypothesis of \\cref{extended:cor:contractive_defect_completion} holds on every bounded time interval. Suppose every \\(P_{m,n}\\) is an orthogonal projection on a Hilbert space and \\(T_n(t)=\\e^{tA_n}\\) is contractive. If, for every fixed \\(m\\),\n\\[\n\\norm{(I-P_{m,n})A_nP_{m,n}}\n\\longrightarrow0,\n\\]\nthen the limiting family \\(U_m\\) is a semigroup.\n\\end{corollary}","lary}[A usable closure condition for refinement towers]\\label{extended:record:027}\\label{extended:cor:generator_closure_tower}\nAssume the contractive summability hypothesis of \\cref{extended:cor:contractive_defect_completion} holds on every bounded time interval. Suppose every \\(P_{m,n}\\) is an orthogonal projection on a Hilbert space and \\(T_n(t)=\\e^{tA_n}\\) is contractive. If, for every fixed \\(m\\),\n\\[\n\\norm{(I-P_{m,n})A_nP_{m,n}}\n\\longrightarrow0,\n\\]\nthen the limiting family \\(U_m\\) is a semigroup.\n\\end{corollary}","usable closure condition for refinement towers]\\label{extended:record:027}\\label{extended:cor:generator_closure_tower}\nAssume the contractive summability hypothesis of \\cref{extended:cor:contractive_defect_completion} holds on every bounded time interval. Suppose every \\(P_{m,n}\\) is an orthogonal projection on a Hilbert space and \\(T_n(t)=\\e^{tA_n}\\) is contractive. If, for every fixed \\(m\\),\n\\[\n\\norm{(I-P_{m,n})A_nP_{m,n}}\n\\longrightarrow0,\n\\]\nthen the limiting family \\(U_m\\) is a semigroup.\n\\end{corollary}","bel{extended:cor:generator_closure_tower}\nAssume the contractive summability hypothesis of \\cref{extended:cor:contractive_defect_completion} holds on every bounded time interval. Suppose every \\(P_{m,n}\\) is an orthogonal projection on a Hilbert space and \\(T_n(t)=\\e^{tA_n}\\) is contractive. If, for every fixed \\(m\\),\n\\[\n\\norm{(I-P_{m,n})A_nP_{m,n}}\n\\longrightarrow0,\n\\]\nthen the limiting family \\(U_m\\) is a semigroup.\n\\end{corollary}","\\begin{proof}\nA bounded generator of a contraction semigroup on a Hilbert space is dissipative. If \\(Q_{m,n}=I-P_{m,n}\\) and \\(y\\in Q_{m,n}X_n\\), then\n\\[\n\\operatorname{Re}\\ip{Q_{m,n}A_nQ_{m,n}y}{y}\n=\n\\operatorname{Re}\\ip{A_ny}{y}\n\\leq0.\n\\]\nHence the compressed hidden semigroup is contractive. Apply \\cref{extended:thm:generator_range_leakage} to \\(P_{m,n}\\). Since \\(P_{m,n}R_{n,m}=R_{n,m}\\),\n\\[\n\\eta_{m,n}(T)\n\\leq\nT\\norm{(I-P_{m,n})A_nP_{m,n}}\n\\norm{R_{n,m}}.\n\\]\nThe right side tends to zero. Use \\cref{extended:prop:range_leakage}.\n\\end{proof}","Probability vectors are columns. A column stochastic matrix \\(K\\) has nonnegative entries and column sums one. A continuous time generator \\(Q\\) has nonnegative off diagonal entries and column sums zero.","\\begin{theorem}[Finite Markov semigroup]\\label{extended:record:028}\\label{extended:thm:finite_markov}\nIf \\(Q\\) is a finite Markov generator, then \\(\\e^{tQ}\\) is column stochastic for every \\(t\\geq0\\).\n\\end{theorem}","\\begin{proof}\nChoose \\(a\\geq\\max_i(-Q_{ii})\\). The matrix \\(Q+aI\\) is nonnegative, so its exponential is nonnegative. Hence\n\\[\n\\e^{tQ}=\\e^{-at}\\e^{t(Q+aI)}\n\\]\nis nonnegative. Since \\(\\one^TQ=0\\), differentiation gives\n\\[\n\\one^T\\e^{tQ}=\\one^T.\n\\]\n\\end{proof}","Let \\(\\{B_1,\\ldots,B_m\\}\\) be a partition of the fine state set. Its aggregation matrix is\n\\[\nC_{\\alpha i}=\\one_{\\{i\\in B_\\alpha\\}}.\n\\]","\\begin{definition}[Canonical lumping obstruction]\\label{extended:record:029}\nFor a linear fine transition operator \\(K\\), define\n\\[\n\\Ob_C(K)\n=\n(CK)|_{\\Ker C}\n\\in\n\\mathcal B(\\Ker C,\\RR^m).\n\\]\n\\end{definition}","\\begin{theorem}[Descent through an aggregation map]\\label{extended:record:030}\\label{extended:thm:canonical_lumping_obstruction}\nLet \\(C:X_f\\to X_c\\) be a surjective linear map between finite dimensional spaces. A unique operator \\(\\overline K:X_c\\to X_c\\) satisfying\n\\[\nCK=\\overline KC\n\\]\nexists if and only if\n\\[\n\\Ob_C(K)=0.\n\\]\nFor any right inverse \\(R\\) of \\(C\\), the only possible coarse operator is\n\\[\n\\overline K=CKR,\n\\]\nand\n\\[\nCK-\\overline KC\n=\nCK(I-RC).\n\\]\n\\end{theorem}","theorem}[Descent through an aggregation map]\\label{extended:record:030}\\label{extended:thm:canonical_lumping_obstruction}\nLet \\(C:X_f\\to X_c\\) be a surjective linear map between finite dimensional spaces. A unique operator \\(\\overline K:X_c\\to X_c\\) satisfying\n\\[\nCK=\\overline KC\n\\]\nexists if and only if\n\\[\n\\Ob_C(K)=0.\n\\]\nFor any right inverse \\(R\\) of \\(C\\), the only possible coarse operator is\n\\[\n\\overline K=CKR,\n\\]\nand\n\\[\nCK-\\overline KC\n=\nCK(I-RC).\n\\]\n\\end{theorem}","aggregation map]\\label{extended:record:030}\\label{extended:thm:canonical_lumping_obstruction}\nLet \\(C:X_f\\to X_c\\) be a surjective linear map between finite dimensional spaces. A unique operator \\(\\overline K:X_c\\to X_c\\) satisfying\n\\[\nCK=\\overline KC\n\\]\nexists if and only if\n\\[\n\\Ob_C(K)=0.\n\\]\nFor any right inverse \\(R\\) of \\(C\\), the only possible coarse operator is\n\\[\n\\overline K=CKR,\n\\]\nand\n\\[\nCK-\\overline KC\n=\nCK(I-RC).\n\\]\n\\end{theorem}","\\begin{proof}\nIf \\(CK=\\overline KC\\), then \\(CK\\) vanishes on \\(\\Ker C\\). Conversely, suppose the obstruction vanishes and define\n\\[\n\\overline K(Cx)=CKx.\n\\]\nIf \\(Cx=Cy\\), then \\(x-y\\in\\Ker C\\), so the value does not depend on the representative. Surjectivity defines \\(\\overline K\\) on all of \\(X_c\\), and the intertwining relation follows from the definition. It also gives uniqueness. If \\(R\\) is a right inverse, then\n\\[\n\\overline K=\\overline KCR=CKR.\n\\]\nSubstitution gives the defect identity.\n\\end{proof}","overline KC\\), then \\(CK\\) vanishes on \\(\\Ker C\\). Conversely, suppose the obstruction vanishes and define\n\\[\n\\overline K(Cx)=CKx.\n\\]\nIf \\(Cx=Cy\\), then \\(x-y\\in\\Ker C\\), so the value does not depend on the representative. Surjectivity defines \\(\\overline K\\) on all of \\(X_c\\), and the intertwining relation follows from the definition. It also gives uniqueness. If \\(R\\) is a right inverse, then\n\\[\n\\overline K=\\overline KCR=CKR.\n\\]\nSubstitution gives the defect identity.\n\\end{proof}","\\begin{corollary}[Coordinate form of the lumping obstruction]\\label{extended:record:031}\\label{extended:cor:lumping_obstruction_coordinates}\nFor the partition aggregation above, \\(\\Ker C\\) is spanned by differences \\(e_i-e_j\\) with \\(i\\) and \\(j\\) in one source block. Also,\n\\[\n\\Ob_C(K)(e_i-e_j)\n=\n\\left(\n\\sum_{r\\in B_\\alpha}(K_{ri}-K_{rj})\n\\right)_{\\alpha=1}^m.\n\\]\n\\end{corollary}","\\begin{proof}\nInside each block, the vectors with coordinate sum zero are spanned by differences of coordinate vectors. Applying \\(CK\\) gives the displayed aggregated column differences.\n\\end{proof}","\\begin{theorem}[Characterization of exact discrete lumping]\\label{extended:record:032}\\label{extended:thm:lumping_characterization}\nLet \\(K\\) be column stochastic. A coarse column stochastic matrix \\(\\overline K\\) satisfying\n\\[\nCK=\\overline KC\n\\]\nexists if and only if, for every source block \\(B_\\beta\\), every pair \\(i,j\\in B_\\beta\\), and every target block \\(B_\\alpha\\),\n\\[\n\\sum_{r\\in B_\\alpha}K_{ri}\n=\n\\sum_{r\\in B_\\alpha}K_{rj}.\n\\]\nWhen it exists, \\(\\overline K\\) is unique and\n\\[\n\\overline K_{\\alpha\\beta}\n=\n\\sum_{r\\in B_\\alpha}K_{ri}\n\\]\nfor any \\(i\\in B_\\beta\\).\n\\end{theorem}","ing]\\label{extended:record:032}\\label{extended:thm:lumping_characterization}\nLet \\(K\\) be column stochastic. A coarse column stochastic matrix \\(\\overline K\\) satisfying\n\\[\nCK=\\overline KC\n\\]\nexists if and only if, for every source block \\(B_\\beta\\), every pair \\(i,j\\in B_\\beta\\), and every target block \\(B_\\alpha\\),\n\\[\n\\sum_{r\\in B_\\alpha}K_{ri}\n=\n\\sum_{r\\in B_\\alpha}K_{rj}.\n\\]\nWhen it exists, \\(\\overline K\\) is unique and\n\\[\n\\overline K_{\\alpha\\beta}\n=\n\\sum_{r\\in B_\\alpha}K_{ri}\n\\]\nfor any \\(i\\in B_\\beta\\).\n\\end{theorem}","}\\label{extended:thm:lumping_characterization}\nLet \\(K\\) be column stochastic. A coarse column stochastic matrix \\(\\overline K\\) satisfying\n\\[\nCK=\\overline KC\n\\]\nexists if and only if, for every source block \\(B_\\beta\\), every pair \\(i,j\\in B_\\beta\\), and every target block \\(B_\\alpha\\),\n\\[\n\\sum_{r\\in B_\\alpha}K_{ri}\n=\n\\sum_{r\\in B_\\alpha}K_{rj}.\n\\]\nWhen it exists, \\(\\overline K\\) is unique and\n\\[\n\\overline K_{\\alpha\\beta}\n=\n\\sum_{r\\in B_\\alpha}K_{ri}\n\\]\nfor any \\(i\\in B_\\beta\\).\n\\end{theorem}","ochastic. A coarse column stochastic matrix \\(\\overline K\\) satisfying\n\\[\nCK=\\overline KC\n\\]\nexists if and only if, for every source block \\(B_\\beta\\), every pair \\(i,j\\in B_\\beta\\), and every target block \\(B_\\alpha\\),\n\\[\n\\sum_{r\\in B_\\alpha}K_{ri}\n=\n\\sum_{r\\in B_\\alpha}K_{rj}.\n\\]\nWhen it exists, \\(\\overline K\\) is unique and\n\\[\n\\overline K_{\\alpha\\beta}\n=\n\\sum_{r\\in B_\\alpha}K_{ri}\n\\]\nfor any \\(i\\in B_\\beta\\).\n\\end{theorem}","\\begin{proof}\nThe \\(j\\)th column of \\(CK\\) records the transition mass from state \\(j\\) into every target block. The \\(j\\)th column of \\(\\overline KC\\) is the column of \\(\\overline K\\) indexed by the source block containing \\(j\\). Equality therefore holds exactly under the stated condition. The formula gives uniqueness. Summing over \\(\\alpha\\) gives one, so the coarse matrix is column stochastic.\n\\end{proof}","\\begin{corollary}[Continuous time lumping]\\label{extended:record:033}\\label{extended:cor:continuous_lumping}\nA coarse Markov generator \\(\\overline Q\\) satisfying\n\\[\nCQ=\\overline QC\n\\]\nexists if and only if the aggregated rates\n\\[\n\\sum_{r\\in B_\\alpha}Q_{ri}\n\\]\ndepend only on the source block of \\(i\\). In that case\n\\[\nC\\e^{tQ}=\\e^{t\\overline Q}C\n\\]\nfor every \\(t\\geq0\\).\n\\end{corollary}","\\begin{proof}\nThe characterization is the same column argument as in \\cref{extended:thm:lumping_characterization}. The off diagonal entries of \\(\\overline Q\\) are nonnegative and its column sums are zero. Exact semigroup intertwining follows from \\cref{extended:prop:infinitesimal_defect}.\n\\end{proof}","\\begin{theorem}[Finite horizon lumping defect]\\label{extended:record:034}\\label{extended:thm:finite_horizon_lumping}\nFor arbitrary fine and coarse operators \\(K\\) and \\(\\overline K\\), with the fixed aggregation map \\(C\\), set\n\\[\n\\Delta_K=CK-\\overline KC.\n\\]\nThen\n\\[\nCK^n-\\overline K^nC\n=\n\\sum_{j=0}^{n-1}\n\\overline K^{n-1-j}\\Delta_KK^j.\n\\]\nIf \\(K\\) and \\(\\overline K\\) are column stochastic, then\n\\[\n\\norm{CK^n-\\overline K^nC}_{1\\to1}\n\\leq\nn\\norm{\\Delta_K}_{1\\to1}.\n\\]\nFor every probability vector \\(p\\), the total variation discrepancy is at most\n\\[\n\\frac n2\\norm{\\Delta_K}_{1\\to1}.\n\\]\n\\end{theorem}","horizon_lumping}\nFor arbitrary fine and coarse operators \\(K\\) and \\(\\overline K\\), with the fixed aggregation map \\(C\\), set\n\\[\n\\Delta_K=CK-\\overline KC.\n\\]\nThen\n\\[\nCK^n-\\overline K^nC\n=\n\\sum_{j=0}^{n-1}\n\\overline K^{n-1-j}\\Delta_KK^j.\n\\]\nIf \\(K\\) and \\(\\overline K\\) are column stochastic, then\n\\[\n\\norm{CK^n-\\overline K^nC}_{1\\to1}\n\\leq\nn\\norm{\\Delta_K}_{1\\to1}.\n\\]\nFor every probability vector \\(p\\), the total variation discrepancy is at most\n\\[\n\\frac n2\\norm{\\Delta_K}_{1\\to1}.\n\\]\n\\end{theorem}","rators \\(K\\) and \\(\\overline K\\), with the fixed aggregation map \\(C\\), set\n\\[\n\\Delta_K=CK-\\overline KC.\n\\]\nThen\n\\[\nCK^n-\\overline K^nC\n=\n\\sum_{j=0}^{n-1}\n\\overline K^{n-1-j}\\Delta_KK^j.\n\\]\nIf \\(K\\) and \\(\\overline K\\) are column stochastic, then\n\\[\n\\norm{CK^n-\\overline K^nC}_{1\\to1}\n\\leq\nn\\norm{\\Delta_K}_{1\\to1}.\n\\]\nFor every probability vector \\(p\\), the total variation discrepancy is at most\n\\[\n\\frac n2\\norm{\\Delta_K}_{1\\to1}.\n\\]\n\\end{theorem}","\\begin{proof}\nSum the telescoping identities\n\\[\n\\overline K^{n-1-j}CK^{j+1}\n-\n\\overline K^{n-j}CK^j\n=\n\\overline K^{n-1-j}\\Delta_KK^j.\n\\]\nStochastic matrices are contractions in the \\(\\ell^1\\) norm. The last statement uses total variation equal to one half of the \\(\\ell^1\\) distance.\n\\end{proof}","\\begin{corollary}[Stationary mass descends]\\label{extended:record:035}\\label{extended:cor:stationary_descends}\nIf \\(CK=\\overline KC\\) and \\(Kp=p\\), then\n\\[\n\\overline K(Cp)=Cp.\n\\]\n\\end{corollary}","\\begin{proof}\nApply the intertwining identity to \\(p\\).\n\\end{proof}","Let \\(\\Omega=\\{1,\\ldots,n\\}\\) have strictly positive probabilities \\(\\pi_i\\). For a partition \\(\\mathcal Q\\), let \\(B(i)\\) denote the block containing \\(i\\). Define\n\\[\n(P_\\mathcal Qf)_i\n=\n\\frac{\\sum_{j\\in B(i)}\\pi_jf_j}\n{\\sum_{j\\in B(i)}\\pi_j}.\n\\]\nEquip \\(\\CC^n\\) with\n\\[\n\\ip{f}{g}_\\pi\n=\n\\sum_i\\pi_if_i\\overline{g_i}.\n\\]","\\begin{theorem}[Finite conditional expectation]\\label{extended:record:036}\\label{extended:thm:finite_conditional_expectation}\nThe operator \\(P_\\mathcal Q\\) is the orthogonal projector onto the functions that are constant on every block of \\(\\mathcal Q\\). In particular,\n\\[\nP_\\mathcal Q^2=P_\\mathcal Q,\n\\qquad\nP_\\mathcal Q^*=P_\\mathcal Q.\n\\]\n\\end{theorem}","\\begin{proof}\nOne application produces a block constant function, so a second application changes nothing. Expanding both weighted inner products block by block gives\n\\[\n\\ip{P_\\mathcal Qf}{g}_\\pi\n=\n\\sum_{B\\in\\mathcal Q}\n\\frac{\n\\left(\\sum_{i\\in B}\\pi_if_i\\right)\n\\overline{\\left(\\sum_{j\\in B}\\pi_jg_j\\right)}\n}{\\pi(B)}\n=\n\\ip{f}{P_\\mathcal Qg}_\\pi.\n\\]\nThe range is exactly the block constant subspace.\n\\end{proof}","\\begin{theorem}[Tower law and martingale packets]\\label{extended:record:037}\\label{extended:thm:martingale_packets}\nLet $P_0,P_1,\\ldots,P_r$ come from increasingly fine partitions, so that\n\\[\n\\Ran P_0\\subseteq\\Ran P_1\\subseteq\\cdots\\subseteq\\Ran P_r.\n\\]\nDefine\n\\[\nD_0=P_0,\n\\qquad\nD_j=P_j-P_{j-1}\n\\quad(1\\leq j\\leq r).\n\\]\nThen\n\\[\nP_iP_j=P_{\\min(i,j)},\n\\qquad\nD_iD_j=\\delta_{ij}D_j,\n\\qquad\nP_r=\\sum_{j=0}^rD_j.\n\\]\nEvery \\(D_j\\) is an orthogonal projector, and\n\\[\n\\norm{P_rf}_\\pi^2\n=\n\\sum_{j=0}^r\\norm{D_jf}_\\pi^2\n\\]\nand\n\\[\n\\norm f_\\pi^2\n=\n\\sum_{j=0}^r\\norm{D_jf}_\\pi^2\n+\n\\norm{f-P_rf}_\\pi^2.\n\\]\n\\end{theorem}","r law and martingale packets]\\label{extended:record:037}\\label{extended:thm:martingale_packets}\nLet $P_0,P_1,\\ldots,P_r$ come from increasingly fine partitions, so that\n\\[\n\\Ran P_0\\subseteq\\Ran P_1\\subseteq\\cdots\\subseteq\\Ran P_r.\n\\]\nDefine\n\\[\nD_0=P_0,\n\\qquad\nD_j=P_j-P_{j-1}\n\\quad(1\\leq j\\leq r).\n\\]\nThen\n\\[\nP_iP_j=P_{\\min(i,j)},\n\\qquad\nD_iD_j=\\delta_{ij}D_j,\n\\qquad\nP_r=\\sum_{j=0}^rD_j.\n\\]\nEvery \\(D_j\\) is an orthogonal projector, and\n\\[\n\\norm{P_rf}_\\pi^2\n=\n\\sum_{j=0}^r\\norm{D_jf}_\\pi^2\n\\]\nand\n\\[\n\\norm f_\\pi^2\n=\n\\sum_{j=0}^r\\norm{D_jf}_\\pi^2\n+\n\\norm{f-P_rf}_\\pi^2.\n\\]\n\\end{theorem}","nded:record:037}\\label{extended:thm:martingale_packets}\nLet $P_0,P_1,\\ldots,P_r$ come from increasingly fine partitions, so that\n\\[\n\\Ran P_0\\subseteq\\Ran P_1\\subseteq\\cdots\\subseteq\\Ran P_r.\n\\]\nDefine\n\\[\nD_0=P_0,\n\\qquad\nD_j=P_j-P_{j-1}\n\\quad(1\\leq j\\leq r).\n\\]\nThen\n\\[\nP_iP_j=P_{\\min(i,j)},\n\\qquad\nD_iD_j=\\delta_{ij}D_j,\n\\qquad\nP_r=\\sum_{j=0}^rD_j.\n\\]\nEvery \\(D_j\\) is an orthogonal projector, and\n\\[\n\\norm{P_rf}_\\pi^2\n=\n\\sum_{j=0}^r\\norm{D_jf}_\\pi^2\n\\]\nand\n\\[\n\\norm f_\\pi^2\n=\n\\sum_{j=0}^r\\norm{D_jf}_\\pi^2\n+\n\\norm{f-P_rf}_\\pi^2.\n\\]\n\\end{theorem}","$P_0,P_1,\\ldots,P_r$ come from increasingly fine partitions, so that\n\\[\n\\Ran P_0\\subseteq\\Ran P_1\\subseteq\\cdots\\subseteq\\Ran P_r.\n\\]\nDefine\n\\[\nD_0=P_0,\n\\qquad\nD_j=P_j-P_{j-1}\n\\quad(1\\leq j\\leq r).\n\\]\nThen\n\\[\nP_iP_j=P_{\\min(i,j)},\n\\qquad\nD_iD_j=\\delta_{ij}D_j,\n\\qquad\nP_r=\\sum_{j=0}^rD_j.\n\\]\nEvery \\(D_j\\) is an orthogonal projector, and\n\\[\n\\norm{P_rf}_\\pi^2\n=\n\\sum_{j=0}^r\\norm{D_jf}_\\pi^2\n\\]\nand\n\\[\n\\norm f_\\pi^2\n=\n\\sum_{j=0}^r\\norm{D_jf}_\\pi^2\n+\n\\norm{f-P_rf}_\\pi^2.\n\\]\n\\end{theorem}","\\begin{proof}\nA coarse average of a fine average is the coarse average. A fine average leaves a coarse constant function unchanged. Hence \\(P_iP_j=P_{\\min(i,j)}\\). Expanding products of the differences gives the orthogonality relations. The sum telescopes to \\(P_r\\). Orthogonal Pythagorean identities give the norm formulas.\n\\end{proof}","\\begin{corollary}[Law of total variance]\\label{extended:record:038}\\label{extended:cor:total_variance}\nFor a real valued \\(f\\), let \\(P_0f=\\Exp_\\pi[f]\\one\\). Then\n\\[\n\\Var_\\pi(f)\n=\n\\Var_\\pi(P_\\mathcal Qf)\n+\n\\norm{f-P_\\mathcal Qf}_\\pi^2.\n\\]\n\\end{corollary}","\\begin{proof}\nSubtract the mean and apply the orthogonal decomposition into the range of \\(P_\\mathcal Q\\) and its orthogonal complement.\n\\end{proof}","\\begin{standardfact}[General conditional expectations]\\label{extended:record:039}\nOn a probability space, conditional expectation onto a sub sigma algebra is the orthogonal projector in \\(L^2\\). Nested sigma algebras satisfy the same tower law. Under the usual hypotheses, martingale convergence gives a limit in \\(L^2\\) and almost surely. Standard treatments include \\cite{Williams1991}.\n\\end{standardfact}","Let \\(\\gamma\\) be the standard Gaussian probability measure on \\(\\RR\\). For \\(t\\geq0\\), define the Ornstein Uhlenbeck transfer\n\\[\n(\\Ocal_t f)(x)\n=\n\\int_\\RR\nf\\left(\n\\e^{-t}x+\\sqrt{1-\\e^{-2t}}\\,y\n\\right)\n\\gamma(\\dd y).\n\\]\nLet \\(H_n\\) be the probabilists' Hermite polynomial defined by\n\\[\n\\exp\\left(ux-\\frac{u^2}{2}\\right)\n=\n\\sum_{n=0}^\\infty H_n(x)\\frac{u^n}{n!},\n\\]\nand set \\(h_n=H_n/\\sqrt{n!}\\).","\\begin{theorem}[Ornstein Uhlenbeck semigroup]\\label{extended:record:040}\\label{extended:thm:ou_semigroup}\nThe operators \\((\\Ocal_t)_{t\\geq0}\\) form a positive Markov contraction semigroup on \\(L^p(\\gamma)\\) for every \\(1\\leq p\\leq\\infty\\). On \\(L^2(\\gamma)\\), the semigroup is self adjoint and\n\\[\n\\Ocal_t h_n=\\e^{-nt}h_n.\n\\]\nThe family \\((h_n)_{n\\geq0}\\) is an orthonormal basis of \\(L^2(\\gamma)\\).\n\\end{theorem}","\\begin{proof}\nPositivity and preservation of constants follow from the integral formula. If \\(X\\) and \\(Y\\) are independent standard Gaussian variables, then\n\\[\n\\e^{-s}\n\\left(\n\\e^{-t}x+\\sqrt{1-\\e^{-2t}}X\n\\right)\n+\n\\sqrt{1-\\e^{-2s}}Y\n\\]\nhas the same distribution as\n\\[\n\\e^{-(s+t)}x\n+\n\\sqrt{1-\\e^{-2(s+t)}}Z\n\\]\nwith \\(Z\\) standard Gaussian. This proves the semigroup law. Jensen's inequality and invariance of \\(\\gamma\\) give the \\(L^p\\) contraction. The joint Gaussian pair\n\\[\n\\left(\nX,\n\\e^{-t}X+\\sqrt{1-\\e^{-2t}}Y\n\\right)\n\\]\nis symmetric under interchange, which proves self adjointness.","of}\nPositivity and preservation of constants follow from the integral formula. If \\(X\\) and \\(Y\\) are independent standard Gaussian variables, then\n\\[\n\\e^{-s}\n\\left(\n\\e^{-t}x+\\sqrt{1-\\e^{-2t}}X\n\\right)\n+\n\\sqrt{1-\\e^{-2s}}Y\n\\]\nhas the same distribution as\n\\[\n\\e^{-(s+t)}x\n+\n\\sqrt{1-\\e^{-2(s+t)}}Z\n\\]\nwith \\(Z\\) standard Gaussian. This proves the semigroup law. Jensen's inequality and invariance of \\(\\gamma\\) give the \\(L^p\\) contraction. The joint Gaussian pair\n\\[\n\\left(\nX,\n\\e^{-t}X+\\sqrt{1-\\e^{-2t}}Y\n\\right)\n\\]\nis symmetric under interchange, which proves self adjointness.","reservation of constants follow from the integral formula. If \\(X\\) and \\(Y\\) are independent standard Gaussian variables, then\n\\[\n\\e^{-s}\n\\left(\n\\e^{-t}x+\\sqrt{1-\\e^{-2t}}X\n\\right)\n+\n\\sqrt{1-\\e^{-2s}}Y\n\\]\nhas the same distribution as\n\\[\n\\e^{-(s+t)}x\n+\n\\sqrt{1-\\e^{-2(s+t)}}Z\n\\]\nwith \\(Z\\) standard Gaussian. This proves the semigroup law. Jensen's inequality and invariance of \\(\\gamma\\) give the \\(L^p\\) contraction. The joint Gaussian pair\n\\[\n\\left(\nX,\n\\e^{-t}X+\\sqrt{1-\\e^{-2t}}Y\n\\right)\n\\]\nis symmetric under interchange, which proves self adjointness.","gral formula. If \\(X\\) and \\(Y\\) are independent standard Gaussian variables, then\n\\[\n\\e^{-s}\n\\left(\n\\e^{-t}x+\\sqrt{1-\\e^{-2t}}X\n\\right)\n+\n\\sqrt{1-\\e^{-2s}}Y\n\\]\nhas the same distribution as\n\\[\n\\e^{-(s+t)}x\n+\n\\sqrt{1-\\e^{-2(s+t)}}Z\n\\]\nwith \\(Z\\) standard Gaussian. This proves the semigroup law. Jensen's inequality and invariance of \\(\\gamma\\) give the \\(L^p\\) contraction. The joint Gaussian pair\n\\[\n\\left(\nX,\n\\e^{-t}X+\\sqrt{1-\\e^{-2t}}Y\n\\right)\n\\]\nis symmetric under interchange, which proves self adjointness.","Apply \\(\\Ocal_t\\) to the Hermite generating function. Gaussian integration gives\n\\[\n\\begin{aligned}\n\\Ocal_t\n\\exp\\left(ux-\\frac{u^2}{2}\\right)\n&=\n\\exp\\left(\n\\e^{-t}ux-\n\\frac{\\e^{-2t}u^2}{2}\n\\right)","Apply \\(\\Ocal_t\\) to the Hermite generating function. Gaussian integration gives\n\\[\n\\begin{aligned}\n\\Ocal_t\n\\exp\\left(ux-\\frac{u^2}{2}\\right)\n&=\n\\exp\\left(\n\\e^{-t}ux-\n\\frac{\\e^{-2t}u^2}{2}\n\\right)\\\\\n&=\n\\sum_{n=0}^\\infty\n\\e^{-nt}H_n(x)\\frac{u^n}{n!}.\n\\end{aligned}\n\\]\nComparison of coefficients proves the eigenvalue formula. The same generating function shows\n\\[\n\\int H_mH_n\\dd\\gamma=n!\\delta_{mn}.\n\\]\nThe Hermite polynomials span all polynomials, and polynomials are dense in \\(L^2(\\gamma)\\). Hence the normalized family is an orthonormal basis. See \\cite{Janson1997} for the general Gaussian Hilbert space setting.\n\\end{proof}","&=\n\\sum_{n=0}^\\infty\n\\e^{-nt}H_n(x)\\frac{u^n}{n!}.\n\\end{aligned}\n\\]\nComparison of coefficients proves the eigenvalue formula. The same generating function shows\n\\[\n\\int H_mH_n\\dd\\gamma=n!\\delta_{mn}.\n\\]\nThe Hermite polynomials span all polynomials, and polynomials are dense in \\(L^2(\\gamma)\\). Hence the normalized family is an orthonormal basis. See \\cite{Janson1997} for the general Gaussian Hilbert space setting.\n\\end{proof}","\\begin{corollary}[Hermite packet reconstruction]\\label{extended:record:041}\\label{extended:cor:hermite_packets}\nLet \\(P_n\\) be the orthogonal projection onto \\(\\CC h_n\\). Then\n\\[\nI=\\sum_{n=0}^\\infty P_n\n\\]\nin the strong operator topology and\n\\[\n\\Ocal_t\n=\n\\sum_{n=0}^\\infty\\e^{-nt}P_n.\n\\]\nFor \\(f\\in L^2(\\gamma)\\),\n\\[\n\\norm{\\Ocal_t f}_2^2\n=\n\\sum_{n=0}^\\infty\n\\e^{-2nt}\\norm{P_nf}_2^2.\n\\]\n\\end{corollary}","\\begin{proposition}[Generator and Dirichlet form]\\label{extended:record:042}\\label{extended:prop:ou_generator}\nOn the polynomial subspace, the infinitesimal generator of \\((\\Ocal_t)\\) is the differential operator\n\\[\n\\Lcal f=f''-xf'.\n\\]\nIt satisfies\n\\[\n\\Lcal h_n=-nh_n.\n\\]\nFor every polynomial \\(f\\),\n\\[\n\\ip{-\\Lcal f}{f}_{L^2(\\gamma)}\n=\n\\int_\\RR\\abs{f'(x)}^2\\gamma(\\dd x).\n\\]\n\\end{proposition}","\\begin{proof}\nWrite\n\\[\n\\rho_t=\\e^{-t}=1-t+O(t^2),\n\\qquad\n\\sigma_t^2=1-\\e^{-2t}=2t+O(t^2).\n\\]\nFor a polynomial \\(f\\), Taylor expansion at \\(x\\) and Gaussian integration give\n\\[\n\\begin{aligned}\n\\Ocal_tf(x)\n={}&f(x)\n+f'(x)(\\rho_t-1)x\n+\\frac12f''(x)\\sigma_t^2\n+o(t)","\\begin{proof}\nWrite\n\\[\n\\rho_t=\\e^{-t}=1-t+O(t^2),\n\\qquad\n\\sigma_t^2=1-\\e^{-2t}=2t+O(t^2).\n\\]\nFor a polynomial \\(f\\), Taylor expansion at \\(x\\) and Gaussian integration give\n\\[\n\\begin{aligned}\n\\Ocal_tf(x)\n={}&f(x)\n+f'(x)(\\rho_t-1)x\n+\\frac12f''(x)\\sigma_t^2\n+o(t)\\\\\n={}&f(x)+t\\bigl(f''(x)-xf'(x)\\bigr)+o(t).\n\\end{aligned}\n\\]\nEvery omitted term is a polynomial in the Gaussian variable multiplied by a quantity that is \\(o(t)\\), so the expansion may be integrated term by term. Therefore\n\\[\n\\left.\\frac{\\dd}{\\dd t}\\Ocal_tf\\right|_{t=0}\n=f''-xf'.\n\\]\nThe Hermite eigenvalue identity also follows by differentiating\n\\(\n\\Ocal_th_n=\\e^{-nt}h_n\n\\)\nat zero. Gaussian integration by parts gives\n\\[\n\\int_\\RR f'(x)\\overline{g(x)}\\gamma(\\dd x)\n=\n\\int_\\RR f(x)\\overline{xg(x)-g'(x)}\\gamma(\\dd x).\n\\]\nSet \\(g=f'\\) and rearrange.\n\\end{proof}","={}&f(x)+t\\bigl(f''(x)-xf'(x)\\bigr)+o(t).\n\\end{aligned}\n\\]\nEvery omitted term is a polynomial in the Gaussian variable multiplied by a quantity that is \\(o(t)\\), so the expansion may be integrated term by term. Therefore\n\\[\n\\left.\\frac{\\dd}{\\dd t}\\Ocal_tf\\right|_{t=0}\n=f''-xf'.\n\\]\nThe Hermite eigenvalue identity also follows by differentiating\n\\(\n\\Ocal_th_n=\\e^{-nt}h_n\n\\)\nat zero. Gaussian integration by parts gives\n\\[\n\\int_\\RR f'(x)\\overline{g(x)}\\gamma(\\dd x)\n=\n\\int_\\RR f(x)\\overline{xg(x)-g'(x)}\\gamma(\\dd x).\n\\]\nSet \\(g=f'\\) and rearrange.\n\\end{proof}","-xf'(x)\\bigr)+o(t).\n\\end{aligned}\n\\]\nEvery omitted term is a polynomial in the Gaussian variable multiplied by a quantity that is \\(o(t)\\), so the expansion may be integrated term by term. Therefore\n\\[\n\\left.\\frac{\\dd}{\\dd t}\\Ocal_tf\\right|_{t=0}\n=f''-xf'.\n\\]\nThe Hermite eigenvalue identity also follows by differentiating\n\\(\n\\Ocal_th_n=\\e^{-nt}h_n\n\\)\nat zero. Gaussian integration by parts gives\n\\[\n\\int_\\RR f'(x)\\overline{g(x)}\\gamma(\\dd x)\n=\n\\int_\\RR f(x)\\overline{xg(x)-g'(x)}\\gamma(\\dd x).\n\\]\nSet \\(g=f'\\) and rearrange.\n\\end{proof}","ed by a quantity that is \\(o(t)\\), so the expansion may be integrated term by term. Therefore\n\\[\n\\left.\\frac{\\dd}{\\dd t}\\Ocal_tf\\right|_{t=0}\n=f''-xf'.\n\\]\nThe Hermite eigenvalue identity also follows by differentiating\n\\(\n\\Ocal_th_n=\\e^{-nt}h_n\n\\)\nat zero. Gaussian integration by parts gives\n\\[\n\\int_\\RR f'(x)\\overline{g(x)}\\gamma(\\dd x)\n=\n\\int_\\RR f(x)\\overline{xg(x)-g'(x)}\\gamma(\\dd x).\n\\]\nSet \\(g=f'\\) and rearrange.\n\\end{proof}","\\begin{corollary}[Gaussian spectral gap]\\label{extended:record:043}\\label{extended:cor:ou_spectral_gap}\nIf \\(f\\in L^2(\\gamma)\\) has mean zero, then\n\\[\n\\norm{\\Ocal_tf}_2\n\\leq\n\\e^{-t}\\norm f_2.\n\\]\nFor every polynomial \\(f\\),\n\\[\n\\Var_\\gamma(f)\n\\leq\n\\int_\\RR\\abs{f'(x)}^2\\gamma(\\dd x).\n\\]\n\\end{corollary}","\\begin{proof}\nA mean zero function has no \\(h_0\\) component, so every surviving Hermite multiplier is at most \\(\\e^{-t}\\). For a polynomial\n\\[\nf-\\Exp_\\gamma f=\\sum_{n\\geq1}c_nh_n.\n\\]\nThen\n\\[\n\\Var_\\gamma(f)=\\sum_{n\\geq1}\\abs{c_n}^2\n\\leq\n\\sum_{n\\geq1}n\\abs{c_n}^2\n=\n\\ip{-\\Lcal f}{f}.\n\\]\nApply \\cref{extended:prop:ou_generator}.\n\\end{proof}","\\begin{corollary}[Fractional Hermite transfer]\\label{extended:record:044}\\label{extended:cor:fractional_hermite}\nLet \\(-\\Lcal\\) be the nonnegative Ornstein Uhlenbeck number operator with\n\\[\n-\\Lcal h_n=nh_n.\n\\]\nFor every \\(\\alpha>0\\),\n\\[\n\\e^{-t(-\\Lcal)^\\alpha}f\n=\n\\sum_{n=0}^\\infty\n\\e^{-tn^\\alpha}P_nf.\n\\]\nThis is a contraction semigroup on \\(L^2(\\gamma)\\).\n\\end{corollary}","Let\n\\[\n\\Delta_{n-1}\n=\n\\left\\{p\\in\\RR^n:\np_i\\geq0,\n\\quad\n\\sum_ip_i=1\n\\right\\}\n\\]\nbe the probability simplex. Let \\(\\lambda\\in[0,\\infty)^n\\) have positive evidence\n\\[\nZ_\\lambda(p)=\\sum_i\\lambda_ip_i.\n\\]\nThe Bayesian transfer is\n\\[\n\\Bcal_\\lambda(p)\n=\n\\frac{\\lambda\\odot p}{Z_\\lambda(p)}.\n\\]","\\begin{proposition}[Scale invariance and composition]\\label{extended:record:045}\\label{extended:prop:bayes_composition}\nFor every \\(c>0\\),\n\\[\n\\Bcal_{c\\lambda}=\\Bcal_\\lambda.\n\\]\nIf all relevant evidences are positive, then\n\\[\n\\Bcal_\\mu(\\Bcal_\\lambda(p))\n=\n\\Bcal_{\\lambda\\odot\\mu}(p).\n\\]\n\\end{proposition}","Let \\(I=I_1\\sqcup\\cdots\\sqcup I_m\\) be a partition. Define packet evidences\n\\[\nZ_\\alpha\n=\n\\sum_{i\\in I_\\alpha}\\lambda_ip_i.\n\\]\nFor \\(Z_\\alpha>0\\), define\n\\[\nm_\\alpha=\\frac{Z_\\alpha}{\\sum_\\beta Z_\\beta},\n\\qquad\nq_{\\alpha,i}=\\frac{\\lambda_ip_i}{Z_\\alpha}\n\\quad(i\\in I_\\alpha).\n\\]\nPackets with zero evidence receive posterior mass zero.","\\begin{theorem}[Bayesian packet reconstruction]\\label{extended:record:046}\\label{extended:thm:bayesian_packet_reconstruction}\nThe posterior is the disjoint packet synthesis\n\\[\n\\Bcal_\\lambda(p)\n=\n\\sum_{\\alpha:Z_\\alpha>0}\nm_\\alpha q_\\alpha.\n\\]\nThe local shape \\(q_\\alpha\\) depends only on the prior and likelihood inside packet \\(I_\\alpha\\). Coupling between different packets occurs through the normalization of the packet evidences.\n\\end{theorem}","\\begin{proof}\nFor \\(i\\in I_\\alpha\\),\n\\[\nm_\\alpha q_{\\alpha,i}\n=\n\\frac{Z_\\alpha}{\\sum_\\beta Z_\\beta}\n\\frac{\\lambda_ip_i}{Z_\\alpha}\n=\n\\frac{\\lambda_ip_i}{Z_\\lambda(p)}.\n\\]\nThe packet supports are disjoint.\n\\end{proof}","\\begin{theorem}[Likelihood perturbation]\\label{extended:record:047}\\label{extended:thm:likelihood_perturbation}\nLet \\(\\lambda,\\mu\\geq0\\) have positive evidences under a fixed prior \\(p\\). Set\n\\[\nd_p(\\lambda,\\mu)\n=\n\\sum_i\\abs{\\lambda_i-\\mu_i}p_i.\n\\]\nThen\n\\[\n\\norm{\\Bcal_\\lambda(p)-\\Bcal_\\mu(p)}_1\n\\leq\n\\frac{2d_p(\\lambda,\\mu)}\n{\\max\\{Z_\\lambda(p),Z_\\mu(p)\\}}.\n\\]\n\\end{theorem}","\\begin{proof}\nAdd and subtract \\(\\mu\\odot p/Z_\\lambda(p)\\). This gives\n\\[\n\\norm{\\Bcal_\\lambda(p)-\\Bcal_\\mu(p)}_1\n\\leq\n\\frac{2d_p(\\lambda,\\mu)}{Z_\\lambda(p)}.\n\\]\nInterchanging \\(\\lambda\\) and \\(\\mu\\) gives the same estimate with \\(Z_\\mu(p)\\). Take the smaller of the two upper bounds.\n\\end{proof}","\\begin{theorem}[Stability with respect to the prior]\\label{extended:record:048}\\label{extended:thm:prior_stability}\nAssume\n\\[\n0<a\\leq\\lambda_i\\leq b\n\\]\nfor every \\(i\\). Then\n\\[\n\\norm{\\Bcal_\\lambda(p)-\\Bcal_\\lambda(q)}_1\n\\leq\n\\frac{2b}{a}\\norm{p-q}_1\n\\]\nfor all priors \\(p,q\\in\\Delta_{n-1}\\).\n\\end{theorem}","\\begin{proof}\nThe evidences are at least \\(a\\). Add and subtract \\(\\lambda\\odot q/Z_\\lambda(p)\\). The numerator difference is at most \\(b\\norm{p-q}_1\\), and the evidence difference has the same bound. The two terms give the stated constant.\n\\end{proof}","Let \\(C\\) send a fine probability vector to its block masses\n\\[\n(Cp)_\\alpha\n=\n\\sum_{i\\in I_\\alpha}p_i.\n\\]\nAssume first that every block mass is positive. Write\n\\[\nr_{\\alpha,i}\n=\n\\frac{p_i}{(Cp)_\\alpha}\n\\]\nfor the prior conditional distribution inside packet \\(I_\\alpha\\). Define the prior dependent packet average likelihood\n\\[\n\\overline\\lambda_\\alpha(p)\n=\n\\sum_{i\\in I_\\alpha}\\lambda_ir_{\\alpha,i}.\n\\]","\\begin{theorem}[Exact two stage Bayesian update]\\label{extended:record:049}\\label{extended:thm:two_stage_bayes}\nFor every strictly positive prior,\n\\[\nC\\Bcal_\\lambda(p)\n=\n\\Bcal_{\\overline\\lambda(p)}(Cp).\n\\]\nFor every packet with \\(\\overline\\lambda_\\alpha(p)>0\\), the posterior conditional distribution is\n\\[\nq_{\\alpha,i}\n=\n\\frac{\\lambda_ir_{\\alpha,i}}\n{\\overline\\lambda_\\alpha(p)}.\n\\]\nA packet with zero average likelihood has zero posterior mass. Thus a fine Bayesian update consists of a coarse update by packet average likelihoods followed by independent local updates inside the packets that survive.\n\\end{theorem}","\\begin{proof}\nThe posterior mass of packet \\(I_\\alpha\\) is\n\\[\n\\frac{\\sum_{i\\in I_\\alpha}\\lambda_ip_i}\n{\\sum_j\\lambda_jp_j}\n=\n\\frac{(Cp)_\\alpha\\overline\\lambda_\\alpha(p)}\n{\\sum_\\beta(Cp)_\\beta\\overline\\lambda_\\beta(p)}.\n\\]\nThis is the coarse Bayesian formula. Dividing the fine posterior coordinates by the packet mass gives the local formula.\n\\end{proof}","\\begin{theorem}[Characterization of fixed coarse sufficiency]\\label{extended:record:050}\\label{extended:thm:bayesian_sufficiency}\nAssume \\(\\lambda_i>0\\) for every state. The following conditions are equivalent.\n\\begin{enumerate}\n\\item There is a positive coarse likelihood \\(\\ell\\in(0,\\infty)^m\\) such that\n\\[\nC\\Bcal_\\lambda(p)\n=\n\\Bcal_\\ell(Cp)\n\\]\nfor every strictly positive prior \\(p\\), and the posterior conditional distribution inside every packet equals its prior conditional distribution.\n\\item The likelihood is constant on every packet.\n\\end{enumerate}\nWhen these conditions hold, \\(\\ell_\\alpha\\) can be chosen as the common value of \\(\\lambda\\) on \\(I_\\alpha\\), and it is unique up to a common positive scale.\n\\end{theorem}","ons are equivalent.\n\\begin{enumerate}\n\\item There is a positive coarse likelihood \\(\\ell\\in(0,\\infty)^m\\) such that\n\\[\nC\\Bcal_\\lambda(p)\n=\n\\Bcal_\\ell(Cp)\n\\]\nfor every strictly positive prior \\(p\\), and the posterior conditional distribution inside every packet equals its prior conditional distribution.\n\\item The likelihood is constant on every packet.\n\\end{enumerate}\nWhen these conditions hold, \\(\\ell_\\alpha\\) can be chosen as the common value of \\(\\lambda\\) on \\(I_\\alpha\\), and it is unique up to a common positive scale.\n\\end{theorem}","a positive coarse likelihood \\(\\ell\\in(0,\\infty)^m\\) such that\n\\[\nC\\Bcal_\\lambda(p)\n=\n\\Bcal_\\ell(Cp)\n\\]\nfor every strictly positive prior \\(p\\), and the posterior conditional distribution inside every packet equals its prior conditional distribution.\n\\item The likelihood is constant on every packet.\n\\end{enumerate}\nWhen these conditions hold, \\(\\ell_\\alpha\\) can be chosen as the common value of \\(\\lambda\\) on \\(I_\\alpha\\), and it is unique up to a common positive scale.\n\\end{theorem}","se likelihood \\(\\ell\\in(0,\\infty)^m\\) such that\n\\[\nC\\Bcal_\\lambda(p)\n=\n\\Bcal_\\ell(Cp)\n\\]\nfor every strictly positive prior \\(p\\), and the posterior conditional distribution inside every packet equals its prior conditional distribution.\n\\item The likelihood is constant on every packet.\n\\end{enumerate}\nWhen these conditions hold, \\(\\ell_\\alpha\\) can be chosen as the common value of \\(\\lambda\\) on \\(I_\\alpha\\), and it is unique up to a common positive scale.\n\\end{theorem}","\\begin{proof}\nIf \\(\\lambda\\) is constant on each packet, \\cref{extended:thm:two_stage_bayes} has packet averages independent of the prior and its local posterior formula gives \\(q_\\alpha=r_\\alpha\\). This proves the first condition.","Conversely, fix any strictly positive prior. Equality of the local posterior and prior conditionals gives\n\\[\n\\frac{\\lambda_i r_{\\alpha,i}}\n{\\overline\\lambda_\\alpha(p)}\n=\nr_{\\alpha,i}\n\\]\nfor every \\(i\\in I_\\alpha\\). Positivity of the conditional prior implies\n\\[\n\\lambda_i=\\overline\\lambda_\\alpha(p)\n\\]\nfor every state in the packet. Thus the likelihood is constant on every packet. If two positive coarse likelihoods define the same update for every positive coarse prior, comparison of any two posterior odds shows that their coordinate ratios agree. Hence they differ only by a common positive factor.\n\\end{proof}","\\begin{theorem}[Coarsest fixed sufficient partition]\\label{extended:record:051}\\label{extended:thm:coarsest_bayesian_partition}\nLet \\(\\mathcal L\\) be a nonempty family of strictly positive likelihood vectors. Define\n\\[\ni\\sim_{\\mathcal L}j\n\\quad\\Longleftrightarrow\\quad\n\\lambda_i=\\lambda_j\n\\quad\\hbox{for every }\\lambda\\in\\mathcal L.\n\\]\nLet \\(\\mathcal Q_{\\mathcal L}\\) be the partition into equivalence classes. A partition \\(\\mathcal Q\\) gives a fixed coarse sufficient description in the sense of \\cref{extended:thm:bayesian_sufficiency} for every \\(\\lambda\\in\\mathcal L\\) if and only if every block of \\(\\mathcal Q\\) is contained in a block of \\(\\mathcal Q_{\\mathcal L}\\). Hence \\(\\mathcal Q_{\\mathcal L}\\) is the unique coarsest partition with this property.\n\\end{theorem}","fixed sufficient partition]\\label{extended:record:051}\\label{extended:thm:coarsest_bayesian_partition}\nLet \\(\\mathcal L\\) be a nonempty family of strictly positive likelihood vectors. Define\n\\[\ni\\sim_{\\mathcal L}j\n\\quad\\Longleftrightarrow\\quad\n\\lambda_i=\\lambda_j\n\\quad\\hbox{for every }\\lambda\\in\\mathcal L.\n\\]\nLet \\(\\mathcal Q_{\\mathcal L}\\) be the partition into equivalence classes. A partition \\(\\mathcal Q\\) gives a fixed coarse sufficient description in the sense of \\cref{extended:thm:bayesian_sufficiency} for every \\(\\lambda\\in\\mathcal L\\) if and only if every block of \\(\\mathcal Q\\) is contained in a block of \\(\\mathcal Q_{\\mathcal L}\\). Hence \\(\\mathcal Q_{\\mathcal L}\\) is the unique coarsest partition with this property.\n\\end{theorem}","nonempty family of strictly positive likelihood vectors. Define\n\\[\ni\\sim_{\\mathcal L}j\n\\quad\\Longleftrightarrow\\quad\n\\lambda_i=\\lambda_j\n\\quad\\hbox{for every }\\lambda\\in\\mathcal L.\n\\]\nLet \\(\\mathcal Q_{\\mathcal L}\\) be the partition into equivalence classes. A partition \\(\\mathcal Q\\) gives a fixed coarse sufficient description in the sense of \\cref{extended:thm:bayesian_sufficiency} for every \\(\\lambda\\in\\mathcal L\\) if and only if every block of \\(\\mathcal Q\\) is contained in a block of \\(\\mathcal Q_{\\mathcal L}\\). Hence \\(\\mathcal Q_{\\mathcal L}\\) is the unique coarsest partition with this property.\n\\end{theorem}","s. Define\n\\[\ni\\sim_{\\mathcal L}j\n\\quad\\Longleftrightarrow\\quad\n\\lambda_i=\\lambda_j\n\\quad\\hbox{for every }\\lambda\\in\\mathcal L.\n\\]\nLet \\(\\mathcal Q_{\\mathcal L}\\) be the partition into equivalence classes. A partition \\(\\mathcal Q\\) gives a fixed coarse sufficient description in the sense of \\cref{extended:thm:bayesian_sufficiency} for every \\(\\lambda\\in\\mathcal L\\) if and only if every block of \\(\\mathcal Q\\) is contained in a block of \\(\\mathcal Q_{\\mathcal L}\\). Hence \\(\\mathcal Q_{\\mathcal L}\\) is the unique coarsest partition with this property.\n\\end{theorem}","gleftrightarrow\\quad\n\\lambda_i=\\lambda_j\n\\quad\\hbox{for every }\\lambda\\in\\mathcal L.\n\\]\nLet \\(\\mathcal Q_{\\mathcal L}\\) be the partition into equivalence classes. A partition \\(\\mathcal Q\\) gives a fixed coarse sufficient description in the sense of \\cref{extended:thm:bayesian_sufficiency} for every \\(\\lambda\\in\\mathcal L\\) if and only if every block of \\(\\mathcal Q\\) is contained in a block of \\(\\mathcal Q_{\\mathcal L}\\). Hence \\(\\mathcal Q_{\\mathcal L}\\) is the unique coarsest partition with this property.\n\\end{theorem}","bda_j\n\\quad\\hbox{for every }\\lambda\\in\\mathcal L.\n\\]\nLet \\(\\mathcal Q_{\\mathcal L}\\) be the partition into equivalence classes. A partition \\(\\mathcal Q\\) gives a fixed coarse sufficient description in the sense of \\cref{extended:thm:bayesian_sufficiency} for every \\(\\lambda\\in\\mathcal L\\) if and only if every block of \\(\\mathcal Q\\) is contained in a block of \\(\\mathcal Q_{\\mathcal L}\\). Hence \\(\\mathcal Q_{\\mathcal L}\\) is the unique coarsest partition with this property.\n\\end{theorem}","\\begin{proof}\nApply \\cref{extended:thm:bayesian_sufficiency} to every likelihood in the family. A fixed sufficient description exists exactly when each likelihood is constant on every block of \\(\\mathcal Q\\). This is equivalent to every block of \\(\\mathcal Q\\) being contained in one common level set of the family. The equivalence class partition is itself admissible, and every other admissible partition refines it.\n\\end{proof}","\\begin{definition}[Within packet information loss]\\label{extended:record:052}\nFor a prior \\(p\\), define\n\\[\n\\eta_i(p)\n=\n\\lambda_i-\\overline\\lambda_\\alpha(p)\n\\quad\\hbox{when }i\\in I_\\alpha.\n\\]\nThen\n\\[\n\\sum_{i\\in I_\\alpha}r_{\\alpha,i}\\eta_i(p)=0.\n\\]\n\\end{definition}","\\begin{proposition}[Exact local distortion]\\label{extended:record:053}\\label{extended:prop:local_bayes_distortion}\nFor every packet with positive mass and evidence,\n\\[\nq_{\\alpha,i}-r_{\\alpha,i}\n=\n\\frac{r_{\\alpha,i}\\eta_i(p)}\n{\\overline\\lambda_\\alpha(p)}.\n\\]\nConsequently,\n\\[\n\\norm{q_\\alpha-r_\\alpha}_1\n=\n\\frac{\n\\sum_{i\\in I_\\alpha}r_{\\alpha,i}\\abs{\\eta_i(p)}\n}{\\overline\\lambda_\\alpha(p)}.\n\\]\nIf \\(\\lambda\\) is positive on the packet, then\n\\[\n\\norm{q_\\alpha-r_\\alpha}_1\n\\leq\n\\frac{\\osc_{I_\\alpha}(\\lambda)}\n{\\min_{i\\in I_\\alpha}\\lambda_i}.\n\\]\n\\end{proposition}","]\\label{extended:record:053}\\label{extended:prop:local_bayes_distortion}\nFor every packet with positive mass and evidence,\n\\[\nq_{\\alpha,i}-r_{\\alpha,i}\n=\n\\frac{r_{\\alpha,i}\\eta_i(p)}\n{\\overline\\lambda_\\alpha(p)}.\n\\]\nConsequently,\n\\[\n\\norm{q_\\alpha-r_\\alpha}_1\n=\n\\frac{\n\\sum_{i\\in I_\\alpha}r_{\\alpha,i}\\abs{\\eta_i(p)}\n}{\\overline\\lambda_\\alpha(p)}.\n\\]\nIf \\(\\lambda\\) is positive on the packet, then\n\\[\n\\norm{q_\\alpha-r_\\alpha}_1\n\\leq\n\\frac{\\osc_{I_\\alpha}(\\lambda)}\n{\\min_{i\\in I_\\alpha}\\lambda_i}.\n\\]\n\\end{proposition}","\\begin{proof}\nSubtract \\(r_{\\alpha,i}\\) from the local posterior formula in \\cref{extended:thm:two_stage_bayes}. The norm identity follows because the factor multiplying \\(\\eta_i\\) is nonnegative. The last estimate uses\n\\[\n\\sum_i r_{\\alpha,i}\\abs{\\eta_i}\n\\leq\n\\osc_{I_\\alpha}(\\lambda)\n\\]\nand \\(\\overline\\lambda_\\alpha(p)\\geq\\min_{I_\\alpha}\\lambda\\).\n\\end{proof}","\\begin{theorem}[Approximate coarse Bayesian rule]\\label{extended:record:054}\\label{extended:thm:approximate_coarse_bayes}\nChoose one positive number \\(\\ell_\\alpha\\) for every packet and define the block constant fine likelihood\n\\[\n\\widetilde\\lambda_i=\\ell_\\alpha\n\\quad(i\\in I_\\alpha).\n\\]\nThen\n\\[\n\\norm{\nC\\Bcal_\\lambda(p)-\\Bcal_\\ell(Cp)\n}_1\n\\leq\n\\frac{\n2\\sum_i p_i\\abs{\\lambda_i-\\widetilde\\lambda_i}\n}{\n\\max\\{Z_\\lambda(p),Z_{\\widetilde\\lambda}(p)\\}\n}.\n\\]\n\\end{theorem}","\\begin{proof}\nBlock constancy gives\n\\[\nC\\Bcal_{\\widetilde\\lambda}(p)\n=\n\\Bcal_\\ell(Cp).\n\\]\nAggregation is an \\(\\ell^1\\) contraction. Apply \\cref{extended:thm:likelihood_perturbation} to the two fine posteriors and then aggregate.\n\\end{proof}","\\begin{definition}[Relative information]\\label{extended:record:055}\nFor probability vectors \\(q\\) and \\(p\\), define\n\\[\nD(q\\Vert p)\n=\n\\sum_iq_i\\log\\frac{q_i}{p_i}.\n\\]\nThe convention is \\(0\\log0=0\\). A term with \\(q_i>0\\) and \\(p_i=0\\) has value \\(+\\infty\\).\n\\end{definition}","\\begin{theorem}[Exact information decomposition]\\label{extended:record:056}\\label{extended:thm:bayesian_information_chain}\nLet \\(p\\) and \\(q\\) be probability vectors on a finite partition \\((I_\\alpha)_\\alpha\\), and assume \\(q_i=0\\) whenever \\(p_i=0\\). Let\n\\[\nm=Cp,\n\\qquad\nn=Cq,\n\\]\nand let \\(p_\\alpha\\) and \\(q_\\alpha\\) be the conditional distributions inside packet \\(I_\\alpha\\) whenever the corresponding mass is positive. Then\n\\[\nD(q\\Vert p)\n=\nD(n\\Vert m)\n+\n\\sum_{\\alpha:n_\\alpha>0}\n n_\\alpha D(q_\\alpha\\Vert p_\\alpha).\n\\]\nThe first term is the information retained by the coarse masses. The second term is the exact information carried by changes inside the packets.\n\\end{theorem}","on decomposition]\\label{extended:record:056}\\label{extended:thm:bayesian_information_chain}\nLet \\(p\\) and \\(q\\) be probability vectors on a finite partition \\((I_\\alpha)_\\alpha\\), and assume \\(q_i=0\\) whenever \\(p_i=0\\). Let\n\\[\nm=Cp,\n\\qquad\nn=Cq,\n\\]\nand let \\(p_\\alpha\\) and \\(q_\\alpha\\) be the conditional distributions inside packet \\(I_\\alpha\\) whenever the corresponding mass is positive. Then\n\\[\nD(q\\Vert p)\n=\nD(n\\Vert m)\n+\n\\sum_{\\alpha:n_\\alpha>0}\n n_\\alpha D(q_\\alpha\\Vert p_\\alpha).\n\\]\nThe first term is the information retained by the coarse masses. The second term is the exact information carried by changes inside the packets.\n\\end{theorem}","\\begin{proof}\nFor \\(i\\in I_\\alpha\\), write\n\\[\np_i=m_\\alpha p_{\\alpha,i},\n\\qquad\nq_i=n_\\alpha q_{\\alpha,i}.\n\\]\nThen\n\\[\n\\log\\frac{q_i}{p_i}\n=\n\\log\\frac{n_\\alpha}{m_\\alpha}\n+\n\\log\\frac{q_{\\alpha,i}}{p_{\\alpha,i}}.\n\\]\nMultiply by \\(q_i\\), sum first inside each packet, and then sum over the packets. This is the finite relative entropy chain rule. Standard information theoretic treatments include \\cite{CoverThomas2006}.\n\\end{proof}","\\begin{corollary}[Coarse data processing and equality]\\label{extended:record:057}\\label{extended:cor:coarse_data_processing}\nUnder the assumptions of \\cref{extended:thm:bayesian_information_chain},\n\\[\nD(Cq\\Vert Cp)\n\\leq\nD(q\\Vert p).\n\\]\nEquality holds if and only if\n\\[\nq_\\alpha=p_\\alpha\n\\]\nfor every packet with positive posterior mass.\n\\end{corollary}","\\begin{corollary}[Information created by a Bayesian update]\\label{extended:record:058}\\label{extended:cor:bayesian_information_gain}\nLet \\(q=\\Bcal_\\lambda(p)\\) with positive likelihood and positive prior. Then\n\\[\nD(q\\Vert p)\n=\n\\sum_iq_i\\log\\lambda_i\n-\n\\log Z_\\lambda(p).\n\\]\nInside a packet,\n\\[\nD(q_\\alpha\\Vert r_\\alpha)\n=\n\\sum_{i\\in I_\\alpha}\nq_{\\alpha,i}\n\\log\\frac{\\lambda_i}{\\overline\\lambda_\\alpha(p)}.\n\\]\nHence\n\\[\nD(q_\\alpha\\Vert r_\\alpha)\n\\leq\n\\log\n\\frac{\\max_{i\\in I_\\alpha}\\lambda_i}\n{\\min_{i\\in I_\\alpha}\\lambda_i}.\n\\]\n\\end{corollary}","[Information created by a Bayesian update]\\label{extended:record:058}\\label{extended:cor:bayesian_information_gain}\nLet \\(q=\\Bcal_\\lambda(p)\\) with positive likelihood and positive prior. Then\n\\[\nD(q\\Vert p)\n=\n\\sum_iq_i\\log\\lambda_i\n-\n\\log Z_\\lambda(p).\n\\]\nInside a packet,\n\\[\nD(q_\\alpha\\Vert r_\\alpha)\n=\n\\sum_{i\\in I_\\alpha}\nq_{\\alpha,i}\n\\log\\frac{\\lambda_i}{\\overline\\lambda_\\alpha(p)}.\n\\]\nHence\n\\[\nD(q_\\alpha\\Vert r_\\alpha)\n\\leq\n\\log\n\\frac{\\max_{i\\in I_\\alpha}\\lambda_i}\n{\\min_{i\\in I_\\alpha}\\lambda_i}.\n\\]\n\\end{corollary}","\\begin{proof}\nThe posterior ratio is\n\\[\n\\frac{q_i}{p_i}\n=\n\\frac{\\lambda_i}{Z_\\lambda(p)}.\n\\]\nThe local ratio is\n\\[\n\\frac{q_{\\alpha,i}}{r_{\\alpha,i}}\n=\n\\frac{\\lambda_i}{\\overline\\lambda_\\alpha(p)}.\n\\]\nInsert these identities into the definition of relative entropy. The final estimate follows because both \\(\\lambda_i\\) and the packet average lie between the packet minimum and maximum.\n\\end{proof}","\\begin{proof}\nIf the likelihood is block constant, every local posterior equals its local prior. Conversely, choose any positive prior. Vanishing relative entropy gives \\(q_\\alpha=r_\\alpha\\). The local posterior formula then forces all likelihood values in that packet to equal their packet average. Repeat for every packet.\n\\end{proof}","Fix a score vector \\(g\\in\\RR^n\\). For \\(t\\geq0\\), define\n\\[\n\\Phi_t(p)_i\n=\n\\frac{\\e^{tg_i}p_i}\n{\\sum_j\\e^{tg_j}p_j}.\n\\]","\\begin{theorem}[Bayesian flow and replicator equation]\\label{extended:record:061}\\label{extended:thm:bayesian_flow}\nThe maps \\(\\Phi_t\\) form a nonlinear semigroup:\n\\[\n\\Phi_{s+t}=\\Phi_s\\circ\\Phi_t.\n\\]\nEvery coordinate satisfies\n\\[\n\\frac{\\dd}{\\dd t}p_i(t)\n=\n\\bigl(g_i-\\ip{g}{p(t)}\\bigr)p_i(t).\n\\]\nThe support of \\(p(t)\\) is independent of \\(t\\), and for any two coordinates in the support,\n\\[\n\\frac{p_i(t)}{p_j(t)}\n=\n\\e^{t(g_i-g_j)}\n\\frac{p_i(0)}{p_j(0)}.\n\\]\n\\end{theorem}","\\begin{proof}\nThe semigroup law follows from multiplication of the likelihoods \\(\\e^{sg}\\) and \\(\\e^{tg}\\). Differentiating the normalized formula gives the replicator equation. A coordinate that starts at zero remains zero, while a positive coordinate remains positive. Dividing two explicit coordinate formulas gives the odds identity.\n\\end{proof}","\\begin{theorem}[Evidence cumulants and score monotonicity]\\label{extended:record:062}\\label{extended:thm:evidence_cumulants}\nSet\n\\[\n\\psi(t)\n=\n\\log\\left(\n\\sum_i\\e^{tg_i}p_i(0)\n\\right).\n\\]\nThen\n\\[\n\\psi'(t)=\\ip{g}{p(t)},\n\\qquad\n\\psi''(t)=\\Var_{p(t)}(g)\\geq0.\n\\]\nConsequently,\n\\[\n\\frac{\\dd}{\\dd t}\\ip{g}{p(t)}\n=\n\\Var_{p(t)}(g).\n\\]\n\\end{theorem}","\\begin{corollary}[Concentration on the maximal score packet]\\label{extended:record:063}\\label{extended:cor:bayes_flow_concentration}\nLet\n\\[\nG=\\max\\{g_i:p_i(0)>0\\}\n\\]\nand let \\(M\\) be the set of supported indices with score \\(G\\). If \\(M\\) is the full support, then the conclusion below is immediate. Otherwise, suppose\n\\[\n\\delta\n=\n\\min\\{G-g_i:p_i(0)>0,\n\\quad i\\notin M\\}>0.\n\\]\nThen\n\\[\n1-\\sum_{i\\in M}p_i(t)\n\\leq\n\\frac{1-p_M(0)}{p_M(0)}\\e^{-\\delta t},\n\\]\nwhere \\(p_M(0)=\\sum_{i\\in M}p_i(0)\\).\n\\end{corollary}","abel{extended:record:063}\\label{extended:cor:bayes_flow_concentration}\nLet\n\\[\nG=\\max\\{g_i:p_i(0)>0\\}\n\\]\nand let \\(M\\) be the set of supported indices with score \\(G\\). If \\(M\\) is the full support, then the conclusion below is immediate. Otherwise, suppose\n\\[\n\\delta\n=\n\\min\\{G-g_i:p_i(0)>0,\n\\quad i\\notin M\\}>0.\n\\]\nThen\n\\[\n1-\\sum_{i\\in M}p_i(t)\n\\leq\n\\frac{1-p_M(0)}{p_M(0)}\\e^{-\\delta t},\n\\]\nwhere \\(p_M(0)=\\sum_{i\\in M}p_i(0)\\).\n\\end{corollary}","\\begin{proof}\nFactor \\(\\e^{tG}\\) from the normalizing constant. The total weight outside \\(M\\) is at most \\((1-p_M(0))\\e^{-\\delta t}\\), while the weight on \\(M\\) is \\(p_M(0)\\).\n\\end{proof}","Let \\(K\\) be column stochastic and let \\(\\lambda^{(r)}\\) be the likelihood of the \\(r\\)th observation. The hidden Markov filter is\n\\[\n\\widehat p_r=Kp_{r-1},\n\\qquad\np_r=\\Bcal_{\\lambda^{(r)}}(\\widehat p_r).\n\\]\nA standard account of finite hidden Markov models appears in \\cite{Rabiner1989}.","\\begin{proposition}[Simplex preservation]\\label{extended:record:064}\\label{extended:prop:hmm_preservation}\nSuppose \\(p_{r-1}\\) is a probability vector and the new evidence is positive. Prediction produces the probability vector \\(\\widehat p_r\\), and normalization produces the probability vector \\(p_r\\).\n\\end{proposition}","\\begin{theorem}[One step filter stability]\\label{extended:record:065}\\label{extended:thm:hmm_stability}\nLet \\(K\\) and \\(\\widetilde K\\) be column stochastic. Assume\n\\[\n0<a\\leq\\lambda_i\\leq b.\n\\]\nThen\n\\[\n\\norm{\n\\Bcal_\\lambda(Kp)-\\Bcal_\\lambda(\\widetilde Kq)\n}_1\n\\leq\n\\frac{2b}{a}\n\\left(\n\\norm{p-q}_1\n+\n\\norm{K-\\widetilde K}_{1\\to1}\n\\right).\n\\]\n\\end{theorem}","\\begin{proof}\nApply \\cref{extended:thm:prior_stability} to \\(Kp\\) and \\(\\widetilde Kq\\). Then use\n\\[\n\\norm{Kp-\\widetilde Kq}_1\n\\leq\n\\norm{K(p-q)}_1\n+\n\\norm{(K-\\widetilde K)q}_1\n\\]\nand the stochastic contraction property.\n\\end{proof}","\\begin{example}[Beta binomial transfer]\\label{extended:record:066}\nA beta prior with parameters \\((a,b)\\) and binomial data containing \\(s\\) successes in \\(n\\) trials gives the posterior parameters\n\\[\n(a+s,b+n-s).\n\\]\nThe update is additive in the two exponent coordinates. This is a finite dimensional conjugate realization of likelihood composition. Standard Bayesian references include \\cite{BernardoSmith1994}.\n\\end{example}","Let \\(H\\) be a finite dimensional Hilbert space and let \\(L=L^*\\geq0\\). Write its spectral resolution as\n\\[\nL=\\sum_j\\lambda_jP_j.\n\\]\nFor \\(\\alpha>0\\), define\n\\[\nL^\\alpha\n=\n\\sum_j\\lambda_j^\\alpha P_j\n\\]\nand\n\\[\nT_\\alpha(t)\n=\n\\e^{-tL^\\alpha}\n=\n\\sum_j\\e^{-t\\lambda_j^\\alpha}P_j.\n\\]","\\begin{theorem}[Fractional packet calculus]\\label{extended:record:067}\\label{extended:thm:fractional_packet_calculus}\nFor \\(\\alpha,\\beta>0\\),\n\\[\nL^\\alpha L^\\beta=L^{\\alpha+\\beta}.\n\\]\nFor \\(s,t\\geq0\\),\n\\[\nT_\\alpha(s)T_\\alpha(t)=T_\\alpha(s+t).\n\\]\nEvery spectral packet \\(P_jH\\) reduces \\(L^\\alpha\\) and \\(T_\\alpha(t)\\).\n\\end{theorem}","\\begin{theorem}[Balakrishnan formula in finite dimension]\\label{extended:record:068}\\label{extended:thm:balakrishnan}\nFor \\(0<\\alpha<1\\),\n\\[\nL^\\alpha\n=\n\\frac{\\sin(\\pi\\alpha)}{\\pi}\n\\int_0^\\infty\ns^{\\alpha-1}L(sI+L)^{-1}\n\\dd s.\n\\]\nThe integral converges in operator norm.\n\\end{theorem}","\\begin{proof}\nIt is enough to prove the formula on an eigenvector with eigenvalue \\(x\\geq0\\). For \\(x>0\\), the substitution \\(s=xu\\) gives\n\\[\n\\frac{\\sin(\\pi\\alpha)}{\\pi}\n\\int_0^\\infty\ns^{\\alpha-1}\\frac{x}{s+x}\\dd s\n=x^\\alpha\n\\frac{\\sin(\\pi\\alpha)}{\\pi}\n\\int_0^\\infty\\frac{u^{\\alpha-1}}{1+u}\\dd u\n=x^\\alpha.\n\\]\nThe value at \\(x=0\\) is zero. There are finitely many eigenvalues, so the operator integral converges. This is the finite spectral form of the fractional power construction in \\cite{Balakrishnan1960}.\n\\end{proof}","\\begin{theorem}[Exact fractional naturality]\\label{extended:record:069}\\label{extended:thm:exact_fractional_naturality}\nLet \\(L_f\\) and \\(L_c\\) be finite dimensional positive self adjoint operators and let \\(C\\) satisfy\n\\[\nCL_f=L_cC.\n\\]\nThen, for every \\(\\alpha>0\\) and \\(t\\geq0\\),\n\\[\nCL_f^\\alpha=L_c^\\alpha C\n\\]\nand\n\\[\nC\\e^{-tL_f^\\alpha}\n=\n\\e^{-tL_c^\\alpha}C.\n\\]\n\\end{theorem}","\\begin{proof}\nThe union of the two finite spectra is a finite set. There is a polynomial that agrees with \\(x^\\alpha\\) on that set, and another polynomial that agrees with \\(\\e^{-tx^\\alpha}\\). Exact intertwining passes to every polynomial by induction.\n\\end{proof}","\\begin{theorem}[Fractional defect bound]\\label{extended:record:070}\\label{extended:thm:fractional_defect}\nLet \\(0<\\alpha<1\\). Suppose\n\\[\nL_f\\geq mI,\n\\qquad\nL_c\\geq mI\n\\]\nfor some \\(m>0\\). Define\n\\[\n\\Delta_L=CL_f-L_cC.\n\\]\nThen\n\\[\n\\norm{CL_f^\\alpha-L_c^\\alpha C}\n\\leq\n\\alpha m^{\\alpha-1}\\norm{\\Delta_L}.\n\\]\nConsequently,\n\\[\n\\norm{\nC\\e^{-tL_f^\\alpha}\n-\n\\e^{-tL_c^\\alpha}C\n}\n\\leq\n\\alpha t m^{\\alpha-1}\\norm{\\Delta_L}.\n\\]\n\\end{theorem}","\\begin{proof}\nSet\n\\[\nR_f(s)=(sI+L_f)^{-1},\n\\qquad\nR_c(s)=(sI+L_c)^{-1}.\n\\]\nThe resolvent identity gives\n\\[\nCR_f(s)-R_c(s)C\n=\n-R_c(s)\\Delta_LR_f(s).\n\\]\nSince \\(LR=I-sR\\),\n\\[\nCL_fR_f(s)-L_cR_c(s)C\n=\nsR_c(s)\\Delta_LR_f(s).\n\\]\nInsert this into \\cref{extended:thm:balakrishnan}. The spectral lower bound gives\n\\[\n\\norm{R_f(s)},\\norm{R_c(s)}\n\\leq\n\\frac{1}{s+m}.\n\\]\nTherefore\n\\[\n\\begin{aligned}\n\\norm{CL_f^\\alpha-L_c^\\alpha C}\n&\\leq\n\\frac{\\sin(\\pi\\alpha)}{\\pi}\n\\norm{\\Delta_L}\n\\int_0^\\infty\n\\frac{s^\\alpha}{(s+m)^2}\\dd s","\\begin{proof}\nSet\n\\[\nR_f(s)=(sI+L_f)^{-1},\n\\qquad\nR_c(s)=(sI+L_c)^{-1}.\n\\]\nThe resolvent identity gives\n\\[\nCR_f(s)-R_c(s)C\n=\n-R_c(s)\\Delta_LR_f(s).\n\\]\nSince \\(LR=I-sR\\),\n\\[\nCL_fR_f(s)-L_cR_c(s)C\n=\nsR_c(s)\\Delta_LR_f(s).\n\\]\nInsert this into \\cref{extended:thm:balakrishnan}. The spectral lower bound gives\n\\[\n\\norm{R_f(s)},\\norm{R_c(s)}\n\\leq\n\\frac{1}{s+m}.\n\\]\nTherefore\n\\[\n\\begin{aligned}\n\\norm{CL_f^\\alpha-L_c^\\alpha C}\n&\\leq\n\\frac{\\sin(\\pi\\alpha)}{\\pi}\n\\norm{\\Delta_L}\n\\int_0^\\infty\n\\frac{s^\\alpha}{(s+m)^2}\\dd s\\\\\n&=\n\\alpha m^{\\alpha-1}\\norm{\\Delta_L}.\n\\end{aligned}\n\\]\nThe semigroup estimate follows from \\cref{extended:thm:duhamel_defect}, applied to the contraction generators \\(-L_f^\\alpha\\) and \\(-L_c^\\alpha\\).\n\\end{proof}","&=\n\\alpha m^{\\alpha-1}\\norm{\\Delta_L}.\n\\end{aligned}\n\\]\nThe semigroup estimate follows from \\cref{extended:thm:duhamel_defect}, applied to the contraction generators \\(-L_f^\\alpha\\) and \\(-L_c^\\alpha\\).\n\\end{proof}","\\begin{remark}\\label{extended:record:071}\nThe factor \\(m^{\\alpha-1}\\) records the singular behavior of \\(x^\\alpha\\) near zero. Without a common positive lower bound, a uniform Lipschitz estimate of this form cannot be expected.\n\\end{remark}","\\begin{standardfact}[Stable subordination]\\label{extended:record:072}\\label{extended:fact:stable_subordination}\nLet \\(0<\\alpha<1\\) and \\(t>0\\). There is a probability measure \\(\\eta_{\\alpha,t}\\) on \\([0,\\infty)\\) such that\n\\[\n\\int_0^\\infty \\e^{-s\\lambda}\\eta_{\\alpha,t}(\\dd s)\n=\n\\e^{-t\\lambda^\\alpha}\n\\]\nfor every \\(\\lambda\\geq0\\). Consequently, for a finite dimensional positive self adjoint operator,\n\\[\n\\e^{-tL^\\alpha}\n=\n\\int_0^\\infty\n\\e^{-sL}\\eta_{\\alpha,t}(\\dd s)\n\\]\nin operator norm. The measures satisfy the convolution law\n\\[\n\\eta_{\\alpha,s}*\\eta_{\\alpha,t}\n=\n\\eta_{\\alpha,s+t}.\n\\]\n\\end{standardfact}","n{standardfact}[Stable subordination]\\label{extended:record:072}\\label{extended:fact:stable_subordination}\nLet \\(0<\\alpha<1\\) and \\(t>0\\). There is a probability measure \\(\\eta_{\\alpha,t}\\) on \\([0,\\infty)\\) such that\n\\[\n\\int_0^\\infty \\e^{-s\\lambda}\\eta_{\\alpha,t}(\\dd s)\n=\n\\e^{-t\\lambda^\\alpha}\n\\]\nfor every \\(\\lambda\\geq0\\). Consequently, for a finite dimensional positive self adjoint operator,\n\\[\n\\e^{-tL^\\alpha}\n=\n\\int_0^\\infty\n\\e^{-sL}\\eta_{\\alpha,t}(\\dd s)\n\\]\nin operator norm. The measures satisfy the convolution law\n\\[\n\\eta_{\\alpha,s}*\\eta_{\\alpha,t}\n=\n\\eta_{\\alpha,s+t}.\n\\]\n\\end{standardfact}","d \\(t>0\\). There is a probability measure \\(\\eta_{\\alpha,t}\\) on \\([0,\\infty)\\) such that\n\\[\n\\int_0^\\infty \\e^{-s\\lambda}\\eta_{\\alpha,t}(\\dd s)\n=\n\\e^{-t\\lambda^\\alpha}\n\\]\nfor every \\(\\lambda\\geq0\\). Consequently, for a finite dimensional positive self adjoint operator,\n\\[\n\\e^{-tL^\\alpha}\n=\n\\int_0^\\infty\n\\e^{-sL}\\eta_{\\alpha,t}(\\dd s)\n\\]\nin operator norm. The measures satisfy the convolution law\n\\[\n\\eta_{\\alpha,s}*\\eta_{\\alpha,t}\n=\n\\eta_{\\alpha,s+t}.\n\\]\n\\end{standardfact}","The existence of \\(\\eta_{\\alpha,t}\\) follows from complete monotonicity of \\(\\e^{-t\\lambda^\\alpha}\\). The operator identity is obtained by applying the scalar Laplace transform on every spectral packet. This is the finite spectral form of Bochner subordination. See \\cite{SchillingSongVondracek2012}.","\\begin{theorem}[Poisson subordination]\\label{extended:record:073}\\label{extended:thm:poisson_subordination}\nFor \\(t>0\\),\n\\[\n\\e^{-t\\sqrt L}\n=\n\\frac{t}{2\\sqrt\\pi}\n\\int_0^\\infty\ns^{-3/2}\n\\e^{-t^2/(4s)}\n\\e^{-sL}\n\\dd s.\n\\]\nThe integral converges in operator norm.\n\\end{theorem}","\\begin{proof}\nOn a spectral packet with eigenvalue \\(\\lambda\\), the formula reduces to\n\\[\n\\e^{-t\\sqrt\\lambda}\n=\n\\frac{t}{2\\sqrt\\pi}\n\\int_0^\\infty\ns^{-3/2}\\e^{-t^2/(4s)-\\lambda s}\\dd s.\n\\]\nThis scalar identity follows from the Gaussian integral. Recombine the finitely many spectral packets.\n\\end{proof}","Consider\n\\[\n-u''=f\n\\quad\\hbox{on }(0,1),\n\\qquad\nu(0)=u(1)=0.\n\\]\nThe continuum Green kernel is\n\\[\nG(x,y)=\\min(x,y)-xy.\n\\]\nThus\n\\[\nu(x)=\\int_0^1G(x,y)f(y)\\dd y.\n\\]","For \\(n\\) interior nodes, let\n\\[\nh=\\frac{1}{n+1},\n\\qquad\nx_i=ih,\n\\]\nand define\n\\[\nL_h\n=\n\\frac{1}{h^2}\n\\begin{pmatrix}\n2&-1&&","For \\(n\\) interior nodes, let\n\\[\nh=\\frac{1}{n+1},\n\\qquad\nx_i=ih,\n\\]\nand define\n\\[\nL_h\n=\n\\frac{1}{h^2}\n\\begin{pmatrix}\n2&-1&&\\\\\n-1&2&\\ddots&\\\\\n&\\ddots&\\ddots&-1\\\\\n&&-1&2\n\\end{pmatrix}.\n\\]","\\begin{theorem}[Exact nodal Green reconstruction]\\label{extended:record:075}\\label{extended:thm:nodal_green}\nFor \\(1\\leq i,j\\leq n\\),\n\\[\n\\frac{1}{h}(L_h^{-1})_{ij}\n=\nG(x_i,x_j).\n\\]\n\\end{theorem}","\\begin{proof}\nThe inverse of the unscaled tridiagonal matrix has entry\n\\[\n\\frac{\\min(i,j)(n+1-\\max(i,j))}{n+1}.\n\\]\nMultiplying by \\(h^2\\) and dividing by \\(h\\) gives\n\\[\nh\\min(i,j)\n\\left(1-\\frac{\\max(i,j)}{n+1}\\right)\n=\n\\min(x_i,x_j)-x_ix_j.\n\\]\n\\end{proof}","\\begin{corollary}[Discrete Green quadrature]\\label{extended:record:076}\\label{extended:cor:discrete_green_quadrature}\nIf \\(u^h\\) solves\n\\[\nL_hu^h=f^h,\n\\]\nthen every nodal value has the exact representation\n\\[\nu_i^h\n=\nh\\sum_{j=1}^nG(x_i,x_j)f_j^h.\n\\]\nThus the discrete Green operator is the nodal rectangle quadrature of the continuum Green kernel.\n\\end{corollary}","\\begin{proof}\nBy \\cref{extended:thm:nodal_green},\n\\[\n(L_h^{-1})_{ij}=hG(x_i,x_j).\n\\]\nMultiply the inverse matrix by the source vector.\n\\end{proof}","\\begin{proposition}[Discrete maximum bound]\\label{extended:record:077}\\label{extended:prop:discrete_inverse_bound}\nThe inverse \\(L_h^{-1}\\) has nonnegative entries and\n\\[\n\\norm{L_h^{-1}}_{\\infty\\to\\infty}\n=\n\\max_{1\\leq i\\leq n}\n\\frac{x_i(1-x_i)}{2}\n\\leq\n\\frac18.\n\\]\n\\end{proposition}","\\begin{proof}\nNonnegativity follows from the entry formula in \\cref{extended:thm:nodal_green}. Let \\(v_i=x_i(1-x_i)/2\\). Since the second difference is exact on quadratics,\n\\[\nL_hv=\\one.\n\\]\nHence the \\(i\\)th row sum of \\(L_h^{-1}\\) is \\(v_i\\). For a nonnegative matrix, the infinity operator norm is the largest row sum.\n\\end{proof}","\\begin{theorem}[Second order Poisson solution error]\\label{extended:record:078}\\label{extended:thm:poisson_error}\nSuppose \\(u\\in C^4[0,1]\\) solves the Dirichlet problem above. Let\n\\[\nf^h_i=f(x_i)\n\\]\nand let \\(u^h\\) solve\n\\[\nL_hu^h=f^h.\n\\]\nThen\n\\[\n\\max_i\\abs{u^h_i-u(x_i)}\n\\leq\n\\frac{h^2}{96}\n\\norm{u^{(4)}}_\\infty.\n\\]\n\\end{theorem}","\\begin{proof}\nTaylor expansion with remainder gives\n\\[\n\\norm{L_h(u(x_i))_i-f^h}_\\infty\n\\leq\n\\frac{h^2}{12}\\norm{u^{(4)}}_\\infty.\n\\]\nApply \\(L_h^{-1}\\) and use \\cref{extended:prop:discrete_inverse_bound}.\n\\end{proof}","The discrete eigenpairs are\n\\[\n\\lambda_{k,h}\n=\n\\frac{4}{h^2}\n\\sin^2\\left(\\frac{k\\pi h}{2}\\right),\n\\qquad\nv_{k,h}(i)=\\sin(k\\pi x_i).\n\\]","\\begin{theorem}[Low mode eigenvalue control]\\label{extended:record:079}\\label{extended:thm:low_mode}\nFor \\(kh\\leq1\\),\n\\[\n0\\leq(k\\pi)^2-\\lambda_{k,h}\n\\leq\n\\frac{(k\\pi)^4}{12}h^2.\n\\]\nConsequently, for every fixed \\(k\\),\n\\[\n\\lambda_{k,h}=(k\\pi)^2+O(h^2).\n\\]\n\\end{theorem}","\\begin{proof}\nSet \\(x=k\\pi h/2\\). The condition \\(kh\\leq1\\) gives \\(0\\leq x\\leq\\pi/2\\). On this interval,\n\\[\n\\sin x\\geq x-\\frac{x^3}{6},\n\\]\nso\n\\[\n0\\leq x^2-\\sin^2x\\leq\\frac{x^4}{3}.\n\\]\nMultiply by \\(4/h^2\\).\n\\end{proof}","\\begin{corollary}[Heat and fractional mode transfer]\\label{extended:record:080}\\label{extended:cor:mode_transfer}\nLet \\(\\lambda_k=(k\\pi)^2\\). For \\(t\\geq0\\),\n\\[\n0\\leq\n\\e^{-t\\lambda_{k,h}}-\\e^{-t\\lambda_k}\n\\leq\n\\frac{t(k\\pi)^4}{12}h^2\\e^{-t\\lambda_{k,h}}.\n\\]\nFor \\(0<\\alpha\\leq1\\) and fixed \\(k\\),\n\\[\n\\lambda_{k,h}^\\alpha\n=\n\\lambda_k^\\alpha+O(h^2)\n\\]\nand\n\\[\n\\e^{-t\\lambda_{k,h}^\\alpha}\n=\n\\e^{-t\\lambda_k^\\alpha}+O(th^2)\n\\]\nuniformly for \\(t\\) in bounded intervals.\n\\end{corollary}","mode transfer]\\label{extended:record:080}\\label{extended:cor:mode_transfer}\nLet \\(\\lambda_k=(k\\pi)^2\\). For \\(t\\geq0\\),\n\\[\n0\\leq\n\\e^{-t\\lambda_{k,h}}-\\e^{-t\\lambda_k}\n\\leq\n\\frac{t(k\\pi)^4}{12}h^2\\e^{-t\\lambda_{k,h}}.\n\\]\nFor \\(0<\\alpha\\leq1\\) and fixed \\(k\\),\n\\[\n\\lambda_{k,h}^\\alpha\n=\n\\lambda_k^\\alpha+O(h^2)\n\\]\nand\n\\[\n\\e^{-t\\lambda_{k,h}^\\alpha}\n=\n\\e^{-t\\lambda_k^\\alpha}+O(th^2)\n\\]\nuniformly for \\(t\\) in bounded intervals.\n\\end{corollary}","\\begin{proof}\nThe ordinary heat estimate follows from the mean value theorem and \\cref{extended:thm:low_mode}. For fixed \\(k\\), the eigenvalues stay uniformly away from zero once \\(h\\) is small. The function \\(x^\\alpha\\) is then Lipschitz on a common compact interval containing both eigenvalues. Apply the mean value theorem twice.\n\\end{proof}","Let \\(K\\) be column stochastic. Let \\(V\\) be a real potential on the states and let\n\\[\nD_V=\\diag(V).\n\\]\nFor functions written as columns, the backward Markov operator is \\(K^T\\).","\\begin{theorem}[Discrete Feynman Kac formula]\\label{extended:record:081}\\label{extended:thm:discrete_feynman_kac}\nDefine\n\\[\nM_V=\\diag(\\e^{-V})K^T.\n\\]\nFor terminal data \\(f\\),\n\\[\n(M_V^nf)_i\n=\n\\Exp_i\\left[\n\\exp\\left(\n-\\sum_{r=0}^{n-1}V(X_r)\n\\right)\nf(X_n)\n\\right].\n\\]\n\\end{theorem}","\\begin{proof}\nFor one step,\n\\[\n(M_Vf)_i\n=\n\\e^{-V(i)}\\sum_jK_{ji}f_j.\n\\]\nIterating expands the matrix product over all paths. The product of transition entries is the path probability, and the diagonal factors give the exponential potential weight.\n\\end{proof}","\\begin{theorem}[Continuous time finite state formula]\\label{extended:record:082}\\label{extended:thm:continuous_feynman_kac}\nLet \\(Q\\) be a finite Markov generator and let \\((X_t)_{t\\geq0}\\) be the corresponding chain. Then\n\\[\n\\left(\n\\e^{t(Q^T-D_V)}f\n\\right)_i\n=\n\\Exp_i\\left[\n\\exp\\left(\n-\\int_0^tV(X_s)\\dd s\n\\right)\nf(X_t)\n\\right].\n\\]\nIf \\(V\\geq v_0\\), then\n\\[\n\\norm{\\e^{t(Q^T-D_V)}}_{\\infty\\to\\infty}\n\\leq\n\\e^{-v_0t}.\n\\]\n\\end{theorem}","\\begin{proof}\nConditioning over the first small time interval gives the backward equation\n\\[\n\\dot u=(Q^T-D_V)u,\n\\qquad\nu(0)=f.\n\\]\nThe finite system has a unique solution, namely the displayed matrix exponential. The expectation bound follows from\n\\[\n\\exp\\left(-\\int_0^tV(X_s)\\dd s\\right)\n\\leq\n\\e^{-v_0t}.\n\\]\nThis finite argument is the matrix form of the Feynman Kac principle originating in \\cite{Kac1949}.\n\\end{proof}","\\begin{theorem}[Alternating propagation and killing]\\label{extended:record:083}\\label{extended:thm:feynman_kac_product}\nFor every \\(t\\geq0\\),\n\\[\n\\e^{t(Q^T-D_V)}\n=\n\\lim_{n\\to\\infty}\n\\left(\n\\e^{-tD_V/n}\\e^{tQ^T/n}\n\\right)^n\n\\]\nin operator norm. Each finite product alternates one Markov propagation step with one diagonal survival weight.\n\\end{theorem}","\\begin{proof}\nSet \\(B=-D_V\\) and \\(C=Q^T\\). For bounded matrices,\n\\[\n\\e^{tB/n}\\e^{tC/n}\n=\nI+\\frac{t}{n}(B+C)+O(n^{-2})\n\\]\nuniformly in operator norm. The standard telescoping comparison with\n\\(\n(I+t(B+C)/n)^n\n\\)\nshows that the product converges to \\(\\e^{t(B+C)}\\). This is the displayed factor order, so the limit follows.\n\\end{proof}","\\begin{corollary}[Positivity and composition]\\label{extended:record:084}\\label{extended:cor:feynman_kac_semigroup}\nThe operators\n\\[\nF_V(t)=\\e^{t(Q^T-D_V)}\n\\]\nform a positive semigroup. If \\(f\\geq0\\), then \\(F_V(t)f\\geq0\\).\n\\end{corollary}","Let \\(X=X_P\\oplus X_Q\\) and write\n\\[\nA=\n\\begin{pmatrix}\nA_{PP}&A_{PQ}","Let \\(X=X_P\\oplus X_Q\\) and write\n\\[\nA=\n\\begin{pmatrix}\nA_{PP}&A_{PQ}\\\\\nA_{QP}&A_{QQ}\n\\end{pmatrix}.\n\\]\nFor \\(zI-A_{QQ}\\) invertible, define\n\\[\n\\Sigma_P(z)\n=\nA_{PQ}(zI-A_{QQ})^{-1}A_{QP}\n\\]\nand\n\\[\nF_P(z)\n=\nzI-A_{PP}-\\Sigma_P(z).\n\\]","A_{QP}&A_{QQ}\n\\end{pmatrix}.\n\\]\nFor \\(zI-A_{QQ}\\) invertible, define\n\\[\n\\Sigma_P(z)\n=\nA_{PQ}(zI-A_{QQ})^{-1}A_{QP}\n\\]\nand\n\\[\nF_P(z)\n=\nzI-A_{PP}-\\Sigma_P(z).\n\\]","\\begin{theorem}[Schur Feshbach resolvent]\\label{extended:record:085}\\label{extended:thm:feshbach_resolvent}\nAssume \\(zI-A_{QQ}\\) is invertible. Then \\(zI-A\\) is invertible if and only if \\(F_P(z)\\) is invertible. In that case,\n\\[\nP(zI-A)^{-1}P\n=\nF_P(z)^{-1}.\n\\]\nIn finite dimension,\n\\[\n\\det(zI-A)\n=\n\\det(zI-A_{QQ})\\det F_P(z).\n\\]\n\\end{theorem}","\\begin{proof}\nSet\n\\[\nL_z=\n\\begin{pmatrix}\nI&-A_{PQ}(zI-A_{QQ})^{-1}\\\\\n0&I\n\\end{pmatrix}\n\\]\nand\n\\[\nS_z=\n\\begin{pmatrix}\nF_P(z)&0","0&I\n\\end{pmatrix}\n\\]\nand\n\\[\nS_z=\n\\begin{pmatrix}\nF_P(z)&0\\\\\n-A_{QP}&zI-A_{QQ}\n\\end{pmatrix}.\n\\]\nDirect multiplication gives \\(zI-A=L_zS_z\\). The first factor is invertible. The claims follow from triangular inversion and the determinant of the second factor. This is the finite block form of the Feshbach reduction \\cite{Feshbach1958}.\n\\end{proof}","-A_{QP}&zI-A_{QQ}\n\\end{pmatrix}.\n\\]\nDirect multiplication gives \\(zI-A=L_zS_z\\). The first factor is invertible. The claims follow from triangular inversion and the determinant of the second factor. This is the finite block form of the Feshbach reduction \\cite{Feshbach1958}.\n\\end{proof}","For the homogeneous equation \\(\\dot x=Ax\\), variation of constants gives\n\\[\nx_Q(t)\n=\n\\e^{tA_{QQ}}x_Q(0)\n+\n\\int_0^t\n\\e^{(t-s)A_{QQ}}A_{QP}x_P(s)\n\\dd s.\n\\]\nSubstitution into the retained equation gives\n\\[\n\\begin{aligned}\n\\dot x_P(t)\n={}&\nA_{PP}x_P(t)\n+\nA_{PQ}\\e^{tA_{QQ}}x_Q(0)","For the homogeneous equation \\(\\dot x=Ax\\), variation of constants gives\n\\[\nx_Q(t)\n=\n\\e^{tA_{QQ}}x_Q(0)\n+\n\\int_0^t\n\\e^{(t-s)A_{QQ}}A_{QP}x_P(s)\n\\dd s.\n\\]\nSubstitution into the retained equation gives\n\\[\n\\begin{aligned}\n\\dot x_P(t)\n={}&\nA_{PP}x_P(t)\n+\nA_{PQ}\\e^{tA_{QQ}}x_Q(0)\\\\\n&+\n\\int_0^t\n\\Kcal(t-s)x_P(s)\n\\dd s,\n\\end{aligned}\n\\]\nwhere\n\\[\n\\Kcal(t)\n=\nA_{PQ}\\e^{tA_{QQ}}A_{QP}.\n\\]","&+\n\\int_0^t\n\\Kcal(t-s)x_P(s)\n\\dd s,\n\\end{aligned}\n\\]\nwhere\n\\[\n\\Kcal(t)\n=\nA_{PQ}\\e^{tA_{QQ}}A_{QP}.\n\\]","\\begin{theorem}[Memory and self energy]\\label{extended:record:086}\\label{extended:thm:memory_self_energy}\nIf \\(z\\) lies to the right of the spectrum of \\(A_{QQ}\\), then\n\\[\n\\int_0^\\infty\n\\e^{-zt}\\Kcal(t)\\dd t\n=\n\\Sigma_P(z).\n\\]\nIf, for constants \\(M\\geq0\\) and \\(\\gamma>0\\),\n\\[\n\\norm{\\e^{tA_{QQ}}}\n\\leq\nM\\e^{-\\gamma t},\n\\]\nthen\n\\[\n\\norm{\\Kcal(t)}\n\\leq\nM\\norm{A_{PQ}}\\norm{A_{QP}}\\e^{-\\gamma t}\n\\]\nand\n\\[\n\\norm{\\Kcal}_{L^1(0,\\infty)}\n\\leq\n\\frac{M}{\\gamma}\n\\norm{A_{PQ}}\\norm{A_{QP}}.\n\\]\n\\end{theorem}","begin{theorem}[Memory and self energy]\\label{extended:record:086}\\label{extended:thm:memory_self_energy}\nIf \\(z\\) lies to the right of the spectrum of \\(A_{QQ}\\), then\n\\[\n\\int_0^\\infty\n\\e^{-zt}\\Kcal(t)\\dd t\n=\n\\Sigma_P(z).\n\\]\nIf, for constants \\(M\\geq0\\) and \\(\\gamma>0\\),\n\\[\n\\norm{\\e^{tA_{QQ}}}\n\\leq\nM\\e^{-\\gamma t},\n\\]\nthen\n\\[\n\\norm{\\Kcal(t)}\n\\leq\nM\\norm{A_{PQ}}\\norm{A_{QP}}\\e^{-\\gamma t}\n\\]\nand\n\\[\n\\norm{\\Kcal}_{L^1(0,\\infty)}\n\\leq\n\\frac{M}{\\gamma}\n\\norm{A_{PQ}}\\norm{A_{QP}}.\n\\]\n\\end{theorem}","\\begin{proof}\nIf \\(X_Q=\\{0\\}\\), the memory kernel and self-energy are both zero, so the identity is immediate. Otherwise write \\(B=A_{QQ}\\), which is a bounded operator on the hidden Banach space, and put \\(s(B)=\\max\\operatorname{Re}\\Spec(B)\\). Choose a real \\(\\beta\\) with \\(s(B)<\\beta<\\operatorname{Re}z\\). Spectral mapping and the spectral-radius formula give \\(r(\\e^B)=\\e^{s(B)}<\\e^\\beta\\), hence \\(\\norm{\\e^{nB}}\\leq M_\\beta\\e^{\\beta n}\\) for every integer \\(n\\geq0\\), after increasing \\(M_\\beta\\) to cover the finitely many initial values. Write \\(t=n+r\\), where \\(0\\leq r<1\\). Then\n\\[\n\\norm{\\e^{tB}}\\leq M_\\beta\\exp(\\norm B+\\max(0,-\\beta))\\e^{\\beta t}.\n\\]\nThus \\(\\e^{-zt}\\e^{tB}\\) is Bochner integrable and tends to zero.","\\begin{proof}\nIf \\(X_Q=\\{0\\}\\), the memory kernel and self-energy are both zero, so the identity is immediate. Otherwise write \\(B=A_{QQ}\\), which is a bounded operator on the hidden Banach space, and put \\(s(B)=\\max\\operatorname{Re}\\Spec(B)\\). Choose a real \\(\\beta\\) with \\(s(B)<\\beta<\\operatorname{Re}z\\). Spectral mapping and the spectral-radius formula give \\(r(\\e^B)=\\e^{s(B)}<\\e^\\beta\\), hence \\(\\norm{\\e^{nB}}\\leq M_\\beta\\e^{\\beta n}\\) for every integer \\(n\\geq0\\), after increasing \\(M_\\beta\\) to cover the finitely many initial values. Write \\(t=n+r\\), where \\(0\\leq r<1\\). Then\n\\[\n\\norm{\\e^{tB}}\\leq M_\\beta\\exp(\\norm B+\\max(0,-\\beta))\\e^{\\beta t}.\n\\]\nThus \\(\\e^{-zt}\\e^{tB}\\) is Bochner integrable and tends to zero. Integrating its derivative gives both inverse identities for \\(R=\\int_0^\\infty\\e^{-zt}\\e^{tB}\\dd t\\), so \\(R","\\begin{proof}\nIf \\(X_Q=\\{0\\}\\), the memory kernel and self-energy are both zero, so the identity is immediate. Otherwise write \\(B=A_{QQ}\\), which is a bounded operator on the hidden Banach space, and put \\(s(B)=\\max\\operatorname{Re}\\Spec(B)\\). Choose a real \\(\\beta\\) with \\(s(B)<\\beta<\\operatorname{Re}z\\). Spectral mapping and the spectral-radius formula give \\(r(\\e^B)=\\e^{s(B)}<\\e^\\beta\\), hence \\(\\norm{\\e^{nB}}\\leq M_\\beta\\e^{\\beta n}\\) for every integer \\(n\\geq0\\), after increasing \\(M_\\beta\\) to cover the finitely many initial values. Write \\(t=n+r\\), where \\(0\\leq r<1\\). Then\n\\[\n\\norm{\\e^{tB}}\\leq M_\\beta\\exp(\\norm B+\\max(0,-\\beta))\\e^{\\beta t}.\n\\]\nThus \\(\\e^{-zt}\\e^{tB}\\) is Bochner integrable and tends to zero. Integrating its derivative gives both inverse identities for \\(R=\\int_0^\\infty\\e^{-zt}\\e^{tB}\\dd t\\), so \\(R=(zI-B)^{-1}\\). This is the bounded-generator growth principle in \\cite[Chapter IV, Corollary 3.12]{Enge","\\begin{proof}\nIf \\(X_Q=\\{0\\}\\), the memory kernel and self-energy are both zero, so the identity is immediate. Otherwise write \\(B=A_{QQ}\\), which is a bounded operator on the hidden Banach space, and put \\(s(B)=\\max\\operatorname{Re}\\Spec(B)\\). Choose a real \\(\\beta\\) with \\(s(B)<\\beta<\\operatorname{Re}z\\). Spectral mapping and the spectral-radius formula give \\(r(\\e^B)=\\e^{s(B)}<\\e^\\beta\\), hence \\(\\norm{\\e^{nB}}\\leq M_\\beta\\e^{\\beta n}\\) for every integer \\(n\\geq0\\), after increasing \\(M_\\beta\\) to cover the finitely many initial values. Write \\(t=n+r\\), where \\(0\\leq r<1\\). Then\n\\[\n\\norm{\\e^{tB}}\\leq M_\\beta\\exp(\\norm B+\\max(0,-\\beta))\\e^{\\beta t}.\n\\]\nThus \\(\\e^{-zt}\\e^{tB}\\) is Bochner integrable and tends to zero. Integrating its derivative gives both inverse identities for \\(R=\\int_0^\\infty\\e^{-zt}\\e^{tB}\\dd t\\), so \\(R=(zI-B)^{-1}\\). This is the bounded-generator growth principle in \\cite[Chapter IV, Corollary 3.12]{EngelNagel2000}. Multiplicati","\\begin{proof}\nIf \\(X_Q=\\{0\\}\\), the memory kernel and self-energy are both zero, so the identity is immediate. Otherwise write \\(B=A_{QQ}\\), which is a bounded operator on the hidden Banach space, and put \\(s(B)=\\max\\operatorname{Re}\\Spec(B)\\). Choose a real \\(\\beta\\) with \\(s(B)<\\beta<\\operatorname{Re}z\\). Spectral mapping and the spectral-radius formula give \\(r(\\e^B)=\\e^{s(B)}<\\e^\\beta\\), hence \\(\\norm{\\e^{nB}}\\leq M_\\beta\\e^{\\beta n}\\) for every integer \\(n\\geq0\\), after increasing \\(M_\\beta\\) to cover the finitely many initial values. Write \\(t=n+r\\), where \\(0\\leq r<1\\). Then\n\\[\n\\norm{\\e^{tB}}\\leq M_\\beta\\exp(\\norm B+\\max(0,-\\beta))\\e^{\\beta t}.\n\\]\nThus \\(\\e^{-zt}\\e^{tB}\\) is Bochner integrable and tends to zero. Integrating its derivative gives both inverse identities for \\(R=\\int_0^\\infty\\e^{-zt}\\e^{tB}\\dd t\\), so \\(R=(zI-B)^{-1}\\). This is the bounded-generator growth principle in \\cite[Chapter IV, Corollary 3.12]{EngelNagel2000}. Multiplication by the coupling blocks gives the fir","\\begin{proof}\nIf \\(X_Q=\\{0\\}\\), the memory kernel and self-energy are both zero, so the identity is immediate. Otherwise write \\(B=A_{QQ}\\), which is a bounded operator on the hidden Banach space, and put \\(s(B)=\\max\\operatorname{Re}\\Spec(B)\\). Choose a real \\(\\beta\\) with \\(s(B)<\\beta<\\operatorname{Re}z\\). Spectral mapping and the spectral-radius formula give \\(r(\\e^B)=\\e^{s(B)}<\\e^\\beta\\), hence \\(\\norm{\\e^{nB}}\\leq M_\\beta\\e^{\\beta n}\\) for every integer \\(n\\geq0\\), after increasing \\(M_\\beta\\) to cover the finitely many initial values. Write \\(t=n+r\\), where \\(0\\leq r<1\\). Then\n\\[\n\\norm{\\e^{tB}}\\leq M_\\beta\\exp(\\norm B+\\max(0,-\\beta))\\e^{\\beta t}.\n\\]\nThus \\(\\e^{-zt}\\e^{tB}\\) is Bochner integrable and tends to zero. Integrating its derivative gives both inverse identities for \\(R=\\int_0^\\infty\\e^{-zt}\\e^{tB}\\dd t\\), so \\(R=(zI-B)^{-1}\\). This is the bounded-generator growth principle in \\cite[Chapter IV, Corollary 3.12]{EngelNagel2000}. Multiplication by the coupling blocks gives the first identity. The remaining estimates follow directly from the assumed positive decay r","gin{proof}\nIf \\(X_Q=\\{0\\}\\), the memory kernel and self-energy are both zero, so the identity is immediate. Otherwise write \\(B=A_{QQ}\\), which is a bounded operator on the hidden Banach space, and put \\(s(B)=\\max\\operatorname{Re}\\Spec(B)\\). Choose a real \\(\\beta\\) with \\(s(B)<\\beta<\\operatorname{Re}z\\). Spectral mapping and the spectral-radius formula give \\(r(\\e^B)=\\e^{s(B)}<\\e^\\beta\\), hence \\(\\norm{\\e^{nB}}\\leq M_\\beta\\e^{\\beta n}\\) for every integer \\(n\\geq0\\), after increasing \\(M_\\beta\\) to cover the finitely many initial values. Write \\(t=n+r\\), where \\(0\\leq r<1\\). Then\n\\[\n\\norm{\\e^{tB}}\\leq M_\\beta\\exp(\\norm B+\\max(0,-\\beta))\\e^{\\beta t}.\n\\]\nThus \\(\\e^{-zt}\\e^{tB}\\) is Bochner integrable and tends to zero. Integrating its derivative gives both inverse identities for \\(R=\\int_0^\\infty\\e^{-zt}\\e^{tB}\\dd t\\), so \\(R=(zI-B)^{-1}\\). This is the bounded-generator growth principle in \\cite[Chapter IV, Corollary 3.12]{EngelNagel2000}. Multiplication by the coupling blocks gives the first identity. The remaining estimates follow directly from the assumed positive decay rate.\n\\end{pr…","y are both zero, so the identity is immediate. Otherwise write \\(B=A_{QQ}\\), which is a bounded operator on the hidden Banach space, and put \\(s(B)=\\max\\operatorname{Re}\\Spec(B)\\). Choose a real \\(\\beta\\) with \\(s(B)<\\beta<\\operatorname{Re}z\\). Spectral mapping and the spectral-radius formula give \\(r(\\e^B)=\\e^{s(B)}<\\e^\\beta\\), hence \\(\\norm{\\e^{nB}}\\leq M_\\beta\\e^{\\beta n}\\) for every integer \\(n\\geq0\\), after increasing \\(M_\\beta\\) to cover the finitely many initial values. Write \\(t=n+r\\), where \\(0\\leq r<1\\). Then\n\\[\n\\norm{\\e^{tB}}\\leq M_\\beta\\exp(\\norm B+\\max(0,-\\beta))\\e^{\\beta t}.\n\\]\nThus \\(\\e^{-zt}\\e^{tB}\\) is Bochner integrable and tends to zero. Integrating its derivative gives both inverse identities for \\(R=\\int_0^\\infty\\e^{-zt}\\e^{tB}\\dd t\\), so \\(R=(zI-B)^{-1}\\). This is the bounded-generator growth principle in \\cite[Chapter IV, Corollary 3.12]{EngelNagel2000}. Multiplication by the coupling blocks gives the first identity. The remaining estimates follow directly from the assumed positive decay rate.\n\\end{proof}","ity is immediate. Otherwise write \\(B=A_{QQ}\\), which is a bounded operator on the hidden Banach space, and put \\(s(B)=\\max\\operatorname{Re}\\Spec(B)\\). Choose a real \\(\\beta\\) with \\(s(B)<\\beta<\\operatorname{Re}z\\). Spectral mapping and the spectral-radius formula give \\(r(\\e^B)=\\e^{s(B)}<\\e^\\beta\\), hence \\(\\norm{\\e^{nB}}\\leq M_\\beta\\e^{\\beta n}\\) for every integer \\(n\\geq0\\), after increasing \\(M_\\beta\\) to cover the finitely many initial values. Write \\(t=n+r\\), where \\(0\\leq r<1\\). Then\n\\[\n\\norm{\\e^{tB}}\\leq M_\\beta\\exp(\\norm B+\\max(0,-\\beta))\\e^{\\beta t}.\n\\]\nThus \\(\\e^{-zt}\\e^{tB}\\) is Bochner integrable and tends to zero. Integrating its derivative gives both inverse identities for \\(R=\\int_0^\\infty\\e^{-zt}\\e^{tB}\\dd t\\), so \\(R=(zI-B)^{-1}\\). This is the bounded-generator growth principle in \\cite[Chapter IV, Corollary 3.12]{EngelNagel2000}. Multiplication by the coupling blocks gives the first identity. The remaining estimates follow directly from the assumed positive decay rate.\n\\end{proof}","unded operator on the hidden Banach space, and put \\(s(B)=\\max\\operatorname{Re}\\Spec(B)\\). Choose a real \\(\\beta\\) with \\(s(B)<\\beta<\\operatorname{Re}z\\). Spectral mapping and the spectral-radius formula give \\(r(\\e^B)=\\e^{s(B)}<\\e^\\beta\\), hence \\(\\norm{\\e^{nB}}\\leq M_\\beta\\e^{\\beta n}\\) for every integer \\(n\\geq0\\), after increasing \\(M_\\beta\\) to cover the finitely many initial values. Write \\(t=n+r\\), where \\(0\\leq r<1\\). Then\n\\[\n\\norm{\\e^{tB}}\\leq M_\\beta\\exp(\\norm B+\\max(0,-\\beta))\\e^{\\beta t}.\n\\]\nThus \\(\\e^{-zt}\\e^{tB}\\) is Bochner integrable and tends to zero. Integrating its derivative gives both inverse identities for \\(R=\\int_0^\\infty\\e^{-zt}\\e^{tB}\\dd t\\), so \\(R=(zI-B)^{-1}\\). This is the bounded-generator growth principle in \\cite[Chapter IV, Corollary 3.12]{EngelNagel2000}. Multiplication by the coupling blocks gives the first identity. The remaining estimates follow directly from the assumed positive decay rate.\n\\end{proof}","the hidden Banach space, and put \\(s(B)=\\max\\operatorname{Re}\\Spec(B)\\). Choose a real \\(\\beta\\) with \\(s(B)<\\beta<\\operatorname{Re}z\\). Spectral mapping and the spectral-radius formula give \\(r(\\e^B)=\\e^{s(B)}<\\e^\\beta\\), hence \\(\\norm{\\e^{nB}}\\leq M_\\beta\\e^{\\beta n}\\) for every integer \\(n\\geq0\\), after increasing \\(M_\\beta\\) to cover the finitely many initial values. Write \\(t=n+r\\), where \\(0\\leq r<1\\). Then\n\\[\n\\norm{\\e^{tB}}\\leq M_\\beta\\exp(\\norm B+\\max(0,-\\beta))\\e^{\\beta t}.\n\\]\nThus \\(\\e^{-zt}\\e^{tB}\\) is Bochner integrable and tends to zero. Integrating its derivative gives both inverse identities for \\(R=\\int_0^\\infty\\e^{-zt}\\e^{tB}\\dd t\\), so \\(R=(zI-B)^{-1}\\). This is the bounded-generator growth principle in \\cite[Chapter IV, Corollary 3.12]{EngelNagel2000}. Multiplication by the coupling blocks gives the first identity. The remaining estimates follow directly from the assumed positive decay rate.\n\\end{proof}","pace, and put \\(s(B)=\\max\\operatorname{Re}\\Spec(B)\\). Choose a real \\(\\beta\\) with \\(s(B)<\\beta<\\operatorname{Re}z\\). Spectral mapping and the spectral-radius formula give \\(r(\\e^B)=\\e^{s(B)}<\\e^\\beta\\), hence \\(\\norm{\\e^{nB}}\\leq M_\\beta\\e^{\\beta n}\\) for every integer \\(n\\geq0\\), after increasing \\(M_\\beta\\) to cover the finitely many initial values. Write \\(t=n+r\\), where \\(0\\leq r<1\\). Then\n\\[\n\\norm{\\e^{tB}}\\leq M_\\beta\\exp(\\norm B+\\max(0,-\\beta))\\e^{\\beta t}.\n\\]\nThus \\(\\e^{-zt}\\e^{tB}\\) is Bochner integrable and tends to zero. Integrating its derivative gives both inverse identities for \\(R=\\int_0^\\infty\\e^{-zt}\\e^{tB}\\dd t\\), so \\(R=(zI-B)^{-1}\\). This is the bounded-generator growth principle in \\cite[Chapter IV, Corollary 3.12]{EngelNagel2000}. Multiplication by the coupling blocks gives the first identity. The remaining estimates follow directly from the assumed positive decay rate.\n\\end{proof}","ith \\(s(B)<\\beta<\\operatorname{Re}z\\). Spectral mapping and the spectral-radius formula give \\(r(\\e^B)=\\e^{s(B)}<\\e^\\beta\\), hence \\(\\norm{\\e^{nB}}\\leq M_\\beta\\e^{\\beta n}\\) for every integer \\(n\\geq0\\), after increasing \\(M_\\beta\\) to cover the finitely many initial values. Write \\(t=n+r\\), where \\(0\\leq r<1\\). Then\n\\[\n\\norm{\\e^{tB}}\\leq M_\\beta\\exp(\\norm B+\\max(0,-\\beta))\\e^{\\beta t}.\n\\]\nThus \\(\\e^{-zt}\\e^{tB}\\) is Bochner integrable and tends to zero. Integrating its derivative gives both inverse identities for \\(R=\\int_0^\\infty\\e^{-zt}\\e^{tB}\\dd t\\), so \\(R=(zI-B)^{-1}\\). This is the bounded-generator growth principle in \\cite[Chapter IV, Corollary 3.12]{EngelNagel2000}. Multiplication by the coupling blocks gives the first identity. The remaining estimates follow directly from the assumed positive decay rate.\n\\end{proof}","), hence \\(\\norm{\\e^{nB}}\\leq M_\\beta\\e^{\\beta n}\\) for every integer \\(n\\geq0\\), after increasing \\(M_\\beta\\) to cover the finitely many initial values. Write \\(t=n+r\\), where \\(0\\leq r<1\\). Then\n\\[\n\\norm{\\e^{tB}}\\leq M_\\beta\\exp(\\norm B+\\max(0,-\\beta))\\e^{\\beta t}.\n\\]\nThus \\(\\e^{-zt}\\e^{tB}\\) is Bochner integrable and tends to zero. Integrating its derivative gives both inverse identities for \\(R=\\int_0^\\infty\\e^{-zt}\\e^{tB}\\dd t\\), so \\(R=(zI-B)^{-1}\\). This is the bounded-generator growth principle in \\cite[Chapter IV, Corollary 3.12]{EngelNagel2000}. Multiplication by the coupling blocks gives the first identity. The remaining estimates follow directly from the assumed positive decay rate.\n\\end{proof}","ta n}\\) for every integer \\(n\\geq0\\), after increasing \\(M_\\beta\\) to cover the finitely many initial values. Write \\(t=n+r\\), where \\(0\\leq r<1\\). Then\n\\[\n\\norm{\\e^{tB}}\\leq M_\\beta\\exp(\\norm B+\\max(0,-\\beta))\\e^{\\beta t}.\n\\]\nThus \\(\\e^{-zt}\\e^{tB}\\) is Bochner integrable and tends to zero. Integrating its derivative gives both inverse identities for \\(R=\\int_0^\\infty\\e^{-zt}\\e^{tB}\\dd t\\), so \\(R=(zI-B)^{-1}\\). This is the bounded-generator growth principle in \\cite[Chapter IV, Corollary 3.12]{EngelNagel2000}. Multiplication by the coupling blocks gives the first identity. The remaining estimates follow directly from the assumed positive decay rate.\n\\end{proof}","\\begin{theorem}[Criterion for exact local closure]\\label{extended:record:087}\\label{extended:thm:local_closure}\nAssume \\(x_Q(0)=0\\) and \\(d=\\dim X_Q<\\infty\\). The following conditions are equivalent.\n\\begin{enumerate}\n\\item For every retained initial state,\n\\[\nP\\e^{tA}P\n=\n\\e^{tA_{PP}}P\n\\]\nfor all \\(t\\geq0\\).\n\\item \\(\\Sigma_P(z)=0\\) for every sufficiently large \\(z\\).\n\\item \\(\\Kcal(t)=0\\) for every \\(t\\geq0\\).\n\\item If \\(d=\\dim X_Q\\), then\n\\[\nA_{PQ}A_{QQ}^kA_{QP}=0\n\\]\nfor \\(0\\leq k<d\\).\n\\end{enumerate}\n\\end{theorem}","gin{theorem}[Criterion for exact local closure]\\label{extended:record:087}\\label{extended:thm:local_closure}\nAssume \\(x_Q(0)=0\\) and \\(d=\\dim X_Q<\\infty\\). The following conditions are equivalent.\n\\begin{enumerate}\n\\item For every retained initial state,\n\\[\nP\\e^{tA}P\n=\n\\e^{tA_{PP}}P\n\\]\nfor all \\(t\\geq0\\).\n\\item \\(\\Sigma_P(z)=0\\) for every sufficiently large \\(z\\).\n\\item \\(\\Kcal(t)=0\\) for every \\(t\\geq0\\).\n\\item If \\(d=\\dim X_Q\\), then\n\\[\nA_{PQ}A_{QQ}^kA_{QP}=0\n\\]\nfor \\(0\\leq k<d\\).\n\\end{enumerate}\n\\end{theorem}","n for exact local closure]\\label{extended:record:087}\\label{extended:thm:local_closure}\nAssume \\(x_Q(0)=0\\) and \\(d=\\dim X_Q<\\infty\\). The following conditions are equivalent.\n\\begin{enumerate}\n\\item For every retained initial state,\n\\[\nP\\e^{tA}P\n=\n\\e^{tA_{PP}}P\n\\]\nfor all \\(t\\geq0\\).\n\\item \\(\\Sigma_P(z)=0\\) for every sufficiently large \\(z\\).\n\\item \\(\\Kcal(t)=0\\) for every \\(t\\geq0\\).\n\\item If \\(d=\\dim X_Q\\), then\n\\[\nA_{PQ}A_{QQ}^kA_{QP}=0\n\\]\nfor \\(0\\leq k<d\\).\n\\end{enumerate}\n\\end{theorem}","sure]\\label{extended:record:087}\\label{extended:thm:local_closure}\nAssume \\(x_Q(0)=0\\) and \\(d=\\dim X_Q<\\infty\\). The following conditions are equivalent.\n\\begin{enumerate}\n\\item For every retained initial state,\n\\[\nP\\e^{tA}P\n=\n\\e^{tA_{PP}}P\n\\]\nfor all \\(t\\geq0\\).\n\\item \\(\\Sigma_P(z)=0\\) for every sufficiently large \\(z\\).\n\\item \\(\\Kcal(t)=0\\) for every \\(t\\geq0\\).\n\\item If \\(d=\\dim X_Q\\), then\n\\[\nA_{PQ}A_{QQ}^kA_{QP}=0\n\\]\nfor \\(0\\leq k<d\\).\n\\end{enumerate}\n\\end{theorem}","label{extended:thm:local_closure}\nAssume \\(x_Q(0)=0\\) and \\(d=\\dim X_Q<\\infty\\). The following conditions are equivalent.\n\\begin{enumerate}\n\\item For every retained initial state,\n\\[\nP\\e^{tA}P\n=\n\\e^{tA_{PP}}P\n\\]\nfor all \\(t\\geq0\\).\n\\item \\(\\Sigma_P(z)=0\\) for every sufficiently large \\(z\\).\n\\item \\(\\Kcal(t)=0\\) for every \\(t\\geq0\\).\n\\item If \\(d=\\dim X_Q\\), then\n\\[\nA_{PQ}A_{QQ}^kA_{QP}=0\n\\]\nfor \\(0\\leq k<d\\).\n\\end{enumerate}\n\\end{theorem}","\\begin{proof}\nBy \\cref{extended:thm:laplace_packet_resolvent}, the Laplace transform of the projected propagator is the projected resolvent. By \\cref{extended:thm:feshbach_resolvent}, the first condition is equivalent to\n\\[\n(zI-A_{PP}-\\Sigma_P(z))^{-1}\n=\n(zI-A_{PP})^{-1}\n\\]\nfor all sufficiently large \\(z\\), which is equivalent to the second condition. The second and third conditions are equivalent by \\cref{extended:thm:memory_self_energy} and uniqueness of the Laplace transform.","For large \\(z\\),\n\\[\n\\Sigma_P(z)\n=\n\\sum_{k=0}^{\\infty}\n\\frac{A_{PQ}A_{QQ}^kA_{QP}}{z^{k+1}}.\n\\]\nThus the second condition implies the moment identities. Conversely, the first \\(d\\) identities imply all later ones by the Cayley Hamilton theorem, so the series vanishes.\n\\end{proof}","\\begin{remark}\\label{extended:record:088}\nA finite hidden block produces a strictly proper rational self energy. It tends to zero as \\(z\\) tends to infinity. Hence a nonzero constant self energy cannot arise from hidden dynamics alone. A constant correction must already be included in the retained generator.\n\\end{remark}","\\begin{theorem}[Polynomial exponential memory of a finite hidden system]\\label{extended:record:089}\\label{extended:thm:modal_memory}\nLet \\(X_Q\\) be a finite dimensional complex space. Let \\(\\mu_1,\\ldots,\\mu_r\\) be the distinct eigenvalues of \\(A_{QQ}\\), let \\(Q_j\\) be the corresponding primary spectral projections, and set\n\\[\nN_j=(A_{QQ}-\\mu_jI)Q_j.\n\\]\nLet \\(\\nu_j\\) be the least positive integer for which \\(N_j^{\\nu_j}=0\\), and define\n\\[\nB_{j,k}\n=\nA_{PQ}N_j^kQ_jA_{QP}\n\\qquad\n(0\\leq k<\\nu_j).\n\\]\nThen\n\\[\n\\Kcal(t)\n=\n\\sum_{j=1}^r\n\\e^{\\mu_jt}\n\\sum_{k=0}^{\\nu_j-1}\n\\frac{t^k}{k!}B_{j,k}\n\\]\nand\n\\[\n\\Sigma_P(z)\n=\n\\sum_{j=1}^r\n\\sum_{k=0}^{\\nu_j-1}\n\\frac{B_{j,k}}{(z-\\mu_j)^{k+1}}.\n\\]\nSuppose \\(\\operatorname{Re}\\mu_j<0\\) for every \\(j\\). If \\(J\\subseteq\\{1,\\ldots,r\\}\\) and \\(\\Kcal_J\\) retains the complete primary blocks indexed by \\(J\\), then","\\begin{theorem}[Polynomial exponential memory of a finite hidden system]\\label{extended:record:089}\\label{extended:thm:modal_memory}\nLet \\(X_Q\\) be a finite dimensional complex space. Let \\(\\mu_1,\\ldots,\\mu_r\\) be the distinct eigenvalues of \\(A_{QQ}\\), let \\(Q_j\\) be the corresponding primary spectral projections, and set\n\\[\nN_j=(A_{QQ}-\\mu_jI)Q_j.\n\\]\nLet \\(\\nu_j\\) be the least positive integer for which \\(N_j^{\\nu_j}=0\\), and define\n\\[\nB_{j,k}\n=\nA_{PQ}N_j^kQ_jA_{QP}\n\\qquad\n(0\\leq k<\\nu_j).\n\\]\nThen\n\\[\n\\Kcal(t)\n=\n\\sum_{j=1}^r\n\\e^{\\mu_jt}\n\\sum_{k=0}^{\\nu_j-1}\n\\frac{t^k}{k!}B_{j,k}\n\\]\nand\n\\[\n\\Sigma_P(z)\n=\n\\sum_{j=1}^r\n\\sum_{k=0}^{\\nu_j-1}\n\\frac{B_{j,k}}{(z-\\mu_j)^{k+1}}.\n\\]\nSuppose \\(\\operatorname{Re}\\mu_j<0\\) for every \\(j\\). If \\(J\\subseteq\\{1,\\ldots,r\\}\\) and \\(\\Kcal_J\\) retains the complete primary blocks indexed by \\(J\\), then\n\\[\n\\norm{\\Kcal-\\Kcal_J}_{L^1(0,\\infty)}\n\\leq\n\\sum_{j\\notin J}\n\\sum","\\begin{theorem}[Polynomial exponential memory of a finite hidden system]\\label{extended:record:089}\\label{extended:thm:modal_memory}\nLet \\(X_Q\\) be a finite dimensional complex space. Let \\(\\mu_1,\\ldots,\\mu_r\\) be the distinct eigenvalues of \\(A_{QQ}\\), let \\(Q_j\\) be the corresponding primary spectral projections, and set\n\\[\nN_j=(A_{QQ}-\\mu_jI)Q_j.\n\\]\nLet \\(\\nu_j\\) be the least positive integer for which \\(N_j^{\\nu_j}=0\\), and define\n\\[\nB_{j,k}\n=\nA_{PQ}N_j^kQ_jA_{QP}\n\\qquad\n(0\\leq k<\\nu_j).\n\\]\nThen\n\\[\n\\Kcal(t)\n=\n\\sum_{j=1}^r\n\\e^{\\mu_jt}\n\\sum_{k=0}^{\\nu_j-1}\n\\frac{t^k}{k!}B_{j,k}\n\\]\nand\n\\[\n\\Sigma_P(z)\n=\n\\sum_{j=1}^r\n\\sum_{k=0}^{\\nu_j-1}\n\\frac{B_{j,k}}{(z-\\mu_j)^{k+1}}.\n\\]\nSuppose \\(\\operatorname{Re}\\mu_j<0\\) for every \\(j\\). If \\(J\\subseteq\\{1,\\ldots,r\\}\\) and \\(\\Kcal_J\\) retains the complete primary blocks indexed by \\(J\\), then\n\\[\n\\norm{\\Kcal-\\Kcal_J}_{L^1(0,\\infty)}\n\\leq\n\\sum_{j\\notin J}\n\\sum_{k=0}^{\\nu_j-1}\n\\frac{\\norm{B_{j,k}}}\n{(-","\\begin{theorem}[Polynomial exponential memory of a finite hidden system]\\label{extended:record:089}\\label{extended:thm:modal_memory}\nLet \\(X_Q\\) be a finite dimensional complex space. Let \\(\\mu_1,\\ldots,\\mu_r\\) be the distinct eigenvalues of \\(A_{QQ}\\), let \\(Q_j\\) be the corresponding primary spectral projections, and set\n\\[\nN_j=(A_{QQ}-\\mu_jI)Q_j.\n\\]\nLet \\(\\nu_j\\) be the least positive integer for which \\(N_j^{\\nu_j}=0\\), and define\n\\[\nB_{j,k}\n=\nA_{PQ}N_j^kQ_jA_{QP}\n\\qquad\n(0\\leq k<\\nu_j).\n\\]\nThen\n\\[\n\\Kcal(t)\n=\n\\sum_{j=1}^r\n\\e^{\\mu_jt}\n\\sum_{k=0}^{\\nu_j-1}\n\\frac{t^k}{k!}B_{j,k}\n\\]\nand\n\\[\n\\Sigma_P(z)\n=\n\\sum_{j=1}^r\n\\sum_{k=0}^{\\nu_j-1}\n\\frac{B_{j,k}}{(z-\\mu_j)^{k+1}}.\n\\]\nSuppose \\(\\operatorname{Re}\\mu_j<0\\) for every \\(j\\). If \\(J\\subseteq\\{1,\\ldots,r\\}\\) and \\(\\Kcal_J\\) retains the complete primary blocks indexed by \\(J\\), then\n\\[\n\\norm{\\Kcal-\\Kcal_J}_{L^1(0,\\infty)}\n\\leq\n\\sum_{j\\notin J}\n\\sum_{k=0}^{\\nu_j-1}\n\\frac{\\norm{B_{j,k}}}\n{(-\\operatorname","\\begin{theorem}[Polynomial exponential memory of a finite hidden system]\\label{extended:record:089}\\label{extended:thm:modal_memory}\nLet \\(X_Q\\) be a finite dimensional complex space. Let \\(\\mu_1,\\ldots,\\mu_r\\) be the distinct eigenvalues of \\(A_{QQ}\\), let \\(Q_j\\) be the corresponding primary spectral projections, and set\n\\[\nN_j=(A_{QQ}-\\mu_jI)Q_j.\n\\]\nLet \\(\\nu_j\\) be the least positive integer for which \\(N_j^{\\nu_j}=0\\), and define\n\\[\nB_{j,k}\n=\nA_{PQ}N_j^kQ_jA_{QP}\n\\qquad\n(0\\leq k<\\nu_j).\n\\]\nThen\n\\[\n\\Kcal(t)\n=\n\\sum_{j=1}^r\n\\e^{\\mu_jt}\n\\sum_{k=0}^{\\nu_j-1}\n\\frac{t^k}{k!}B_{j,k}\n\\]\nand\n\\[\n\\Sigma_P(z)\n=\n\\sum_{j=1}^r\n\\sum_{k=0}^{\\nu_j-1}\n\\frac{B_{j,k}}{(z-\\mu_j)^{k+1}}.\n\\]\nSuppose \\(\\operatorname{Re}\\mu_j<0\\) for every \\(j\\). If \\(J\\subseteq\\{1,\\ldots,r\\}\\) and \\(\\Kcal_J\\) retains the complete primary blocks indexed by \\(J\\), then\n\\[\n\\norm{\\Kcal-\\Kcal_J}_{L^1(0,\\infty)}\n\\leq\n\\sum_{j\\notin J}\n\\sum_{k=0}^{\\nu_j-1}\n\\frac{\\norm{B_{j,k}}}\n{(-\\operatorname{Re}\\mu_j)^{k+1}}.\n\\]\n\\end{theorem}","rem}[Polynomial exponential memory of a finite hidden system]\\label{extended:record:089}\\label{extended:thm:modal_memory}\nLet \\(X_Q\\) be a finite dimensional complex space. Let \\(\\mu_1,\\ldots,\\mu_r\\) be the distinct eigenvalues of \\(A_{QQ}\\), let \\(Q_j\\) be the corresponding primary spectral projections, and set\n\\[\nN_j=(A_{QQ}-\\mu_jI)Q_j.\n\\]\nLet \\(\\nu_j\\) be the least positive integer for which \\(N_j^{\\nu_j}=0\\), and define\n\\[\nB_{j,k}\n=\nA_{PQ}N_j^kQ_jA_{QP}\n\\qquad\n(0\\leq k<\\nu_j).\n\\]\nThen\n\\[\n\\Kcal(t)\n=\n\\sum_{j=1}^r\n\\e^{\\mu_jt}\n\\sum_{k=0}^{\\nu_j-1}\n\\frac{t^k}{k!}B_{j,k}\n\\]\nand\n\\[\n\\Sigma_P(z)\n=\n\\sum_{j=1}^r\n\\sum_{k=0}^{\\nu_j-1}\n\\frac{B_{j,k}}{(z-\\mu_j)^{k+1}}.\n\\]\nSuppose \\(\\operatorname{Re}\\mu_j<0\\) for every \\(j\\). If \\(J\\subseteq\\{1,\\ldots,r\\}\\) and \\(\\Kcal_J\\) retains the complete primary blocks indexed by \\(J\\), then\n\\[\n\\norm{\\Kcal-\\Kcal_J}_{L^1(0,\\infty)}\n\\leq\n\\sum_{j\\notin J}\n\\sum_{k=0}^{\\nu_j-1}\n\\frac{\\norm{B_{j,k}}}\n{(-\\operatorname{Re}\\mu_j)^{k+1}}.\n\\]\n\\end{theorem}","mory of a finite hidden system]\\label{extended:record:089}\\label{extended:thm:modal_memory}\nLet \\(X_Q\\) be a finite dimensional complex space. Let \\(\\mu_1,\\ldots,\\mu_r\\) be the distinct eigenvalues of \\(A_{QQ}\\), let \\(Q_j\\) be the corresponding primary spectral projections, and set\n\\[\nN_j=(A_{QQ}-\\mu_jI)Q_j.\n\\]\nLet \\(\\nu_j\\) be the least positive integer for which \\(N_j^{\\nu_j}=0\\), and define\n\\[\nB_{j,k}\n=\nA_{PQ}N_j^kQ_jA_{QP}\n\\qquad\n(0\\leq k<\\nu_j).\n\\]\nThen\n\\[\n\\Kcal(t)\n=\n\\sum_{j=1}^r\n\\e^{\\mu_jt}\n\\sum_{k=0}^{\\nu_j-1}\n\\frac{t^k}{k!}B_{j,k}\n\\]\nand\n\\[\n\\Sigma_P(z)\n=\n\\sum_{j=1}^r\n\\sum_{k=0}^{\\nu_j-1}\n\\frac{B_{j,k}}{(z-\\mu_j)^{k+1}}.\n\\]\nSuppose \\(\\operatorname{Re}\\mu_j<0\\) for every \\(j\\). If \\(J\\subseteq\\{1,\\ldots,r\\}\\) and \\(\\Kcal_J\\) retains the complete primary blocks indexed by \\(J\\), then\n\\[\n\\norm{\\Kcal-\\Kcal_J}_{L^1(0,\\infty)}\n\\leq\n\\sum_{j\\notin J}\n\\sum_{k=0}^{\\nu_j-1}\n\\frac{\\norm{B_{j,k}}}\n{(-\\operatorname{Re}\\mu_j)^{k+1}}.\n\\]\n\\end{theorem}","xtended:thm:modal_memory}\nLet \\(X_Q\\) be a finite dimensional complex space. Let \\(\\mu_1,\\ldots,\\mu_r\\) be the distinct eigenvalues of \\(A_{QQ}\\), let \\(Q_j\\) be the corresponding primary spectral projections, and set\n\\[\nN_j=(A_{QQ}-\\mu_jI)Q_j.\n\\]\nLet \\(\\nu_j\\) be the least positive integer for which \\(N_j^{\\nu_j}=0\\), and define\n\\[\nB_{j,k}\n=\nA_{PQ}N_j^kQ_jA_{QP}\n\\qquad\n(0\\leq k<\\nu_j).\n\\]\nThen\n\\[\n\\Kcal(t)\n=\n\\sum_{j=1}^r\n\\e^{\\mu_jt}\n\\sum_{k=0}^{\\nu_j-1}\n\\frac{t^k}{k!}B_{j,k}\n\\]\nand\n\\[\n\\Sigma_P(z)\n=\n\\sum_{j=1}^r\n\\sum_{k=0}^{\\nu_j-1}\n\\frac{B_{j,k}}{(z-\\mu_j)^{k+1}}.\n\\]\nSuppose \\(\\operatorname{Re}\\mu_j<0\\) for every \\(j\\). If \\(J\\subseteq\\{1,\\ldots,r\\}\\) and \\(\\Kcal_J\\) retains the complete primary blocks indexed by \\(J\\), then\n\\[\n\\norm{\\Kcal-\\Kcal_J}_{L^1(0,\\infty)}\n\\leq\n\\sum_{j\\notin J}\n\\sum_{k=0}^{\\nu_j-1}\n\\frac{\\norm{B_{j,k}}}\n{(-\\operatorname{Re}\\mu_j)^{k+1}}.\n\\]\n\\end{theorem}","\\ldots,\\mu_r\\) be the distinct eigenvalues of \\(A_{QQ}\\), let \\(Q_j\\) be the corresponding primary spectral projections, and set\n\\[\nN_j=(A_{QQ}-\\mu_jI)Q_j.\n\\]\nLet \\(\\nu_j\\) be the least positive integer for which \\(N_j^{\\nu_j}=0\\), and define\n\\[\nB_{j,k}\n=\nA_{PQ}N_j^kQ_jA_{QP}\n\\qquad\n(0\\leq k<\\nu_j).\n\\]\nThen\n\\[\n\\Kcal(t)\n=\n\\sum_{j=1}^r\n\\e^{\\mu_jt}\n\\sum_{k=0}^{\\nu_j-1}\n\\frac{t^k}{k!}B_{j,k}\n\\]\nand\n\\[\n\\Sigma_P(z)\n=\n\\sum_{j=1}^r\n\\sum_{k=0}^{\\nu_j-1}\n\\frac{B_{j,k}}{(z-\\mu_j)^{k+1}}.\n\\]\nSuppose \\(\\operatorname{Re}\\mu_j<0\\) for every \\(j\\). If \\(J\\subseteq\\{1,\\ldots,r\\}\\) and \\(\\Kcal_J\\) retains the complete primary blocks indexed by \\(J\\), then\n\\[\n\\norm{\\Kcal-\\Kcal_J}_{L^1(0,\\infty)}\n\\leq\n\\sum_{j\\notin J}\n\\sum_{k=0}^{\\nu_j-1}\n\\frac{\\norm{B_{j,k}}}\n{(-\\operatorname{Re}\\mu_j)^{k+1}}.\n\\]\n\\end{theorem}","ary spectral projections, and set\n\\[\nN_j=(A_{QQ}-\\mu_jI)Q_j.\n\\]\nLet \\(\\nu_j\\) be the least positive integer for which \\(N_j^{\\nu_j}=0\\), and define\n\\[\nB_{j,k}\n=\nA_{PQ}N_j^kQ_jA_{QP}\n\\qquad\n(0\\leq k<\\nu_j).\n\\]\nThen\n\\[\n\\Kcal(t)\n=\n\\sum_{j=1}^r\n\\e^{\\mu_jt}\n\\sum_{k=0}^{\\nu_j-1}\n\\frac{t^k}{k!}B_{j,k}\n\\]\nand\n\\[\n\\Sigma_P(z)\n=\n\\sum_{j=1}^r\n\\sum_{k=0}^{\\nu_j-1}\n\\frac{B_{j,k}}{(z-\\mu_j)^{k+1}}.\n\\]\nSuppose \\(\\operatorname{Re}\\mu_j<0\\) for every \\(j\\). If \\(J\\subseteq\\{1,\\ldots,r\\}\\) and \\(\\Kcal_J\\) retains the complete primary blocks indexed by \\(J\\), then\n\\[\n\\norm{\\Kcal-\\Kcal_J}_{L^1(0,\\infty)}\n\\leq\n\\sum_{j\\notin J}\n\\sum_{k=0}^{\\nu_j-1}\n\\frac{\\norm{B_{j,k}}}\n{(-\\operatorname{Re}\\mu_j)^{k+1}}.\n\\]\n\\end{theorem}","j=(A_{QQ}-\\mu_jI)Q_j.\n\\]\nLet \\(\\nu_j\\) be the least positive integer for which \\(N_j^{\\nu_j}=0\\), and define\n\\[\nB_{j,k}\n=\nA_{PQ}N_j^kQ_jA_{QP}\n\\qquad\n(0\\leq k<\\nu_j).\n\\]\nThen\n\\[\n\\Kcal(t)\n=\n\\sum_{j=1}^r\n\\e^{\\mu_jt}\n\\sum_{k=0}^{\\nu_j-1}\n\\frac{t^k}{k!}B_{j,k}\n\\]\nand\n\\[\n\\Sigma_P(z)\n=\n\\sum_{j=1}^r\n\\sum_{k=0}^{\\nu_j-1}\n\\frac{B_{j,k}}{(z-\\mu_j)^{k+1}}.\n\\]\nSuppose \\(\\operatorname{Re}\\mu_j<0\\) for every \\(j\\). If \\(J\\subseteq\\{1,\\ldots,r\\}\\) and \\(\\Kcal_J\\) retains the complete primary blocks indexed by \\(J\\), then\n\\[\n\\norm{\\Kcal-\\Kcal_J}_{L^1(0,\\infty)}\n\\leq\n\\sum_{j\\notin J}\n\\sum_{k=0}^{\\nu_j-1}\n\\frac{\\norm{B_{j,k}}}\n{(-\\operatorname{Re}\\mu_j)^{k+1}}.\n\\]\n\\end{theorem}","\\mu_jI)Q_j.\n\\]\nLet \\(\\nu_j\\) be the least positive integer for which \\(N_j^{\\nu_j}=0\\), and define\n\\[\nB_{j,k}\n=\nA_{PQ}N_j^kQ_jA_{QP}\n\\qquad\n(0\\leq k<\\nu_j).\n\\]\nThen\n\\[\n\\Kcal(t)\n=\n\\sum_{j=1}^r\n\\e^{\\mu_jt}\n\\sum_{k=0}^{\\nu_j-1}\n\\frac{t^k}{k!}B_{j,k}\n\\]\nand\n\\[\n\\Sigma_P(z)\n=\n\\sum_{j=1}^r\n\\sum_{k=0}^{\\nu_j-1}\n\\frac{B_{j,k}}{(z-\\mu_j)^{k+1}}.\n\\]\nSuppose \\(\\operatorname{Re}\\mu_j<0\\) for every \\(j\\). If \\(J\\subseteq\\{1,\\ldots,r\\}\\) and \\(\\Kcal_J\\) retains the complete primary blocks indexed by \\(J\\), then\n\\[\n\\norm{\\Kcal-\\Kcal_J}_{L^1(0,\\infty)}\n\\leq\n\\sum_{j\\notin J}\n\\sum_{k=0}^{\\nu_j-1}\n\\frac{\\norm{B_{j,k}}}\n{(-\\operatorname{Re}\\mu_j)^{k+1}}.\n\\]\n\\end{theorem}","he least positive integer for which \\(N_j^{\\nu_j}=0\\), and define\n\\[\nB_{j,k}\n=\nA_{PQ}N_j^kQ_jA_{QP}\n\\qquad\n(0\\leq k<\\nu_j).\n\\]\nThen\n\\[\n\\Kcal(t)\n=\n\\sum_{j=1}^r\n\\e^{\\mu_jt}\n\\sum_{k=0}^{\\nu_j-1}\n\\frac{t^k}{k!}B_{j,k}\n\\]\nand\n\\[\n\\Sigma_P(z)\n=\n\\sum_{j=1}^r\n\\sum_{k=0}^{\\nu_j-1}\n\\frac{B_{j,k}}{(z-\\mu_j)^{k+1}}.\n\\]\nSuppose \\(\\operatorname{Re}\\mu_j<0\\) for every \\(j\\). If \\(J\\subseteq\\{1,\\ldots,r\\}\\) and \\(\\Kcal_J\\) retains the complete primary blocks indexed by \\(J\\), then\n\\[\n\\norm{\\Kcal-\\Kcal_J}_{L^1(0,\\infty)}\n\\leq\n\\sum_{j\\notin J}\n\\sum_{k=0}^{\\nu_j-1}\n\\frac{\\norm{B_{j,k}}}\n{(-\\operatorname{Re}\\mu_j)^{k+1}}.\n\\]\n\\end{theorem}","define\n\\[\nB_{j,k}\n=\nA_{PQ}N_j^kQ_jA_{QP}\n\\qquad\n(0\\leq k<\\nu_j).\n\\]\nThen\n\\[\n\\Kcal(t)\n=\n\\sum_{j=1}^r\n\\e^{\\mu_jt}\n\\sum_{k=0}^{\\nu_j-1}\n\\frac{t^k}{k!}B_{j,k}\n\\]\nand\n\\[\n\\Sigma_P(z)\n=\n\\sum_{j=1}^r\n\\sum_{k=0}^{\\nu_j-1}\n\\frac{B_{j,k}}{(z-\\mu_j)^{k+1}}.\n\\]\nSuppose \\(\\operatorname{Re}\\mu_j<0\\) for every \\(j\\). If \\(J\\subseteq\\{1,\\ldots,r\\}\\) and \\(\\Kcal_J\\) retains the complete primary blocks indexed by \\(J\\), then\n\\[\n\\norm{\\Kcal-\\Kcal_J}_{L^1(0,\\infty)}\n\\leq\n\\sum_{j\\notin J}\n\\sum_{k=0}^{\\nu_j-1}\n\\frac{\\norm{B_{j,k}}}\n{(-\\operatorname{Re}\\mu_j)^{k+1}}.\n\\]\n\\end{theorem}","{j,k}\n=\nA_{PQ}N_j^kQ_jA_{QP}\n\\qquad\n(0\\leq k<\\nu_j).\n\\]\nThen\n\\[\n\\Kcal(t)\n=\n\\sum_{j=1}^r\n\\e^{\\mu_jt}\n\\sum_{k=0}^{\\nu_j-1}\n\\frac{t^k}{k!}B_{j,k}\n\\]\nand\n\\[\n\\Sigma_P(z)\n=\n\\sum_{j=1}^r\n\\sum_{k=0}^{\\nu_j-1}\n\\frac{B_{j,k}}{(z-\\mu_j)^{k+1}}.\n\\]\nSuppose \\(\\operatorname{Re}\\mu_j<0\\) for every \\(j\\). If \\(J\\subseteq\\{1,\\ldots,r\\}\\) and \\(\\Kcal_J\\) retains the complete primary blocks indexed by \\(J\\), then\n\\[\n\\norm{\\Kcal-\\Kcal_J}_{L^1(0,\\infty)}\n\\leq\n\\sum_{j\\notin J}\n\\sum_{k=0}^{\\nu_j-1}\n\\frac{\\norm{B_{j,k}}}\n{(-\\operatorname{Re}\\mu_j)^{k+1}}.\n\\]\n\\end{theorem}","\\begin{proof}\nThe primary decomposition gives\n\\[\nA_{QQ}Q_j=\\mu_jQ_j+N_j\n\\]\nwith commuting scalar and nilpotent parts on \\(Q_jX_Q\\). Therefore\n\\[\n\\e^{tA_{QQ}}Q_j\n=\n\\e^{\\mu_jt}\n\\sum_{k=0}^{\\nu_j-1}\n\\frac{t^k}{k!}N_j^kQ_j.\n\\]\nMultiplication by \\(A_{PQ}\\) and \\(A_{QP}\\) gives the memory formula. On the \\(j\\)th primary space, the nilpotent resolvent expansion is\n\\[\n(zI-A_{QQ})^{-1}Q_j\n=\n\\sum_{k=0}^{\\nu_j-1}\n\\frac{N_j^kQ_j}{(z-\\mu_j)^{k+1}}.\n\\]\nMultiplication by the coupling blocks gives the self energy formula for every \\(z\\) outside the hidden spectrum. Equivalently, on a right half plane it is the Laplace transform identity\n\\[\n\\int_0^\\infty\n\\e^{-zt}\\e^{\\mu_jt}\n\\frac{t^k}{k!}\n\\dd t\n=\n\\frac{1}{(z-\\mu_j)^{k+1}}.\n\\]\nUnder stability,\n\\[\n\\int_0^\\infty\n\\e^{\\operatorname{R","\\begin{proof}\nThe primary decomposition gives\n\\[\nA_{QQ}Q_j=\\mu_jQ_j+N_j\n\\]\nwith commuting scalar and nilpotent parts on \\(Q_jX_Q\\). Therefore\n\\[\n\\e^{tA_{QQ}}Q_j\n=\n\\e^{\\mu_jt}\n\\sum_{k=0}^{\\nu_j-1}\n\\frac{t^k}{k!}N_j^kQ_j.\n\\]\nMultiplication by \\(A_{PQ}\\) and \\(A_{QP}\\) gives the memory formula. On the \\(j\\)th primary space, the nilpotent resolvent expansion is\n\\[\n(zI-A_{QQ})^{-1}Q_j\n=\n\\sum_{k=0}^{\\nu_j-1}\n\\frac{N_j^kQ_j}{(z-\\mu_j)^{k+1}}.\n\\]\nMultiplication by the coupling blocks gives the self energy formula for every \\(z\\) outside the hidden spectrum. Equivalently, on a right half plane it is the Laplace transform identity\n\\[\n\\int_0^\\infty\n\\e^{-zt}\\e^{\\mu_jt}\n\\frac{t^k}{k!}\n\\dd t\n=\n\\frac{1}{(z-\\mu_j)^{k+1}}.\n\\]\nUnder stability,\n\\[\n\\int_0^\\infty\n\\e^{\\operatorname{Re}\\mu_jt}\n\\frac{t^k}{k!}\n\\dd t\n=\n\\frac{1}{(-\\operatornam","\\begin{proof}\nThe primary decomposition gives\n\\[\nA_{QQ}Q_j=\\mu_jQ_j+N_j\n\\]\nwith commuting scalar and nilpotent parts on \\(Q_jX_Q\\). Therefore\n\\[\n\\e^{tA_{QQ}}Q_j\n=\n\\e^{\\mu_jt}\n\\sum_{k=0}^{\\nu_j-1}\n\\frac{t^k}{k!}N_j^kQ_j.\n\\]\nMultiplication by \\(A_{PQ}\\) and \\(A_{QP}\\) gives the memory formula. On the \\(j\\)th primary space, the nilpotent resolvent expansion is\n\\[\n(zI-A_{QQ})^{-1}Q_j\n=\n\\sum_{k=0}^{\\nu_j-1}\n\\frac{N_j^kQ_j}{(z-\\mu_j)^{k+1}}.\n\\]\nMultiplication by the coupling blocks gives the self energy formula for every \\(z\\) outside the hidden spectrum. Equivalently, on a right half plane it is the Laplace transform identity\n\\[\n\\int_0^\\infty\n\\e^{-zt}\\e^{\\mu_jt}\n\\frac{t^k}{k!}\n\\dd t\n=\n\\frac{1}{(z-\\mu_j)^{k+1}}.\n\\]\nUnder stability,\n\\[\n\\int_0^\\infty\n\\e^{\\operatorname{Re}\\mu_jt}\n\\frac{t^k}{k!}\n\\dd t\n=\n\\frac{1}{(-\\operatorname{Re}\\mu_j)^{k+1}}.\n\\]\nSum the omitted terms.\n\\end{proof}","_jX_Q\\). Therefore\n\\[\n\\e^{tA_{QQ}}Q_j\n=\n\\e^{\\mu_jt}\n\\sum_{k=0}^{\\nu_j-1}\n\\frac{t^k}{k!}N_j^kQ_j.\n\\]\nMultiplication by \\(A_{PQ}\\) and \\(A_{QP}\\) gives the memory formula. On the \\(j\\)th primary space, the nilpotent resolvent expansion is\n\\[\n(zI-A_{QQ})^{-1}Q_j\n=\n\\sum_{k=0}^{\\nu_j-1}\n\\frac{N_j^kQ_j}{(z-\\mu_j)^{k+1}}.\n\\]\nMultiplication by the coupling blocks gives the self energy formula for every \\(z\\) outside the hidden spectrum. Equivalently, on a right half plane it is the Laplace transform identity\n\\[\n\\int_0^\\infty\n\\e^{-zt}\\e^{\\mu_jt}\n\\frac{t^k}{k!}\n\\dd t\n=\n\\frac{1}{(z-\\mu_j)^{k+1}}.\n\\]\nUnder stability,\n\\[\n\\int_0^\\infty\n\\e^{\\operatorname{Re}\\mu_jt}\n\\frac{t^k}{k!}\n\\dd t\n=\n\\frac{1}{(-\\operatorname{Re}\\mu_j)^{k+1}}.\n\\]\nSum the omitted terms.\n\\end{proof}","cation by \\(A_{PQ}\\) and \\(A_{QP}\\) gives the memory formula. On the \\(j\\)th primary space, the nilpotent resolvent expansion is\n\\[\n(zI-A_{QQ})^{-1}Q_j\n=\n\\sum_{k=0}^{\\nu_j-1}\n\\frac{N_j^kQ_j}{(z-\\mu_j)^{k+1}}.\n\\]\nMultiplication by the coupling blocks gives the self energy formula for every \\(z\\) outside the hidden spectrum. Equivalently, on a right half plane it is the Laplace transform identity\n\\[\n\\int_0^\\infty\n\\e^{-zt}\\e^{\\mu_jt}\n\\frac{t^k}{k!}\n\\dd t\n=\n\\frac{1}{(z-\\mu_j)^{k+1}}.\n\\]\nUnder stability,\n\\[\n\\int_0^\\infty\n\\e^{\\operatorname{Re}\\mu_jt}\n\\frac{t^k}{k!}\n\\dd t\n=\n\\frac{1}{(-\\operatorname{Re}\\mu_j)^{k+1}}.\n\\]\nSum the omitted terms.\n\\end{proof}","esolvent expansion is\n\\[\n(zI-A_{QQ})^{-1}Q_j\n=\n\\sum_{k=0}^{\\nu_j-1}\n\\frac{N_j^kQ_j}{(z-\\mu_j)^{k+1}}.\n\\]\nMultiplication by the coupling blocks gives the self energy formula for every \\(z\\) outside the hidden spectrum. Equivalently, on a right half plane it is the Laplace transform identity\n\\[\n\\int_0^\\infty\n\\e^{-zt}\\e^{\\mu_jt}\n\\frac{t^k}{k!}\n\\dd t\n=\n\\frac{1}{(z-\\mu_j)^{k+1}}.\n\\]\nUnder stability,\n\\[\n\\int_0^\\infty\n\\e^{\\operatorname{Re}\\mu_jt}\n\\frac{t^k}{k!}\n\\dd t\n=\n\\frac{1}{(-\\operatorname{Re}\\mu_j)^{k+1}}.\n\\]\nSum the omitted terms.\n\\end{proof}","\\begin{corollary}[Optimal selection for the certified primary block bound]\\label{extended:record:090}\\label{extended:cor:optimal_primary_selection}\nUnder the stable assumptions of \\cref{extended:thm:modal_memory}, define the primary block weights\n\\[\nw_j\n=\n\\sum_{k=0}^{\\nu_j-1}\n\\frac{\\norm{B_{j,k}}}\n{(-\\operatorname{Re}\\mu_j)^{k+1}}.\n\\]\nAmong all sets \\(J\\) containing \\(k\\) primary blocks, the certified upper bound\n\\[\n\\sum_{j\\notin J}w_j\n\\]\nis minimized by retaining the \\(k\\) largest weights. If\n\\[\nw_{(1)}\\geq w_{(2)}\\geq\\cdots\\geq w_{(r)},\n\\]\nthen the least number of retained primary blocks certified to give \\(L^1\\) error at most \\(\\varepsilon\\) is the least \\(k\\) for which\n\\[\n\\sum_{j>k}w_{(j)}\\leq\\varepsilon.\n\\]\n\\end{corollary}","ptimal selection for the certified primary block bound]\\label{extended:record:090}\\label{extended:cor:optimal_primary_selection}\nUnder the stable assumptions of \\cref{extended:thm:modal_memory}, define the primary block weights\n\\[\nw_j\n=\n\\sum_{k=0}^{\\nu_j-1}\n\\frac{\\norm{B_{j,k}}}\n{(-\\operatorname{Re}\\mu_j)^{k+1}}.\n\\]\nAmong all sets \\(J\\) containing \\(k\\) primary blocks, the certified upper bound\n\\[\n\\sum_{j\\notin J}w_j\n\\]\nis minimized by retaining the \\(k\\) largest weights. If\n\\[\nw_{(1)}\\geq w_{(2)}\\geq\\cdots\\geq w_{(r)},\n\\]\nthen the least number of retained primary blocks certified to give \\(L^1\\) error at most \\(\\varepsilon\\) is the least \\(k\\) for which\n\\[\n\\sum_{j>k}w_{(j)}\\leq\\varepsilon.\n\\]\n\\end{corollary}","label{extended:record:090}\\label{extended:cor:optimal_primary_selection}\nUnder the stable assumptions of \\cref{extended:thm:modal_memory}, define the primary block weights\n\\[\nw_j\n=\n\\sum_{k=0}^{\\nu_j-1}\n\\frac{\\norm{B_{j,k}}}\n{(-\\operatorname{Re}\\mu_j)^{k+1}}.\n\\]\nAmong all sets \\(J\\) containing \\(k\\) primary blocks, the certified upper bound\n\\[\n\\sum_{j\\notin J}w_j\n\\]\nis minimized by retaining the \\(k\\) largest weights. If\n\\[\nw_{(1)}\\geq w_{(2)}\\geq\\cdots\\geq w_{(r)},\n\\]\nthen the least number of retained primary blocks certified to give \\(L^1\\) error at most \\(\\varepsilon\\) is the least \\(k\\) for which\n\\[\n\\sum_{j>k}w_{(j)}\\leq\\varepsilon.\n\\]\n\\end{corollary}","\\label{extended:cor:optimal_primary_selection}\nUnder the stable assumptions of \\cref{extended:thm:modal_memory}, define the primary block weights\n\\[\nw_j\n=\n\\sum_{k=0}^{\\nu_j-1}\n\\frac{\\norm{B_{j,k}}}\n{(-\\operatorname{Re}\\mu_j)^{k+1}}.\n\\]\nAmong all sets \\(J\\) containing \\(k\\) primary blocks, the certified upper bound\n\\[\n\\sum_{j\\notin J}w_j\n\\]\nis minimized by retaining the \\(k\\) largest weights. If\n\\[\nw_{(1)}\\geq w_{(2)}\\geq\\cdots\\geq w_{(r)},\n\\]\nthen the least number of retained primary blocks certified to give \\(L^1\\) error at most \\(\\varepsilon\\) is the least \\(k\\) for which\n\\[\n\\sum_{j>k}w_{(j)}\\leq\\varepsilon.\n\\]\n\\end{corollary}","ine the primary block weights\n\\[\nw_j\n=\n\\sum_{k=0}^{\\nu_j-1}\n\\frac{\\norm{B_{j,k}}}\n{(-\\operatorname{Re}\\mu_j)^{k+1}}.\n\\]\nAmong all sets \\(J\\) containing \\(k\\) primary blocks, the certified upper bound\n\\[\n\\sum_{j\\notin J}w_j\n\\]\nis minimized by retaining the \\(k\\) largest weights. If\n\\[\nw_{(1)}\\geq w_{(2)}\\geq\\cdots\\geq w_{(r)},\n\\]\nthen the least number of retained primary blocks certified to give \\(L^1\\) error at most \\(\\varepsilon\\) is the least \\(k\\) for which\n\\[\n\\sum_{j>k}w_{(j)}\\leq\\varepsilon.\n\\]\n\\end{corollary}","weights\n\\[\nw_j\n=\n\\sum_{k=0}^{\\nu_j-1}\n\\frac{\\norm{B_{j,k}}}\n{(-\\operatorname{Re}\\mu_j)^{k+1}}.\n\\]\nAmong all sets \\(J\\) containing \\(k\\) primary blocks, the certified upper bound\n\\[\n\\sum_{j\\notin J}w_j\n\\]\nis minimized by retaining the \\(k\\) largest weights. If\n\\[\nw_{(1)}\\geq w_{(2)}\\geq\\cdots\\geq w_{(r)},\n\\]\nthen the least number of retained primary blocks certified to give \\(L^1\\) error at most \\(\\varepsilon\\) is the least \\(k\\) for which\n\\[\n\\sum_{j>k}w_{(j)}\\leq\\varepsilon.\n\\]\n\\end{corollary}","\\nu_j-1}\n\\frac{\\norm{B_{j,k}}}\n{(-\\operatorname{Re}\\mu_j)^{k+1}}.\n\\]\nAmong all sets \\(J\\) containing \\(k\\) primary blocks, the certified upper bound\n\\[\n\\sum_{j\\notin J}w_j\n\\]\nis minimized by retaining the \\(k\\) largest weights. If\n\\[\nw_{(1)}\\geq w_{(2)}\\geq\\cdots\\geq w_{(r)},\n\\]\nthen the least number of retained primary blocks certified to give \\(L^1\\) error at most \\(\\varepsilon\\) is the least \\(k\\) for which\n\\[\n\\sum_{j>k}w_{(j)}\\leq\\varepsilon.\n\\]\n\\end{corollary}","norm{B_{j,k}}}\n{(-\\operatorname{Re}\\mu_j)^{k+1}}.\n\\]\nAmong all sets \\(J\\) containing \\(k\\) primary blocks, the certified upper bound\n\\[\n\\sum_{j\\notin J}w_j\n\\]\nis minimized by retaining the \\(k\\) largest weights. If\n\\[\nw_{(1)}\\geq w_{(2)}\\geq\\cdots\\geq w_{(r)},\n\\]\nthen the least number of retained primary blocks certified to give \\(L^1\\) error at most \\(\\varepsilon\\) is the least \\(k\\) for which\n\\[\n\\sum_{j>k}w_{(j)}\\leq\\varepsilon.\n\\]\n\\end{corollary}","\\begin{proof}\nIf a retained set contains a block with smaller weight than an omitted block, exchanging them reduces the omitted weight sum. Repeating the exchange gives the set of the \\(k\\) largest weights. The tolerance statement is the corresponding sorted tail condition.\n\\end{proof}","\\begin{corollary}[Normal hidden dynamics]\\label{extended:record:091}\\label{extended:cor:normal_modal_memory}\nSuppose \\(X_Q\\) is a finite dimensional Hilbert space. If \\(A_{QQ}\\) is normal with spectral resolution\n\\[\nA_{QQ}=\\sum_{j=1}^r\\mu_jQ_j,\n\\]\nthen\n\\[\n\\Kcal(t)=\\sum_{j=1}^r\\e^{\\mu_jt}B_j,\n\\qquad\n\\Sigma_P(z)=\\sum_{j=1}^r\\frac{B_j}{z-\\mu_j},\n\\]\nwhere\n\\[\nB_j=A_{PQ}Q_jA_{QP}.\n\\]\nIf \\(\\operatorname{Re}\\mu_j<0\\), then\n\\[\n\\norm{\\Kcal-\\Kcal_J}_{L^1(0,\\infty)}\n\\leq\n\\sum_{j\\notin J}\n\\frac{\\norm{B_j}}{-\\operatorname{Re}\\mu_j}.\n\\]\n\\end{corollary}","mal hidden dynamics]\\label{extended:record:091}\\label{extended:cor:normal_modal_memory}\nSuppose \\(X_Q\\) is a finite dimensional Hilbert space. If \\(A_{QQ}\\) is normal with spectral resolution\n\\[\nA_{QQ}=\\sum_{j=1}^r\\mu_jQ_j,\n\\]\nthen\n\\[\n\\Kcal(t)=\\sum_{j=1}^r\\e^{\\mu_jt}B_j,\n\\qquad\n\\Sigma_P(z)=\\sum_{j=1}^r\\frac{B_j}{z-\\mu_j},\n\\]\nwhere\n\\[\nB_j=A_{PQ}Q_jA_{QP}.\n\\]\nIf \\(\\operatorname{Re}\\mu_j<0\\), then\n\\[\n\\norm{\\Kcal-\\Kcal_J}_{L^1(0,\\infty)}\n\\leq\n\\sum_{j\\notin J}\n\\frac{\\norm{B_j}}{-\\operatorname{Re}\\mu_j}.\n\\]\n\\end{corollary}","\\begin{proof}\nNormality removes every nilpotent part, so \\(\\nu_j=1\\). Apply \\cref{extended:thm:modal_memory}.\n\\end{proof}","\\begin{corollary}[Pole order and hidden Jordan depth]\\label{extended:record:092}\\label{extended:cor:pole_jordan_depth}\nEvery finite hidden block produces a strictly proper rational self energy. At an eigenvalue \\(\\mu_j\\), its pole order is at most the size of the largest Jordan block of \\(A_{QQ}\\) at \\(\\mu_j\\). Equality holds when the corresponding highest coefficient \\(B_{j,\\nu_j-1}\\) is nonzero.\n\\end{corollary}","\\begin{corollary}[Finite auxiliary state realization]\\label{extended:record:093}\\label{extended:cor:finite_auxiliary_realization}\nEvery kernel of the form\n\\[\n\\Kcal(t)=B\\e^{tD}C\n\\]\ncan be represented exactly by adjoining the finite auxiliary state\n\\[\n\\dot y=Dy+Cx_P,\n\\qquad\n\\dot x_P=A_{PP}x_P+By,\n\\qquad\ny(0)=0.\n\\]\nA nonzero auxiliary initial state adds the explicit forcing \\(B\\e^{tD}y(0)\\). The self energy of the augmented system is\n\\[\nB(zI-D)^{-1}C.\n\\]\n\\end{corollary}","ate realization]\\label{extended:record:093}\\label{extended:cor:finite_auxiliary_realization}\nEvery kernel of the form\n\\[\n\\Kcal(t)=B\\e^{tD}C\n\\]\ncan be represented exactly by adjoining the finite auxiliary state\n\\[\n\\dot y=Dy+Cx_P,\n\\qquad\n\\dot x_P=A_{PP}x_P+By,\n\\qquad\ny(0)=0.\n\\]\nA nonzero auxiliary initial state adds the explicit forcing \\(B\\e^{tD}y(0)\\). The self energy of the augmented system is\n\\[\nB(zI-D)^{-1}C.\n\\]\n\\end{corollary}","\\begin{proof}\nEliminate \\(y\\) by variation of constants or apply \\cref{extended:thm:feshbach_resolvent} to the augmented block matrix.\n\\end{proof}","\\begin{theorem}[Volterra stability under kernel compression]\\label{extended:record:094}\\label{extended:thm:volterra_compression}\nLet \\(S(t)\\) be a strongly continuous semigroup on a Banach space \\(X\\), fix \\(T>0\\), and assume\n\\[\n\\norm{S(t)}\\leq M_T\n\\quad(0\\leq t\\leq T).\n\\]\nLet \\(K,\\widetilde K\\in L^1((0,T),\\mathcal B(X))\\), let \\(g\\in L^1((0,T),X)\\), and let \\(x,\\widetilde x\\in C([0,T],X)\\). Suppose these are mild solutions with the same initial state and forcing, except that \\(K\\) is replaced by \\(\\widetilde K\\):\n\\[\n\\begin{aligned}\nx(t)\n={}&S(t)x_0+\n\\int_0^tS(t-s)g(s)\\dd s","094}\\label{extended:thm:volterra_compression}\nLet \\(S(t)\\) be a strongly continuous semigroup on a Banach space \\(X\\), fix \\(T>0\\), and assume\n\\[\n\\norm{S(t)}\\leq M_T\n\\quad(0\\leq t\\leq T).\n\\]\nLet \\(K,\\widetilde K\\in L^1((0,T),\\mathcal B(X))\\), let \\(g\\in L^1((0,T),X)\\), and let \\(x,\\widetilde x\\in C([0,T],X)\\). Suppose these are mild solutions with the same initial state and forcing, except that \\(K\\) is replaced by \\(\\widetilde K\\):\n\\[\n\\begin{aligned}\nx(t)\n={}&S(t)x_0+\n\\int_0^tS(t-s)g(s)\\dd s","hm:volterra_compression}\nLet \\(S(t)\\) be a strongly continuous semigroup on a Banach space \\(X\\), fix \\(T>0\\), and assume\n\\[\n\\norm{S(t)}\\leq M_T\n\\quad(0\\leq t\\leq T).\n\\]\nLet \\(K,\\widetilde K\\in L^1((0,T),\\mathcal B(X))\\), let \\(g\\in L^1((0,T),X)\\), and let \\(x,\\widetilde x\\in C([0,T],X)\\). Suppose these are mild solutions with the same initial state and forcing, except that \\(K\\) is replaced by \\(\\widetilde K\\):\n\\[\n\\begin{aligned}\nx(t)\n={}&S(t)x_0+\n\\int_0^tS(t-s)g(s)\\dd s","ssion}\nLet \\(S(t)\\) be a strongly continuous semigroup on a Banach space \\(X\\), fix \\(T>0\\), and assume\n\\[\n\\norm{S(t)}\\leq M_T\n\\quad(0\\leq t\\leq T).\n\\]\nLet \\(K,\\widetilde K\\in L^1((0,T),\\mathcal B(X))\\), let \\(g\\in L^1((0,T),X)\\), and let \\(x,\\widetilde x\\in C([0,T],X)\\). Suppose these are mild solutions with the same initial state and forcing, except that \\(K\\) is replaced by \\(\\widetilde K\\):\n\\[\n\\begin{aligned}\nx(t)\n={}&S(t)x_0+\n\\int_0^tS(t-s)g(s)\\dd s\\\\\n&+\n\\int_0^tS(t-s)\n\\int_0^sK(s-r)x(r)\\dd r\\dd s.\n\\end{aligned}\n\\]\nAssume\n\\[\n\\sup_{0\\leq t\\leq T}\\norm{\\widetilde x(t)}\\leq B.\n\\]\nSet\n\\[\nL=\\norm K_{L^1(0,T)},\n\\qquad\n\\varepsilon=\\norm{K-\\widetilde K}_{L^1(0,T)}.\n\\]\nThen\n\\[\n\\sup_{0\\leq t\\leq T}\n\\norm{x(t)-\\widetilde x(t)}\n\\leq\nM_TBT\\varepsilon\n\\exp(M_TLT).\n\\]\n\\end{theorem}","&+\n\\int_0^tS(t-s)\n\\int_0^sK(s-r)x(r)\\dd r\\dd s.\n\\end{aligned}\n\\]\nAssume\n\\[\n\\sup_{0\\leq t\\leq T}\\norm{\\widetilde x(t)}\\leq B.\n\\]\nSet\n\\[\nL=\\norm K_{L^1(0,T)},\n\\qquad\n\\varepsilon=\\norm{K-\\widetilde K}_{L^1(0,T)}.\n\\]\nThen\n\\[\n\\sup_{0\\leq t\\leq T}\n\\norm{x(t)-\\widetilde x(t)}\n\\leq\nM_TBT\\varepsilon\n\\exp(M_TLT).\n\\]\n\\end{theorem}","\\begin{proof}\nLet\n\\[\nE(t)=\\sup_{0\\leq r\\leq t}\\norm{x(r)-\\widetilde x(r)}.\n\\]\nSubtract the two mild equations. Fubini's theorem and the triangle inequality give\n\\[\nE(t)\n\\leq\nM_TL\\int_0^tE(r)\\dd r\n+\nM_TBt\\varepsilon.\n\\]\nFor \\(t\\leq T\\), replace the last factor \\(t\\) by \\(T\\) and apply Gronwall's inequality.\n\\end{proof}","\\begin{corollary}[Trajectory bound for modal truncation]\\label{extended:record:095}\\label{extended:cor:modal_trajectory_bound}\nUnder the assumptions of \\cref{extended:thm:modal_memory,extended:thm:volterra_compression}, replacing \\(\\Kcal\\) by \\(\\Kcal_J\\) changes the retained trajectory by at most\n\\[\nM_TBT\n\\exp(M_TLT)\n\\sum_{j\\notin J}\n\\sum_{k=0}^{\\nu_j-1}\n\\frac{\\norm{B_{j,k}}}\n{(-\\operatorname{Re}\\mu_j)^{k+1}}.\n\\]\n\\end{corollary}","\\begin{proof}\nInsert the \\(L^1\\) error from \\cref{extended:thm:modal_memory} into \\cref{extended:thm:volterra_compression}.\n\\end{proof}","\\begin{remark}\\label{extended:record:096}\nA finite hidden realization is exact for rational self energy. A general memory kernel may require infinitely many modes or a continuum of modes. The compression theorem gives a stability estimate once an \\(L^1\\) approximation has been supplied. The largest weight rule is optimal for the certified blockwise bound. It need not minimize the true kernel error when different modal terms cancel.\n\\end{remark}","Let \\(A\\) be a transient state subgenerator. Its off diagonal entries are nonnegative and its column sums are nonpositive. Let \\(p_0\\) be a transient probability vector. Define\n\\[\np(t)=\\e^{tA}p_0,\n\\qquad\nS(t)=\\one^Tp(t),\n\\qquad\na^T=-\\one^TA.\n\\]\nThe row vector \\(a^T\\) contains the exit rates.","\\begin{theorem}[State dependent killing as survival]\\label{extended:record:097}\\label{extended:thm:killing_survival}\nLet \\(Q\\) be a finite Markov generator, let \\(V\\geq0\\), and set\n\\[\nA=Q-D_V.\n\\]\nFor an initial distribution \\(p_0\\), define\n\\[\nS_V(t)=\\one^T\\e^{tA}p_0.\n\\]\nThen\n\\[\nS_V(t)\n=\n\\Exp_{p_0}\n\\left[\n\\exp\\left(\n-\\int_0^tV(X_s)\\dd s\n\\right)\n\\right].\n\\]\nAlso,\n\\[\n-S_V'(t)\n=\n\\one^TD_V\\e^{tA}p_0.\n\\]\nThe matrix \\(A=Q-D_V\\) has nonnegative off diagonal entries and nonpositive column sums, so it is the subgenerator of the killed, possibly nontransient chain. Thus the potential in the finite Feynman Kac formula is a state dependent killing rate in the forward sub-Markov semigroup. Almost-sure absorption follows under the additional spectral stability hypothesis used below.\n\\end{theorem}","[State dependent killing as survival]\\label{extended:record:097}\\label{extended:thm:killing_survival}\nLet \\(Q\\) be a finite Markov generator, let \\(V\\geq0\\), and set\n\\[\nA=Q-D_V.\n\\]\nFor an initial distribution \\(p_0\\), define\n\\[\nS_V(t)=\\one^T\\e^{tA}p_0.\n\\]\nThen\n\\[\nS_V(t)\n=\n\\Exp_{p_0}\n\\left[\n\\exp\\left(\n-\\int_0^tV(X_s)\\dd s\n\\right)\n\\right].\n\\]\nAlso,\n\\[\n-S_V'(t)\n=\n\\one^TD_V\\e^{tA}p_0.\n\\]\nThe matrix \\(A=Q-D_V\\) has nonnegative off diagonal entries and nonpositive column sums, so it is the subgenerator of the killed, possibly nontransient chain. Thus the potential in the finite Feynman Kac formula is a state dependent killing rate in the forward sub-Markov semigroup. Almost-sure absorption follows under the additional spectral stability hypothesis used below.\n\\end{theorem}","\\begin{proof}\nAverage \\cref{extended:thm:continuous_feynman_kac} with terminal function \\(f=\\one\\) against the initial distribution. Transposition gives\n\\[\np_0^T\\e^{t(Q^T-D_V)}\\one\n=\n\\one^T\\e^{t(Q-D_V)}p_0.\n\\]\nSince \\(\\one^TQ=0\\), differentiation gives\n\\[\nS_V'(t)\n=\n-\\one^TD_V\\e^{tA}p_0.\n\\]\n\\end{proof}","\\begin{theorem}[Survival and hazard identities]\\label{extended:record:098}\\label{extended:thm:survival_hazard}\nThe event density and hazard are\n\\[\nf(t)=a^T\\e^{tA}p_0=-S'(t),\n\\qquad\nh(t)=\\frac{f(t)}{S(t)}\n\\]\nwhenever \\(S(t)>0\\). The survival function is nonincreasing and\n\\[\nS(t)\n=\n\\exp\\left(\n-\\int_0^th(s)\\dd s\n\\right).\n\\]\n\\end{theorem}","\\begin{proof}\nBecause \\(A\\) is a Metzler matrix, \\(\\e^{tA}\\) is nonnegative. The column sum assumption makes \\(a^T\\) nonnegative. Differentiate \\(S(t)\\):\n\\[\nS'(t)=\\one^TA\\e^{tA}p_0=-a^T\\e^{tA}p_0.\n\\]\nThe remaining identity follows by integrating \\((\\log S)'=-h\\).\n\\end{proof}","\\begin{theorem}[Laplace transform and moments]\\label{extended:record:099}\\label{extended:thm:survival_moments}\nAssume the spectrum of \\(A\\) lies in the open left half plane. Then absorption occurs with probability one. For \\(s\\geq0\\),\n\\[\n\\Exp[\\e^{-s\\tau}]\n=\na^T(sI-A)^{-1}p_0.\n\\] For every integer \\(m\\geq1\\),\n\\[\n\\Exp[\\tau^m]\n=\nm!\\one^T(-A)^{-m}p_0.\n\\]\n\\end{theorem}","\\begin{proof}\nThe Laplace transform follows by integrating the density in \\cref{extended:thm:survival_hazard}. Stability of \\(A\\) gives\n\\[\n\\int_0^\\infty\n\\e^{-st}\\e^{tA}\\dd t\n=\n(sI-A)^{-1}.\n\\]\nAt \\(s=0\\),\n\\[\na^T(-A)^{-1}p_0\n=\n\\one^Tp_0=1.\n\\]\nFor the moments,\n\\[\n\\int_0^\\infty t^m\\e^{tA}\\dd t\n=\nm!(-A)^{-m-1}.\n\\]\nMultiplying by \\(a^T=\\one^T(-A)\\) gives the result. These are the standard matrix formulas for phase type laws \\cite{Neuts1981,AsmussenNermanOlsson1996}.\n\\end{proof}","\\begin{proposition}[Residual life as a restarted semigroup]\\label{extended:record:100}\\label{extended:prop:residual_life}\nFor \\(S(t)>0\\), define the conditional transient distribution\n\\[\np_t\n=\n\\frac{\\e^{tA}p_0}{S(t)}.\n\\]\nThen\n\\[\n\\frac{S(t+s)}{S(t)}\n=\n\\one^T\\e^{sA}p_t.\n\\]\n\\end{proposition}","\\begin{proof}\nUse the semigroup law in the numerator \\(S(t+s)=\\one^T\\e^{sA}\\e^{tA}p_0\\).\n\\end{proof}","\\begin{theorem}[Competing risk packets]\\label{extended:record:101}\\label{extended:thm:competing_risks}\nSuppose the exit row splits into nonnegative cause rows\n\\[\na^T\n=\n\\sum_{r=1}^q a_r^T.\n\\]\nDefine\n\\[\nf_r(t)=a_r^T\\e^{tA}p_0.\n\\]\nThen\n\\[\nf(t)=\\sum_{r=1}^qf_r(t).\n\\]\nThe cause specific hazard is\n\\[\nh_r(t)=\\frac{f_r(t)}{S(t)},\n\\]\nand\n\\[\nh(t)=\\sum_{r=1}^qh_r(t).\n\\]\nIf the spectrum of \\(A\\) lies in the open left half-plane, the eventual probability of cause \\(r\\) is\n\\[\n\\int_0^\\infty f_r(t)\\dd t\n=\na_r^T(-A)^{-1}p_0.\n\\]\n\\end{theorem}","cord:101}\\label{extended:thm:competing_risks}\nSuppose the exit row splits into nonnegative cause rows\n\\[\na^T\n=\n\\sum_{r=1}^q a_r^T.\n\\]\nDefine\n\\[\nf_r(t)=a_r^T\\e^{tA}p_0.\n\\]\nThen\n\\[\nf(t)=\\sum_{r=1}^qf_r(t).\n\\]\nThe cause specific hazard is\n\\[\nh_r(t)=\\frac{f_r(t)}{S(t)},\n\\]\nand\n\\[\nh(t)=\\sum_{r=1}^qh_r(t).\n\\]\nIf the spectrum of \\(A\\) lies in the open left half-plane, the eventual probability of cause \\(r\\) is\n\\[\n\\int_0^\\infty f_r(t)\\dd t\n=\na_r^T(-A)^{-1}p_0.\n\\]\n\\end{theorem}","\\label{extended:thm:competing_risks}\nSuppose the exit row splits into nonnegative cause rows\n\\[\na^T\n=\n\\sum_{r=1}^q a_r^T.\n\\]\nDefine\n\\[\nf_r(t)=a_r^T\\e^{tA}p_0.\n\\]\nThen\n\\[\nf(t)=\\sum_{r=1}^qf_r(t).\n\\]\nThe cause specific hazard is\n\\[\nh_r(t)=\\frac{f_r(t)}{S(t)},\n\\]\nand\n\\[\nh(t)=\\sum_{r=1}^qh_r(t).\n\\]\nIf the spectrum of \\(A\\) lies in the open left half-plane, the eventual probability of cause \\(r\\) is\n\\[\n\\int_0^\\infty f_r(t)\\dd t\n=\na_r^T(-A)^{-1}p_0.\n\\]\n\\end{theorem}","\\midrule\nPacket block convolution & below \\(3\\times10^{-14}\\)","Duhamel defect & below \\(3\\times10^{-12}\\)","Markov lumping & below \\(3\\times10^{-14}\\)","Bayesian reconstruction & below \\(3\\times10^{-15}\\)","Fractional spectral law & below \\(3\\times10^{-12}\\)","Nodal Green identity & below \\(3\\times10^{-13}\\)","Feynman Kac enumeration & below \\(3\\times10^{-14}\\)","Schur Feshbach resolvent & below \\(3\\times10^{-12}\\)","Memory Laplace identity & below \\(3\\times10^{-11}\\)","Survival moment formulas & below \\(3\\times10^{-11}\\)","The article therefore separates four operations that are often compressed into one informal idea of coarse dynamics:\n\\[\n\\begin{aligned}\n\\text{intertwining}\n&\\longrightarrow\n\\text{refinement control}\\\\\n&\\longrightarrow\n\\text{statistical sufficiency}\\\\\n&\\longrightarrow\n\\text{memory closure}.\n\\end{aligned}\n\\]\nEach arrow has its own hypotheses and its own obstruction.","figure}{0}\\setcounter{table}{0}\n\\renewcommand{\\thesection}{R\\arabic{section}}\n\\renewcommand{\\theHsection}{R.\\arabic{section}}\n\\renewcommand{\\thefigure}{R\\arabic{figure}}\n\\renewcommand{\\theHfigure}{R.\\arabic{figure}}\n\\renewcommand{\\thetable}{R\\arabic{table}}\n\\renewcommand{\\theHtable}{R.\\arabic{table}}\n\\FloatBarrier\n\\section{Transfer paths and product-topology limits}\\label{earlier:section:01}\nLet $T(t)$ be an arbitrary strongly continuous semigroup on a complex Banach space $X$, with generator $A$, and suppose $\\norm{T(t)}\\leq M e^{\\omega t}$ for $M\\geq1$ and $\\omega\\in\\mathbb R$. Let $(P_\\alpha)_{\\alpha\\in I_0}$ be a finite family of bounded pairwise orthogonal idempotents summing to $I$, and put $T_{\\beta\\alpha}(t)=P_\\beta T(t)P_\\alpha$. The generator need not be bounded. In the following resolvent statement the integral is strong: apply the integrand to each $x\\in X$ and take its Bochner integral in $X$. It is not asserted to be a Bochner integral in the operator-norm space $\\mathcal B(X)$. The general Laplace formula is \\cite[Chapter II, Theorem 1.10]{EngelNagel2000}.\n\\begin{theor…","renewcommand{\\theHsection}{R.\\arabic{section}}\n\\renewcommand{\\thefigure}{R\\arabic{figure}}\n\\renewcommand{\\theHfigure}{R.\\arabic{figure}}\n\\renewcommand{\\thetable}{R\\arabic{table}}\n\\renewcommand{\\theHtable}{R.\\arabic{table}}\n\\FloatBarrier\n\\section{Transfer paths and product-topology limits}\\label{earlier:section:01}\nLet $T(t)$ be an arbitrary strongly continuous semigroup on a complex Banach space $X$, with generator $A$, and suppose $\\norm{T(t)}\\leq M e^{\\omega t}$ for $M\\geq1$ and $\\omega\\in\\mathbb R$. Let $(P_\\alpha)_{\\alpha\\in I_0}$ be a finite family of bounded pairwise orthogonal idempotents summing to $I$, and put $T_{\\beta\\alpha}(t)=P_\\beta T(t)P_\\alpha$. The generator need not be bounded. In the following resolvent statement the integral is strong: apply the integrand to each $x\\in X$ and take its Bochner integral in $X$. It is not asserted to be a Bochner integral in the operator-norm space $\\mathcal B(X)$. The general Laplace formula is \\cite[Chapter II, Theorem 1.10]{EngelNagel2000}.\n\\begin{theorem}[Packet convolution]\\label{earlier:record:001}\\label{earlier:thm:packet_conv…","ction}{R.\\arabic{section}}\n\\renewcommand{\\thefigure}{R\\arabic{figure}}\n\\renewcommand{\\theHfigure}{R.\\arabic{figure}}\n\\renewcommand{\\thetable}{R\\arabic{table}}\n\\renewcommand{\\theHtable}{R.\\arabic{table}}\n\\FloatBarrier\n\\section{Transfer paths and product-topology limits}\\label{earlier:section:01}\nLet $T(t)$ be an arbitrary strongly continuous semigroup on a complex Banach space $X$, with generator $A$, and suppose $\\norm{T(t)}\\leq M e^{\\omega t}$ for $M\\geq1$ and $\\omega\\in\\mathbb R$. Let $(P_\\alpha)_{\\alpha\\in I_0}$ be a finite family of bounded pairwise orthogonal idempotents summing to $I$, and put $T_{\\beta\\alpha}(t)=P_\\beta T(t)P_\\alpha$. The generator need not be bounded. In the following resolvent statement the integral is strong: apply the integrand to each $x\\in X$ and take its Bochner integral in $X$. It is not asserted to be a Bochner integral in the operator-norm space $\\mathcal B(X)$. The general Laplace formula is \\cite[Chapter II, Theorem 1.10]{EngelNagel2000}.\n\\begin{theorem}[Packet convolution]\\label{earlier:record:001}\\label{earlier:thm:packet_convolution}\nFor all \\(s…","section}}\n\\renewcommand{\\thefigure}{R\\arabic{figure}}\n\\renewcommand{\\theHfigure}{R.\\arabic{figure}}\n\\renewcommand{\\thetable}{R\\arabic{table}}\n\\renewcommand{\\theHtable}{R.\\arabic{table}}\n\\FloatBarrier\n\\section{Transfer paths and product-topology limits}\\label{earlier:section:01}\nLet $T(t)$ be an arbitrary strongly continuous semigroup on a complex Banach space $X$, with generator $A$, and suppose $\\norm{T(t)}\\leq M e^{\\omega t}$ for $M\\geq1$ and $\\omega\\in\\mathbb R$. Let $(P_\\alpha)_{\\alpha\\in I_0}$ be a finite family of bounded pairwise orthogonal idempotents summing to $I$, and put $T_{\\beta\\alpha}(t)=P_\\beta T(t)P_\\alpha$. The generator need not be bounded. In the following resolvent statement the integral is strong: apply the integrand to each $x\\in X$ and take its Bochner integral in $X$. It is not asserted to be a Bochner integral in the operator-norm space $\\mathcal B(X)$. The general Laplace formula is \\cite[Chapter II, Theorem 1.10]{EngelNagel2000}.\n\\begin{theorem}[Packet convolution]\\label{earlier:record:001}\\label{earlier:thm:packet_convolution}\nFor all \\(s,t\\geq0\\) and all…","\\arabic{figure}}\n\\renewcommand{\\theHfigure}{R.\\arabic{figure}}\n\\renewcommand{\\thetable}{R\\arabic{table}}\n\\renewcommand{\\theHtable}{R.\\arabic{table}}\n\\FloatBarrier\n\\section{Transfer paths and product-topology limits}\\label{earlier:section:01}\nLet $T(t)$ be an arbitrary strongly continuous semigroup on a complex Banach space $X$, with generator $A$, and suppose $\\norm{T(t)}\\leq M e^{\\omega t}$ for $M\\geq1$ and $\\omega\\in\\mathbb R$. Let $(P_\\alpha)_{\\alpha\\in I_0}$ be a finite family of bounded pairwise orthogonal idempotents summing to $I$, and put $T_{\\beta\\alpha}(t)=P_\\beta T(t)P_\\alpha$. The generator need not be bounded. In the following resolvent statement the integral is strong: apply the integrand to each $x\\in X$ and take its Bochner integral in $X$. It is not asserted to be a Bochner integral in the operator-norm space $\\mathcal B(X)$. The general Laplace formula is \\cite[Chapter II, Theorem 1.10]{EngelNagel2000}.\n\\begin{theorem}[Packet convolution]\\label{earlier:record:001}\\label{earlier:thm:packet_convolution}\nFor all \\(s,t\\geq0\\) and all packets \\(\\alpha,\\beta\\),\n\\[\n\\boxed{…","e}}\n\\renewcommand{\\theHfigure}{R.\\arabic{figure}}\n\\renewcommand{\\thetable}{R\\arabic{table}}\n\\renewcommand{\\theHtable}{R.\\arabic{table}}\n\\FloatBarrier\n\\section{Transfer paths and product-topology limits}\\label{earlier:section:01}\nLet $T(t)$ be an arbitrary strongly continuous semigroup on a complex Banach space $X$, with generator $A$, and suppose $\\norm{T(t)}\\leq M e^{\\omega t}$ for $M\\geq1$ and $\\omega\\in\\mathbb R$. Let $(P_\\alpha)_{\\alpha\\in I_0}$ be a finite family of bounded pairwise orthogonal idempotents summing to $I$, and put $T_{\\beta\\alpha}(t)=P_\\beta T(t)P_\\alpha$. The generator need not be bounded. In the following resolvent statement the integral is strong: apply the integrand to each $x\\in X$ and take its Bochner integral in $X$. It is not asserted to be a Bochner integral in the operator-norm space $\\mathcal B(X)$. The general Laplace formula is \\cite[Chapter II, Theorem 1.10]{EngelNagel2000}.\n\\begin{theorem}[Packet convolution]\\label{earlier:record:001}\\label{earlier:thm:packet_convolution}\nFor all \\(s,t\\geq0\\) and all packets \\(\\alpha,\\beta\\),\n\\[\n\\boxed{\nT_{\\beta\\alp…","ure}{R.\\arabic{figure}}\n\\renewcommand{\\thetable}{R\\arabic{table}}\n\\renewcommand{\\theHtable}{R.\\arabic{table}}\n\\FloatBarrier\n\\section{Transfer paths and product-topology limits}\\label{earlier:section:01}\nLet $T(t)$ be an arbitrary strongly continuous semigroup on a complex Banach space $X$, with generator $A$, and suppose $\\norm{T(t)}\\leq M e^{\\omega t}$ for $M\\geq1$ and $\\omega\\in\\mathbb R$. Let $(P_\\alpha)_{\\alpha\\in I_0}$ be a finite family of bounded pairwise orthogonal idempotents summing to $I$, and put $T_{\\beta\\alpha}(t)=P_\\beta T(t)P_\\alpha$. The generator need not be bounded. In the following resolvent statement the integral is strong: apply the integrand to each $x\\in X$ and take its Bochner integral in $X$. It is not asserted to be a Bochner integral in the operator-norm space $\\mathcal B(X)$. The general Laplace formula is \\cite[Chapter II, Theorem 1.10]{EngelNagel2000}.\n\\begin{theorem}[Packet convolution]\\label{earlier:record:001}\\label{earlier:thm:packet_convolution}\nFor all \\(s,t\\geq0\\) and all packets \\(\\alpha,\\beta\\),\n\\[\n\\boxed{\nT_{\\beta\\alpha}(s+t)\n=\n\\sum_{\\gamma\\in…","table}}\n\\FloatBarrier\n\\section{Transfer paths and product-topology limits}\\label{earlier:section:01}\nLet $T(t)$ be an arbitrary strongly continuous semigroup on a complex Banach space $X$, with generator $A$, and suppose $\\norm{T(t)}\\leq M e^{\\omega t}$ for $M\\geq1$ and $\\omega\\in\\mathbb R$. Let $(P_\\alpha)_{\\alpha\\in I_0}$ be a finite family of bounded pairwise orthogonal idempotents summing to $I$, and put $T_{\\beta\\alpha}(t)=P_\\beta T(t)P_\\alpha$. The generator need not be bounded. In the following resolvent statement the integral is strong: apply the integrand to each $x\\in X$ and take its Bochner integral in $X$. It is not asserted to be a Bochner integral in the operator-norm space $\\mathcal B(X)$. The general Laplace formula is \\cite[Chapter II, Theorem 1.10]{EngelNagel2000}.\n\\begin{theorem}[Packet convolution]\\label{earlier:record:001}\\label{earlier:thm:packet_convolution}\nFor all \\(s,t\\geq0\\) and all packets \\(\\alpha,\\beta\\),\n\\[\n\\boxed{\nT_{\\beta\\alpha}(s+t)\n=\n\\sum_{\\gamma\\in I_0}\nT_{\\beta\\gamma}(s)T_{\\gamma\\alpha}(t).\n}\n\\]\nThe original semigroup is reconstructed from its blo…","tBarrier\n\\section{Transfer paths and product-topology limits}\\label{earlier:section:01}\nLet $T(t)$ be an arbitrary strongly continuous semigroup on a complex Banach space $X$, with generator $A$, and suppose $\\norm{T(t)}\\leq M e^{\\omega t}$ for $M\\geq1$ and $\\omega\\in\\mathbb R$. Let $(P_\\alpha)_{\\alpha\\in I_0}$ be a finite family of bounded pairwise orthogonal idempotents summing to $I$, and put $T_{\\beta\\alpha}(t)=P_\\beta T(t)P_\\alpha$. The generator need not be bounded. In the following resolvent statement the integral is strong: apply the integrand to each $x\\in X$ and take its Bochner integral in $X$. It is not asserted to be a Bochner integral in the operator-norm space $\\mathcal B(X)$. The general Laplace formula is \\cite[Chapter II, Theorem 1.10]{EngelNagel2000}.\n\\begin{theorem}[Packet convolution]\\label{earlier:record:001}\\label{earlier:thm:packet_convolution}\nFor all \\(s,t\\geq0\\) and all packets \\(\\alpha,\\beta\\),\n\\[\n\\boxed{\nT_{\\beta\\alpha}(s+t)\n=\n\\sum_{\\gamma\\in I_0}\nT_{\\beta\\gamma}(s)T_{\\gamma\\alpha}(t).\n}\n\\]\nThe original semigroup is reconstructed from its blocks by\n\\[\nT(t…","ace $X$, with generator $A$, and suppose $\\norm{T(t)}\\leq M e^{\\omega t}$ for $M\\geq1$ and $\\omega\\in\\mathbb R$. Let $(P_\\alpha)_{\\alpha\\in I_0}$ be a finite family of bounded pairwise orthogonal idempotents summing to $I$, and put $T_{\\beta\\alpha}(t)=P_\\beta T(t)P_\\alpha$. The generator need not be bounded. In the following resolvent statement the integral is strong: apply the integrand to each $x\\in X$ and take its Bochner integral in $X$. It is not asserted to be a Bochner integral in the operator-norm space $\\mathcal B(X)$. The general Laplace formula is \\cite[Chapter II, Theorem 1.10]{EngelNagel2000}.\n\\begin{theorem}[Packet convolution]\\label{earlier:record:001}\\label{earlier:thm:packet_convolution}\nFor all \\(s,t\\geq0\\) and all packets \\(\\alpha,\\beta\\),\n\\[\n\\boxed{\nT_{\\beta\\alpha}(s+t)\n=\n\\sum_{\\gamma\\in I_0}\nT_{\\beta\\gamma}(s)T_{\\gamma\\alpha}(t).\n}\n\\]\nThe original semigroup is reconstructed from its blocks by\n\\[\nT(t)=\\sum_{\\alpha,\\beta\\in I_0}T_{\\beta\\alpha}(t).\n\\]\n\\end{theorem}","\\norm{T(t)}\\leq M e^{\\omega t}$ for $M\\geq1$ and $\\omega\\in\\mathbb R$. Let $(P_\\alpha)_{\\alpha\\in I_0}$ be a finite family of bounded pairwise orthogonal idempotents summing to $I$, and put $T_{\\beta\\alpha}(t)=P_\\beta T(t)P_\\alpha$. The generator need not be bounded. In the following resolvent statement the integral is strong: apply the integrand to each $x\\in X$ and take its Bochner integral in $X$. It is not asserted to be a Bochner integral in the operator-norm space $\\mathcal B(X)$. The general Laplace formula is \\cite[Chapter II, Theorem 1.10]{EngelNagel2000}.\n\\begin{theorem}[Packet convolution]\\label{earlier:record:001}\\label{earlier:thm:packet_convolution}\nFor all \\(s,t\\geq0\\) and all packets \\(\\alpha,\\beta\\),\n\\[\n\\boxed{\nT_{\\beta\\alpha}(s+t)\n=\n\\sum_{\\gamma\\in I_0}\nT_{\\beta\\gamma}(s)T_{\\gamma\\alpha}(t).\n}\n\\]\nThe original semigroup is reconstructed from its blocks by\n\\[\nT(t)=\\sum_{\\alpha,\\beta\\in I_0}T_{\\beta\\alpha}(t).\n\\]\n\\end{theorem}","(P_\\alpha)_{\\alpha\\in I_0}$ be a finite family of bounded pairwise orthogonal idempotents summing to $I$, and put $T_{\\beta\\alpha}(t)=P_\\beta T(t)P_\\alpha$. The generator need not be bounded. In the following resolvent statement the integral is strong: apply the integrand to each $x\\in X$ and take its Bochner integral in $X$. It is not asserted to be a Bochner integral in the operator-norm space $\\mathcal B(X)$. The general Laplace formula is \\cite[Chapter II, Theorem 1.10]{EngelNagel2000}.\n\\begin{theorem}[Packet convolution]\\label{earlier:record:001}\\label{earlier:thm:packet_convolution}\nFor all \\(s,t\\geq0\\) and all packets \\(\\alpha,\\beta\\),\n\\[\n\\boxed{\nT_{\\beta\\alpha}(s+t)\n=\n\\sum_{\\gamma\\in I_0}\nT_{\\beta\\gamma}(s)T_{\\gamma\\alpha}(t).\n}\n\\]\nThe original semigroup is reconstructed from its blocks by\n\\[\nT(t)=\\sum_{\\alpha,\\beta\\in I_0}T_{\\beta\\alpha}(t).\n\\]\n\\end{theorem}","ng resolvent statement the integral is strong: apply the integrand to each $x\\in X$ and take its Bochner integral in $X$. It is not asserted to be a Bochner integral in the operator-norm space $\\mathcal B(X)$. The general Laplace formula is \\cite[Chapter II, Theorem 1.10]{EngelNagel2000}.\n\\begin{theorem}[Packet convolution]\\label{earlier:record:001}\\label{earlier:thm:packet_convolution}\nFor all \\(s,t\\geq0\\) and all packets \\(\\alpha,\\beta\\),\n\\[\n\\boxed{\nT_{\\beta\\alpha}(s+t)\n=\n\\sum_{\\gamma\\in I_0}\nT_{\\beta\\gamma}(s)T_{\\gamma\\alpha}(t).\n}\n\\]\nThe original semigroup is reconstructed from its blocks by\n\\[\nT(t)=\\sum_{\\alpha,\\beta\\in I_0}T_{\\beta\\alpha}(t).\n\\]\n\\end{theorem}","tegral is strong: apply the integrand to each $x\\in X$ and take its Bochner integral in $X$. It is not asserted to be a Bochner integral in the operator-norm space $\\mathcal B(X)$. The general Laplace formula is \\cite[Chapter II, Theorem 1.10]{EngelNagel2000}.\n\\begin{theorem}[Packet convolution]\\label{earlier:record:001}\\label{earlier:thm:packet_convolution}\nFor all \\(s,t\\geq0\\) and all packets \\(\\alpha,\\beta\\),\n\\[\n\\boxed{\nT_{\\beta\\alpha}(s+t)\n=\n\\sum_{\\gamma\\in I_0}\nT_{\\beta\\gamma}(s)T_{\\gamma\\alpha}(t).\n}\n\\]\nThe original semigroup is reconstructed from its blocks by\n\\[\nT(t)=\\sum_{\\alpha,\\beta\\in I_0}T_{\\beta\\alpha}(t).\n\\]\n\\end{theorem}","apply the integrand to each $x\\in X$ and take its Bochner integral in $X$. It is not asserted to be a Bochner integral in the operator-norm space $\\mathcal B(X)$. The general Laplace formula is \\cite[Chapter II, Theorem 1.10]{EngelNagel2000}.\n\\begin{theorem}[Packet convolution]\\label{earlier:record:001}\\label{earlier:thm:packet_convolution}\nFor all \\(s,t\\geq0\\) and all packets \\(\\alpha,\\beta\\),\n\\[\n\\boxed{\nT_{\\beta\\alpha}(s+t)\n=\n\\sum_{\\gamma\\in I_0}\nT_{\\beta\\gamma}(s)T_{\\gamma\\alpha}(t).\n}\n\\]\nThe original semigroup is reconstructed from its blocks by\n\\[\nT(t)=\\sum_{\\alpha,\\beta\\in I_0}T_{\\beta\\alpha}(t).\n\\]\n\\end{theorem}","X)$. The general Laplace formula is \\cite[Chapter II, Theorem 1.10]{EngelNagel2000}.\n\\begin{theorem}[Packet convolution]\\label{earlier:record:001}\\label{earlier:thm:packet_convolution}\nFor all \\(s,t\\geq0\\) and all packets \\(\\alpha,\\beta\\),\n\\[\n\\boxed{\nT_{\\beta\\alpha}(s+t)\n=\n\\sum_{\\gamma\\in I_0}\nT_{\\beta\\gamma}(s)T_{\\gamma\\alpha}(t).\n}\n\\]\nThe original semigroup is reconstructed from its blocks by\n\\[\nT(t)=\\sum_{\\alpha,\\beta\\in I_0}T_{\\beta\\alpha}(t).\n\\]\n\\end{theorem}","\\begin{proof}\nInsert the identity between the two semigroup factors:\n\\[\n\\begin{aligned}\nP_\\beta T(s+t)P_\\alpha\n&=P_\\beta T(s)\\left(\\sum_\\gamma P_\\gamma\\right)T(t)P_\\alpha\\\\\n&=\\sum_\\gamma\nP_\\beta T(s)P_\\gamma T(t)P_\\alpha.\n\\end{aligned}\n\\]\nThe reconstruction formula follows by inserting the packet identity on both sides of \\(T(t)\\).\n\\end{proof}","&=\\sum_\\gamma\nP_\\beta T(s)P_\\gamma T(t)P_\\alpha.\n\\end{aligned}\n\\]\nThe reconstruction formula follows by inserting the packet identity on both sides of \\(T(t)\\).\n\\end{proof}","\\begin{proposition}[Packet resolvent convolution]\\label{earlier:record:002}\\label{earlier:prop:packet_resolvent}\nLet \\(z\\) belong to the resolvent set of \\(A\\). If \\(\\operatorname{Re}z\\) is larger than a growth bound for the semigroup, then\n\\[\nP_\\beta(zI-A)^{-1}P_\\alpha\n=\n\\int_0^{\\infty}\\e^{-zt}T_{\\beta\\alpha}(t)\\dd t.\n\\]\n\\end{proposition}","\\begin{proof}\nThe standard Laplace representation of the resolvent is\n\\[\n(zI-A)^{-1}\n=\n\\int_0^{\\infty}\\e^{-zt}T(t)\\dd t.\n\\]\nMultiplication by the bounded packet projections may be passed through the Bochner integral.\n\\end{proof}","For the path-order statement, now assume $A$ is bounded as in Part II. A directed edge from $\\alpha$ to $\\beta$ means that $P_\\beta A P_\\alpha$ is nonzero, with $\\alpha\\ne\\beta$; a packet always has the length-zero path to itself.\n\\begin{corollary}[Minimum transfer order]\\label{earlier:record:003}\\label{earlier:cor:minimum_transfer_order}\nSuppose every directed packet path from \\(\\alpha\\) to \\(\\beta\\) uses at least \\(m_0\\) off diagonal transfers. Then\n\\[\nP_\\beta\\e^{tA}P_\\alpha\n=O(t^{m_0})\n\\quad\\text{as }t\\downarrow0.\n\\]\nIf there is no directed path from \\(\\alpha\\) to \\(\\beta\\), the block vanishes for all \\(t\\).\n\\end{corollary}","atement, now assume $A$ is bounded as in Part II. A directed edge from $\\alpha$ to $\\beta$ means that $P_\\beta A P_\\alpha$ is nonzero, with $\\alpha\\ne\\beta$; a packet always has the length-zero path to itself.\n\\begin{corollary}[Minimum transfer order]\\label{earlier:record:003}\\label{earlier:cor:minimum_transfer_order}\nSuppose every directed packet path from \\(\\alpha\\) to \\(\\beta\\) uses at least \\(m_0\\) off diagonal transfers. Then\n\\[\nP_\\beta\\e^{tA}P_\\alpha\n=O(t^{m_0})\n\\quad\\text{as }t\\downarrow0.\n\\]\nIf there is no directed path from \\(\\alpha\\) to \\(\\beta\\), the block vanishes for all \\(t\\).\n\\end{corollary}","in Part II. A directed edge from $\\alpha$ to $\\beta$ means that $P_\\beta A P_\\alpha$ is nonzero, with $\\alpha\\ne\\beta$; a packet always has the length-zero path to itself.\n\\begin{corollary}[Minimum transfer order]\\label{earlier:record:003}\\label{earlier:cor:minimum_transfer_order}\nSuppose every directed packet path from \\(\\alpha\\) to \\(\\beta\\) uses at least \\(m_0\\) off diagonal transfers. Then\n\\[\nP_\\beta\\e^{tA}P_\\alpha\n=O(t^{m_0})\n\\quad\\text{as }t\\downarrow0.\n\\]\nIf there is no directed path from \\(\\alpha\\) to \\(\\beta\\), the block vanishes for all \\(t\\).\n\\end{corollary}","\\alpha\\ne\\beta$; a packet always has the length-zero path to itself.\n\\begin{corollary}[Minimum transfer order]\\label{earlier:record:003}\\label{earlier:cor:minimum_transfer_order}\nSuppose every directed packet path from \\(\\alpha\\) to \\(\\beta\\) uses at least \\(m_0\\) off diagonal transfers. Then\n\\[\nP_\\beta\\e^{tA}P_\\alpha\n=O(t^{m_0})\n\\quad\\text{as }t\\downarrow0.\n\\]\nIf there is no directed path from \\(\\alpha\\) to \\(\\beta\\), the block vanishes for all \\(t\\).\n\\end{corollary}","$; a packet always has the length-zero path to itself.\n\\begin{corollary}[Minimum transfer order]\\label{earlier:record:003}\\label{earlier:cor:minimum_transfer_order}\nSuppose every directed packet path from \\(\\alpha\\) to \\(\\beta\\) uses at least \\(m_0\\) off diagonal transfers. Then\n\\[\nP_\\beta\\e^{tA}P_\\alpha\n=O(t^{m_0})\n\\quad\\text{as }t\\downarrow0.\n\\]\nIf there is no directed path from \\(\\alpha\\) to \\(\\beta\\), the block vanishes for all \\(t\\).\n\\end{corollary}","path to itself.\n\\begin{corollary}[Minimum transfer order]\\label{earlier:record:003}\\label{earlier:cor:minimum_transfer_order}\nSuppose every directed packet path from \\(\\alpha\\) to \\(\\beta\\) uses at least \\(m_0\\) off diagonal transfers. Then\n\\[\nP_\\beta\\e^{tA}P_\\alpha\n=O(t^{m_0})\n\\quad\\text{as }t\\downarrow0.\n\\]\nIf there is no directed path from \\(\\alpha\\) to \\(\\beta\\), the block vanishes for all \\(t\\).\n\\end{corollary}","\\begin{proof}\nEvery nonzero term in \\cref{extended:thm:dyson_paths} contains at least \\(m_0\\) time integrations. The simplex \\(0<r_1<\\cdots<r_m<t\\) has volume \\(t^m/m!\\), so the first possible order is \\(t^{m_0}\\). If there is no path, every packet term vanishes.\n\\end{proof}","Let $C_j:X_{j+1}\\to X_j$ be bounded linear maps, and put $C_{\\ell,m}=C_\\ell\\cdots C_{m-1}$ for $\\ell<m$, with $C_{j,j}=I$. No common ambient space is assumed.\n\\begin{theorem}[Several level telescoping identity]\\label{earlier:record:004}\\label{earlier:thm:multilevel_telescoping}\nLet \\(T_j\\) be bounded operators on \\(X_j\\), and set\n\\[\nE_j=C_jT_{j+1}-T_jC_j.\n\\]\nThen\n\\[\n\\boxed{\nC_{\\ell,m}T_m-T_\\ell C_{\\ell,m}\n=\n\\sum_{j=\\ell}^{m-1}\nC_{\\ell,j}E_jC_{j+1,m}.\n}\n\\]\n\\end{theorem}","\\begin{proof}\nExpand each summand as\n\\[\nC_{\\ell,j+1}T_{j+1}C_{j+1,m}\n-\nC_{\\ell,j}T_jC_{j,m}.\n\\]\nThe intermediate terms cancel. The first surviving term is \\(C_{\\ell,m}T_m\\), and the last is \\(-T_\\ell C_{\\ell,m}\\).\n\\end{proof}","\\begin{corollary}[Summable generator defects]\\label{earlier:record:005}\\label{earlier:cor:summable_defects}\nSuppose \\(C_j\\) are contractions and \\((T_j(t))\\) are contraction semigroups with bounded generators \\(A_j\\). Set\n\\[\n\\delta_j=\\norm{C_jA_{j+1}-A_jC_j}.\n\\]\nThen\n\\[\n\\norm{C_{\\ell,m}T_m(t)-T_\\ell(t)C_{\\ell,m}}\n\\leq\n t\\sum_{j=\\ell}^{m-1}\\delta_j.\n\\]\nIf \\(\\sum_{j\\geq0}\\delta_j<\\infty\\), the right side tends to zero as \\(\\ell\\to\\infty\\), uniformly in \\(m>\\ell\\) and uniformly for \\(t\\) in a fixed bounded interval.\n\\end{corollary}","e generator defects]\\label{earlier:record:005}\\label{earlier:cor:summable_defects}\nSuppose \\(C_j\\) are contractions and \\((T_j(t))\\) are contraction semigroups with bounded generators \\(A_j\\). Set\n\\[\n\\delta_j=\\norm{C_jA_{j+1}-A_jC_j}.\n\\]\nThen\n\\[\n\\norm{C_{\\ell,m}T_m(t)-T_\\ell(t)C_{\\ell,m}}\n\\leq\n t\\sum_{j=\\ell}^{m-1}\\delta_j.\n\\]\nIf \\(\\sum_{j\\geq0}\\delta_j<\\infty\\), the right side tends to zero as \\(\\ell\\to\\infty\\), uniformly in \\(m>\\ell\\) and uniformly for \\(t\\) in a fixed bounded interval.\n\\end{corollary}","er:record:005}\\label{earlier:cor:summable_defects}\nSuppose \\(C_j\\) are contractions and \\((T_j(t))\\) are contraction semigroups with bounded generators \\(A_j\\). Set\n\\[\n\\delta_j=\\norm{C_jA_{j+1}-A_jC_j}.\n\\]\nThen\n\\[\n\\norm{C_{\\ell,m}T_m(t)-T_\\ell(t)C_{\\ell,m}}\n\\leq\n t\\sum_{j=\\ell}^{m-1}\\delta_j.\n\\]\nIf \\(\\sum_{j\\geq0}\\delta_j<\\infty\\), the right side tends to zero as \\(\\ell\\to\\infty\\), uniformly in \\(m>\\ell\\) and uniformly for \\(t\\) in a fixed bounded interval.\n\\end{corollary}","cor:summable_defects}\nSuppose \\(C_j\\) are contractions and \\((T_j(t))\\) are contraction semigroups with bounded generators \\(A_j\\). Set\n\\[\n\\delta_j=\\norm{C_jA_{j+1}-A_jC_j}.\n\\]\nThen\n\\[\n\\norm{C_{\\ell,m}T_m(t)-T_\\ell(t)C_{\\ell,m}}\n\\leq\n t\\sum_{j=\\ell}^{m-1}\\delta_j.\n\\]\nIf \\(\\sum_{j\\geq0}\\delta_j<\\infty\\), the right side tends to zero as \\(\\ell\\to\\infty\\), uniformly in \\(m>\\ell\\) and uniformly for \\(t\\) in a fixed bounded interval.\n\\end{corollary}","\\begin{proof}\nBy \\cref{extended:thm:duhamel_defect}, the interlevel semigroup defect has norm at most \\(t\\delta_j\\). Apply \\cref{earlier:thm:multilevel_telescoping} and use the contraction bounds on all coarsening maps.\n\\end{proof}","\\begin{definition}[Projective limit space]\\label{earlier:record:006}\\label{earlier:def:projective_limit}\nLet \\((X_j,C_j)\\) be a sequence of Banach spaces and bounded linear maps \\(C_j:X_{j+1}\\to X_j\\). Its projective limit is\n\\[\nX_\\infty\n=\n\\left\\{\n(x_j)_{j\\geq0}\\in\\prod_{j\\geq0}X_j:\nC_jx_{j+1}=x_j\\text{ for every }j\n\\right\\}.\n\\]\nWe equip \\(X_\\infty\\) with the subspace topology inherited from the product topology.\n\\end{definition}","\\begin{theorem}[Exact projective limit semigroup]\\label{earlier:record:007}\\label{earlier:thm:projective_limit_semigroup}\nSuppose each \\(X_j\\) carries a strongly continuous semigroup \\((T_j(t))_{t\\geq0}\\), and\n\\[\nC_jT_{j+1}(t)=T_j(t)C_j\n\\]\nfor every \\(j\\) and every \\(t\\geq0\\). Then\n\\[\nT_\\infty(t)(x_j)_{j\\geq0}\n=\n(T_j(t)x_j)_{j\\geq0}\n\\]\ndefines a strongly continuous semigroup on \\(X_\\infty\\).\n\\end{theorem}","\\begin{proof}\nExact intertwining gives\n\\[\nC_jT_{j+1}(t)x_{j+1}\n=\nT_j(t)C_jx_{j+1}\n=\nT_j(t)x_j.\n\\]\nThus the coordinatewise evolution remains in the projective limit. The semigroup law holds in every coordinate. Strong continuity in the product topology follows because every coordinate map\n\\[\nt\\longmapsto T_j(t)x_j\n\\]\nis continuous.\n\\end{proof}","\\begin{corollary}[Coordinate generators]\\label{earlier:record:008}\\label{earlier:cor:projective_generators}\nAssume the setting of \\cref{earlier:thm:projective_limit_semigroup}. If \\(x=(x_j)\\) satisfies \\(x_j\\in\\operatorname{Dom}(A_j)\\) and the sequence \\((A_jx_j)\\) belongs to \\(X_\\infty\\), then \\(x\\) lies in the generator domain of \\(T_\\infty\\) and\n\\[\nA_\\infty x=(A_jx_j)_{j\\geq0}.\n\\]\n\\end{corollary}","\\begin{proof}\nThe difference quotients converge in every coordinate to \\(A_jx_j\\). The assumed compatibility places the coordinatewise limit in \\(X_\\infty\\), which is exactly convergence in the product topology.\n\\end{proof}","The product topology used here differs from the bounded inverse-limit norm of Part II. No uniform boundedness of coordinate generators is required by the coordinatewise statement.\n\\FloatBarrier\n\\section{Reversible coarse dynamics}\\label{earlier:section:02}\nLet $Q$ be a finite Markov generator acting on column probability vectors, let $(B_a)$ be a partition into nonempty blocks, and let $C_{ai}=\\one_{i\\in B_a}$. Strong lumpability means $CQ=\\overline Q C$ for a coarse Markov generator. For observables, the backward generator is $Q^T$; weighted $L^2$ inner products use the stated stationary probability.\n\\begin{theorem}[Reversibility descends]\\label{earlier:record:009}\\label{earlier:thm:reversibility_descends}\nAssume \\(Q\\) is strongly lumpable and reversible with respect to a positive stationary distribution \\(\\pi\\), so\n\\[\nQ_{ij}\\pi_j=Q_{ji}\\pi_i.\n\\]\nDefine\n\\[\n\\overline\\pi_a=\\sum_{i\\in B_a}\\pi_i.\n\\]\nThen \\(\\overline\\pi\\) is stationary for \\(\\overline Q","The product topology used here differs from the bounded inverse-limit norm of Part II. No uniform boundedness of coordinate generators is required by the coordinatewise statement.\n\\FloatBarrier\n\\section{Reversible coarse dynamics}\\label{earlier:section:02}\nLet $Q$ be a finite Markov generator acting on column probability vectors, let $(B_a)$ be a partition into nonempty blocks, and let $C_{ai}=\\one_{i\\in B_a}$. Strong lumpability means $CQ=\\overline Q C$ for a coarse Markov generator. For observables, the backward generator is $Q^T$; weighted $L^2$ inner products use the stated stationary probability.\n\\begin{theorem}[Reversibility descends]\\label{earlier:record:009}\\label{earlier:thm:reversibility_descends}\nAssume \\(Q\\) is strongly lumpable and reversible with respect to a positive stationary distribution \\(\\pi\\), so\n\\[\nQ_{ij}\\pi_j=Q_{ji}\\pi_i.\n\\]\nDefine\n\\[\n\\overline\\pi_a=\\sum_{i\\in B_a}\\pi_i.\n\\]\nThen \\(\\overline\\pi\\) is stationary for \\(\\overline Q\\), and\n\\[\n\\overline Q_{ab}\\overline\\pi_b\n=\n\\overline Q_{ba}\\overline\\pi_a.\n\\]","The product topology used here differs from the bounded inverse-limit norm of Part II. No uniform boundedness of coordinate generators is required by the coordinatewise statement.\n\\FloatBarrier\n\\section{Reversible coarse dynamics}\\label{earlier:section:02}\nLet $Q$ be a finite Markov generator acting on column probability vectors, let $(B_a)$ be a partition into nonempty blocks, and let $C_{ai}=\\one_{i\\in B_a}$. Strong lumpability means $CQ=\\overline Q C$ for a coarse Markov generator. For observables, the backward generator is $Q^T$; weighted $L^2$ inner products use the stated stationary probability.\n\\begin{theorem}[Reversibility descends]\\label{earlier:record:009}\\label{earlier:thm:reversibility_descends}\nAssume \\(Q\\) is strongly lumpable and reversible with respect to a positive stationary distribution \\(\\pi\\), so\n\\[\nQ_{ij}\\pi_j=Q_{ji}\\pi_i.\n\\]\nDefine\n\\[\n\\overline\\pi_a=\\sum_{i\\in B_a}\\pi_i.\n\\]\nThen \\(\\overline\\pi\\) is stationary for \\(\\overline Q\\), and\n\\[\n\\overline Q_{ab}\\overline\\pi_b\n=\n\\overline Q_{ba}\\overline\\pi_a.\n\\]\nThus the coarse chain is reversible.\n\\end{theorem}","om the bounded inverse-limit norm of Part II. No uniform boundedness of coordinate generators is required by the coordinatewise statement.\n\\FloatBarrier\n\\section{Reversible coarse dynamics}\\label{earlier:section:02}\nLet $Q$ be a finite Markov generator acting on column probability vectors, let $(B_a)$ be a partition into nonempty blocks, and let $C_{ai}=\\one_{i\\in B_a}$. Strong lumpability means $CQ=\\overline Q C$ for a coarse Markov generator. For observables, the backward generator is $Q^T$; weighted $L^2$ inner products use the stated stationary probability.\n\\begin{theorem}[Reversibility descends]\\label{earlier:record:009}\\label{earlier:thm:reversibility_descends}\nAssume \\(Q\\) is strongly lumpable and reversible with respect to a positive stationary distribution \\(\\pi\\), so\n\\[\nQ_{ij}\\pi_j=Q_{ji}\\pi_i.\n\\]\nDefine\n\\[\n\\overline\\pi_a=\\sum_{i\\in B_a}\\pi_i.\n\\]\nThen \\(\\overline\\pi\\) is stationary for \\(\\overline Q\\), and\n\\[\n\\overline Q_{ab}\\overline\\pi_b\n=\n\\overline Q_{ba}\\overline\\pi_a.\n\\]\nThus the coarse chain is reversible.\n\\end{theorem}","is required by the coordinatewise statement.\n\\FloatBarrier\n\\section{Reversible coarse dynamics}\\label{earlier:section:02}\nLet $Q$ be a finite Markov generator acting on column probability vectors, let $(B_a)$ be a partition into nonempty blocks, and let $C_{ai}=\\one_{i\\in B_a}$. Strong lumpability means $CQ=\\overline Q C$ for a coarse Markov generator. For observables, the backward generator is $Q^T$; weighted $L^2$ inner products use the stated stationary probability.\n\\begin{theorem}[Reversibility descends]\\label{earlier:record:009}\\label{earlier:thm:reversibility_descends}\nAssume \\(Q\\) is strongly lumpable and reversible with respect to a positive stationary distribution \\(\\pi\\), so\n\\[\nQ_{ij}\\pi_j=Q_{ji}\\pi_i.\n\\]\nDefine\n\\[\n\\overline\\pi_a=\\sum_{i\\in B_a}\\pi_i.\n\\]\nThen \\(\\overline\\pi\\) is stationary for \\(\\overline Q\\), and\n\\[\n\\overline Q_{ab}\\overline\\pi_b\n=\n\\overline Q_{ba}\\overline\\pi_a.\n\\]\nThus the coarse chain is reversible.\n\\end{theorem}","the coordinatewise statement.\n\\FloatBarrier\n\\section{Reversible coarse dynamics}\\label{earlier:section:02}\nLet $Q$ be a finite Markov generator acting on column probability vectors, let $(B_a)$ be a partition into nonempty blocks, and let $C_{ai}=\\one_{i\\in B_a}$. Strong lumpability means $CQ=\\overline Q C$ for a coarse Markov generator. For observables, the backward generator is $Q^T$; weighted $L^2$ inner products use the stated stationary probability.\n\\begin{theorem}[Reversibility descends]\\label{earlier:record:009}\\label{earlier:thm:reversibility_descends}\nAssume \\(Q\\) is strongly lumpable and reversible with respect to a positive stationary distribution \\(\\pi\\), so\n\\[\nQ_{ij}\\pi_j=Q_{ji}\\pi_i.\n\\]\nDefine\n\\[\n\\overline\\pi_a=\\sum_{i\\in B_a}\\pi_i.\n\\]\nThen \\(\\overline\\pi\\) is stationary for \\(\\overline Q\\), and\n\\[\n\\overline Q_{ab}\\overline\\pi_b\n=\n\\overline Q_{ba}\\overline\\pi_a.\n\\]\nThus the coarse chain is reversible.\n\\end{theorem}","tors, let $(B_a)$ be a partition into nonempty blocks, and let $C_{ai}=\\one_{i\\in B_a}$. Strong lumpability means $CQ=\\overline Q C$ for a coarse Markov generator. For observables, the backward generator is $Q^T$; weighted $L^2$ inner products use the stated stationary probability.\n\\begin{theorem}[Reversibility descends]\\label{earlier:record:009}\\label{earlier:thm:reversibility_descends}\nAssume \\(Q\\) is strongly lumpable and reversible with respect to a positive stationary distribution \\(\\pi\\), so\n\\[\nQ_{ij}\\pi_j=Q_{ji}\\pi_i.\n\\]\nDefine\n\\[\n\\overline\\pi_a=\\sum_{i\\in B_a}\\pi_i.\n\\]\nThen \\(\\overline\\pi\\) is stationary for \\(\\overline Q\\), and\n\\[\n\\overline Q_{ab}\\overline\\pi_b\n=\n\\overline Q_{ba}\\overline\\pi_a.\n\\]\nThus the coarse chain is reversible.\n\\end{theorem}","ng lumpability means $CQ=\\overline Q C$ for a coarse Markov generator. For observables, the backward generator is $Q^T$; weighted $L^2$ inner products use the stated stationary probability.\n\\begin{theorem}[Reversibility descends]\\label{earlier:record:009}\\label{earlier:thm:reversibility_descends}\nAssume \\(Q\\) is strongly lumpable and reversible with respect to a positive stationary distribution \\(\\pi\\), so\n\\[\nQ_{ij}\\pi_j=Q_{ji}\\pi_i.\n\\]\nDefine\n\\[\n\\overline\\pi_a=\\sum_{i\\in B_a}\\pi_i.\n\\]\nThen \\(\\overline\\pi\\) is stationary for \\(\\overline Q\\), and\n\\[\n\\overline Q_{ab}\\overline\\pi_b\n=\n\\overline Q_{ba}\\overline\\pi_a.\n\\]\nThus the coarse chain is reversible.\n\\end{theorem}","ty means $CQ=\\overline Q C$ for a coarse Markov generator. For observables, the backward generator is $Q^T$; weighted $L^2$ inner products use the stated stationary probability.\n\\begin{theorem}[Reversibility descends]\\label{earlier:record:009}\\label{earlier:thm:reversibility_descends}\nAssume \\(Q\\) is strongly lumpable and reversible with respect to a positive stationary distribution \\(\\pi\\), so\n\\[\nQ_{ij}\\pi_j=Q_{ji}\\pi_i.\n\\]\nDefine\n\\[\n\\overline\\pi_a=\\sum_{i\\in B_a}\\pi_i.\n\\]\nThen \\(\\overline\\pi\\) is stationary for \\(\\overline Q\\), and\n\\[\n\\overline Q_{ab}\\overline\\pi_b\n=\n\\overline Q_{ba}\\overline\\pi_a.\n\\]\nThus the coarse chain is reversible.\n\\end{theorem}","se Markov generator. For observables, the backward generator is $Q^T$; weighted $L^2$ inner products use the stated stationary probability.\n\\begin{theorem}[Reversibility descends]\\label{earlier:record:009}\\label{earlier:thm:reversibility_descends}\nAssume \\(Q\\) is strongly lumpable and reversible with respect to a positive stationary distribution \\(\\pi\\), so\n\\[\nQ_{ij}\\pi_j=Q_{ji}\\pi_i.\n\\]\nDefine\n\\[\n\\overline\\pi_a=\\sum_{i\\in B_a}\\pi_i.\n\\]\nThen \\(\\overline\\pi\\) is stationary for \\(\\overline Q\\), and\n\\[\n\\overline Q_{ab}\\overline\\pi_b\n=\n\\overline Q_{ba}\\overline\\pi_a.\n\\]\nThus the coarse chain is reversible.\n\\end{theorem}","rd generator is $Q^T$; weighted $L^2$ inner products use the stated stationary probability.\n\\begin{theorem}[Reversibility descends]\\label{earlier:record:009}\\label{earlier:thm:reversibility_descends}\nAssume \\(Q\\) is strongly lumpable and reversible with respect to a positive stationary distribution \\(\\pi\\), so\n\\[\nQ_{ij}\\pi_j=Q_{ji}\\pi_i.\n\\]\nDefine\n\\[\n\\overline\\pi_a=\\sum_{i\\in B_a}\\pi_i.\n\\]\nThen \\(\\overline\\pi\\) is stationary for \\(\\overline Q\\), and\n\\[\n\\overline Q_{ab}\\overline\\pi_b\n=\n\\overline Q_{ba}\\overline\\pi_a.\n\\]\nThus the coarse chain is reversible.\n\\end{theorem}","2$ inner products use the stated stationary probability.\n\\begin{theorem}[Reversibility descends]\\label{earlier:record:009}\\label{earlier:thm:reversibility_descends}\nAssume \\(Q\\) is strongly lumpable and reversible with respect to a positive stationary distribution \\(\\pi\\), so\n\\[\nQ_{ij}\\pi_j=Q_{ji}\\pi_i.\n\\]\nDefine\n\\[\n\\overline\\pi_a=\\sum_{i\\in B_a}\\pi_i.\n\\]\nThen \\(\\overline\\pi\\) is stationary for \\(\\overline Q\\), and\n\\[\n\\overline Q_{ab}\\overline\\pi_b\n=\n\\overline Q_{ba}\\overline\\pi_a.\n\\]\nThus the coarse chain is reversible.\n\\end{theorem}","the stated stationary probability.\n\\begin{theorem}[Reversibility descends]\\label{earlier:record:009}\\label{earlier:thm:reversibility_descends}\nAssume \\(Q\\) is strongly lumpable and reversible with respect to a positive stationary distribution \\(\\pi\\), so\n\\[\nQ_{ij}\\pi_j=Q_{ji}\\pi_i.\n\\]\nDefine\n\\[\n\\overline\\pi_a=\\sum_{i\\in B_a}\\pi_i.\n\\]\nThen \\(\\overline\\pi\\) is stationary for \\(\\overline Q\\), and\n\\[\n\\overline Q_{ab}\\overline\\pi_b\n=\n\\overline Q_{ba}\\overline\\pi_a.\n\\]\nThus the coarse chain is reversible.\n\\end{theorem}","\\begin{proof}\nStationarity follows from\n\\[\n\\overline Q\\overline\\pi\n=\\overline QC\\pi\n=CQ\\pi=0.\n\\]\nFor reversibility, use the aggregate rate formula:\n\\[\n\\begin{aligned}\n\\overline Q_{ab}\\overline\\pi_b\n&=\\sum_{j\\in B_b}\\pi_j\\sum_{i\\in B_a}Q_{ij}","\\begin{proof}\nStationarity follows from\n\\[\n\\overline Q\\overline\\pi\n=\\overline QC\\pi\n=CQ\\pi=0.\n\\]\nFor reversibility, use the aggregate rate formula:\n\\[\n\\begin{aligned}\n\\overline Q_{ab}\\overline\\pi_b\n&=\\sum_{j\\in B_b}\\pi_j\\sum_{i\\in B_a}Q_{ij}\\\\\n&=\\sum_{i\\in B_a}\\sum_{j\\in B_b}\\pi_iQ_{ji}\\\\\n&=\\overline Q_{ba}\\overline\\pi_a.\n\\end{aligned}\n\\]\n\\end{proof}","\\begin{corollary}[Coarse Dirichlet form]\\label{earlier:record:010}\\label{earlier:cor:coarse_dirichlet_form}\nUnder the hypotheses of \\cref{earlier:thm:reversibility_descends}, let \\(Jg\\) be the fine function that is constant with value \\(g_a\\) on \\(B_a\\). Then\n\\[\n-\\ip{Jg}{Q^TJg}_{L^2(\\pi)}\n=\n-\\ip{g}{\\overline Q^Tg}_{L^2(\\overline\\pi)}.\n\\]\n\\end{corollary}","\\FloatBarrier\n\\section{Prior-restricted sufficiency and filtering}\\label{earlier:section:03}\nLet $L$ be a strictly positive finite likelihood and let $\\Bcal_L(p)=L\\odot p/\\ip{L}{p}$ for a probability vector $p$. Fix probability vectors $r^{(a)}$ supported on $B_a$, and restrict the priors to $p_i=q_a r_i^{(a)}$ for $i\\in B_a$, with $q$ a coarse probability vector.\n\\begin{proposition}[Sufficiency relative to a prior disintegration]\\label{earlier:record:011}\\label{earlier:prop:relative_sufficiency}\nDefine\n\\[\n\\overline L_a\n=\n\\sum_{i\\in B_a}L_ir_i^{(a)}.\n\\]\nFor every prior in the declared family,\n\\[\nC\\Bcal_L(p)\n=\n\\Bcal_{\\overline L}(q).\n\\]\n\\end{proposition}","tly positive finite likelihood and let $\\Bcal_L(p)=L\\odot p/\\ip{L}{p}$ for a probability vector $p$. Fix probability vectors $r^{(a)}$ supported on $B_a$, and restrict the priors to $p_i=q_a r_i^{(a)}$ for $i\\in B_a$, with $q$ a coarse probability vector.\n\\begin{proposition}[Sufficiency relative to a prior disintegration]\\label{earlier:record:011}\\label{earlier:prop:relative_sufficiency}\nDefine\n\\[\n\\overline L_a\n=\n\\sum_{i\\in B_a}L_ir_i^{(a)}.\n\\]\nFor every prior in the declared family,\n\\[\nC\\Bcal_L(p)\n=\n\\Bcal_{\\overline L}(q).\n\\]\n\\end{proposition}","ctor $p$. Fix probability vectors $r^{(a)}$ supported on $B_a$, and restrict the priors to $p_i=q_a r_i^{(a)}$ for $i\\in B_a$, with $q$ a coarse probability vector.\n\\begin{proposition}[Sufficiency relative to a prior disintegration]\\label{earlier:record:011}\\label{earlier:prop:relative_sufficiency}\nDefine\n\\[\n\\overline L_a\n=\n\\sum_{i\\in B_a}L_ir_i^{(a)}.\n\\]\nFor every prior in the declared family,\n\\[\nC\\Bcal_L(p)\n=\n\\Bcal_{\\overline L}(q).\n\\]\n\\end{proposition}","\\begin{proof}\nThe evidence in block \\(B_a\\) is\n\\[\nZ_a\n=q_a\\sum_{i\\in B_a}L_ir_i^{(a)}\n=q_a\\overline L_a.\n\\]\nApply \\cref{extended:thm:bayesian_packet_reconstruction}.\n\\end{proof}","\\begin{lemma}[Normalization bound]\\label{earlier:record:012}\\label{earlier:lem:normalization_bound}\nLet \\(a,b\\in[0,\\infty)^m\\) be nonzero, and set \\(A=\\norm a_1\\) and \\(B=\\norm b_1\\). Then\n\\[\n\\boxed{\n\\norm{a/A-b/B}_1\n\\leq\n\\frac{2\\norm{a-b}_1}{\\max\\{A,B\\}}.\n}\n\\]\n\\end{lemma}","\\begin{proof}\nUsing \\(A\\) in the denominator gives\n\\[\n\\norm{a/A-b/B}_1\n\\leq\n\\frac{\\norm{a-b}_1}{A}\n+\n\\norm b_1\\abs{1/A-1/B}.\n\\]\nThe second term equals \\(\\abs{B-A}/A\\), which is at most \\(\\norm{a-b}_1/A\\). Interchanging \\(a\\) and \\(b\\) gives the analogous bound with \\(B\\). Taking the smaller of the two bounds gives the result.\n\\end{proof}","For an arbitrary prior $p$, put $q=Cp$ and $Z(p,L)=\\ip{L}{p}$. Define $\\ell_a(p)=\\sum_{i\\in B_a}L_ip_i/q_a$ when $q_a>0$; choose any positive value when $q_a=0$. Only the product $q_a\\ell_a(p)$ enters the formulas.\n\\begin{theorem}[Exact information loss formula]\\label{earlier:record:013}\\label{earlier:thm:information_loss}\nFor every prior,\n\\[\n\\boxed{\nC\\Bcal_L(p)\n=\n\\Bcal_{\\ell(p)}(q).\n}\n\\]\nLet \\(\\overline L\\) be any fixed positive coarse likelihood and set\n\\[\nd(p\\mid L,\\overline L)\n=\n\\sum_{a=1}^mq_a\\abs{\\ell_a(p)-\\overline L_a}.\n\\]\nThen\n\\[\n\\norm{\nC\\Bcal_L(p)-\\Bcal_{\\overline L}(q)\n}_1\n\\leq\n\\frac{2d(p\\mid L,\\overline L)}\n{\\max\\{Z(p,L),\\ip{\\overline L}{q}\\}}.\n\\]\n\\end{theorem}","Define $\\ell_a(p)=\\sum_{i\\in B_a}L_ip_i/q_a$ when $q_a>0$; choose any positive value when $q_a=0$. Only the product $q_a\\ell_a(p)$ enters the formulas.\n\\begin{theorem}[Exact information loss formula]\\label{earlier:record:013}\\label{earlier:thm:information_loss}\nFor every prior,\n\\[\n\\boxed{\nC\\Bcal_L(p)\n=\n\\Bcal_{\\ell(p)}(q).\n}\n\\]\nLet \\(\\overline L\\) be any fixed positive coarse likelihood and set\n\\[\nd(p\\mid L,\\overline L)\n=\n\\sum_{a=1}^mq_a\\abs{\\ell_a(p)-\\overline L_a}.\n\\]\nThen\n\\[\n\\norm{\nC\\Bcal_L(p)-\\Bcal_{\\overline L}(q)\n}_1\n\\leq\n\\frac{2d(p\\mid L,\\overline L)}\n{\\max\\{Z(p,L),\\ip{\\overline L}{q}\\}}.\n\\]\n\\end{theorem}","lue when $q_a=0$. Only the product $q_a\\ell_a(p)$ enters the formulas.\n\\begin{theorem}[Exact information loss formula]\\label{earlier:record:013}\\label{earlier:thm:information_loss}\nFor every prior,\n\\[\n\\boxed{\nC\\Bcal_L(p)\n=\n\\Bcal_{\\ell(p)}(q).\n}\n\\]\nLet \\(\\overline L\\) be any fixed positive coarse likelihood and set\n\\[\nd(p\\mid L,\\overline L)\n=\n\\sum_{a=1}^mq_a\\abs{\\ell_a(p)-\\overline L_a}.\n\\]\nThen\n\\[\n\\norm{\nC\\Bcal_L(p)-\\Bcal_{\\overline L}(q)\n}_1\n\\leq\n\\frac{2d(p\\mid L,\\overline L)}\n{\\max\\{Z(p,L),\\ip{\\overline L}{q}\\}}.\n\\]\n\\end{theorem}","\\begin{proof}\nThe packet evidence in block \\(a\\) is \\(q_a\\ell_a(p)\\), which proves the exact identity. Apply \\cref{earlier:lem:normalization_bound} to\n\\[\na=q\\odot\\ell(p),\n\\qquad\nb=q\\odot\\overline L.\n\\]\nTheir \\(\\ell^1\\) norms are the two evidences, and \\(\\norm{a-b}_1=d(p\\mid L,\\overline L)\\).\n\\end{proof}","\\begin{corollary}[Oscillation bound]\\label{earlier:record:014}\\label{earlier:cor:bayes_oscillation}\nSuppose \\(\\overline L_a\\) is any representative likelihood for block \\(B_a\\), and set\n\\[\n\\omega_a\n=\n\\max_{i\\in B_a}\\abs{L_i-\\overline L_a}.\n\\]\nThen\n\\[\nd(p\\mid L,\\overline L)\n\\leq\n\\sum_aq_a\\omega_a.\n\\]\nThus small within block likelihood oscillation gives a small coarse posterior defect whenever the evidence is bounded away from zero.\n\\end{corollary}","\\begin{proof}\nThe conditional average \\(\\ell_a(p)\\) lies in the convex hull of the likelihood values inside \\(B_a\\). Hence\n\\[\n\\abs{\\ell_a(p)-\\overline L_a}\\leq\\omega_a.\n\\]\nMultiply by \\(q_a\\) and sum.\n\\end{proof}","Along the Bayesian flow $p_i(t)=e^{tg_i}p_i(0)/\\sum_j e^{tg_j}p_j(0)$, the following statement holds for every real observable $h$.\n\\begin{corollary}[Covariance identity]\\label{earlier:record:015}\\label{earlier:cor:bayesian_covariance}\nFor any observable \\(h\\in\\RR^n\\),\n\\[\n\\frac{\\dd}{\\dd t}\\EE_{p(t)}[h]\n=\n\\Cov_{p(t)}(h,g).\n\\]\nIn particular,\n\\[\n\\frac{\\dd}{\\dd t}\\EE_{p(t)}[g]\n=\n\\Var_{p(t)}(g)\\geq0.\n\\]\n\\end{corollary}","\\begin{proof}\nDifferentiate \\(\\sum_ih_ip_i(t)\\) and insert the replicator equation.\n\\end{proof}","For stochastic matrices $K,\\overline K$ and positive likelihoods $L_y,\\overline L_y$, put $\\Phi_y=\\Bcal_{L_y}\\circ K$ and $\\overline\\Phi_y=\\Bcal_{\\overline L_y}\\circ\\overline K$.\n\\begin{theorem}[Exact coarse hidden Markov filter]\\label{earlier:record:016}\\label{earlier:thm:exact_coarse_filter}\nAssume\n\\[\nCK=\\overline KC\n\\]\nand every likelihood \\(L_y\\) is constant on each coarse block with block values \\(\\overline L_y\\). Then\n\\[\nC\\Phi_y(p)=\\overline\\Phi_y(Cp)\n\\]\nfor every prior. For any observation sequence \\(y_1,\\ldots,y_r\\), the complete filtering recursion commutes with aggregation:\n\\[\nC\\Phi_{y_r}\\circ\\cdots\\circ\\Phi_{y_1}\n=\n\\overline\\Phi_{y_r}\\circ\\cdots\\circ\\overline\\Phi_{y_1}C.\n\\]\n\\end{theorem}","ochastic matrices $K,\\overline K$ and positive likelihoods $L_y,\\overline L_y$, put $\\Phi_y=\\Bcal_{L_y}\\circ K$ and $\\overline\\Phi_y=\\Bcal_{\\overline L_y}\\circ\\overline K$.\n\\begin{theorem}[Exact coarse hidden Markov filter]\\label{earlier:record:016}\\label{earlier:thm:exact_coarse_filter}\nAssume\n\\[\nCK=\\overline KC\n\\]\nand every likelihood \\(L_y\\) is constant on each coarse block with block values \\(\\overline L_y\\). Then\n\\[\nC\\Phi_y(p)=\\overline\\Phi_y(Cp)\n\\]\nfor every prior. For any observation sequence \\(y_1,\\ldots,y_r\\), the complete filtering recursion commutes with aggregation:\n\\[\nC\\Phi_{y_r}\\circ\\cdots\\circ\\Phi_{y_1}\n=\n\\overline\\Phi_{y_r}\\circ\\cdots\\circ\\overline\\Phi_{y_1}C.\n\\]\n\\end{theorem}","erline K$ and positive likelihoods $L_y,\\overline L_y$, put $\\Phi_y=\\Bcal_{L_y}\\circ K$ and $\\overline\\Phi_y=\\Bcal_{\\overline L_y}\\circ\\overline K$.\n\\begin{theorem}[Exact coarse hidden Markov filter]\\label{earlier:record:016}\\label{earlier:thm:exact_coarse_filter}\nAssume\n\\[\nCK=\\overline KC\n\\]\nand every likelihood \\(L_y\\) is constant on each coarse block with block values \\(\\overline L_y\\). Then\n\\[\nC\\Phi_y(p)=\\overline\\Phi_y(Cp)\n\\]\nfor every prior. For any observation sequence \\(y_1,\\ldots,y_r\\), the complete filtering recursion commutes with aggregation:\n\\[\nC\\Phi_{y_r}\\circ\\cdots\\circ\\Phi_{y_1}\n=\n\\overline\\Phi_{y_r}\\circ\\cdots\\circ\\overline\\Phi_{y_1}C.\n\\]\n\\end{theorem}","c K$ and $\\overline\\Phi_y=\\Bcal_{\\overline L_y}\\circ\\overline K$.\n\\begin{theorem}[Exact coarse hidden Markov filter]\\label{earlier:record:016}\\label{earlier:thm:exact_coarse_filter}\nAssume\n\\[\nCK=\\overline KC\n\\]\nand every likelihood \\(L_y\\) is constant on each coarse block with block values \\(\\overline L_y\\). Then\n\\[\nC\\Phi_y(p)=\\overline\\Phi_y(Cp)\n\\]\nfor every prior. For any observation sequence \\(y_1,\\ldots,y_r\\), the complete filtering recursion commutes with aggregation:\n\\[\nC\\Phi_{y_r}\\circ\\cdots\\circ\\Phi_{y_1}\n=\n\\overline\\Phi_{y_r}\\circ\\cdots\\circ\\overline\\Phi_{y_1}C.\n\\]\n\\end{theorem}","}[Exact coarse hidden Markov filter]\\label{earlier:record:016}\\label{earlier:thm:exact_coarse_filter}\nAssume\n\\[\nCK=\\overline KC\n\\]\nand every likelihood \\(L_y\\) is constant on each coarse block with block values \\(\\overline L_y\\). Then\n\\[\nC\\Phi_y(p)=\\overline\\Phi_y(Cp)\n\\]\nfor every prior. For any observation sequence \\(y_1,\\ldots,y_r\\), the complete filtering recursion commutes with aggregation:\n\\[\nC\\Phi_{y_r}\\circ\\cdots\\circ\\Phi_{y_1}\n=\n\\overline\\Phi_{y_r}\\circ\\cdots\\circ\\overline\\Phi_{y_1}C.\n\\]\n\\end{theorem}","\\begin{proof}\nExact Markov lumping gives \\(CKp=\\overline KCp\\). The Bayesian sufficiency criterion gives exact commutation of the update. Their composition proves the one step identity. Iterate it over the observation sequence.\n\\end{proof}","\\begin{theorem}[One step filter defect]\\label{earlier:record:017}\\label{earlier:thm:filter_defect}\nLet \\(r=Kp\\), \\(q=Cr\\), and let \\(\\ell_y(r)\\) be the exact coarse likelihood obtained by averaging \\(L_y\\) inside each block with respect to \\(r\\). Assume\n\\[\n0<\\overline m\\leq(\\overline L_y)_a\\leq\\overline M.\n\\]\nThen\n\\[\n\\begin{aligned}\n\\norm{C\\Phi_y(p)-\\overline\\Phi_y(Cp)}_1\n\\leq{}&\n\\frac{2d(r\\mid L_y,\\overline L_y)}\n{\\max\\{\\ip{L_y}{r},\\ip{\\overline L_y}{q}\\}}","\\begin{theorem}[One step filter defect]\\label{earlier:record:017}\\label{earlier:thm:filter_defect}\nLet \\(r=Kp\\), \\(q=Cr\\), and let \\(\\ell_y(r)\\) be the exact coarse likelihood obtained by averaging \\(L_y\\) inside each block with respect to \\(r\\). Assume\n\\[\n0<\\overline m\\leq(\\overline L_y)_a\\leq\\overline M.\n\\]\nThen\n\\[\n\\begin{aligned}\n\\norm{C\\Phi_y(p)-\\overline\\Phi_y(Cp)}_1\n\\leq{}&\n\\frac{2d(r\\mid L_y,\\overline L_y)}\n{\\max\\{\\ip{L_y}{r},\\ip{\\overline L_y}{q}\\}}\\\\\n&+\n\\frac{2\\overline M}{\\overline m}\n\\norm{CKp-\\overline KCp}_1.\n\\end{aligned}\n\\]\n\\end{theorem}","\\begin{proof}\nInsert the intermediate posterior \\(\\Bcal_{\\overline L_y}(CKp)\\). The first difference is bounded by \\cref{earlier:thm:information_loss}. The second is bounded by \\cref{extended:thm:prior_stability} applied to the coarse likelihood.\n\\end{proof}","\\FloatBarrier\n\\section{Fractional modes and killed transfer}\\label{earlier:section:04}\nFor the one-dimensional Dirichlet difference operator, let $h=1/M$ with an integer $M\\geq2$, let $1\\leq k\\leq M-1$, and set $\\lambda_{k,h}=4h^{-2}\\sin^2(k\\pi h/2)$. The retained explicit coefficient below is valid for every discrete mode.\n\\begin{corollary}[Fractional low mode dispersion]\\label{earlier:record:018}\\label{earlier:cor:fractional_low_mode_dispersion}\nLet \\(0<\\alpha\\leq1\\). Then\n\\[\n0\n\\leq\n(k\\pi)^{2\\alpha}-\\lambda_{k,h}^\\alpha\n\\leq\n\\frac{\\alpha(4k^2)^{\\alpha-1}(k\\pi)^4}{12}h^2.\n\\]\nFor every \\(t\\geq0\\),\n\\[\n\\abs{\n\\e^{-t\\lambda_{k,h}^\\alpha}\n-\n\\e^{-t(k\\pi)^{2\\alpha}}\n}\n\\leq\n\\frac{t\\alpha(4k^2)^{\\alpha-1}(k\\pi)^4}{12}h^2.\n\\]\n\\end{corollary}","r}\\label{earlier:section:04}\nFor the one-dimensional Dirichlet difference operator, let $h=1/M$ with an integer $M\\geq2$, let $1\\leq k\\leq M-1$, and set $\\lambda_{k,h}=4h^{-2}\\sin^2(k\\pi h/2)$. The retained explicit coefficient below is valid for every discrete mode.\n\\begin{corollary}[Fractional low mode dispersion]\\label{earlier:record:018}\\label{earlier:cor:fractional_low_mode_dispersion}\nLet \\(0<\\alpha\\leq1\\). Then\n\\[\n0\n\\leq\n(k\\pi)^{2\\alpha}-\\lambda_{k,h}^\\alpha\n\\leq\n\\frac{\\alpha(4k^2)^{\\alpha-1}(k\\pi)^4}{12}h^2.\n\\]\nFor every \\(t\\geq0\\),\n\\[\n\\abs{\n\\e^{-t\\lambda_{k,h}^\\alpha}\n-\n\\e^{-t(k\\pi)^{2\\alpha}}\n}\n\\leq\n\\frac{t\\alpha(4k^2)^{\\alpha-1}(k\\pi)^4}{12}h^2.\n\\]\n\\end{corollary}",":04}\nFor the one-dimensional Dirichlet difference operator, let $h=1/M$ with an integer $M\\geq2$, let $1\\leq k\\leq M-1$, and set $\\lambda_{k,h}=4h^{-2}\\sin^2(k\\pi h/2)$. The retained explicit coefficient below is valid for every discrete mode.\n\\begin{corollary}[Fractional low mode dispersion]\\label{earlier:record:018}\\label{earlier:cor:fractional_low_mode_dispersion}\nLet \\(0<\\alpha\\leq1\\). Then\n\\[\n0\n\\leq\n(k\\pi)^{2\\alpha}-\\lambda_{k,h}^\\alpha\n\\leq\n\\frac{\\alpha(4k^2)^{\\alpha-1}(k\\pi)^4}{12}h^2.\n\\]\nFor every \\(t\\geq0\\),\n\\[\n\\abs{\n\\e^{-t\\lambda_{k,h}^\\alpha}\n-\n\\e^{-t(k\\pi)^{2\\alpha}}\n}\n\\leq\n\\frac{t\\alpha(4k^2)^{\\alpha-1}(k\\pi)^4}{12}h^2.\n\\]\n\\end{corollary}","q M-1$, and set $\\lambda_{k,h}=4h^{-2}\\sin^2(k\\pi h/2)$. The retained explicit coefficient below is valid for every discrete mode.\n\\begin{corollary}[Fractional low mode dispersion]\\label{earlier:record:018}\\label{earlier:cor:fractional_low_mode_dispersion}\nLet \\(0<\\alpha\\leq1\\). Then\n\\[\n0\n\\leq\n(k\\pi)^{2\\alpha}-\\lambda_{k,h}^\\alpha\n\\leq\n\\frac{\\alpha(4k^2)^{\\alpha-1}(k\\pi)^4}{12}h^2.\n\\]\nFor every \\(t\\geq0\\),\n\\[\n\\abs{\n\\e^{-t\\lambda_{k,h}^\\alpha}\n-\n\\e^{-t(k\\pi)^{2\\alpha}}\n}\n\\leq\n\\frac{t\\alpha(4k^2)^{\\alpha-1}(k\\pi)^4}{12}h^2.\n\\]\n\\end{corollary}","set $\\lambda_{k,h}=4h^{-2}\\sin^2(k\\pi h/2)$. The retained explicit coefficient below is valid for every discrete mode.\n\\begin{corollary}[Fractional low mode dispersion]\\label{earlier:record:018}\\label{earlier:cor:fractional_low_mode_dispersion}\nLet \\(0<\\alpha\\leq1\\). Then\n\\[\n0\n\\leq\n(k\\pi)^{2\\alpha}-\\lambda_{k,h}^\\alpha\n\\leq\n\\frac{\\alpha(4k^2)^{\\alpha-1}(k\\pi)^4}{12}h^2.\n\\]\nFor every \\(t\\geq0\\),\n\\[\n\\abs{\n\\e^{-t\\lambda_{k,h}^\\alpha}\n-\n\\e^{-t(k\\pi)^{2\\alpha}}\n}\n\\leq\n\\frac{t\\alpha(4k^2)^{\\alpha-1}(k\\pi)^4}{12}h^2.\n\\]\n\\end{corollary}","\\begin{proof}\nThe chord bound \\(\\sin x\\geq2x/\\pi\\) on \\([0,\\pi/2]\\) gives \\(\\lambda_{k,h}\\geq4k^2\\). Since \\(x\\mapsto x^\\alpha\\) is increasing and has decreasing derivative,\n\\[\n(k\\pi)^{2\\alpha}-\\lambda_{k,h}^\\alpha\n\\leq\n\\alpha\\lambda_{k,h}^{\\alpha-1}\n\\bigl((k\\pi)^2-\\lambda_{k,h}\\bigr).\n\\]\nUse \\cref{extended:thm:low_mode} and the lower bound for \\(\\lambda_{k,h}\\). The exponential estimate follows from\n\\[\n\\abs{\\e^{-ta}-\\e^{-tb}}\n\\leq t\\abs{a-b}\n\\]\nfor nonnegative \\(a\\) and \\(b\\).\n\\end{proof}","{proof}\nThe chord bound \\(\\sin x\\geq2x/\\pi\\) on \\([0,\\pi/2]\\) gives \\(\\lambda_{k,h}\\geq4k^2\\). Since \\(x\\mapsto x^\\alpha\\) is increasing and has decreasing derivative,\n\\[\n(k\\pi)^{2\\alpha}-\\lambda_{k,h}^\\alpha\n\\leq\n\\alpha\\lambda_{k,h}^{\\alpha-1}\n\\bigl((k\\pi)^2-\\lambda_{k,h}\\bigr).\n\\]\nUse \\cref{extended:thm:low_mode} and the lower bound for \\(\\lambda_{k,h}\\). The exponential estimate follows from\n\\[\n\\abs{\\e^{-ta}-\\e^{-tb}}\n\\leq t\\abs{a-b}\n\\]\nfor nonnegative \\(a\\) and \\(b\\).\n\\end{proof}","gives \\(\\lambda_{k,h}\\geq4k^2\\). Since \\(x\\mapsto x^\\alpha\\) is increasing and has decreasing derivative,\n\\[\n(k\\pi)^{2\\alpha}-\\lambda_{k,h}^\\alpha\n\\leq\n\\alpha\\lambda_{k,h}^{\\alpha-1}\n\\bigl((k\\pi)^2-\\lambda_{k,h}\\bigr).\n\\]\nUse \\cref{extended:thm:low_mode} and the lower bound for \\(\\lambda_{k,h}\\). The exponential estimate follows from\n\\[\n\\abs{\\e^{-ta}-\\e^{-tb}}\n\\leq t\\abs{a-b}\n\\]\nfor nonnegative \\(a\\) and \\(b\\).\n\\end{proof}","ambda_{k,h}\\geq4k^2\\). Since \\(x\\mapsto x^\\alpha\\) is increasing and has decreasing derivative,\n\\[\n(k\\pi)^{2\\alpha}-\\lambda_{k,h}^\\alpha\n\\leq\n\\alpha\\lambda_{k,h}^{\\alpha-1}\n\\bigl((k\\pi)^2-\\lambda_{k,h}\\bigr).\n\\]\nUse \\cref{extended:thm:low_mode} and the lower bound for \\(\\lambda_{k,h}\\). The exponential estimate follows from\n\\[\n\\abs{\\e^{-ta}-\\e^{-tb}}\n\\leq t\\abs{a-b}\n\\]\nfor nonnegative \\(a\\) and \\(b\\).\n\\end{proof}","Let $Q$ and $\\overline Q$ be finite Markov generators in the column convention. All times in this section satisfy $t\\geq0$.\n\\begin{theorem}[Potential perturbation bound]\\label{earlier:record:019}\\label{earlier:thm:feynman_kac_potential_bound}\nLet \\(V\\) and \\(W\\) be nonnegative diagonal matrices. In the supremum norm,\n\\[\n\\boxed{\n\\norm{\n\\e^{t(Q^T-V)}\n-\n\\e^{t(Q^T-W)}\n}_{\\infty\\to\\infty}\n\\leq\n t\\norm{V-W}_{\\infty\\to\\infty}.\n}\n\\]\n\\end{theorem}","\\begin{proof}\nBoth killed Markov semigroups are contractions in the supremum norm. Apply the Duhamel identity with generator defect \\(W-V\\).\n\\end{proof}","\\begin{theorem}[Coarse Feynman Kac transfer]\\label{earlier:record:020}\\label{earlier:thm:coarse_feynman_kac}\nAssume \\(CQ=\\overline QC\\), and let \\(J=C^T\\) lift a coarse function to the corresponding block constant fine function. Suppose the potential is block constant, so\n\\[\nVJ=J\\overline V.\n\\]\nThen\n\\[\n\\boxed{\n\\e^{t(Q^T-V)}J\n=\nJ\\e^{t(\\overline Q^T-\\overline V)}\n}\n\\]\nfor every \\(t\\geq0\\).\n\\end{theorem}","\\begin{proof}\nTranspose the Markov intertwining relation to obtain\n\\[\nQ^TJ=J\\overline Q^T.\n\\]\nTogether with \\(VJ=J\\overline V\\), this gives\n\\[\n(Q^T-V)J=J(\\overline Q^T-\\overline V).\n\\]\nThe exponential intertwining follows from the power series.\n\\end{proof}","\\FloatBarrier\n\\section{Memory compression, rational realization, and phase-type consequences}\\label{earlier:section:05}\nFix $T>0$. For the next estimate let $A$ be a bounded operator on a Banach space $X$, let $\\Kcal,\\widetilde\\Kcal\\in L^1((0,T),\\mathcal B(X))$, and let $f$ be continuous. Take continuously differentiable solutions of the displayed equations. This direct bounded-generator estimate retains a sharper exponential constant than the general mild-semigroup estimate in Part II.\n\\begin{theorem}[Stability of memory compression]\\label{earlier:record:021}\\label{earlier:thm:memory_compression_stability}\nLet \\(x\\) and \\(\\widetilde x\\) solve\n\\[\n\\dot x(t)\n=Ax(t)+\\int_0^t\\Kcal(t-s)x(s)\\dd s+f(t)\n\\]\nand\n\\[\n\\dot{\\widetilde x}(t)\n=A\\widetilde x(t)\n+\\int_0^t\\widetilde\\Kcal(t-s)\\widetilde x(s)\\dd s\n+f(t)\n\\]\nwith the same","\\FloatBarrier\n\\section{Memory compression, rational realization, and phase-type consequences}\\label{earlier:section:05}\nFix $T>0$. For the next estimate let $A$ be a bounded operator on a Banach space $X$, let $\\Kcal,\\widetilde\\Kcal\\in L^1((0,T),\\mathcal B(X))$, and let $f$ be continuous. Take continuously differentiable solutions of the displayed equations. This direct bounded-generator estimate retains a sharper exponential constant than the general mild-semigroup estimate in Part II.\n\\begin{theorem}[Stability of memory compression]\\label{earlier:record:021}\\label{earlier:thm:memory_compression_stability}\nLet \\(x\\) and \\(\\widetilde x\\) solve\n\\[\n\\dot x(t)\n=Ax(t)+\\int_0^t\\Kcal(t-s)x(s)\\dd s+f(t)\n\\]\nand\n\\[\n\\dot{\\widetilde x}(t)\n=A\\widetilde x(t)\n+\\int_0^t\\widetilde\\Kcal(t-s)\\widetilde x(s)\\dd s\n+f(t)\n\\]\nwith the same initial value. Fix \\(T>0\\), and","\\FloatBarrier\n\\section{Memory compression, rational realization, and phase-type consequences}\\label{earlier:section:05}\nFix $T>0$. For the next estimate let $A$ be a bounded operator on a Banach space $X$, let $\\Kcal,\\widetilde\\Kcal\\in L^1((0,T),\\mathcal B(X))$, and let $f$ be continuous. Take continuously differentiable solutions of the displayed equations. This direct bounded-generator estimate retains a sharper exponential constant than the general mild-semigroup estimate in Part II.\n\\begin{theorem}[Stability of memory compression]\\label{earlier:record:021}\\label{earlier:thm:memory_compression_stability}\nLet \\(x\\) and \\(\\widetilde x\\) solve\n\\[\n\\dot x(t)\n=Ax(t)+\\int_0^t\\Kcal(t-s)x(s)\\dd s+f(t)\n\\]\nand\n\\[\n\\dot{\\widetilde x}(t)\n=A\\widetilde x(t)\n+\\int_0^t\\widetilde\\Kcal(t-s)\\widetilde x(s)\\dd s\n+f(t)\n\\]\nwith the same initial value. Fix \\(T>0\\), and assume\n\\[\n\\sup_{0\\leq t\\leq T}\\norm{\\wideti","\\FloatBarrier\n\\section{Memory compression, rational realization, and phase-type consequences}\\label{earlier:section:05}\nFix $T>0$. For the next estimate let $A$ be a bounded operator on a Banach space $X$, let $\\Kcal,\\widetilde\\Kcal\\in L^1((0,T),\\mathcal B(X))$, and let $f$ be continuous. Take continuously differentiable solutions of the displayed equations. This direct bounded-generator estimate retains a sharper exponential constant than the general mild-semigroup estimate in Part II.\n\\begin{theorem}[Stability of memory compression]\\label{earlier:record:021}\\label{earlier:thm:memory_compression_stability}\nLet \\(x\\) and \\(\\widetilde x\\) solve\n\\[\n\\dot x(t)\n=Ax(t)+\\int_0^t\\Kcal(t-s)x(s)\\dd s+f(t)\n\\]\nand\n\\[\n\\dot{\\widetilde x}(t)\n=A\\widetilde x(t)\n+\\int_0^t\\widetilde\\Kcal(t-s)\\widetilde x(s)\\dd s\n+f(t)\n\\]\nwith the same initial value. Fix \\(T>0\\), and assume\n\\[\n\\sup_{0\\leq t\\leq T}\\norm{\\widetilde x(t)}\\leq M.\n\\]\nSet\n\\[\na=\\norm A,\n\\qquad\nk=\\norm{\\Kca","\\FloatBarrier\n\\section{Memory compression, rational realization, and phase-type consequences}\\label{earlier:section:05}\nFix $T>0$. For the next estimate let $A$ be a bounded operator on a Banach space $X$, let $\\Kcal,\\widetilde\\Kcal\\in L^1((0,T),\\mathcal B(X))$, and let $f$ be continuous. Take continuously differentiable solutions of the displayed equations. This direct bounded-generator estimate retains a sharper exponential constant than the general mild-semigroup estimate in Part II.\n\\begin{theorem}[Stability of memory compression]\\label{earlier:record:021}\\label{earlier:thm:memory_compression_stability}\nLet \\(x\\) and \\(\\widetilde x\\) solve\n\\[\n\\dot x(t)\n=Ax(t)+\\int_0^t\\Kcal(t-s)x(s)\\dd s+f(t)\n\\]\nand\n\\[\n\\dot{\\widetilde x}(t)\n=A\\widetilde x(t)\n+\\int_0^t\\widetilde\\Kcal(t-s)\\widetilde x(s)\\dd s\n+f(t)\n\\]\nwith the same initial value. Fix \\(T>0\\), and assume\n\\[\n\\sup_{0\\leq t\\leq T}\\norm{\\widetilde x(t)}\\leq M.\n\\]\nSet\n\\[\na=\\norm A,\n\\qquad\nk=\\norm{\\Kcal}_{L^1(0,T)}","etilde\\Kcal\\in L^1((0,T),\\mathcal B(X))$, and let $f$ be continuous. Take continuously differentiable solutions of the displayed equations. This direct bounded-generator estimate retains a sharper exponential constant than the general mild-semigroup estimate in Part II.\n\\begin{theorem}[Stability of memory compression]\\label{earlier:record:021}\\label{earlier:thm:memory_compression_stability}\nLet \\(x\\) and \\(\\widetilde x\\) solve\n\\[\n\\dot x(t)\n=Ax(t)+\\int_0^t\\Kcal(t-s)x(s)\\dd s+f(t)\n\\]\nand\n\\[\n\\dot{\\widetilde x}(t)\n=A\\widetilde x(t)\n+\\int_0^t\\widetilde\\Kcal(t-s)\\widetilde x(s)\\dd s\n+f(t)\n\\]\nwith the same initial value. Fix \\(T>0\\), and assume\n\\[\n\\sup_{0\\leq t\\leq T}\\norm{\\widetilde x(t)}\\leq M.\n\\]\nSet\n\\[\na=\\norm A,\n\\qquad\nk=\\norm{\\Kcal}_{L^1(0,T)},\n\\qquad\n\\delta=\\norm{\\Kcal-\\widetilde\\Kcal}_{L^1(0,T)}.\n\\]\nThen\n\\[\n\\boxed{\n\\sup_{0\\leq t\\leq T}\n\\norm{x(t)-\\widetilde x(t)}\n\\leq\nMT\\delta\\e^{(a+k)T}.\n}\n\\]\n\\end{theorem}","l\\in L^1((0,T),\\mathcal B(X))$, and let $f$ be continuous. Take continuously differentiable solutions of the displayed equations. This direct bounded-generator estimate retains a sharper exponential constant than the general mild-semigroup estimate in Part II.\n\\begin{theorem}[Stability of memory compression]\\label{earlier:record:021}\\label{earlier:thm:memory_compression_stability}\nLet \\(x\\) and \\(\\widetilde x\\) solve\n\\[\n\\dot x(t)\n=Ax(t)+\\int_0^t\\Kcal(t-s)x(s)\\dd s+f(t)\n\\]\nand\n\\[\n\\dot{\\widetilde x}(t)\n=A\\widetilde x(t)\n+\\int_0^t\\widetilde\\Kcal(t-s)\\widetilde x(s)\\dd s\n+f(t)\n\\]\nwith the same initial value. Fix \\(T>0\\), and assume\n\\[\n\\sup_{0\\leq t\\leq T}\\norm{\\widetilde x(t)}\\leq M.\n\\]\nSet\n\\[\na=\\norm A,\n\\qquad\nk=\\norm{\\Kcal}_{L^1(0,T)},\n\\qquad\n\\delta=\\norm{\\Kcal-\\widetilde\\Kcal}_{L^1(0,T)}.\n\\]\nThen\n\\[\n\\boxed{\n\\sup_{0\\leq t\\leq T}\n\\norm{x(t)-\\widetilde x(t)}\n\\leq\nMT\\delta\\e^{(a+k)T}.\n}\n\\]\n\\end{theorem}","B(X))$, and let $f$ be continuous. Take continuously differentiable solutions of the displayed equations. This direct bounded-generator estimate retains a sharper exponential constant than the general mild-semigroup estimate in Part II.\n\\begin{theorem}[Stability of memory compression]\\label{earlier:record:021}\\label{earlier:thm:memory_compression_stability}\nLet \\(x\\) and \\(\\widetilde x\\) solve\n\\[\n\\dot x(t)\n=Ax(t)+\\int_0^t\\Kcal(t-s)x(s)\\dd s+f(t)\n\\]\nand\n\\[\n\\dot{\\widetilde x}(t)\n=A\\widetilde x(t)\n+\\int_0^t\\widetilde\\Kcal(t-s)\\widetilde x(s)\\dd s\n+f(t)\n\\]\nwith the same initial value. Fix \\(T>0\\), and assume\n\\[\n\\sup_{0\\leq t\\leq T}\\norm{\\widetilde x(t)}\\leq M.\n\\]\nSet\n\\[\na=\\norm A,\n\\qquad\nk=\\norm{\\Kcal}_{L^1(0,T)},\n\\qquad\n\\delta=\\norm{\\Kcal-\\widetilde\\Kcal}_{L^1(0,T)}.\n\\]\nThen\n\\[\n\\boxed{\n\\sup_{0\\leq t\\leq T}\n\\norm{x(t)-\\widetilde x(t)}\n\\leq\nMT\\delta\\e^{(a+k)T}.\n}\n\\]\n\\end{theorem}","entiable solutions of the displayed equations. This direct bounded-generator estimate retains a sharper exponential constant than the general mild-semigroup estimate in Part II.\n\\begin{theorem}[Stability of memory compression]\\label{earlier:record:021}\\label{earlier:thm:memory_compression_stability}\nLet \\(x\\) and \\(\\widetilde x\\) solve\n\\[\n\\dot x(t)\n=Ax(t)+\\int_0^t\\Kcal(t-s)x(s)\\dd s+f(t)\n\\]\nand\n\\[\n\\dot{\\widetilde x}(t)\n=A\\widetilde x(t)\n+\\int_0^t\\widetilde\\Kcal(t-s)\\widetilde x(s)\\dd s\n+f(t)\n\\]\nwith the same initial value. Fix \\(T>0\\), and assume\n\\[\n\\sup_{0\\leq t\\leq T}\\norm{\\widetilde x(t)}\\leq M.\n\\]\nSet\n\\[\na=\\norm A,\n\\qquad\nk=\\norm{\\Kcal}_{L^1(0,T)},\n\\qquad\n\\delta=\\norm{\\Kcal-\\widetilde\\Kcal}_{L^1(0,T)}.\n\\]\nThen\n\\[\n\\boxed{\n\\sup_{0\\leq t\\leq T}\n\\norm{x(t)-\\widetilde x(t)}\n\\leq\nMT\\delta\\e^{(a+k)T}.\n}\n\\]\n\\end{theorem}","eneral mild-semigroup estimate in Part II.\n\\begin{theorem}[Stability of memory compression]\\label{earlier:record:021}\\label{earlier:thm:memory_compression_stability}\nLet \\(x\\) and \\(\\widetilde x\\) solve\n\\[\n\\dot x(t)\n=Ax(t)+\\int_0^t\\Kcal(t-s)x(s)\\dd s+f(t)\n\\]\nand\n\\[\n\\dot{\\widetilde x}(t)\n=A\\widetilde x(t)\n+\\int_0^t\\widetilde\\Kcal(t-s)\\widetilde x(s)\\dd s\n+f(t)\n\\]\nwith the same initial value. Fix \\(T>0\\), and assume\n\\[\n\\sup_{0\\leq t\\leq T}\\norm{\\widetilde x(t)}\\leq M.\n\\]\nSet\n\\[\na=\\norm A,\n\\qquad\nk=\\norm{\\Kcal}_{L^1(0,T)},\n\\qquad\n\\delta=\\norm{\\Kcal-\\widetilde\\Kcal}_{L^1(0,T)}.\n\\]\nThen\n\\[\n\\boxed{\n\\sup_{0\\leq t\\leq T}\n\\norm{x(t)-\\widetilde x(t)}\n\\leq\nMT\\delta\\e^{(a+k)T}.\n}\n\\]\n\\end{theorem}","p estimate in Part II.\n\\begin{theorem}[Stability of memory compression]\\label{earlier:record:021}\\label{earlier:thm:memory_compression_stability}\nLet \\(x\\) and \\(\\widetilde x\\) solve\n\\[\n\\dot x(t)\n=Ax(t)+\\int_0^t\\Kcal(t-s)x(s)\\dd s+f(t)\n\\]\nand\n\\[\n\\dot{\\widetilde x}(t)\n=A\\widetilde x(t)\n+\\int_0^t\\widetilde\\Kcal(t-s)\\widetilde x(s)\\dd s\n+f(t)\n\\]\nwith the same initial value. Fix \\(T>0\\), and assume\n\\[\n\\sup_{0\\leq t\\leq T}\\norm{\\widetilde x(t)}\\leq M.\n\\]\nSet\n\\[\na=\\norm A,\n\\qquad\nk=\\norm{\\Kcal}_{L^1(0,T)},\n\\qquad\n\\delta=\\norm{\\Kcal-\\widetilde\\Kcal}_{L^1(0,T)}.\n\\]\nThen\n\\[\n\\boxed{\n\\sup_{0\\leq t\\leq T}\n\\norm{x(t)-\\widetilde x(t)}\n\\leq\nMT\\delta\\e^{(a+k)T}.\n}\n\\]\n\\end{theorem}","ompression]\\label{earlier:record:021}\\label{earlier:thm:memory_compression_stability}\nLet \\(x\\) and \\(\\widetilde x\\) solve\n\\[\n\\dot x(t)\n=Ax(t)+\\int_0^t\\Kcal(t-s)x(s)\\dd s+f(t)\n\\]\nand\n\\[\n\\dot{\\widetilde x}(t)\n=A\\widetilde x(t)\n+\\int_0^t\\widetilde\\Kcal(t-s)\\widetilde x(s)\\dd s\n+f(t)\n\\]\nwith the same initial value. Fix \\(T>0\\), and assume\n\\[\n\\sup_{0\\leq t\\leq T}\\norm{\\widetilde x(t)}\\leq M.\n\\]\nSet\n\\[\na=\\norm A,\n\\qquad\nk=\\norm{\\Kcal}_{L^1(0,T)},\n\\qquad\n\\delta=\\norm{\\Kcal-\\widetilde\\Kcal}_{L^1(0,T)}.\n\\]\nThen\n\\[\n\\boxed{\n\\sup_{0\\leq t\\leq T}\n\\norm{x(t)-\\widetilde x(t)}\n\\leq\nMT\\delta\\e^{(a+k)T}.\n}\n\\]\n\\end{theorem}","e x\\) solve\n\\[\n\\dot x(t)\n=Ax(t)+\\int_0^t\\Kcal(t-s)x(s)\\dd s+f(t)\n\\]\nand\n\\[\n\\dot{\\widetilde x}(t)\n=A\\widetilde x(t)\n+\\int_0^t\\widetilde\\Kcal(t-s)\\widetilde x(s)\\dd s\n+f(t)\n\\]\nwith the same initial value. Fix \\(T>0\\), and assume\n\\[\n\\sup_{0\\leq t\\leq T}\\norm{\\widetilde x(t)}\\leq M.\n\\]\nSet\n\\[\na=\\norm A,\n\\qquad\nk=\\norm{\\Kcal}_{L^1(0,T)},\n\\qquad\n\\delta=\\norm{\\Kcal-\\widetilde\\Kcal}_{L^1(0,T)}.\n\\]\nThen\n\\[\n\\boxed{\n\\sup_{0\\leq t\\leq T}\n\\norm{x(t)-\\widetilde x(t)}\n\\leq\nMT\\delta\\e^{(a+k)T}.\n}\n\\]\n\\end{theorem}","\\begin{proof}\nLet \\(e=x-\\widetilde x\\). Integrating the difference equation and changing the order of integration gives\n\\[\n\\norm{e(t)}\n\\leq\n(a+k)\\int_0^t\\norm{e(r)}\\dd r\n+Mt\\delta.\n\\]\nSince \\(t\\leq T\\), Gronwall's inequality gives\n\\[\n\\norm{e(t)}\n\\leq\nMT\\delta\\e^{(a+k)t}.\n\\]\nTake the supremum over \\([0,T]\\).\n\\end{proof}","For the exact closure criterion of Part II, allowing an arbitrary bounded local generator $B$ does not enlarge the class: differentiating $P e^{tA}P=e^{tB}P$ at $t=0$ forces $B=A_{PP}$ on the retained space. Thus its zero-memory criterion excludes every alternative bounded local generator.","A finite auxiliary realization of a matrix kernel means $\\Kcal(t)=B e^{tD}C$, with finite matrices of compatible sizes. Laplace transforms are taken in a right half-plane of absolute convergence. The rational characterization concerns continuous kernels of exponential order, so uniqueness of the ordinary Laplace transform applies.\n\\begin{theorem}[Rational memory realization]\\label{earlier:record:022}\\label{earlier:thm:rational_memory_realization}\nA matrix valued memory transform has a finite auxiliary realization if and only if it is a strictly proper rational matrix function. More precisely,\n\\[\n\\boxed{\n\\widehat\\Kcal(z)=B(zI-D)^{-1}C\n}\n\\]\nif and only if\n\\[\n\\boxed{\n\\Kcal(t)=B\\e^{tD}C.\n}\n\\]\n\\end{theorem}","tional characterization concerns continuous kernels of exponential order, so uniqueness of the ordinary Laplace transform applies.\n\\begin{theorem}[Rational memory realization]\\label{earlier:record:022}\\label{earlier:thm:rational_memory_realization}\nA matrix valued memory transform has a finite auxiliary realization if and only if it is a strictly proper rational matrix function. More precisely,\n\\[\n\\boxed{\n\\widehat\\Kcal(z)=B(zI-D)^{-1}C\n}\n\\]\nif and only if\n\\[\n\\boxed{\n\\Kcal(t)=B\\e^{tD}C.\n}\n\\]\n\\end{theorem}","tial order, so uniqueness of the ordinary Laplace transform applies.\n\\begin{theorem}[Rational memory realization]\\label{earlier:record:022}\\label{earlier:thm:rational_memory_realization}\nA matrix valued memory transform has a finite auxiliary realization if and only if it is a strictly proper rational matrix function. More precisely,\n\\[\n\\boxed{\n\\widehat\\Kcal(z)=B(zI-D)^{-1}C\n}\n\\]\nif and only if\n\\[\n\\boxed{\n\\Kcal(t)=B\\e^{tD}C.\n}\n\\]\n\\end{theorem}","\\begin{proof}\nThe forward implication follows by taking the Laplace transform. The resulting matrix is rational and tends to zero as \\(z\\) tends to infinity.","Conversely, decompose every entry of a strictly proper rational matrix into partial fractions. A term\n\\[\n\\frac{c}{(z-\\lambda)^r}\n\\]\nis an entry of the resolvent of an \\(r\\) by \\(r\\) Jordan block with eigenvalue \\(\\lambda\\). Taking a direct sum of these blocks and choosing the input and output maps entry by entry gives matrices \\(B,D,C\\) with the required resolvent. Inverse Laplace transformation gives the time domain formula.\n\\end{proof}","\\begin{corollary}[Exponential sum memory]\\label{earlier:record:023}\\label{earlier:cor:exponential_sum_memory}\nIf\n\\[\n\\Kcal(t)=\\sum_{j=1}^{m}B_j\\e^{\\lambda_jt}C_j,\n\\]\nthen \\(\\Kcal\\) has a finite auxiliary realization with one auxiliary block for each exponential. Repeated poles produce polynomial factors in \\(t\\) and are realized by Jordan blocks.\n\\end{corollary}","\\begin{proof}\nIf \\(B_j\\) and \\(C_j\\) have an intermediate space of dimension \\(d_j\\), use a block diagonal matrix \\(D\\) with blocks \\(\\lambda_j I_{d_j}\\), or the required Jordan blocks when poles repeat. Assemble the matrices \\(B_j\\) and \\(C_j\\) into the output and input maps.\n\\end{proof}","For the survival consequences, $A$ is a finite Metzler matrix with nonpositive column sums and spectrum in the open left half-plane, $p_0$ is a probability vector, $S(t)=\\one^T e^{tA}p_0$, and $f(t)=-S'(t)$. The one-phase model has $\\lambda>0$. The survival Laplace identity of Part II also holds for every complex $s$ with $\\operatorname{Re}s\\geq0$: stability gives absolute convergence of $\\int_0^\\infty e^{-st}e^{tA}\\,dt=(sI-A)^{-1}$, and multiplying by the exit row and $p_0$ gives the same transform.\n\\begin{corollary}[Phase type rationality]\\label{earlier:record:024}\\label{earlier:cor:phase_type_rationality}\nThe Laplace transform of every finite phase type distribution is rational. If \\(A\\) is diagonalizable, then \\(S(t)\\)","For the survival consequences, $A$ is a finite Metzler matrix with nonpositive column sums and spectrum in the open left half-plane, $p_0$ is a probability vector, $S(t)=\\one^T e^{tA}p_0$, and $f(t)=-S'(t)$. The one-phase model has $\\lambda>0$. The survival Laplace identity of Part II also holds for every complex $s$ with $\\operatorname{Re}s\\geq0$: stability gives absolute convergence of $\\int_0^\\infty e^{-st}e^{tA}\\,dt=(sI-A)^{-1}$, and multiplying by the exit row and $p_0$ gives the same transform.\n\\begin{corollary}[Phase type rationality]\\label{earlier:record:024}\\label{earlier:cor:phase_type_rationality}\nThe Laplace transform of every finite phase type distribution is rational. If \\(A\\) is diagonalizable, then \\(S(t)\\) and \\(f(t)\\) are finite linear combinations of exponential functions. Jordan blocks add polynomial facto","For the survival consequences, $A$ is a finite Metzler matrix with nonpositive column sums and spectrum in the open left half-plane, $p_0$ is a probability vector, $S(t)=\\one^T e^{tA}p_0$, and $f(t)=-S'(t)$. The one-phase model has $\\lambda>0$. The survival Laplace identity of Part II also holds for every complex $s$ with $\\operatorname{Re}s\\geq0$: stability gives absolute convergence of $\\int_0^\\infty e^{-st}e^{tA}\\,dt=(sI-A)^{-1}$, and multiplying by the exit row and $p_0$ gives the same transform.\n\\begin{corollary}[Phase type rationality]\\label{earlier:record:024}\\label{earlier:cor:phase_type_rationality}\nThe Laplace transform of every finite phase type distribution is rational. If \\(A\\) is diagonalizable, then \\(S(t)\\) and \\(f(t)\\) are finite linear combinations of exponential functions. Jordan blocks add polynomial factors in \\(t\\).\n\\end{corollary}","ive column sums and spectrum in the open left half-plane, $p_0$ is a probability vector, $S(t)=\\one^T e^{tA}p_0$, and $f(t)=-S'(t)$. The one-phase model has $\\lambda>0$. The survival Laplace identity of Part II also holds for every complex $s$ with $\\operatorname{Re}s\\geq0$: stability gives absolute convergence of $\\int_0^\\infty e^{-st}e^{tA}\\,dt=(sI-A)^{-1}$, and multiplying by the exit row and $p_0$ gives the same transform.\n\\begin{corollary}[Phase type rationality]\\label{earlier:record:024}\\label{earlier:cor:phase_type_rationality}\nThe Laplace transform of every finite phase type distribution is rational. If \\(A\\) is diagonalizable, then \\(S(t)\\) and \\(f(t)\\) are finite linear combinations of exponential functions. Jordan blocks add polynomial factors in \\(t\\).\n\\end{corollary}","for every complex $s$ with $\\operatorname{Re}s\\geq0$: stability gives absolute convergence of $\\int_0^\\infty e^{-st}e^{tA}\\,dt=(sI-A)^{-1}$, and multiplying by the exit row and $p_0$ gives the same transform.\n\\begin{corollary}[Phase type rationality]\\label{earlier:record:024}\\label{earlier:cor:phase_type_rationality}\nThe Laplace transform of every finite phase type distribution is rational. If \\(A\\) is diagonalizable, then \\(S(t)\\) and \\(f(t)\\) are finite linear combinations of exponential functions. Jordan blocks add polynomial factors in \\(t\\).\n\\end{corollary}","operatorname{Re}s\\geq0$: stability gives absolute convergence of $\\int_0^\\infty e^{-st}e^{tA}\\,dt=(sI-A)^{-1}$, and multiplying by the exit row and $p_0$ gives the same transform.\n\\begin{corollary}[Phase type rationality]\\label{earlier:record:024}\\label{earlier:cor:phase_type_rationality}\nThe Laplace transform of every finite phase type distribution is rational. If \\(A\\) is diagonalizable, then \\(S(t)\\) and \\(f(t)\\) are finite linear combinations of exponential functions. Jordan blocks add polynomial factors in \\(t\\).\n\\end{corollary}","Re}s\\geq0$: stability gives absolute convergence of $\\int_0^\\infty e^{-st}e^{tA}\\,dt=(sI-A)^{-1}$, and multiplying by the exit row and $p_0$ gives the same transform.\n\\begin{corollary}[Phase type rationality]\\label{earlier:record:024}\\label{earlier:cor:phase_type_rationality}\nThe Laplace transform of every finite phase type distribution is rational. If \\(A\\) is diagonalizable, then \\(S(t)\\) and \\(f(t)\\) are finite linear combinations of exponential functions. Jordan blocks add polynomial factors in \\(t\\).\n\\end{corollary}","tiplying by the exit row and $p_0$ gives the same transform.\n\\begin{corollary}[Phase type rationality]\\label{earlier:record:024}\\label{earlier:cor:phase_type_rationality}\nThe Laplace transform of every finite phase type distribution is rational. If \\(A\\) is diagonalizable, then \\(S(t)\\) and \\(f(t)\\) are finite linear combinations of exponential functions. Jordan blocks add polynomial factors in \\(t\\).\n\\end{corollary}","\\begin{proof}\nThe resolvent \\((sI-A)^{-1}\\) is rational. The time domain statement follows from the Jordan form of \\(A\\).\n\\end{proof}","\\begin{corollary}[One phase model]\\label{earlier:record:025}\\label{earlier:cor:one_phase_model}\nIf \\(A=(-\\lambda)\\) and \\(p_0=1\\), then\n\\[\nS(t)=\\e^{-\\lambda t},\n\\qquad\nf(t)=\\lambda\\e^{-\\lambda t},\n\\qquad\nh(t)=\\lambda.\n\\]\n\\end{corollary}","\\FloatBarrier\n\\section{Earlier exact examples}\\label{earlier:section:06}\nThese five short examples are retained from the earlier human edition. Their numerical parameters are independent of the later visual guide's worked models.\n\\begin{example}[One transfer between two packets]\\label{earlier:record:026}\\label{earlier:ex:two_packet_transfer}\nLet\n\\[\nA=\n\\begin{pmatrix}\na&0\\\\\nc&b\n\\end{pmatrix}\n\\]\nand let the two coordinate axes be the packet spaces. If \\(a\\neq b\\), then\n\\[\n\\e^{tA}\n=\n\\begin{pmatrix}\n\\e^{at}&0","c&b\n\\end{pmatrix}\n\\]\nand let the two coordinate axes be the packet spaces. If \\(a\\neq b\\), then\n\\[\n\\e^{tA}\n=\n\\begin{pmatrix}\n\\e^{at}&0","c&b\n\\end{pmatrix}\n\\]\nand let the two coordinate axes be the packet spaces. If \\(a\\neq b\\), then\n\\[\n\\e^{tA}\n=\n\\begin{pmatrix}\n\\e^{at}&0\\\\\n\\displaystyle c\\frac{\\e^{at}-\\e^{bt}}{a-b}&\\e^{bt}\n\\end{pmatrix}.\n\\]\nThe lower left block is produced by one transfer from the first packet to the second. Its expansion begins with \\(ct\\), in agreement with \\cref{earlier:cor:minimum_transfer_order}. When \\(a=b\\), the same block is \\(ct\\e^{at}\\).\n\\end{example}","\\displaystyle c\\frac{\\e^{at}-\\e^{bt}}{a-b}&\\e^{bt}\n\\end{pmatrix}.\n\\]\nThe lower left block is produced by one transfer from the first packet to the second. Its expansion begins with \\(ct\\), in agreement with \\cref{earlier:cor:minimum_transfer_order}. When \\(a=b\\), the same block is \\(ct\\e^{at}\\).\n\\end{example}","\\begin{example}[A four state lumpable chain]\\label{earlier:record:027}\\label{earlier:ex:four_state_lumping}\nPartition four states into \\(B_1=\\{1,2\\}\\) and \\(B_2=\\{3,4\\}\\). For positive rates \\(a,b,u,v\\), set\n\\[\nQ=\n\\begin{pmatrix}\n-u-a&u&b/2&b/2","\\begin{example}[A four state lumpable chain]\\label{earlier:record:027}\\label{earlier:ex:four_state_lumping}\nPartition four states into \\(B_1=\\{1,2\\}\\) and \\(B_2=\\{3,4\\}\\). For positive rates \\(a,b,u,v\\), set\n\\[\nQ=\n\\begin{pmatrix}\n-u-a&u&b/2&b/2\\\\\nu&-u-a&b/2&b/2\\\\\na/2&a/2&-v-b&v\\\\\na/2&a/2&v&-v-b\n\\end{pmatrix}.\n\\]\nThe rates \\(u\\) and \\(v\\) move inside the coarse blocks. Every state in \\(B_1\\) leaves for \\(B_2\\) at total rate \\(a\\), and every state in \\(B_2\\) leaves for \\(B_1\\) at total rate \\(b\\). Hence\n\\[\n\\overline Q=\n\\begin{pmatrix}\n-a&b","a/2&a/2&v&-v-b\n\\end{pmatrix}.\n\\]\nThe rates \\(u\\) and \\(v\\) move inside the coarse blocks. Every state in \\(B_1\\) leaves for \\(B_2\\) at total rate \\(a\\), and every state in \\(B_2\\) leaves for \\(B_1\\) at total rate \\(b\\). Hence\n\\[\n\\overline Q=\n\\begin{pmatrix}\n-a&b","a/2&a/2&v&-v-b\n\\end{pmatrix}.\n\\]\nThe rates \\(u\\) and \\(v\\) move inside the coarse blocks. Every state in \\(B_1\\) leaves for \\(B_2\\) at total rate \\(a\\), and every state in \\(B_2\\) leaves for \\(B_1\\) at total rate \\(b\\). Hence\n\\[\n\\overline Q=\n\\begin{pmatrix}\n-a&b\\\\\na&-b\n\\end{pmatrix}\n\\]\nand \\(CQ=\\overline QC\\). The internal rates disappear from the coarse generator even though they remain present in the fine evolution.\n\\end{example}","a&-b\n\\end{pmatrix}\n\\]\nand \\(CQ=\\overline QC\\). The internal rates disappear from the coarse generator even though they remain present in the fine evolution.\n\\end{example}","\\begin{example}[A visible likelihood loss]\\label{earlier:record:028}\\label{earlier:ex:bayes_information_loss}\nLet \\(B_1=\\{1,2\\}\\), let \\(B_2=\\{3\\}\\), and take\n\\[\np=(3/10,2/10,5/10),\n\\qquad\nL=(1,3,2).\n\\]\nThe coarse prior is \\((1/2,1/2)\\). The exact block likelihoods are\n\\[\n\\ell_1(p)=\\frac{(3/10)+3(2/10)}{1/2}=\\frac95,\n\\qquad\n\\ell_2(p)=2.\n\\]\nThus the exact coarse posterior is \\((9/19,10/19)\\). If the first block is replaced by the fixed representative value \\(2\\), the approximate coarse posterior is \\((1/2,1/2)\\). The \\(\\ell^1\\) defect is \\(1/19\\), while \\cref{earlier:thm:information_loss} gives the valid upper bound \\(1/10\\).\n\\end{example}","d:028}\\label{earlier:ex:bayes_information_loss}\nLet \\(B_1=\\{1,2\\}\\), let \\(B_2=\\{3\\}\\), and take\n\\[\np=(3/10,2/10,5/10),\n\\qquad\nL=(1,3,2).\n\\]\nThe coarse prior is \\((1/2,1/2)\\). The exact block likelihoods are\n\\[\n\\ell_1(p)=\\frac{(3/10)+3(2/10)}{1/2}=\\frac95,\n\\qquad\n\\ell_2(p)=2.\n\\]\nThus the exact coarse posterior is \\((9/19,10/19)\\). If the first block is replaced by the fixed representative value \\(2\\), the approximate coarse posterior is \\((1/2,1/2)\\). The \\(\\ell^1\\) defect is \\(1/19\\), while \\cref{earlier:thm:information_loss} gives the valid upper bound \\(1/10\\).\n\\end{example}","oss}\nLet \\(B_1=\\{1,2\\}\\), let \\(B_2=\\{3\\}\\), and take\n\\[\np=(3/10,2/10,5/10),\n\\qquad\nL=(1,3,2).\n\\]\nThe coarse prior is \\((1/2,1/2)\\). The exact block likelihoods are\n\\[\n\\ell_1(p)=\\frac{(3/10)+3(2/10)}{1/2}=\\frac95,\n\\qquad\n\\ell_2(p)=2.\n\\]\nThus the exact coarse posterior is \\((9/19,10/19)\\). If the first block is replaced by the fixed representative value \\(2\\), the approximate coarse posterior is \\((1/2,1/2)\\). The \\(\\ell^1\\) defect is \\(1/19\\), while \\cref{earlier:thm:information_loss} gives the valid upper bound \\(1/10\\).\n\\end{example}","2\\}\\), let \\(B_2=\\{3\\}\\), and take\n\\[\np=(3/10,2/10,5/10),\n\\qquad\nL=(1,3,2).\n\\]\nThe coarse prior is \\((1/2,1/2)\\). The exact block likelihoods are\n\\[\n\\ell_1(p)=\\frac{(3/10)+3(2/10)}{1/2}=\\frac95,\n\\qquad\n\\ell_2(p)=2.\n\\]\nThus the exact coarse posterior is \\((9/19,10/19)\\). If the first block is replaced by the fixed representative value \\(2\\), the approximate coarse posterior is \\((1/2,1/2)\\). The \\(\\ell^1\\) defect is \\(1/19\\), while \\cref{earlier:thm:information_loss} gives the valid upper bound \\(1/10\\).\n\\end{example}","}\\), and take\n\\[\np=(3/10,2/10,5/10),\n\\qquad\nL=(1,3,2).\n\\]\nThe coarse prior is \\((1/2,1/2)\\). The exact block likelihoods are\n\\[\n\\ell_1(p)=\\frac{(3/10)+3(2/10)}{1/2}=\\frac95,\n\\qquad\n\\ell_2(p)=2.\n\\]\nThus the exact coarse posterior is \\((9/19,10/19)\\). If the first block is replaced by the fixed representative value \\(2\\), the approximate coarse posterior is \\((1/2,1/2)\\). The \\(\\ell^1\\) defect is \\(1/19\\), while \\cref{earlier:thm:information_loss} gives the valid upper bound \\(1/10\\).\n\\end{example}","(1/2,1/2)\\). The exact block likelihoods are\n\\[\n\\ell_1(p)=\\frac{(3/10)+3(2/10)}{1/2}=\\frac95,\n\\qquad\n\\ell_2(p)=2.\n\\]\nThus the exact coarse posterior is \\((9/19,10/19)\\). If the first block is replaced by the fixed representative value \\(2\\), the approximate coarse posterior is \\((1/2,1/2)\\). The \\(\\ell^1\\) defect is \\(1/19\\), while \\cref{earlier:thm:information_loss} gives the valid upper bound \\(1/10\\).\n\\end{example}","\\begin{example}[One hidden mode]\\label{earlier:record:029}\\label{earlier:ex:one_hidden_mode}\nConsider\n\\[\nA=\n\\begin{pmatrix}\n-a&c\\\\\nd&-b\n\\end{pmatrix},\n\\qquad\na,b>0,\n\\]\nand retain the first coordinate. The memory kernel and self energy are\n\\[\n\\Kcal(t)=cd\\e^{-bt},\n\\qquad\n\\Sigma(z)=\\frac{cd}{z+b}.\n\\]\nThe retained coordinate has an exact time local closure for every retained initial value precisely when \\(cd=0\\). When \\(cd\\neq0\\), one auxiliary scalar state realizes the memory exactly.\n\\end{example}","d&-b\n\\end{pmatrix},\n\\qquad\na,b>0,\n\\]\nand retain the first coordinate. The memory kernel and self energy are\n\\[\n\\Kcal(t)=cd\\e^{-bt},\n\\qquad\n\\Sigma(z)=\\frac{cd}{z+b}.\n\\]\nThe retained coordinate has an exact time local closure for every retained initial value precisely when \\(cd=0\\). When \\(cd\\neq0\\), one auxiliary scalar state realizes the memory exactly.\n\\end{example}","\\begin{example}[Two sequential phases]\\label{earlier:record:030}\\label{earlier:ex:two_phase_survival}\nLet\n\\[\nA=\n\\begin{pmatrix}\n-\\lambda&0\\\\\n\\lambda&-\\mu\n\\end{pmatrix},\n\\qquad\np_0=\n\\begin{pmatrix}1\\\\0\\end{pmatrix},\n\\]\nwhere \\(\\lambda\\neq\\mu\\) are positive. The first phase moves to the second at rate \\(\\lambda\\), and the second is absorbed at rate \\(\\mu\\). Direct calculation gives\n\\[\nS(t)=\n\\frac{\\mu\\e^{-\\lambda t}-\\lambda\\e^{-\\mu t}}{\\mu-\\lambda},\n\\]\n\\[\nf(t)=\n\\frac{\\lambda\\mu}{\\mu-\\lambda}\n\\left(\\e^{-\\lambda t}-\\e^{-\\mu t}\\right).\n\\]\nThe ratio \\(f(t)/S(t)\\) is generally not constant. A time local Markov model on two transient phases can therefore produce a nonconstant hazard after the phase coordinate is hidden.\n\\end{example}","0\\end{pmatrix},\n\\]\nwhere \\(\\lambda\\neq\\mu\\) are positive. The first phase moves to the second at rate \\(\\lambda\\), and the second is absorbed at rate \\(\\mu\\). Direct calculation gives\n\\[\nS(t)=\n\\frac{\\mu\\e^{-\\lambda t}-\\lambda\\e^{-\\mu t}}{\\mu-\\lambda},\n\\]\n\\[\nf(t)=\n\\frac{\\lambda\\mu}{\\mu-\\lambda}\n\\left(\\e^{-\\lambda t}-\\e^{-\\mu t}\\right).\n\\]\nThe ratio \\(f(t)/S(t)\\) is generally not constant. A time local Markov model on two transient phases can therefore produce a nonconstant hazard after the phase coordinate is hidden.\n\\end{example}","This section receives the operator-substitution result deliberately moved out of Paper IV. Fix an integer $N\\geq1$ and the entire scalar function\n\\[\n\\Dfun_N(z)=\\sum_{m=0}^{\\infty}\\frac{(-1)^m z^{Nm}}{(Nm)!}.\n\\]\nMore generally, the result applies to any specified holomorphic scalar function that is nonzero on the declared domain. On a simply connected zero-free domain $U$, choose $g=\\log\\Dfun_N$ once and define $\\Sfun_N^{(r)}(z)=\\exp(-r g(z))$ for complex $r$. The bounded Banach-space result below uses classical holomorphic functional calculus; it is separate from the finite canonical-jet calculus formalized in Part I.","ceives the operator-substitution result deliberately moved out of Paper IV. Fix an integer $N\\geq1$ and the entire scalar function\n\\[\n\\Dfun_N(z)=\\sum_{m=0}^{\\infty}\\frac{(-1)^m z^{Nm}}{(Nm)!}.\n\\]\nMore generally, the result applies to any specified holomorphic scalar function that is nonzero on the declared domain. On a simply connected zero-free domain $U$, choose $g=\\log\\Dfun_N$ once and define $\\Sfun_N^{(r)}(z)=\\exp(-r g(z))$ for complex $r$. The bounded Banach-space result below uses classical holomorphic functional calculus; it is separate from the finite canonical-jet calculus formalized in Part I.","erately moved out of Paper IV. Fix an integer $N\\geq1$ and the entire scalar function\n\\[\n\\Dfun_N(z)=\\sum_{m=0}^{\\infty}\\frac{(-1)^m z^{Nm}}{(Nm)!}.\n\\]\nMore generally, the result applies to any specified holomorphic scalar function that is nonzero on the declared domain. On a simply connected zero-free domain $U$, choose $g=\\log\\Dfun_N$ once and define $\\Sfun_N^{(r)}(z)=\\exp(-r g(z))$ for complex $r$. The bounded Banach-space result below uses classical holomorphic functional calculus; it is separate from the finite canonical-jet calculus formalized in Part I.","\\begin{standardfact}[Holomorphic functional calculus]\\label{migration:record:001}\\label{migration:sf:holomorphic_calculus}\nLet $A$ be a bounded operator on a complex Banach space.  If $f$ is holomorphic on a neighborhood of $\\Spec(A)$, then\n\\[\nf(A)\n=\n\\frac1{2\\pi i}\\int_\\Gamma f(z)(zI-A)^{-1}\\,dz\n\\]\nfor an admissible positively oriented contour $\\Gamma$ lying in the holomorphy domain and the resolvent set, with winding number one about the spectrum and zero outside the domain. The calculus preserves sums, products, and composition.  We use the standard form from \\cite{DunfordSchwartz1958}; an accessible proof is given in \\cite[Appendix A]{TaylorBanachAlgebras}.\n\\end{standardfact}","\\begin{proposition}[Operator reciprocal family]\\label{migration:record:002}\\label{migration:prop:operator_reciprocal}\nLet $A$ be bounded and suppose $\\Spec(A)$ lies in a simply connected region on which $\\Dfun_N$ has no zeros.  Fix a holomorphic branch $g=\\log\\Dfun_N$ there.  Then\n\\[\n\\Sfun_N^{(r)}(A)\n=\n\\exp\\bigl(-r g(A)\\bigr)\n=\n\\bigl(\\Dfun_N^{-r}\\bigr)(A)\n\\]\nis well defined and\n\\[\n\\Sfun_N^{(r)}(A)\\Sfun_N^{(s)}(A)\n=\n\\Sfun_N^{(r+s)}(A).\n\\]\nIf $(P_i)$ is a finite complete family of bounded idempotents, then\n\\[\nP_i f(A)P_j\n=\n\\frac1{2\\pi i}\n\\int_\\Gamma\nf(z)P_i(zI-A)^{-1}P_j\\,dz\n\\]\nfor every holomorphic $f$ on a neighborhood of the spectrum.\n\\end{proposition}","]\\label{migration:record:002}\\label{migration:prop:operator_reciprocal}\nLet $A$ be bounded and suppose $\\Spec(A)$ lies in a simply connected region on which $\\Dfun_N$ has no zeros.  Fix a holomorphic branch $g=\\log\\Dfun_N$ there.  Then\n\\[\n\\Sfun_N^{(r)}(A)\n=\n\\exp\\bigl(-r g(A)\\bigr)\n=\n\\bigl(\\Dfun_N^{-r}\\bigr)(A)\n\\]\nis well defined and\n\\[\n\\Sfun_N^{(r)}(A)\\Sfun_N^{(s)}(A)\n=\n\\Sfun_N^{(r+s)}(A).\n\\]\nIf $(P_i)$ is a finite complete family of bounded idempotents, then\n\\[\nP_i f(A)P_j\n=\n\\frac1{2\\pi i}\n\\int_\\Gamma\nf(z)P_i(zI-A)^{-1}P_j\\,dz\n\\]\nfor every holomorphic $f$ on a neighborhood of the spectrum.\n\\end{proposition}","ocal}\nLet $A$ be bounded and suppose $\\Spec(A)$ lies in a simply connected region on which $\\Dfun_N$ has no zeros.  Fix a holomorphic branch $g=\\log\\Dfun_N$ there.  Then\n\\[\n\\Sfun_N^{(r)}(A)\n=\n\\exp\\bigl(-r g(A)\\bigr)\n=\n\\bigl(\\Dfun_N^{-r}\\bigr)(A)\n\\]\nis well defined and\n\\[\n\\Sfun_N^{(r)}(A)\\Sfun_N^{(s)}(A)\n=\n\\Sfun_N^{(r+s)}(A).\n\\]\nIf $(P_i)$ is a finite complete family of bounded idempotents, then\n\\[\nP_i f(A)P_j\n=\n\\frac1{2\\pi i}\n\\int_\\Gamma\nf(z)P_i(zI-A)^{-1}P_j\\,dz\n\\]\nfor every holomorphic $f$ on a neighborhood of the spectrum.\n\\end{proposition}","fun_N$ has no zeros.  Fix a holomorphic branch $g=\\log\\Dfun_N$ there.  Then\n\\[\n\\Sfun_N^{(r)}(A)\n=\n\\exp\\bigl(-r g(A)\\bigr)\n=\n\\bigl(\\Dfun_N^{-r}\\bigr)(A)\n\\]\nis well defined and\n\\[\n\\Sfun_N^{(r)}(A)\\Sfun_N^{(s)}(A)\n=\n\\Sfun_N^{(r+s)}(A).\n\\]\nIf $(P_i)$ is a finite complete family of bounded idempotents, then\n\\[\nP_i f(A)P_j\n=\n\\frac1{2\\pi i}\n\\int_\\Gamma\nf(z)P_i(zI-A)^{-1}P_j\\,dz\n\\]\nfor every holomorphic $f$ on a neighborhood of the spectrum.\n\\end{proposition}","\\begin{proof}\nThe spectral mapping theorem makes $\\Dfun_N(A)$ invertible.  The product identity follows from multiplicativity of the functional calculus and the fixed logarithm.  Applying $P_i$ and $P_j$ to the Cauchy integral formula gives the packet formula.\n\\end{proof}","aper's language, not by pretending that their state spaces are identical.\n\\vfig{three_views}\n\\platecaption{Left and center use one probability vector and its two block masses. The right panel is a separate retained/hidden model, developed on \\pagerefV{22}. Every curve is drawn from an explicit formula.}\n\\head{Coordinates retain; aggregation forgets; elimination remembers}\nKeeping every component $P_a x$ is an exact change of representation when $\\sum_aP_a=I$. Reconstruct by summing the components. In contrast, replacing $(p_0,p_1,p_2,p_3)$ by $(p_0+p_1,p_2+p_3)$ forgets how mass was distributed inside either pair. That information can matter for later observations.","e spaces are identical.\n\\vfig{three_views}\n\\platecaption{Left and center use one probability vector and its two block masses. The right panel is a separate retained/hidden model, developed on \\pagerefV{22}. Every curve is drawn from an explicit formula.}\n\\head{Coordinates retain; aggregation forgets; elimination remembers}\nKeeping every component $P_a x$ is an exact change of representation when $\\sum_aP_a=I$. Reconstruct by summing the components. In contrast, replacing $(p_0,p_1,p_2,p_3)$ by $(p_0+p_1,p_2+p_3)$ forgets how mass was distributed inside either pair. That information can matter for later observations.","one probability vector and its two block masses. The right panel is a separate retained/hidden model, developed on \\pagerefV{22}. Every curve is drawn from an explicit formula.}\n\\head{Coordinates retain; aggregation forgets; elimination remembers}\nKeeping every component $P_a x$ is an exact change of representation when $\\sum_aP_a=I$. Reconstruct by summing the components. In contrast, replacing $(p_0,p_1,p_2,p_3)$ by $(p_0+p_1,p_2+p_3)$ forgets how mass was distributed inside either pair. That information can matter for later observations.","and its two block masses. The right panel is a separate retained/hidden model, developed on \\pagerefV{22}. Every curve is drawn from an explicit formula.}\n\\head{Coordinates retain; aggregation forgets; elimination remembers}\nKeeping every component $P_a x$ is an exact change of representation when $\\sum_aP_a=I$. Reconstruct by summing the components. In contrast, replacing $(p_0,p_1,p_2,p_3)$ by $(p_0+p_1,p_2+p_3)$ forgets how mass was distributed inside either pair. That information can matter for later observations.","\\vpage{2}{A dictionary with orientations}{READ BEFORE MULTIPLYING / CONVENTIONS}{Traditional Part I: Sections 2, 4, 6, 7, and 12. Column probabilities and backward observables.}\nAn apparent disagreement between two transfer formulas often comes from acting on different objects. Throughout this guide, probabilities are columns and observables are functions written as columns. Their pairing is $f^{\\mathsf T}p$. The transpose distinguishes the forward action on distributions from the backward action on observations.\n\\begin{center}\\small\n\\begin{tabularx}{\\linewidth}{@{}lYY@{}}\\toprule\nObject & Exact meaning & What to check","midrule\n$p$ & A nonnegative column with $\\ones^{\\mathsf T}p=1$. & It is not a vector of function values.","$Q$ & Continuous Markov generator. & Off-diagonal entries are nonnegative and columns sum to zero.","$K$ & One discrete stochastic step. & $\\ones^{\\mathsf T}K=\\ones^{\\mathsf T}$ and $K_{ij}\\ge0$.","$K^{\\mathsf T}f$ & Expected next observable given the current state. & The same dynamics, with the opposite action on the dual object.","$P_a$ & A coordinate packet projector. & All packets together reconstruct the full state.","$C$ & Sums fine probability masses into coarse blocks. & Usually rectangular and not invertible.","$E_\\pi$ & Conditional expectation of observables. & A weighted average, self-adjoint for the $\\pi$ inner product.","$\\mathcal B_\\lambda$ & $\\operatorname{diag}(\\lambda)p/(\\lambda^{\\mathsf T}p)$. & Nonnegative likelihood and positive total evidence.","$L$ & A nonnegative self-adjoint operator. & Its evolution is $e^{-tL}$, not $e^{tL}$.","$R(z)$ & $(zI-A)^{-1}$. & $z$ belongs to the resolvent set.","$\\mathcal K(t)$ & The memory kernel after hidden-state elimination. & It is not the transition matrix $K$.","bottomrule\n\\end{tabularx}\\end{center}\n\\head{One convention test}\nFor\n\\[\nK_2=\\begin{pmatrix}3/4&1/4\\\\1/4&3/4\\end{pmatrix},\n\\qquad p=e_0,\\qquad f=(2,1)^{\\mathsf T},\n\\]\nwe obtain $K_2p=(3/4,1/4)^{\\mathsf T}$ and\n\\[\nf^{\\mathsf T}K_2p=7/4=(K_2^{\\mathsf T}f)_0.\n\\]\nWith departure weights $w_i=e^{-V(i)}$, the backward Feynman--Kac operator is $\\operatorname{diag}(w)K_2^{\\mathsf T}$. Its forward dual is $K_2\\operatorname{diag}(w)$. Interchanging these factors changes when the weight is applied.\n\\takeaway{Always name the object being transported. The transpose and the position of the likelihood or potential are mathematical choices, not typographic details.}","1/4&3/4\\end{pmatrix},\n\\qquad p=e_0,\\qquad f=(2,1)^{\\mathsf T},\n\\]\nwe obtain $K_2p=(3/4,1/4)^{\\mathsf T}$ and\n\\[\nf^{\\mathsf T}K_2p=7/4=(K_2^{\\mathsf T}f)_0.\n\\]\nWith departure weights $w_i=e^{-V(i)}$, the backward Feynman--Kac operator is $\\operatorname{diag}(w)K_2^{\\mathsf T}$. Its forward dual is $K_2\\operatorname{diag}(w)$. Interchanging these factors changes when the weight is applied.\n\\takeaway{Always name the object being transported. The transpose and the position of the likelihood or potential are mathematical choices, not typographic details.}","\\vpage{3}{A clock generated by one matrix}{CONTINUOUS TRANSFER / THE ENTIRE SOLUTION}{Traditional Part I: Theorems 1.2 and 4.1. Finite matrix exponentials and Markov positivity.}\nTake the generator\n\\[\nG=\\begin{pmatrix}-1&1\\\\1&-1\\end{pmatrix}.\n\\]\nThe total-mass direction $(1,1)$ has eigenvalue zero. The contrast $(1,-1)$ has eigenvalue $-2$. Resolving those two directions gives the full transfer, not only its first-order approximation:\n\\[\nT(t)=e^{tG}=\\frac12\\begin{pmatrix}1+e^{-2t}&1-e^{-2t}","1&-1\\end{pmatrix}.\n\\]\nThe total-mass direction $(1,1)$ has eigenvalue zero. The contrast $(1,-1)$ has eigenvalue $-2$. Resolving those two directions gives the full transfer, not only its first-order approximation:\n\\[\nT(t)=e^{tG}=\\frac12\\begin{pmatrix}1+e^{-2t}&1-e^{-2t}","1&-1\\end{pmatrix}.\n\\]\nThe total-mass direction $(1,1)$ has eigenvalue zero. The contrast $(1,-1)$ has eigenvalue $-2$. Resolving those two directions gives the full transfer, not only its first-order approximation:\n\\[\nT(t)=e^{tG}=\\frac12\\begin{pmatrix}1+e^{-2t}&1-e^{-2t}\\\\1-e^{-2t}&1+e^{-2t}\\end{pmatrix}.\n\\]\n\\vfig{clock}\n\\platecaption{A point seed at state zero approaches the uniform distribution. The displayed times were selected so that every bar label is rational. This is continuous-time evolution, not a simulated trajectory.}\n\\head{Two checks explain the mechanism}\nThe constant mode is unchanged, while each elapsed interval multiplies the contrast by $e^{-2t}$. Consequently two intervals multiply their contrast factors:\n\\[\ne^{-2s}e^{-2t}=e^{-2(s+t)},\\qquad T(s)T(t)=T(s+t).\n\\]\nFor $t\\ge0$, the entries are nonnegative and every column sums to one. Differentiation gives $T'(t)=GT(t)$. These three facts separately check the initial condition, the evolution equation, and preservation of probab","1-e^{-2t}&1+e^{-2t}\\end{pmatrix}.\n\\]\n\\vfig{clock}\n\\platecaption{A point seed at state zero approaches the uniform distribution. The displayed times were selected so that every bar label is rational. This is continuous-time evolution, not a simulated trajectory.}\n\\head{Two checks explain the mechanism}\nThe constant mode is unchanged, while each elapsed interval multiplies the contrast by $e^{-2t}$. Consequently two intervals multiply their contrast factors:\n\\[\ne^{-2s}e^{-2t}=e^{-2(s+t)},\\qquad T(s)T(t)=T(s+t).\n\\]\nFor $t\\ge0$, the entries are nonnegative and every column sums to one. Differentiation gives $T'(t)=GT(t)$. These three facts separately check the initial condition, the evolution equation, and preservation of probability.","{A point seed at state zero approaches the uniform distribution. The displayed times were selected so that every bar label is rational. This is continuous-time evolution, not a simulated trajectory.}\n\\head{Two checks explain the mechanism}\nThe constant mode is unchanged, while each elapsed interval multiplies the contrast by $e^{-2t}$. Consequently two intervals multiply their contrast factors:\n\\[\ne^{-2s}e^{-2t}=e^{-2(s+t)},\\qquad T(s)T(t)=T(s+t).\n\\]\nFor $t\\ge0$, the entries are nonnegative and every column sums to one. Differentiation gives $T'(t)=GT(t)$. These three facts separately check the initial condition, the evolution equation, and preservation of probability.","ion. The displayed times were selected so that every bar label is rational. This is continuous-time evolution, not a simulated trajectory.}\n\\head{Two checks explain the mechanism}\nThe constant mode is unchanged, while each elapsed interval multiplies the contrast by $e^{-2t}$. Consequently two intervals multiply their contrast factors:\n\\[\ne^{-2s}e^{-2t}=e^{-2(s+t)},\\qquad T(s)T(t)=T(s+t).\n\\]\nFor $t\\ge0$, the entries are nonnegative and every column sums to one. Differentiation gives $T'(t)=GT(t)$. These three facts separately check the initial condition, the evolution equation, and preservation of probability.","nuous-time evolution, not a simulated trajectory.}\n\\head{Two checks explain the mechanism}\nThe constant mode is unchanged, while each elapsed interval multiplies the contrast by $e^{-2t}$. Consequently two intervals multiply their contrast factors:\n\\[\ne^{-2s}e^{-2t}=e^{-2(s+t)},\\qquad T(s)T(t)=T(s+t).\n\\]\nFor $t\\ge0$, the entries are nonnegative and every column sums to one. Differentiation gives $T'(t)=GT(t)$. These three facts separately check the initial condition, the evolution equation, and preservation of probability.","At $t=\\log 2$, the retained first coordinate is $5/8$. It would instead be $1/2$ under the isolated scalar equation $\\dot x=-x$. The discrepancy is the mass that can leave state zero and return. On \\pagerefV{22}, we remove the second coordinate correctly and recover exactly the same $5/8$ using a memory integral.\n\\takeaway{The semigroup lives on the complete state. One diagonal entry of a semigroup need not itself satisfy a scalar semigroup law.}","\\vpage{4}{One evolution, two packet choices}{FOUR-STATE MODEL / COORDINATES AND MODES}{Traditional Part I: Theorem 2.1, Corollary 2.2, and the packet path expansion.}\nOur main finite model has generator\n\\[\nQ=\\begin{pmatrix}-3&1&2&0\\\\1&-3&0&2\\\\2&0&-3&1\\\\0&2&1&-3\\end{pmatrix}.\n\\]\nThe graph shows rates between actual states. The four normalized sign vectors on the right form an orthonormal basis $u_j$. Their projectors $P_j=u_ju_j^{\\mathsf T}$ sum to the identity and satisfy $P_iP_j=0$ for $i\\ne j$.\n\\vfig{generator_modes}\n\\platecaption{State coordinates have off-diagonal transfers. Spectral packets reduce the generator and evolve independently. The two dashed state groups are the coarse partition used next.}\n\\head{The same evolution has a closed formula}\nWriting $q=e^{-2t}$ gives\n\\[\ne^{tQ}=P_0+qP_1+q^2P_2+q^3P_3.\n\\]\nAt $q=1/2$, the point seed $e_0$ becomes $(15,5,9,3)^{\\mathsf T}/32$. The polynomial identity $T(q)T(r)=T(qr)$ checks composition exactly, without nume","0&2&1&-3\\end{pmatrix}.\n\\]\nThe graph shows rates between actual states. The four normalized sign vectors on the right form an orthonormal basis $u_j$. Their projectors $P_j=u_ju_j^{\\mathsf T}$ sum to the identity and satisfy $P_iP_j=0$ for $i\\ne j$.\n\\vfig{generator_modes}\n\\platecaption{State coordinates have off-diagonal transfers. Spectral packets reduce the generator and evolve independently. The two dashed state groups are the coarse partition used next.}\n\\head{The same evolution has a closed formula}\nWriting $q=e^{-2t}$ gives\n\\[\ne^{tQ}=P_0+qP_1+q^2P_2+q^3P_3.\n\\]\nAt $q=1/2$, the point seed $e_0$ becomes $(15,5,9,3)^{\\mathsf T}/32$. The polynomial identity $T(q)T(r)=T(qr)$ checks composition exactly, without numerical matrix exponentiation.","orm an orthonormal basis $u_j$. Their projectors $P_j=u_ju_j^{\\mathsf T}$ sum to the identity and satisfy $P_iP_j=0$ for $i\\ne j$.\n\\vfig{generator_modes}\n\\platecaption{State coordinates have off-diagonal transfers. Spectral packets reduce the generator and evolve independently. The two dashed state groups are the coarse partition used next.}\n\\head{The same evolution has a closed formula}\nWriting $q=e^{-2t}$ gives\n\\[\ne^{tQ}=P_0+qP_1+q^2P_2+q^3P_3.\n\\]\nAt $q=1/2$, the point seed $e_0$ becomes $(15,5,9,3)^{\\mathsf T}/32$. The polynomial identity $T(q)T(r)=T(qr)$ checks composition exactly, without numerical matrix exponentiation.","basis $u_j$. Their projectors $P_j=u_ju_j^{\\mathsf T}$ sum to the identity and satisfy $P_iP_j=0$ for $i\\ne j$.\n\\vfig{generator_modes}\n\\platecaption{State coordinates have off-diagonal transfers. Spectral packets reduce the generator and evolve independently. The two dashed state groups are the coarse partition used next.}\n\\head{The same evolution has a closed formula}\nWriting $q=e^{-2t}$ gives\n\\[\ne^{tQ}=P_0+qP_1+q^2P_2+q^3P_3.\n\\]\nAt $q=1/2$, the point seed $e_0$ becomes $(15,5,9,3)^{\\mathsf T}/32$. The polynomial identity $T(q)T(r)=T(qr)$ checks composition exactly, without numerical matrix exponentiation.","_j^{\\mathsf T}$ sum to the identity and satisfy $P_iP_j=0$ for $i\\ne j$.\n\\vfig{generator_modes}\n\\platecaption{State coordinates have off-diagonal transfers. Spectral packets reduce the generator and evolve independently. The two dashed state groups are the coarse partition used next.}\n\\head{The same evolution has a closed formula}\nWriting $q=e^{-2t}$ gives\n\\[\ne^{tQ}=P_0+qP_1+q^2P_2+q^3P_3.\n\\]\nAt $q=1/2$, the point seed $e_0$ becomes $(15,5,9,3)^{\\mathsf T}/32$. The polynomial identity $T(q)T(r)=T(qr)$ checks composition exactly, without numerical matrix exponentiation.","he identity and satisfy $P_iP_j=0$ for $i\\ne j$.\n\\vfig{generator_modes}\n\\platecaption{State coordinates have off-diagonal transfers. Spectral packets reduce the generator and evolve independently. The two dashed state groups are the coarse partition used next.}\n\\head{The same evolution has a closed formula}\nWriting $q=e^{-2t}$ gives\n\\[\ne^{tQ}=P_0+qP_1+q^2P_2+q^3P_3.\n\\]\nAt $q=1/2$, the point seed $e_0$ becomes $(15,5,9,3)^{\\mathsf T}/32$. The polynomial identity $T(q)T(r)=T(qr)$ checks composition exactly, without numerical matrix exponentiation.","d satisfy $P_iP_j=0$ for $i\\ne j$.\n\\vfig{generator_modes}\n\\platecaption{State coordinates have off-diagonal transfers. Spectral packets reduce the generator and evolve independently. The two dashed state groups are the coarse partition used next.}\n\\head{The same evolution has a closed formula}\nWriting $q=e^{-2t}$ gives\n\\[\ne^{tQ}=P_0+qP_1+q^2P_2+q^3P_3.\n\\]\nAt $q=1/2$, the point seed $e_0$ becomes $(15,5,9,3)^{\\mathsf T}/32$. The polynomial identity $T(q)T(r)=T(qr)$ checks composition exactly, without numerical matrix exponentiation.","des}\n\\platecaption{State coordinates have off-diagonal transfers. Spectral packets reduce the generator and evolve independently. The two dashed state groups are the coarse partition used next.}\n\\head{The same evolution has a closed formula}\nWriting $q=e^{-2t}$ gives\n\\[\ne^{tQ}=P_0+qP_1+q^2P_2+q^3P_3.\n\\]\nAt $q=1/2$, the point seed $e_0$ becomes $(15,5,9,3)^{\\mathsf T}/32$. The polynomial identity $T(q)T(r)=T(qr)$ checks composition exactly, without numerical matrix exponentiation.","Any other complete packet family still has exact block convolution. For the coordinate groups A and B,\n\\[\nP_b e^{(s+t)Q}P_a=\\sum_{c\\in\\{\\mathrm A,\\mathrm B\\}}(P_b e^{sQ}P_c)(P_c e^{tQ}P_a).\n\\]\nThe intermediate packet cannot generally be omitted. Only a reducing family eliminates all cross terms.","For later discrete calculations define $K=I+Q/8$. This matrix is stochastic, but \\emph{it is not} $e^{Q/8}$. It is the step kernel in the uniformization identity $e^{tQ}=e^{-8t}\\sum_{n\\ge0}(8t)^nK^n/n!$.\n\\takeaway{Changing packet coordinates does not change the evolution. Replacing continuous time by one discrete step does.}","\\vpage{5}{When two states are truly equivalent}{EXACT LUMPABILITY / AGGREGATE COLUMNS}{Traditional Part I: Theorem 5.1. The hypothesis is an intertwining identity.}\nGroup states $0,1$ into A and states $2,3$ into B. The aggregation map and the candidate coarse kernel are\n\\[\nC=\\begin{pmatrix}1&1&0&0","\\vpage{5}{When two states are truly equivalent}{EXACT LUMPABILITY / AGGREGATE COLUMNS}{Traditional Part I: Theorem 5.1. The hypothesis is an intertwining identity.}\nGroup states $0,1$ into A and states $2,3$ into B. The aggregation map and the candidate coarse kernel are\n\\[\nC=\\begin{pmatrix}1&1&0&0\\\\0&0&1&1\\end{pmatrix},\\qquad\n\\bar K=\\begin{pmatrix}3/4&1/4\\\\1/4&3/4\\end{pmatrix}.\n\\]\nA fine state in A has probability $1/4$ of entering B in one step, regardless of which fine state in A was occupied. The analogous statement holds for B.\n\\vfig{lumpability}\n\\platecaption{The fine columns are visibly different. Their sums over each target packet agree within each source packet. That equality is the reason a two-state description works.}\n\\head{A one-step calculation controls every discrete horizon}\nThe displayed columns give $CK=\\bar KC$. Multiplication then yields\n\\[\nCK^n=\\bar K^nC\\qquad(n\\ge0).\n\\]\nFor example, $e_0$ and $e_1$ have the same coarse initial vector $(1,0)^{\\mathsf T}$. Their fine evolutions differ, but their coarse vectors agree at every step. No choice of th","1/4&3/4\\end{pmatrix}.\n\\]\nA fine state in A has probability $1/4$ of entering B in one step, regardless of which fine state in A was occupied. The analogous statement holds for B.\n\\vfig{lumpability}\n\\platecaption{The fine columns are visibly different. Their sums over each target packet agree within each source packet. That equality is the reason a two-state description works.}\n\\head{A one-step calculation controls every discrete horizon}\nThe displayed columns give $CK=\\bar KC$. Multiplication then yields\n\\[\nCK^n=\\bar K^nC\\qquad(n\\ge0).\n\\]\nFor example, $e_0$ and $e_1$ have the same coarse initial vector $(1,0)^{\\mathsf T}$. Their fine evolutions differ, but their coarse vectors agree at every step. No choice of the within-A conditional distribution is ne","tering B in one step, regardless of which fine state in A was occupied. The analogous statement holds for B.\n\\vfig{lumpability}\n\\platecaption{The fine columns are visibly different. Their sums over each target packet agree within each source packet. That equality is the reason a two-state description works.}\n\\head{A one-step calculation controls every discrete horizon}\nThe displayed columns give $CK=\\bar KC$. Multiplication then yields\n\\[\nCK^n=\\bar K^nC\\qquad(n\\ge0).\n\\]\nFor example, $e_0$ and $e_1$ have the same coarse initial vector $(1,0)^{\\mathsf T}$. Their fine evolutions differ, but their coarse vectors agree at every step. No choice of the within-A conditional distribution is needed for the prediction.","fine state in A was occupied. The analogous statement holds for B.\n\\vfig{lumpability}\n\\platecaption{The fine columns are visibly different. Their sums over each target packet agree within each source packet. That equality is the reason a two-state description works.}\n\\head{A one-step calculation controls every discrete horizon}\nThe displayed columns give $CK=\\bar KC$. Multiplication then yields\n\\[\nCK^n=\\bar K^nC\\qquad(n\\ge0).\n\\]\nFor example, $e_0$ and $e_1$ have the same coarse initial vector $(1,0)^{\\mathsf T}$. Their fine evolutions differ, but their coarse vectors agree at every step. No choice of the within-A conditional distribution is needed for the prediction.","tement holds for B.\n\\vfig{lumpability}\n\\platecaption{The fine columns are visibly different. Their sums over each target packet agree within each source packet. That equality is the reason a two-state description works.}\n\\head{A one-step calculation controls every discrete horizon}\nThe displayed columns give $CK=\\bar KC$. Multiplication then yields\n\\[\nCK^n=\\bar K^nC\\qquad(n\\ge0).\n\\]\nFor example, $e_0$ and $e_1$ have the same coarse initial vector $(1,0)^{\\mathsf T}$. Their fine evolutions differ, but their coarse vectors agree at every step. No choice of the within-A conditional distribution is needed for the prediction.","ds for B.\n\\vfig{lumpability}\n\\platecaption{The fine columns are visibly different. Their sums over each target packet agree within each source packet. That equality is the reason a two-state description works.}\n\\head{A one-step calculation controls every discrete horizon}\nThe displayed columns give $CK=\\bar KC$. Multiplication then yields\n\\[\nCK^n=\\bar K^nC\\qquad(n\\ge0).\n\\]\nFor example, $e_0$ and $e_1$ have the same coarse initial vector $(1,0)^{\\mathsf T}$. Their fine evolutions differ, but their coarse vectors agree at every step. No choice of the within-A conditional distribution is needed for the prediction.","{The fine columns are visibly different. Their sums over each target packet agree within each source packet. That equality is the reason a two-state description works.}\n\\head{A one-step calculation controls every discrete horizon}\nThe displayed columns give $CK=\\bar KC$. Multiplication then yields\n\\[\nCK^n=\\bar K^nC\\qquad(n\\ge0).\n\\]\nFor example, $e_0$ and $e_1$ have the same coarse initial vector $(1,0)^{\\mathsf T}$. Their fine evolutions differ, but their coarse vectors agree at every step. No choice of the within-A conditional distribution is needed for the prediction.","The continuous generator also closes:\n\\[\nCQ=\\bar Q C,\\qquad\n\\bar Q=\\begin{pmatrix}-2&2\\\\2&-2\\end{pmatrix}.\n\\]\nThe Duhamel identity therefore gives $Ce^{tQ}=e^{t\\bar Q}C$. This is a second consequence of the corresponding generator identity, not an inference that $K$ is a fixed-time sample of $Q$.","2&-2\\end{pmatrix}.\n\\]\nThe Duhamel identity therefore gives $Ce^{tQ}=e^{t\\bar Q}C$. This is a second consequence of the corresponding generator identity, not an inference that $K$ is a fixed-time sample of $Q$.","\\vpage{6}{A broken lump and its error budget}{DEFECTS / A GUARANTEE WITHOUT A FIT}{Traditional Part I: Theorems 3.1, 3.3, and 5.2. Continuous and discrete defect identities.}\nModify only column zero of $K$: move $1/16$ of its mass from target state zero to target state two. The perturbed matrix $K_\\varepsilon$ remains stochastic, but its aggregate column now differs from column one. Relative to the unchanged coarse kernel,\n\\[\n\\Delta=CK_\\varepsilon-\\bar KC\n=\\begin{pmatrix}-1/16&0&0&0","ts error budget}{DEFECTS / A GUARANTEE WITHOUT A FIT}{Traditional Part I: Theorems 3.1, 3.3, and 5.2. Continuous and discrete defect identities.}\nModify only column zero of $K$: move $1/16$ of its mass from target state zero to target state two. The perturbed matrix $K_\\varepsilon$ remains stochastic, but its aggregate column now differs from column one. Relative to the unchanged coarse kernel,\n\\[\n\\Delta=CK_\\varepsilon-\\bar KC\n=\\begin{pmatrix}-1/16&0&0&0\\\\1/16&0&0&0\\end{pmatrix},\n\\qquad \\|\\Delta\\|_1=1/8.\n\\]\n\\vfig{defect_budget}\n\\platecaption{The solid values are computed with rational matrix powers. The dashed certificate uses only stochasticity and the exact one-step defect. It need not be sharp.}\n\\head{Telescoping tells us where every error originates}\nFor $n\\ge0$,\n\\[\nCK_\\varepsilon^n-\\bar K^nC\n=\\sum_{j=0}^{n-1}\\bar K^{n-1-j}\\Delta K_\\varepsilon^j.\n\\]\nEach summand propagates one defect between an earlier fine evolution and a later coarse evolution. Column-stochastic matrices have induced $\\ell^1$ norm one, so the norm is at most $n/8$. The empty sum at $n=0$ is zero.","1/16&0&0&0\\end{pmatrix},\n\\qquad \\|\\Delta\\|_1=1/8.\n\\]\n\\vfig{defect_budget}\n\\platecaption{The solid values are computed with rational matrix powers. The dashed certificate uses only stochasticity and the exact one-step defect. It need not be sharp.}\n\\head{Telescoping tells us where every error originates}\nFor $n\\ge0$,\n\\[\nCK_\\varepsilon^n-\\bar K^nC\n=\\sum_{j=0}^{n-1}\\bar K^{n-1-j}\\Delta K_\\varepsilon^j.\n\\]\nEach summand propagates one defect between an earlier fine evolution and a later coarse evolution. Column-stochastic matrices have induced $\\ell^1$ norm one, so the norm is at most $n/8$. The empty sum at $n=0$ is zero.","The dashed certificate uses only stochasticity and the exact one-step defect. It need not be sharp.}\n\\head{Telescoping tells us where every error originates}\nFor $n\\ge0$,\n\\[\nCK_\\varepsilon^n-\\bar K^nC\n=\\sum_{j=0}^{n-1}\\bar K^{n-1-j}\\Delta K_\\varepsilon^j.\n\\]\nEach summand propagates one defect between an earlier fine evolution and a later coarse evolution. Column-stochastic matrices have induced $\\ell^1$ norm one, so the norm is at most $n/8$. The empty sum at $n=0$ is zero.","icity and the exact one-step defect. It need not be sharp.}\n\\head{Telescoping tells us where every error originates}\nFor $n\\ge0$,\n\\[\nCK_\\varepsilon^n-\\bar K^nC\n=\\sum_{j=0}^{n-1}\\bar K^{n-1-j}\\Delta K_\\varepsilon^j.\n\\]\nEach summand propagates one defect between an earlier fine evolution and a later coarse evolution. Column-stochastic matrices have induced $\\ell^1$ norm one, so the norm is at most $n/8$. The empty sum at $n=0$ is zero.","step defect. It need not be sharp.}\n\\head{Telescoping tells us where every error originates}\nFor $n\\ge0$,\n\\[\nCK_\\varepsilon^n-\\bar K^nC\n=\\sum_{j=0}^{n-1}\\bar K^{n-1-j}\\Delta K_\\varepsilon^j.\n\\]\nEach summand propagates one defect between an earlier fine evolution and a later coarse evolution. Column-stochastic matrices have induced $\\ell^1$ norm one, so the norm is at most $n/8$. The empty sum at $n=0$ is zero.","The continuous counterpart is equally explicit. With fine generator $-1$, coarse generator $-2$, and $C=1$, the generator defect is $1$ and\n\\[\ne^{-t}-e^{-2t}=\\int_0^t e^{-2(t-r)}e^{-r}\\,dr.\n\\]\nIts resolvent version is $1/(z+1)-1/(z+2)=1/((z+1)(z+2))$ for $z>0$.\n\\takeaway{A small defect does not disappear. The identity specifies how it accumulates, and the norm assumptions turn that identity into a finite-horizon certificate.}","\\vpage{7}{Averaging is not summing}{OBSERVABLES / WEIGHTED ORTHOGONALITY}{Traditional Part I: Theorem 6.1. Strictly positive weights are part of the setting.}\nFor this observable example, choose the fixed probability weights $\\pi=(1,2,3,4)/10$. They are not being identified with the stationary law of the earlier symmetric chain. On the same two-block partition, conditional expectation is\n\\[\nE=\\begin{pmatrix}\n1/3&2/3&0&0","\\vpage{7}{Averaging is not summing}{OBSERVABLES / WEIGHTED ORTHOGONALITY}{Traditional Part I: Theorem 6.1. Strictly positive weights are part of the setting.}\nFor this observable example, choose the fixed probability weights $\\pi=(1,2,3,4)/10$. They are not being identified with the stationary law of the earlier symmetric chain. On the same two-block partition, conditional expectation is\n\\[\nE=\\begin{pmatrix}\n1/3&2/3&0&0\\\\1/3&2/3&0&0\\\\0&0&3/7&4/7\\\\0&0&3/7&4/7\n\\end{pmatrix}.\n\\]\nIt replaces function values by their \\emph{weighted block average}. In particular, $f=(0,3,2,5)^{\\mathsf T}$ becomes $Ef=(2,2,26/7,26/7)^{\\mathsf T}$.\n\\vfig{weighted_projection}\n\\platecaption{The bars are function values, not state probabilities. Their total need not be one. The block masses $3/10$ and $7/10$ determine the denominators in the averaging formula.}\n\\head{Which inner product makes this an orthogonal projection?}\nPut $W=\\operatorname{diag}(\\pi)$. Direct multiplication checks\n\\[\nE^2=E,\\qquad WE=E^{\\mathsf T}W,\\qquad E\\ones=\\ones.\n\\]\nThe middle equation is weighted self-adjointness. Ordinary symmetry f…","0&0&3/7&4/7\n\\end{pmatrix}.\n\\]\nIt replaces function values by their \\emph{weighted block average}. In particular, $f=(0,3,2,5)^{\\mathsf T}$ becomes $Ef=(2,2,26/7,26/7)^{\\mathsf T}$.\n\\vfig{weighted_projection}\n\\platecaption{The bars are function values, not state probabilities. Their total need not be one. The block masses $3/10$ and $7/10$ determine the denominators in the averaging formula.}\n\\head{Which inner product makes this an orthogonal projection?}\nPut $W=\\operatorname{diag}(\\pi)$. Direct multiplication checks\n\\[\nE^2=E,\\qquad WE=E^{\\mathsf T}W,\\qquad E\\ones=\\ones.\n\\]\nThe middle equation is weighted self-adjointness. Ordinary symmetry fails: $E_{01}=2/3$ whereas $E_{10}=1/3$. Testing $E=E^{\\mathsf T}$ would therefore reject a correct conditional-expectation projector.","ir \\emph{weighted block average}. In particular, $f=(0,3,2,5)^{\\mathsf T}$ becomes $Ef=(2,2,26/7,26/7)^{\\mathsf T}$.\n\\vfig{weighted_projection}\n\\platecaption{The bars are function values, not state probabilities. Their total need not be one. The block masses $3/10$ and $7/10$ determine the denominators in the averaging formula.}\n\\head{Which inner product makes this an orthogonal projection?}\nPut $W=\\operatorname{diag}(\\pi)$. Direct multiplication checks\n\\[\nE^2=E,\\qquad WE=E^{\\mathsf T}W,\\qquad E\\ones=\\ones.\n\\]\nThe middle equation is weighted self-adjointness. Ordinary symmetry fails: $E_{01}=2/3$ whereas $E_{10}=1/3$. Testing $E=E^{\\mathsf T}$ would therefore reject a correct conditional-expectation projector.",")^{\\mathsf T}$ becomes $Ef=(2,2,26/7,26/7)^{\\mathsf T}$.\n\\vfig{weighted_projection}\n\\platecaption{The bars are function values, not state probabilities. Their total need not be one. The block masses $3/10$ and $7/10$ determine the denominators in the averaging formula.}\n\\head{Which inner product makes this an orthogonal projection?}\nPut $W=\\operatorname{diag}(\\pi)$. Direct multiplication checks\n\\[\nE^2=E,\\qquad WE=E^{\\mathsf T}W,\\qquad E\\ones=\\ones.\n\\]\nThe middle equation is weighted self-adjointness. Ordinary symmetry fails: $E_{01}=2/3$ whereas $E_{10}=1/3$. Testing $E=E^{\\mathsf T}$ would therefore reject a correct conditional-expectation projector.","state probabilities. Their total need not be one. The block masses $3/10$ and $7/10$ determine the denominators in the averaging formula.}\n\\head{Which inner product makes this an orthogonal projection?}\nPut $W=\\operatorname{diag}(\\pi)$. Direct multiplication checks\n\\[\nE^2=E,\\qquad WE=E^{\\mathsf T}W,\\qquad E\\ones=\\ones.\n\\]\nThe middle equation is weighted self-adjointness. Ordinary symmetry fails: $E_{01}=2/3$ whereas $E_{10}=1/3$. Testing $E=E^{\\mathsf T}$ would therefore reject a correct conditional-expectation projector.","Their total need not be one. The block masses $3/10$ and $7/10$ determine the denominators in the averaging formula.}\n\\head{Which inner product makes this an orthogonal projection?}\nPut $W=\\operatorname{diag}(\\pi)$. Direct multiplication checks\n\\[\nE^2=E,\\qquad WE=E^{\\mathsf T}W,\\qquad E\\ones=\\ones.\n\\]\nThe middle equation is weighted self-adjointness. Ordinary symmetry fails: $E_{01}=2/3$ whereas $E_{10}=1/3$. Testing $E=E^{\\mathsf T}$ would therefore reject a correct conditional-expectation projector.","e one. The block masses $3/10$ and $7/10$ determine the denominators in the averaging formula.}\n\\head{Which inner product makes this an orthogonal projection?}\nPut $W=\\operatorname{diag}(\\pi)$. Direct multiplication checks\n\\[\nE^2=E,\\qquad WE=E^{\\mathsf T}W,\\qquad E\\ones=\\ones.\n\\]\nThe middle equation is weighted self-adjointness. Ordinary symmetry fails: $E_{01}=2/3$ whereas $E_{10}=1/3$. Testing $E=E^{\\mathsf T}$ would therefore reject a correct conditional-expectation projector.","The range consists exactly of functions constant on each block. The error $f-Ef$ has weighted mean zero inside every block, so it is weighted-orthogonal to every block-constant function. This is the finite projection interpretation of conditional expectation.","In contrast, $C$ acts on probability columns by summing masses. The shapes of $C$ and $E$ differ, and their domains have different interpretations. The notation ``coarsening'' alone does not specify which operation is intended.\n\\takeaway{An exact weighted projector can be oblique in the unweighted Euclidean picture. Keep the measure and the inner product in the model.}","\\vpage{8}{Where the variance goes}{MARTINGALE PACKETS / THREE EXACT PIECES}{Traditional Part I: Theorem 6.2. Nested weighted projections and their differences.}\nKeep the weights and observable from \\pagerefV{7}. Let $E_0f$ be the constant global mean, let $E_1=E$ average within A and B, and let $E_2=I$ retain every value. Define\n\\[\nD_0=E_0,\\qquad D_1=E_1-E_0,\\qquad D_2=I-E_1.\n\\]\nThen $D_0f$ is the mean, $D_1f$ describes the contrast between block means, and $D_2f$ is the fluctuation left inside each block.\n\\vfig{variance}\n\\platecaption{The strip lengths are proportional to the three exact weighted energies. The between-block component is small but not zero. The vectors below specify the corresponding directions, not only their sizes.}\n\\head{The energy identity follows from actual orthogonality}\nThe tower identities give $E_0E_1=E_1E_0=E_0$. Together with weighted self-adjointness, these imply\n\\[\nD_iD_j=0\\ (i","\\vpage{8}{Where the variance goes}{MARTINGALE PACKETS / THREE EXACT PIECES}{Traditional Part I: Theorem 6.2. Nested weighted projections and their differences.}\nKeep the weights and observable from \\pagerefV{7}. Let $E_0f$ be the constant global mean, let $E_1=E$ average within A and B, and let $E_2=I$ retain every value. Define\n\\[\nD_0=E_0,\\qquad D_1=E_1-E_0,\\qquad D_2=I-E_1.\n\\]\nThen $D_0f$ is the mean, $D_1f$ describes the contrast between block means, and $D_2f$ is the fluctuation left inside each block.\n\\vfig{variance}\n\\platecaption{The strip lengths are proportional to the three exact weighted energies. The between-block component is small but not zero. The vectors below specify the corresponding directions, not only their sizes.}\n\\head{The energy identity follows from actual orthogonality}\nThe tower identities give $E_0E_1=E_1E_0=E_0$. Together with weighted self-adjointness, these imply\n\\[\nD_iD_j=0\\ (i\\ne j),\\qquad \\sum_{j=0}^2 D_j=I.\n\\]\nFor","\\vpage{8}{Where the variance goes}{MARTINGALE PACKETS / THREE EXACT PIECES}{Traditional Part I: Theorem 6.2. Nested weighted projections and their differences.}\nKeep the weights and observable from \\pagerefV{7}. Let $E_0f$ be the constant global mean, let $E_1=E$ average within A and B, and let $E_2=I$ retain every value. Define\n\\[\nD_0=E_0,\\qquad D_1=E_1-E_0,\\qquad D_2=I-E_1.\n\\]\nThen $D_0f$ is the mean, $D_1f$ describes the contrast between block means, and $D_2f$ is the fluctuation left inside each block.\n\\vfig{variance}\n\\platecaption{The strip lengths are proportional to the three exact weighted energies. The between-block component is small but not zero. The vectors below specify the corresponding directions, not only their sizes.}\n\\head{The energy identity follows from actual orthogonality}\nThe tower identities give $E_0E_1=E_1E_0=E_0$. Together with weighted self-adjointness, these imply\n\\[\nD_iD_j=0\\ (i\\ne j),\\qquad \\sum_{j=0}^2 D_j=I.\n\\]\nFor the chosen $f$, the total weighted energ","\\vpage{8}{Where the variance goes}{MARTINGALE PACKETS / THREE EXACT PIECES}{Traditional Part I: Theorem 6.2. Nested weighted projections and their differences.}\nKeep the weights and observable from \\pagerefV{7}. Let $E_0f$ be the constant global mean, let $E_1=E$ average within A and B, and let $E_2=I$ retain every value. Define\n\\[\nD_0=E_0,\\qquad D_1=E_1-E_0,\\qquad D_2=I-E_1.\n\\]\nThen $D_0f$ is the mean, $D_1f$ describes the contrast between block means, and $D_2f$ is the fluctuation left inside each block.\n\\vfig{variance}\n\\platecaption{The strip lengths are proportional to the three exact weighted energies. The between-block component is small but not zero. The vectors below specify the corresponding directions, not only their sizes.}\n\\head{The energy identity follows from actual orthogonality}\nThe tower identities give $E_0E_1=E_1E_0=E_0$. Together with weighted self-adjointness, these imply\n\\[\nD_iD_j=0\\ (i\\ne j),\\qquad \\sum_{j=0}^2 D_j=I.\n\\]\nFor the chosen $f$, the total weighted energy is $\\|f\\|_\\pi^2=13$ and the mean is $16/5$. Consequently\n\\[\n\\operatorname{V","\\vpage{8}{Where the variance goes}{MARTINGALE PACKETS / THREE EXACT PIECES}{Traditional Part I: Theorem 6.2. Nested weighted projections and their differences.}\nKeep the weights and observable from \\pagerefV{7}. Let $E_0f$ be the constant global mean, let $E_1=E$ average within A and B, and let $E_2=I$ retain every value. Define\n\\[\nD_0=E_0,\\qquad D_1=E_1-E_0,\\qquad D_2=I-E_1.\n\\]\nThen $D_0f$ is the mean, $D_1f$ describes the contrast between block means, and $D_2f$ is the fluctuation left inside each block.\n\\vfig{variance}\n\\platecaption{The strip lengths are proportional to the three exact weighted energies. The between-block component is small but not zero. The vectors below specify the corresponding directions, not only their sizes.}\n\\head{The energy identity follows from actual orthogonality}\nThe tower identities give $E_0E_1=E_1E_0=E_0$. Together with weighted self-adjointness, these imply\n\\[\nD_iD_j=0\\ (i\\ne j),\\qquad \\sum_{j=0}^2 D_j=I.\n\\]\nFor the chosen $f$, the total weighted energy is $\\|f\\|_\\pi^2=13$ and the mean is $16/5$. Consequently\n\\[\n\\operatorname{Var}_\\pi(f)\n=1","}{Where the variance goes}{MARTINGALE PACKETS / THREE EXACT PIECES}{Traditional Part I: Theorem 6.2. Nested weighted projections and their differences.}\nKeep the weights and observable from \\pagerefV{7}. Let $E_0f$ be the constant global mean, let $E_1=E$ average within A and B, and let $E_2=I$ retain every value. Define\n\\[\nD_0=E_0,\\qquad D_1=E_1-E_0,\\qquad D_2=I-E_1.\n\\]\nThen $D_0f$ is the mean, $D_1f$ describes the contrast between block means, and $D_2f$ is the fluctuation left inside each block.\n\\vfig{variance}\n\\platecaption{The strip lengths are proportional to the three exact weighted energies. The between-block component is small but not zero. The vectors below specify the corresponding directions, not only their sizes.}\n\\head{The energy identity follows from actual orthogonality}\nThe tower identities give $E_0E_1=E_1E_0=E_0$. Together with weighted self-adjointness, these imply\n\\[\nD_iD_j=0\\ (i\\ne j),\\qquad \\sum_{j=0}^2 D_j=I.\n\\]\nFor the chosen $f$, the total weighted energy is $\\|f\\|_\\pi^2=13$ and the mean is $16/5$. Consequently\n\\[\n\\operatorname{Var}_\\pi(f)\n=13-\\left(\\frac{16…","XACT PIECES}{Traditional Part I: Theorem 6.2. Nested weighted projections and their differences.}\nKeep the weights and observable from \\pagerefV{7}. Let $E_0f$ be the constant global mean, let $E_1=E$ average within A and B, and let $E_2=I$ retain every value. Define\n\\[\nD_0=E_0,\\qquad D_1=E_1-E_0,\\qquad D_2=I-E_1.\n\\]\nThen $D_0f$ is the mean, $D_1f$ describes the contrast between block means, and $D_2f$ is the fluctuation left inside each block.\n\\vfig{variance}\n\\platecaption{The strip lengths are proportional to the three exact weighted energies. The between-block component is small but not zero. The vectors below specify the corresponding directions, not only their sizes.}\n\\head{The energy identity follows from actual orthogonality}\nThe tower identities give $E_0E_1=E_1E_0=E_0$. Together with weighted self-adjointness, these imply\n\\[\nD_iD_j=0\\ (i\\ne j),\\qquad \\sum_{j=0}^2 D_j=I.\n\\]\nFor the chosen $f$, the total weighted energy is $\\|f\\|_\\pi^2=13$ and the mean is $16/5$. Consequently\n\\[\n\\operatorname{Var}_\\pi(f)\n=13-\\left(\\frac{16}{5}\\right)^2\n=\\frac{69}{25}\n=\\frac{108}{175}+\\frac{15}…","ast between block means, and $D_2f$ is the fluctuation left inside each block.\n\\vfig{variance}\n\\platecaption{The strip lengths are proportional to the three exact weighted energies. The between-block component is small but not zero. The vectors below specify the corresponding directions, not only their sizes.}\n\\head{The energy identity follows from actual orthogonality}\nThe tower identities give $E_0E_1=E_1E_0=E_0$. Together with weighted self-adjointness, these imply\n\\[\nD_iD_j=0\\ (i\\ne j),\\qquad \\sum_{j=0}^2 D_j=I.\n\\]\nFor the chosen $f$, the total weighted energy is $\\|f\\|_\\pi^2=13$ and the mean is $16/5$. Consequently\n\\[\n\\operatorname{Var}_\\pi(f)\n=13-\\left(\\frac{16}{5}\\right)^2\n=\\frac{69}{25}\n=\\frac{108}{175}+\\frac{15}{7}.\n\\]\nThe first term is the energy of the between-block contrast, and the second is the weighted within-block variance. This is a decomposition of one observable in one fixed weighted space.","ck.\n\\vfig{variance}\n\\platecaption{The strip lengths are proportional to the three exact weighted energies. The between-block component is small but not zero. The vectors below specify the corresponding directions, not only their sizes.}\n\\head{The energy identity follows from actual orthogonality}\nThe tower identities give $E_0E_1=E_1E_0=E_0$. Together with weighted self-adjointness, these imply\n\\[\nD_iD_j=0\\ (i\\ne j),\\qquad \\sum_{j=0}^2 D_j=I.\n\\]\nFor the chosen $f$, the total weighted energy is $\\|f\\|_\\pi^2=13$ and the mean is $16/5$. Consequently\n\\[\n\\operatorname{Var}_\\pi(f)\n=13-\\left(\\frac{16}{5}\\right)^2\n=\\frac{69}{25}\n=\\frac{108}{175}+\\frac{15}{7}.\n\\]\nThe first term is the energy of the between-block contrast, and the second is the weighted within-block variance. This is a decomposition of one observable in one fixed weighted space.","al to the three exact weighted energies. The between-block component is small but not zero. The vectors below specify the corresponding directions, not only their sizes.}\n\\head{The energy identity follows from actual orthogonality}\nThe tower identities give $E_0E_1=E_1E_0=E_0$. Together with weighted self-adjointness, these imply\n\\[\nD_iD_j=0\\ (i\\ne j),\\qquad \\sum_{j=0}^2 D_j=I.\n\\]\nFor the chosen $f$, the total weighted energy is $\\|f\\|_\\pi^2=13$ and the mean is $16/5$. Consequently\n\\[\n\\operatorname{Var}_\\pi(f)\n=13-\\left(\\frac{16}{5}\\right)^2\n=\\frac{69}{25}\n=\\frac{108}{175}+\\frac{15}{7}.\n\\]\nThe first term is the energy of the between-block contrast, and the second is the weighted within-block variance. This is a decomposition of one observable in one fixed weighted space.","rgies. The between-block component is small but not zero. The vectors below specify the corresponding directions, not only their sizes.}\n\\head{The energy identity follows from actual orthogonality}\nThe tower identities give $E_0E_1=E_1E_0=E_0$. Together with weighted self-adjointness, these imply\n\\[\nD_iD_j=0\\ (i\\ne j),\\qquad \\sum_{j=0}^2 D_j=I.\n\\]\nFor the chosen $f$, the total weighted energy is $\\|f\\|_\\pi^2=13$ and the mean is $16/5$. Consequently\n\\[\n\\operatorname{Var}_\\pi(f)\n=13-\\left(\\frac{16}{5}\\right)^2\n=\\frac{69}{25}\n=\\frac{108}{175}+\\frac{15}{7}.\n\\]\nThe first term is the energy of the between-block contrast, and the second is the weighted within-block variance. This is a decomposition of one observable in one fixed weighted space.","t is small but not zero. The vectors below specify the corresponding directions, not only their sizes.}\n\\head{The energy identity follows from actual orthogonality}\nThe tower identities give $E_0E_1=E_1E_0=E_0$. Together with weighted self-adjointness, these imply\n\\[\nD_iD_j=0\\ (i\\ne j),\\qquad \\sum_{j=0}^2 D_j=I.\n\\]\nFor the chosen $f$, the total weighted energy is $\\|f\\|_\\pi^2=13$ and the mean is $16/5$. Consequently\n\\[\n\\operatorname{Var}_\\pi(f)\n=13-\\left(\\frac{16}{5}\\right)^2\n=\\frac{69}{25}\n=\\frac{108}{175}+\\frac{15}{7}.\n\\]\nThe first term is the energy of the between-block contrast, and the second is the weighted within-block variance. This is a decomposition of one observable in one fixed weighted space.","o. The vectors below specify the corresponding directions, not only their sizes.}\n\\head{The energy identity follows from actual orthogonality}\nThe tower identities give $E_0E_1=E_1E_0=E_0$. Together with weighted self-adjointness, these imply\n\\[\nD_iD_j=0\\ (i\\ne j),\\qquad \\sum_{j=0}^2 D_j=I.\n\\]\nFor the chosen $f$, the total weighted energy is $\\|f\\|_\\pi^2=13$ and the mean is $16/5$. Consequently\n\\[\n\\operatorname{Var}_\\pi(f)\n=13-\\left(\\frac{16}{5}\\right)^2\n=\\frac{69}{25}\n=\\frac{108}{175}+\\frac{15}{7}.\n\\]\nThe first term is the energy of the between-block contrast, and the second is the weighted within-block variance. This is a decomposition of one observable in one fixed weighted space.","Changing the probability weights between levels would change the inner products and the averaging operators. The displayed orthogonal decomposition would no longer follow merely from nesting the sets. Likewise, choosing unrelated projections can preserve the telescope $\\sum D_j=I$ while losing orthogonality and the energy sum.\n\\takeaway{The differences between nested conditional expectations are not just bookkeeping. They are orthogonal, separately measurable parts of the original signal.}","\\vpage{9}{Evidence changes shape and mass}{BAYESIAN TRANSFER / AN EXACT POSTERIOR}{Traditional Part I: Theorems 7.1--7.3. Normalization, packet reconstruction, and positive evidence.}\nStart with $p=(1,1,1,1)^{\\mathsf T}/4$ and likelihood weights $\\lambda=(4,1,1,1)^{\\mathsf T}$. The unnormalized vector is $(1,1/4,1/4,1/4)^{\\mathsf T}$, its evidence is $Z=7/4$, and\n\\[\n\\mathcal B_\\lambda(p)=\\frac{(4,1,1,1)^{\\mathsf T}}7.\n\\]\nLikelihood weights may be rescaled, so an evidence larger than one is not a probability claim. Multiplying all likelihoods by one common positive factor leaves this posterior unchanged.\n\\vfig{evidence}\n\\platecaption{The A packet becomes heavier and its internal shape changes. The B packet retains an equal internal split but loses total mass. These are different effects of the same likelihood.}\n\\head{Reconstruct from packet mass times packet shape}\nThe unnormalized packet evidences are $Z_{\\m","\\vpage{9}{Evidence changes shape and mass}{BAYESIAN TRANSFER / AN EXACT POSTERIOR}{Traditional Part I: Theorems 7.1--7.3. Normalization, packet reconstruction, and positive evidence.}\nStart with $p=(1,1,1,1)^{\\mathsf T}/4$ and likelihood weights $\\lambda=(4,1,1,1)^{\\mathsf T}$. The unnormalized vector is $(1,1/4,1/4,1/4)^{\\mathsf T}$, its evidence is $Z=7/4$, and\n\\[\n\\mathcal B_\\lambda(p)=\\frac{(4,1,1,1)^{\\mathsf T}}7.\n\\]\nLikelihood weights may be rescaled, so an evidence larger than one is not a probability claim. Multiplying all likelihoods by one common positive factor leaves this posterior unchanged.\n\\vfig{evidence}\n\\platecaption{The A packet becomes heavier and its internal shape changes. The B packet retains an equal internal split but loses total mass. These are different effects of the same likelihood.}\n\\head{Reconstruct from packet mass times packet shape}\nThe unnormalized packet evidences are $Z_{\\mathrm A}=5/4$ and $Z_{\\mathrm B}=1/2$. The resulting ma","\\vpage{9}{Evidence changes shape and mass}{BAYESIAN TRANSFER / AN EXACT POSTERIOR}{Traditional Part I: Theorems 7.1--7.3. Normalization, packet reconstruction, and positive evidence.}\nStart with $p=(1,1,1,1)^{\\mathsf T}/4$ and likelihood weights $\\lambda=(4,1,1,1)^{\\mathsf T}$. The unnormalized vector is $(1,1/4,1/4,1/4)^{\\mathsf T}$, its evidence is $Z=7/4$, and\n\\[\n\\mathcal B_\\lambda(p)=\\frac{(4,1,1,1)^{\\mathsf T}}7.\n\\]\nLikelihood weights may be rescaled, so an evidence larger than one is not a probability claim. Multiplying all likelihoods by one common positive factor leaves this posterior unchanged.\n\\vfig{evidence}\n\\platecaption{The A packet becomes heavier and its internal shape changes. The B packet retains an equal internal split but loses total mass. These are different effects of the same likelihood.}\n\\head{Reconstruct from packet mass times packet shape}\nThe unnormalized packet evidences are $Z_{\\mathrm A}=5/4$ and $Z_{\\mathrm B}=1/2$. The resulting masses and conditional shapes are\n\\[\nm_{\\mathrm A}=5/7,\\quad","\\vpage{9}{Evidence changes shape and mass}{BAYESIAN TRANSFER / AN EXACT POSTERIOR}{Traditional Part I: Theorems 7.1--7.3. Normalization, packet reconstruction, and positive evidence.}\nStart with $p=(1,1,1,1)^{\\mathsf T}/4$ and likelihood weights $\\lambda=(4,1,1,1)^{\\mathsf T}$. The unnormalized vector is $(1,1/4,1/4,1/4)^{\\mathsf T}$, its evidence is $Z=7/4$, and\n\\[\n\\mathcal B_\\lambda(p)=\\frac{(4,1,1,1)^{\\mathsf T}}7.\n\\]\nLikelihood weights may be rescaled, so an evidence larger than one is not a probability claim. Multiplying all likelihoods by one common positive factor leaves this posterior unchanged.\n\\vfig{evidence}\n\\platecaption{The A packet becomes heavier and its internal shape changes. The B packet retains an equal internal split but loses total mass. These are different effects of the same likelihood.}\n\\head{Reconstruct from packet mass times packet shape}\nThe unnormalized packet evidences are $Z_{\\mathrm A}=5/4$ and $Z_{\\mathrm B}=1/2$. The resulting masses and conditional shapes are\n\\[\nm_{\\mathrm A}=5/7,\\quad q_{\\mathrm A}=(4/5,1/5,0","\\vpage{9}{Evidence changes shape and mass}{BAYESIAN TRANSFER / AN EXACT POSTERIOR}{Traditional Part I: Theorems 7.1--7.3. Normalization, packet reconstruction, and positive evidence.}\nStart with $p=(1,1,1,1)^{\\mathsf T}/4$ and likelihood weights $\\lambda=(4,1,1,1)^{\\mathsf T}$. The unnormalized vector is $(1,1/4,1/4,1/4)^{\\mathsf T}$, its evidence is $Z=7/4$, and\n\\[\n\\mathcal B_\\lambda(p)=\\frac{(4,1,1,1)^{\\mathsf T}}7.\n\\]\nLikelihood weights may be rescaled, so an evidence larger than one is not a probability claim. Multiplying all likelihoods by one common positive factor leaves this posterior unchanged.\n\\vfig{evidence}\n\\platecaption{The A packet becomes heavier and its internal shape changes. The B packet retains an equal internal split but loses total mass. These are different effects of the same likelihood.}\n\\head{Reconstruct from packet mass times packet shape}\nThe unnormalized packet evidences are $Z_{\\mathrm A}=5/4$ and $Z_{\\mathrm B}=1/2$. The resulting masses and conditional shapes are\n\\[\nm_{\\mathrm A}=5/7,\\quad q_{\\mathrm A}=(4/5,1/5,0,0)^{\\mathsf T},\n\\qquad m_{\\mathrm B}=2/7…","im. Multiplying all likelihoods by one common positive factor leaves this posterior unchanged.\n\\vfig{evidence}\n\\platecaption{The A packet becomes heavier and its internal shape changes. The B packet retains an equal internal split but loses total mass. These are different effects of the same likelihood.}\n\\head{Reconstruct from packet mass times packet shape}\nThe unnormalized packet evidences are $Z_{\\mathrm A}=5/4$ and $Z_{\\mathrm B}=1/2$. The resulting masses and conditional shapes are\n\\[\nm_{\\mathrm A}=5/7,\\quad q_{\\mathrm A}=(4/5,1/5,0,0)^{\\mathsf T},\n\\qquad m_{\\mathrm B}=2/7,\\quad q_{\\mathrm B}=(0,0,1/2,1/2)^{\\mathsf T}.\n\\]\nThus $m_{\\mathrm A}q_{\\mathrm A}+m_{\\mathrm B}q_{\\mathrm B}$ is exactly the displayed posterior. Keeping only the masses would not reconstruct the fine state.","lihoods by one common positive factor leaves this posterior unchanged.\n\\vfig{evidence}\n\\platecaption{The A packet becomes heavier and its internal shape changes. The B packet retains an equal internal split but loses total mass. These are different effects of the same likelihood.}\n\\head{Reconstruct from packet mass times packet shape}\nThe unnormalized packet evidences are $Z_{\\mathrm A}=5/4$ and $Z_{\\mathrm B}=1/2$. The resulting masses and conditional shapes are\n\\[\nm_{\\mathrm A}=5/7,\\quad q_{\\mathrm A}=(4/5,1/5,0,0)^{\\mathsf T},\n\\qquad m_{\\mathrm B}=2/7,\\quad q_{\\mathrm B}=(0,0,1/2,1/2)^{\\mathsf T}.\n\\]\nThus $m_{\\mathrm A}q_{\\mathrm A}+m_{\\mathrm B}q_{\\mathrm B}$ is exactly the displayed posterior. Keeping only the masses would not reconstruct the fine state.","\\vfig{evidence}\n\\platecaption{The A packet becomes heavier and its internal shape changes. The B packet retains an equal internal split but loses total mass. These are different effects of the same likelihood.}\n\\head{Reconstruct from packet mass times packet shape}\nThe unnormalized packet evidences are $Z_{\\mathrm A}=5/4$ and $Z_{\\mathrm B}=1/2$. The resulting masses and conditional shapes are\n\\[\nm_{\\mathrm A}=5/7,\\quad q_{\\mathrm A}=(4/5,1/5,0,0)^{\\mathsf T},\n\\qquad m_{\\mathrm B}=2/7,\\quad q_{\\mathrm B}=(0,0,1/2,1/2)^{\\mathsf T}.\n\\]\nThus $m_{\\mathrm A}q_{\\mathrm A}+m_{\\mathrm B}q_{\\mathrm B}$ is exactly the displayed posterior. Keeping only the masses would not reconstruct the fine state.","total mass. These are different effects of the same likelihood.}\n\\head{Reconstruct from packet mass times packet shape}\nThe unnormalized packet evidences are $Z_{\\mathrm A}=5/4$ and $Z_{\\mathrm B}=1/2$. The resulting masses and conditional shapes are\n\\[\nm_{\\mathrm A}=5/7,\\quad q_{\\mathrm A}=(4/5,1/5,0,0)^{\\mathsf T},\n\\qquad m_{\\mathrm B}=2/7,\\quad q_{\\mathrm B}=(0,0,1/2,1/2)^{\\mathsf T}.\n\\]\nThus $m_{\\mathrm A}q_{\\mathrm A}+m_{\\mathrm B}q_{\\mathrm B}$ is exactly the displayed posterior. Keeping only the masses would not reconstruct the fine state.","There is no need to divide by a zero packet evidence. For likelihood $(0,0,1,1)$, packet A has mass zero and contributes the zero vector. Only the occupied B packet has a conditional probability vector. If the \\emph{total} evidence is zero, however, the Bayesian update is not defined and the implementation rejects the input.\n\\takeaway{Bayesian reconstruction uses two kinds of data: how much posterior mass a packet carries and how that mass is distributed inside the packet.}","\\vpage{10}{Two priors that fool the coarse model}{SUFFICIENCY / A COMPUTABLE FAILURE WITNESS}{Traditional Part I: Theorem 8.2 and the one-packet exception.}\nKeep the same likelihood $\\lambda=(4,1,1,1)^{\\mathsf T}$ and partition A$=\\{0,1\\}$, B$=\\{2,3\\}$. Compare\n\\[\np^{(a)}=(1/2,0,1/2,0)^{\\mathsf T},\\qquad\np^{(b)}=(0,1/2,1/2,0)^{\\mathsf T}.\n\\]\nBoth have coarse prior $(1/2,1/2)^{\\mathsf T}$. A procedure given only that two-entry vector cannot distinguish them.\n\\vfig{insufficiency}\n\\platecaption{After the identical fine likelihood is applied, the coarse posteriors are $(4/5,1/5)$ and $(1/2,1/2)$. A single deterministic update of the common coarse prior cannot produce both.}\n\\head{Why this witnesses the necessity of blockwise constancy}\nThe A evidence for $p^{(a)}$ is $2$, whereas it is $1/2$ for $p^{(b)}$. Their B evidence is $1/2$ in both cases. The posterior odds therefore distinguish the hidden distr","\\vpage{10}{Two priors that fool the coarse model}{SUFFICIENCY / A COMPUTABLE FAILURE WITNESS}{Traditional Part I: Theorem 8.2 and the one-packet exception.}\nKeep the same likelihood $\\lambda=(4,1,1,1)^{\\mathsf T}$ and partition A$=\\{0,1\\}$, B$=\\{2,3\\}$. Compare\n\\[\np^{(a)}=(1/2,0,1/2,0)^{\\mathsf T},\\qquad\np^{(b)}=(0,1/2,1/2,0)^{\\mathsf T}.\n\\]\nBoth have coarse prior $(1/2,1/2)^{\\mathsf T}$. A procedure given only that two-entry vector cannot distinguish them.\n\\vfig{insufficiency}\n\\platecaption{After the identical fine likelihood is applied, the coarse posteriors are $(4/5,1/5)$ and $(1/2,1/2)$. A single deterministic update of the common coarse prior cannot produce both.}\n\\head{Why this witnesses the necessity of blockwise constancy}\nThe A evidence for $p^{(a)}$ is $2$, whereas it is $1/2$ for $p^{(b)}$. Their B evidence is $1/2$ in both cases. The posterior odds therefore distinguish the hidden distribution inside A.","ikelihood $\\lambda=(4,1,1,1)^{\\mathsf T}$ and partition A$=\\{0,1\\}$, B$=\\{2,3\\}$. Compare\n\\[\np^{(a)}=(1/2,0,1/2,0)^{\\mathsf T},\\qquad\np^{(b)}=(0,1/2,1/2,0)^{\\mathsf T}.\n\\]\nBoth have coarse prior $(1/2,1/2)^{\\mathsf T}$. A procedure given only that two-entry vector cannot distinguish them.\n\\vfig{insufficiency}\n\\platecaption{After the identical fine likelihood is applied, the coarse posteriors are $(4/5,1/5)$ and $(1/2,1/2)$. A single deterministic update of the common coarse prior cannot produce both.}\n\\head{Why this witnesses the necessity of blockwise constancy}\nThe A evidence for $p^{(a)}$ is $2$, whereas it is $1/2$ for $p^{(b)}$. Their B evidence is $1/2$ in both cases. The posterior odds therefore distinguish the hidden distribution inside A.","da=(4,1,1,1)^{\\mathsf T}$ and partition A$=\\{0,1\\}$, B$=\\{2,3\\}$. Compare\n\\[\np^{(a)}=(1/2,0,1/2,0)^{\\mathsf T},\\qquad\np^{(b)}=(0,1/2,1/2,0)^{\\mathsf T}.\n\\]\nBoth have coarse prior $(1/2,1/2)^{\\mathsf T}$. A procedure given only that two-entry vector cannot distinguish them.\n\\vfig{insufficiency}\n\\platecaption{After the identical fine likelihood is applied, the coarse posteriors are $(4/5,1/5)$ and $(1/2,1/2)$. A single deterministic update of the common coarse prior cannot produce both.}\n\\head{Why this witnesses the necessity of blockwise constancy}\nThe A evidence for $p^{(a)}$ is $2$, whereas it is $1/2$ for $p^{(b)}$. Their B evidence is $1/2$ in both cases. The posterior odds therefore distinguish the hidden distribution inside A.","rior $(1/2,1/2)^{\\mathsf T}$. A procedure given only that two-entry vector cannot distinguish them.\n\\vfig{insufficiency}\n\\platecaption{After the identical fine likelihood is applied, the coarse posteriors are $(4/5,1/5)$ and $(1/2,1/2)$. A single deterministic update of the common coarse prior cannot produce both.}\n\\head{Why this witnesses the necessity of blockwise constancy}\nThe A evidence for $p^{(a)}$ is $2$, whereas it is $1/2$ for $p^{(b)}$. Their B evidence is $1/2$ in both cases. The posterior odds therefore distinguish the hidden distribution inside A.","2)^{\\mathsf T}$. A procedure given only that two-entry vector cannot distinguish them.\n\\vfig{insufficiency}\n\\platecaption{After the identical fine likelihood is applied, the coarse posteriors are $(4/5,1/5)$ and $(1/2,1/2)$. A single deterministic update of the common coarse prior cannot produce both.}\n\\head{Why this witnesses the necessity of blockwise constancy}\nThe A evidence for $p^{(a)}$ is $2$, whereas it is $1/2$ for $p^{(b)}$. Their B evidence is $1/2$ in both cases. The posterior odds therefore distinguish the hidden distribution inside A.","procedure given only that two-entry vector cannot distinguish them.\n\\vfig{insufficiency}\n\\platecaption{After the identical fine likelihood is applied, the coarse posteriors are $(4/5,1/5)$ and $(1/2,1/2)$. A single deterministic update of the common coarse prior cannot produce both.}\n\\head{Why this witnesses the necessity of blockwise constancy}\nThe A evidence for $p^{(a)}$ is $2$, whereas it is $1/2$ for $p^{(b)}$. Their B evidence is $1/2$ in both cases. The posterior odds therefore distinguish the hidden distribution inside A.","given only that two-entry vector cannot distinguish them.\n\\vfig{insufficiency}\n\\platecaption{After the identical fine likelihood is applied, the coarse posteriors are $(4/5,1/5)$ and $(1/2,1/2)$. A single deterministic update of the common coarse prior cannot produce both.}\n\\head{Why this witnesses the necessity of blockwise constancy}\nThe A evidence for $p^{(a)}$ is $2$, whereas it is $1/2$ for $p^{(b)}$. Their B evidence is $1/2$ in both cases. The posterior odds therefore distinguish the hidden distribution inside A.","r cannot distinguish them.\n\\vfig{insufficiency}\n\\platecaption{After the identical fine likelihood is applied, the coarse posteriors are $(4/5,1/5)$ and $(1/2,1/2)$. A single deterministic update of the common coarse prior cannot produce both.}\n\\head{Why this witnesses the necessity of blockwise constancy}\nThe A evidence for $p^{(a)}$ is $2$, whereas it is $1/2$ for $p^{(b)}$. Their B evidence is $1/2$ in both cases. The posterior odds therefore distinguish the hidden distribution inside A.","For strictly positive fine and coarse likelihoods, and at least two coarse blocks, the manuscript proves the all-prior criterion\n\\[\nC\\mathcal B_\\lambda(p)=\\mathcal B_{\\bar\\lambda}(Cp)\\quad\\hbox{for every }p\n\\quad\\Longleftrightarrow\\quad\n\\lambda_i=c\\bar\\lambda_a\\quad(i\\in I_a)\n\\]\nfor one global $c>0$. The global scale is invisible after normalization. A different arbitrary scale for each packet is not invisible: it changes posterior odds.","If all states are merged into a single packet, every normalized posterior coarsens to $(1)$. That degenerate commutation holds for every likelihood and cannot identify any within-packet information. It is explicitly excluded from the nontrivial criterion.\n\\takeaway{A prior-dependent effective likelihood can always summarize the update. A fixed likelihood that works for every prior is a much stronger requirement.}","\\vpage{11}{The exact information-loss vector}{BAYESIAN DEFECT / MORE THAN AN ERROR BAR}{Traditional Part I: Theorems 8.1 and 8.4, and Corollary 8.6.}\nFor a prior $p$, let $P=Cp$ and let $r_a(p)$ be the conditional average of the fine likelihood inside packet $a$. For the uniform prior and $\\lambda=(4,1,1,1)$,\n\\[\nP=(1/2,1/2)^{\\mathsf T},\\qquad r=(5/2,1)^{\\mathsf T}.\n\\]\nThe exact coarse posterior is $\\mathcal B_r(P)=(5/7,2/7)^{\\mathsf T}$. Suppose instead that the declared coarse likelihood is $\\bar\\lambda=(1,1)$.\n\\vfig{loss_vector}\n\\platecaption{The posterior error lies along the mass-preserving direction of the two-state simplex. Its entries are opposite because both endpoints have total mass one.}\n\\head{Compute the factorization, including the null direction}\nHere $\\bar b=\\mathcal B_{\\bar\\lambda}(P)=P$, $\\delta=r-\\bar\\lambda=(3/2,0)^{\\mathsf T}$, and","\\vpage{11}{The exact information-loss vector}{BAYESIAN DEFECT / MORE THAN AN ERROR BAR}{Traditional Part I: Theorems 8.1 and 8.4, and Corollary 8.6.}\nFor a prior $p$, let $P=Cp$ and let $r_a(p)$ be the conditional average of the fine likelihood inside packet $a$. For the uniform prior and $\\lambda=(4,1,1,1)$,\n\\[\nP=(1/2,1/2)^{\\mathsf T},\\qquad r=(5/2,1)^{\\mathsf T}.\n\\]\nThe exact coarse posterior is $\\mathcal B_r(P)=(5/7,2/7)^{\\mathsf T}$. Suppose instead that the declared coarse likelihood is $\\bar\\lambda=(1,1)$.\n\\vfig{loss_vector}\n\\platecaption{The posterior error lies along the mass-preserving direction of the two-state simplex. Its entries are opposite because both endpoints have total mass one.}\n\\head{Compute the factorization, including the null direction}\nHere $\\bar b=\\mathcal B_{\\bar\\lambda}(P)=P$, $\\delta=r-\\bar\\lambda=(3/2,0)^{\\mathsf T}$, and $Z_r=7/4$. H","\\vpage{11}{The exact information-loss vector}{BAYESIAN DEFECT / MORE THAN AN ERROR BAR}{Traditional Part I: Theorems 8.1 and 8.4, and Corollary 8.6.}\nFor a prior $p$, let $P=Cp$ and let $r_a(p)$ be the conditional average of the fine likelihood inside packet $a$. For the uniform prior and $\\lambda=(4,1,1,1)$,\n\\[\nP=(1/2,1/2)^{\\mathsf T},\\qquad r=(5/2,1)^{\\mathsf T}.\n\\]\nThe exact coarse posterior is $\\mathcal B_r(P)=(5/7,2/7)^{\\mathsf T}$. Suppose instead that the declared coarse likelihood is $\\bar\\lambda=(1,1)$.\n\\vfig{loss_vector}\n\\platecaption{The posterior error lies along the mass-preserving direction of the two-state simplex. Its entries are opposite because both endpoints have total mass one.}\n\\head{Compute the factorization, including the null direction}\nHere $\\bar b=\\mathcal B_{\\bar\\lambda}(P)=P$, $\\delta=r-\\bar\\lambda=(3/2,0)^{\\mathsf T}$, and $Z_r=7/4$. Hence\n\\[\n\\frac1{Z_","\\vpage{11}{The exact information-loss vector}{BAYESIAN DEFECT / MORE THAN AN ERROR BAR}{Traditional Part I: Theorems 8.1 and 8.4, and Corollary 8.6.}\nFor a prior $p$, let $P=Cp$ and let $r_a(p)$ be the conditional average of the fine likelihood inside packet $a$. For the uniform prior and $\\lambda=(4,1,1,1)$,\n\\[\nP=(1/2,1/2)^{\\mathsf T},\\qquad r=(5/2,1)^{\\mathsf T}.\n\\]\nThe exact coarse posterior is $\\mathcal B_r(P)=(5/7,2/7)^{\\mathsf T}$. Suppose instead that the declared coarse likelihood is $\\bar\\lambda=(1,1)$.\n\\vfig{loss_vector}\n\\platecaption{The posterior error lies along the mass-preserving direction of the two-state simplex. Its entries are opposite because both endpoints have total mass one.}\n\\head{Compute the factorization, including the null direction}\nHere $\\bar b=\\mathcal B_{\\bar\\lambda}(P)=P$, $\\delta=r-\\bar\\lambda=(3/2,0)^{\\mathsf T}$, and $Z_r=7/4$. Hence\n\\[\n\\frac1{Z_r}\\left(\\diag(P)-\\bar bP^{\\mathsf T}\\right)\\delta\n=\\frac4{7}\\begin{p","\\vpage{11}{The exact information-loss vector}{BAYESIAN DEFECT / MORE THAN AN ERROR BAR}{Traditional Part I: Theorems 8.1 and 8.4, and Corollary 8.6.}\nFor a prior $p$, let $P=Cp$ and let $r_a(p)$ be the conditional average of the fine likelihood inside packet $a$. For the uniform prior and $\\lambda=(4,1,1,1)$,\n\\[\nP=(1/2,1/2)^{\\mathsf T},\\qquad r=(5/2,1)^{\\mathsf T}.\n\\]\nThe exact coarse posterior is $\\mathcal B_r(P)=(5/7,2/7)^{\\mathsf T}$. Suppose instead that the declared coarse likelihood is $\\bar\\lambda=(1,1)$.\n\\vfig{loss_vector}\n\\platecaption{The posterior error lies along the mass-preserving direction of the two-state simplex. Its entries are opposite because both endpoints have total mass one.}\n\\head{Compute the factorization, including the null direction}\nHere $\\bar b=\\mathcal B_{\\bar\\lambda}(P)=P$, $\\delta=r-\\bar\\lambda=(3/2,0)^{\\mathsf T}$, and $Z_r=7/4$. Hence\n\\[\n\\frac1{Z_r}\\left(\\diag(P)-\\bar bP^{\\mathsf T}\\right)\\delta\n=\\frac4{7}\\begin{pmatrix}1/4&-1/4","page{11}{The exact information-loss vector}{BAYESIAN DEFECT / MORE THAN AN ERROR BAR}{Traditional Part I: Theorems 8.1 and 8.4, and Corollary 8.6.}\nFor a prior $p$, let $P=Cp$ and let $r_a(p)$ be the conditional average of the fine likelihood inside packet $a$. For the uniform prior and $\\lambda=(4,1,1,1)$,\n\\[\nP=(1/2,1/2)^{\\mathsf T},\\qquad r=(5/2,1)^{\\mathsf T}.\n\\]\nThe exact coarse posterior is $\\mathcal B_r(P)=(5/7,2/7)^{\\mathsf T}$. Suppose instead that the declared coarse likelihood is $\\bar\\lambda=(1,1)$.\n\\vfig{loss_vector}\n\\platecaption{The posterior error lies along the mass-preserving direction of the two-state simplex. Its entries are opposite because both endpoints have total mass one.}\n\\head{Compute the factorization, including the null direction}\nHere $\\bar b=\\mathcal B_{\\bar\\lambda}(P)=P$, $\\delta=r-\\bar\\lambda=(3/2,0)^{\\mathsf T}$, and $Z_r=7/4$. Hence\n\\[\n\\frac1{Z_r}\\left(\\diag(P)-\\bar bP^{\\mathsf T}\\right)\\delta\n=\\frac4{7}\\begin{pmatrix}1/4&-1/4","l Part I: Theorems 8.1 and 8.4, and Corollary 8.6.}\nFor a prior $p$, let $P=Cp$ and let $r_a(p)$ be the conditional average of the fine likelihood inside packet $a$. For the uniform prior and $\\lambda=(4,1,1,1)$,\n\\[\nP=(1/2,1/2)^{\\mathsf T},\\qquad r=(5/2,1)^{\\mathsf T}.\n\\]\nThe exact coarse posterior is $\\mathcal B_r(P)=(5/7,2/7)^{\\mathsf T}$. Suppose instead that the declared coarse likelihood is $\\bar\\lambda=(1,1)$.\n\\vfig{loss_vector}\n\\platecaption{The posterior error lies along the mass-preserving direction of the two-state simplex. Its entries are opposite because both endpoints have total mass one.}\n\\head{Compute the factorization, including the null direction}\nHere $\\bar b=\\mathcal B_{\\bar\\lambda}(P)=P$, $\\delta=r-\\bar\\lambda=(3/2,0)^{\\mathsf T}$, and $Z_r=7/4$. Hence\n\\[\n\\frac1{Z_r}\\left(\\diag(P)-\\bar bP^{\\mathsf T}\\right)\\delta\n=\\frac4{7}\\begin{pmatrix}1/4&-1/4","act coarse posterior is $\\mathcal B_r(P)=(5/7,2/7)^{\\mathsf T}$. Suppose instead that the declared coarse likelihood is $\\bar\\lambda=(1,1)$.\n\\vfig{loss_vector}\n\\platecaption{The posterior error lies along the mass-preserving direction of the two-state simplex. Its entries are opposite because both endpoints have total mass one.}\n\\head{Compute the factorization, including the null direction}\nHere $\\bar b=\\mathcal B_{\\bar\\lambda}(P)=P$, $\\delta=r-\\bar\\lambda=(3/2,0)^{\\mathsf T}$, and $Z_r=7/4$. Hence\n\\[\n\\frac1{Z_r}\\left(\\diag(P)-\\bar bP^{\\mathsf T}\\right)\\delta\n=\\frac4{7}\\begin{pmatrix}1/4&-1/4","=(5/7,2/7)^{\\mathsf T}$. Suppose instead that the declared coarse likelihood is $\\bar\\lambda=(1,1)$.\n\\vfig{loss_vector}\n\\platecaption{The posterior error lies along the mass-preserving direction of the two-state simplex. Its entries are opposite because both endpoints have total mass one.}\n\\head{Compute the factorization, including the null direction}\nHere $\\bar b=\\mathcal B_{\\bar\\lambda}(P)=P$, $\\delta=r-\\bar\\lambda=(3/2,0)^{\\mathsf T}$, and $Z_r=7/4$. Hence\n\\[\n\\frac1{Z_r}\\left(\\diag(P)-\\bar bP^{\\mathsf T}\\right)\\delta\n=\\frac4{7}\\begin{pmatrix}1/4&-1/4","e declared coarse likelihood is $\\bar\\lambda=(1,1)$.\n\\vfig{loss_vector}\n\\platecaption{The posterior error lies along the mass-preserving direction of the two-state simplex. Its entries are opposite because both endpoints have total mass one.}\n\\head{Compute the factorization, including the null direction}\nHere $\\bar b=\\mathcal B_{\\bar\\lambda}(P)=P$, $\\delta=r-\\bar\\lambda=(3/2,0)^{\\mathsf T}$, and $Z_r=7/4$. Hence\n\\[\n\\frac1{Z_r}\\left(\\diag(P)-\\bar bP^{\\mathsf T}\\right)\\delta\n=\\frac4{7}\\begin{pmatrix}1/4&-1/4","likelihood is $\\bar\\lambda=(1,1)$.\n\\vfig{loss_vector}\n\\platecaption{The posterior error lies along the mass-preserving direction of the two-state simplex. Its entries are opposite because both endpoints have total mass one.}\n\\head{Compute the factorization, including the null direction}\nHere $\\bar b=\\mathcal B_{\\bar\\lambda}(P)=P$, $\\delta=r-\\bar\\lambda=(3/2,0)^{\\mathsf T}$, and $Z_r=7/4$. Hence\n\\[\n\\frac1{Z_r}\\left(\\diag(P)-\\bar bP^{\\mathsf T}\\right)\\delta\n=\\frac4{7}\\begin{pmatrix}1/4&-1/4\\\\-1/4&1/4\\end{pmatrix}\n\\begin{pmatrix}3/2\\\\0\\end{pmatrix}\n=\\begin{pmatrix}3/14\\\\-3/14\\end{pmatrix}.\n\\]\nThis is the actual fine-to-coarse discrepancy, not a fitted approximation. The matrix kills $\\bar\\lambda$, recording the common-scale null direction. Its $\\ell^1$ error is $3/7$.","-3/14\\end{pmatrix}.\n\\]\nThis is the actual fine-to-coarse discrepancy, not a fitted approximation. The matrix kills $\\bar\\lambda$, recording the common-scale null direction. Its $\\ell^1$ error is $3/7$.","For a family of likelihoods, states may be grouped by equality of their \\emph{whole likelihood profile}. The family $(3,3,1,1)$ and $(1,2,1,2)$ separates all four states even though either likelihood alone permits a coarser grouping. The common refinement is therefore the discrete partition. The theorem's one-packet exception still remains.\n\\takeaway{The defect factors through exactly the conditional evidence information missing from the declared coarse model. A new observation family can require a finer partition.}","\\vpage{12}{A nonlinear semigroup of odds}{BAYESIAN FLOW / COMPOSITION WITHOUT LINEARITY}{Traditional Part I: Theorems 7.1 and 9.1. Likelihood composition and the replicator field.}\nLet the score vector be $g=(1,0)$ and start at $(1/2,1/2)$. The likelihood at time $t$ is $(e^t,1)$, so the first posterior coordinate is\n\\[\np_0(t)=\\frac{e^t}{1+e^t},\\qquad p_1(t)=\\frac1{1+e^t}.\n\\]\nThe odds $p_0/p_1$ grow exponentially, while the probabilities remain inside the simplex. At $t=\\log 2$ the posterior is exactly $(2/3,1/3)$.\n\\vfig{odds}\n\\platecaption{The solid and dashed curves sum to one. The curvature is caused by the shared evidence denominator, not by a nonlinear likelihood multiplication law.}\n\\head{Two properties that must not be conflated}\nIf $a>0$ multiplies the odds, the first-coordinate map is\n\\[\n\\phi_a(x)=\\frac{ax}{1+(a-1)x}.\n\\]\nDirect algebra gives $\\phi_a(\\phi_b(x))=\\phi_{ab}(x)$. Thus $a=e^t$ giv","\\vpage{12}{A nonlinear semigroup of odds}{BAYESIAN FLOW / COMPOSITION WITHOUT LINEARITY}{Traditional Part I: Theorems 7.1 and 9.1. Likelihood composition and the replicator field.}\nLet the score vector be $g=(1,0)$ and start at $(1/2,1/2)$. The likelihood at time $t$ is $(e^t,1)$, so the first posterior coordinate is\n\\[\np_0(t)=\\frac{e^t}{1+e^t},\\qquad p_1(t)=\\frac1{1+e^t}.\n\\]\nThe odds $p_0/p_1$ grow exponentially, while the probabilities remain inside the simplex. At $t=\\log 2$ the posterior is exactly $(2/3,1/3)$.\n\\vfig{odds}\n\\platecaption{The solid and dashed curves sum to one. The curvature is caused by the shared evidence denominator, not by a nonlinear likelihood multiplication law.}\n\\head{Two properties that must not be conflated}\nIf $a>0$ multiplies the odds, the first-coordinate map is\n\\[\n\\phi_a(x)=\\frac{ax}{1+(a-1)x}.\n\\]\nDirect algebra gives $\\phi_a(\\phi_b(x))=\\phi_{ab}(x)$. Thus $a=e^t$ gives a nonlinear semigroup.","\\vpage{12}{A nonlinear semigroup of odds}{BAYESIAN FLOW / COMPOSITION WITHOUT LINEARITY}{Traditional Part I: Theorems 7.1 and 9.1. Likelihood composition and the replicator field.}\nLet the score vector be $g=(1,0)$ and start at $(1/2,1/2)$. The likelihood at time $t$ is $(e^t,1)$, so the first posterior coordinate is\n\\[\np_0(t)=\\frac{e^t}{1+e^t},\\qquad p_1(t)=\\frac1{1+e^t}.\n\\]\nThe odds $p_0/p_1$ grow exponentially, while the probabilities remain inside the simplex. At $t=\\log 2$ the posterior is exactly $(2/3,1/3)$.\n\\vfig{odds}\n\\platecaption{The solid and dashed curves sum to one. The curvature is caused by the shared evidence denominator, not by a nonlinear likelihood multiplication law.}\n\\head{Two properties that must not be conflated}\nIf $a>0$ multiplies the odds, the first-coordinate map is\n\\[\n\\phi_a(x)=\\frac{ax}{1+(a-1)x}.\n\\]\nDirect algebra gives $\\phi_a(\\phi_b(x))=\\phi_{ab}(x)$. Thus $a=e^t$ gives a nonlinear semigroup. Nevertheless, $\\phi_3(1/2)=","\\vpage{12}{A nonlinear semigroup of odds}{BAYESIAN FLOW / COMPOSITION WITHOUT LINEARITY}{Traditional Part I: Theorems 7.1 and 9.1. Likelihood composition and the replicator field.}\nLet the score vector be $g=(1,0)$ and start at $(1/2,1/2)$. The likelihood at time $t$ is $(e^t,1)$, so the first posterior coordinate is\n\\[\np_0(t)=\\frac{e^t}{1+e^t},\\qquad p_1(t)=\\frac1{1+e^t}.\n\\]\nThe odds $p_0/p_1$ grow exponentially, while the probabilities remain inside the simplex. At $t=\\log 2$ the posterior is exactly $(2/3,1/3)$.\n\\vfig{odds}\n\\platecaption{The solid and dashed curves sum to one. The curvature is caused by the shared evidence denominator, not by a nonlinear likelihood multiplication law.}\n\\head{Two properties that must not be conflated}\nIf $a>0$ multiplies the odds, the first-coordinate map is\n\\[\n\\phi_a(x)=\\frac{ax}{1+(a-1)x}.\n\\]\nDirect algebra gives $\\phi_a(\\phi_b(x))=\\phi_{ab}(x)$. Thus $a=e^t$ gives a nonlinear semigroup. Nevertheless, $\\phi_3(1/2)=3/4$, while t","\\vpage{12}{A nonlinear semigroup of odds}{BAYESIAN FLOW / COMPOSITION WITHOUT LINEARITY}{Traditional Part I: Theorems 7.1 and 9.1. Likelihood composition and the replicator field.}\nLet the score vector be $g=(1,0)$ and start at $(1/2,1/2)$. The likelihood at time $t$ is $(e^t,1)$, so the first posterior coordinate is\n\\[\np_0(t)=\\frac{e^t}{1+e^t},\\qquad p_1(t)=\\frac1{1+e^t}.\n\\]\nThe odds $p_0/p_1$ grow exponentially, while the probabilities remain inside the simplex. At $t=\\log 2$ the posterior is exactly $(2/3,1/3)$.\n\\vfig{odds}\n\\platecaption{The solid and dashed curves sum to one. The curvature is caused by the shared evidence denominator, not by a nonlinear likelihood multiplication law.}\n\\head{Two properties that must not be conflated}\nIf $a>0$ multiplies the odds, the first-coordinate map is\n\\[\n\\phi_a(x)=\\frac{ax}{1+(a-1)x}.\n\\]\nDirect algebra gives $\\phi_a(\\phi_b(x))=\\phi_{ab}(x)$. Thus $a=e^t$ gives a nonlinear semigroup. Nevertheless, $\\phi_3(1/2)=3/4$, while the mean of $\\phi_3(0)$ and $\\phi_3(1)$ is $1/2$. Superposition fails.","HOUT LINEARITY}{Traditional Part I: Theorems 7.1 and 9.1. Likelihood composition and the replicator field.}\nLet the score vector be $g=(1,0)$ and start at $(1/2,1/2)$. The likelihood at time $t$ is $(e^t,1)$, so the first posterior coordinate is\n\\[\np_0(t)=\\frac{e^t}{1+e^t},\\qquad p_1(t)=\\frac1{1+e^t}.\n\\]\nThe odds $p_0/p_1$ grow exponentially, while the probabilities remain inside the simplex. At $t=\\log 2$ the posterior is exactly $(2/3,1/3)$.\n\\vfig{odds}\n\\platecaption{The solid and dashed curves sum to one. The curvature is caused by the shared evidence denominator, not by a nonlinear likelihood multiplication law.}\n\\head{Two properties that must not be conflated}\nIf $a>0$ multiplies the odds, the first-coordinate map is\n\\[\n\\phi_a(x)=\\frac{ax}{1+(a-1)x}.\n\\]\nDirect algebra gives $\\phi_a(\\phi_b(x))=\\phi_{ab}(x)$. Thus $a=e^t$ gives a nonlinear semigroup. Nevertheless, $\\phi_3(1/2)=3/4$, while the mean of $\\phi_3(0)$ and $\\phi_3(1)$ is $1/2$. Superposition fails.","Theorems 7.1 and 9.1. Likelihood composition and the replicator field.}\nLet the score vector be $g=(1,0)$ and start at $(1/2,1/2)$. The likelihood at time $t$ is $(e^t,1)$, so the first posterior coordinate is\n\\[\np_0(t)=\\frac{e^t}{1+e^t},\\qquad p_1(t)=\\frac1{1+e^t}.\n\\]\nThe odds $p_0/p_1$ grow exponentially, while the probabilities remain inside the simplex. At $t=\\log 2$ the posterior is exactly $(2/3,1/3)$.\n\\vfig{odds}\n\\platecaption{The solid and dashed curves sum to one. The curvature is caused by the shared evidence denominator, not by a nonlinear likelihood multiplication law.}\n\\head{Two properties that must not be conflated}\nIf $a>0$ multiplies the odds, the first-coordinate map is\n\\[\n\\phi_a(x)=\\frac{ax}{1+(a-1)x}.\n\\]\nDirect algebra gives $\\phi_a(\\phi_b(x))=\\phi_{ab}(x)$. Thus $a=e^t$ gives a nonlinear semigroup. Nevertheless, $\\phi_3(1/2)=3/4$, while the mean of $\\phi_3(0)$ and $\\phi_3(1)$ is $1/2$. Superposition fails.","ad p_1(t)=\\frac1{1+e^t}.\n\\]\nThe odds $p_0/p_1$ grow exponentially, while the probabilities remain inside the simplex. At $t=\\log 2$ the posterior is exactly $(2/3,1/3)$.\n\\vfig{odds}\n\\platecaption{The solid and dashed curves sum to one. The curvature is caused by the shared evidence denominator, not by a nonlinear likelihood multiplication law.}\n\\head{Two properties that must not be conflated}\nIf $a>0$ multiplies the odds, the first-coordinate map is\n\\[\n\\phi_a(x)=\\frac{ax}{1+(a-1)x}.\n\\]\nDirect algebra gives $\\phi_a(\\phi_b(x))=\\phi_{ab}(x)$. Thus $a=e^t$ gives a nonlinear semigroup. Nevertheless, $\\phi_3(1/2)=3/4$, while the mean of $\\phi_3(0)$ and $\\phi_3(1)$ is $1/2$. Superposition fails.","nentially, while the probabilities remain inside the simplex. At $t=\\log 2$ the posterior is exactly $(2/3,1/3)$.\n\\vfig{odds}\n\\platecaption{The solid and dashed curves sum to one. The curvature is caused by the shared evidence denominator, not by a nonlinear likelihood multiplication law.}\n\\head{Two properties that must not be conflated}\nIf $a>0$ multiplies the odds, the first-coordinate map is\n\\[\n\\phi_a(x)=\\frac{ax}{1+(a-1)x}.\n\\]\nDirect algebra gives $\\phi_a(\\phi_b(x))=\\phi_{ab}(x)$. Thus $a=e^t$ gives a nonlinear semigroup. Nevertheless, $\\phi_3(1/2)=3/4$, while the mean of $\\phi_3(0)$ and $\\phi_3(1)$ is $1/2$. Superposition fails.","lex. At $t=\\log 2$ the posterior is exactly $(2/3,1/3)$.\n\\vfig{odds}\n\\platecaption{The solid and dashed curves sum to one. The curvature is caused by the shared evidence denominator, not by a nonlinear likelihood multiplication law.}\n\\head{Two properties that must not be conflated}\nIf $a>0$ multiplies the odds, the first-coordinate map is\n\\[\n\\phi_a(x)=\\frac{ax}{1+(a-1)x}.\n\\]\nDirect algebra gives $\\phi_a(\\phi_b(x))=\\phi_{ab}(x)$. Thus $a=e^t$ gives a nonlinear semigroup. Nevertheless, $\\phi_3(1/2)=3/4$, while the mean of $\\phi_3(0)$ and $\\phi_3(1)$ is $1/2$. Superposition fails.","ctly $(2/3,1/3)$.\n\\vfig{odds}\n\\platecaption{The solid and dashed curves sum to one. The curvature is caused by the shared evidence denominator, not by a nonlinear likelihood multiplication law.}\n\\head{Two properties that must not be conflated}\nIf $a>0$ multiplies the odds, the first-coordinate map is\n\\[\n\\phi_a(x)=\\frac{ax}{1+(a-1)x}.\n\\]\nDirect algebra gives $\\phi_a(\\phi_b(x))=\\phi_{ab}(x)$. Thus $a=e^t$ gives a nonlinear semigroup. Nevertheless, $\\phi_3(1/2)=3/4$, while the mean of $\\phi_3(0)$ and $\\phi_3(1)$ is $1/2$. Superposition fails.","id and dashed curves sum to one. The curvature is caused by the shared evidence denominator, not by a nonlinear likelihood multiplication law.}\n\\head{Two properties that must not be conflated}\nIf $a>0$ multiplies the odds, the first-coordinate map is\n\\[\n\\phi_a(x)=\\frac{ax}{1+(a-1)x}.\n\\]\nDirect algebra gives $\\phi_a(\\phi_b(x))=\\phi_{ab}(x)$. Thus $a=e^t$ gives a nonlinear semigroup. Nevertheless, $\\phi_3(1/2)=3/4$, while the mean of $\\phi_3(0)$ and $\\phi_3(1)$ is $1/2$. Superposition fails.","curvature is caused by the shared evidence denominator, not by a nonlinear likelihood multiplication law.}\n\\head{Two properties that must not be conflated}\nIf $a>0$ multiplies the odds, the first-coordinate map is\n\\[\n\\phi_a(x)=\\frac{ax}{1+(a-1)x}.\n\\]\nDirect algebra gives $\\phi_a(\\phi_b(x))=\\phi_{ab}(x)$. Thus $a=e^t$ gives a nonlinear semigroup. Nevertheless, $\\phi_3(1/2)=3/4$, while the mean of $\\phi_3(0)$ and $\\phi_3(1)$ is $1/2$. Superposition fails.","sed by the shared evidence denominator, not by a nonlinear likelihood multiplication law.}\n\\head{Two properties that must not be conflated}\nIf $a>0$ multiplies the odds, the first-coordinate map is\n\\[\n\\phi_a(x)=\\frac{ax}{1+(a-1)x}.\n\\]\nDirect algebra gives $\\phi_a(\\phi_b(x))=\\phi_{ab}(x)$. Thus $a=e^t$ gives a nonlinear semigroup. Nevertheless, $\\phi_3(1/2)=3/4$, while the mean of $\\phi_3(0)$ and $\\phi_3(1)$ is $1/2$. Superposition fails.","ed evidence denominator, not by a nonlinear likelihood multiplication law.}\n\\head{Two properties that must not be conflated}\nIf $a>0$ multiplies the odds, the first-coordinate map is\n\\[\n\\phi_a(x)=\\frac{ax}{1+(a-1)x}.\n\\]\nDirect algebra gives $\\phi_a(\\phi_b(x))=\\phi_{ab}(x)$. Thus $a=e^t$ gives a nonlinear semigroup. Nevertheless, $\\phi_3(1/2)=3/4$, while the mean of $\\phi_3(0)$ and $\\phi_3(1)$ is $1/2$. Superposition fails.","Differentiation of the normalized flow yields the centered score equation\n\\[\n\\dot p_i=\\bigl(g_i-\\langle g,p\\rangle\\bigr)p_i.\n\\]\nFor the displayed model, $\\dot p_0=p_0(1-p_0)$. The negative average-score term is what keeps the total derivative zero. Omitting the derivative of the normalizer would not preserve mass.","\\vpage{13}{Prediction and observation do not commute}{HIDDEN MARKOV FILTERING / TIME ORDER}{Traditional Part I: Proposition 9.3. Linear prediction followed by a positive-evidence update.}\nUse the two-state kernel $K_2$ from \\pagerefV{2}, the initial distribution $e_0$, and likelihood $(1,1/2)$. Predicting first gives $(3/4,1/4)$. Its likelihood-weighted vector is $(3/4,1/8)$, with evidence $7/8$, so the posterior is $(6/7,1/7)$.","and observation do not commute}{HIDDEN MARKOV FILTERING / TIME ORDER}{Traditional Part I: Proposition 9.3. Linear prediction followed by a positive-evidence update.}\nUse the two-state kernel $K_2$ from \\pagerefV{2}, the initial distribution $e_0$, and likelihood $(1,1/2)$. Predicting first gives $(3/4,1/4)$. Its likelihood-weighted vector is $(3/4,1/8)$, with evidence $7/8$, so the posterior is $(6/7,1/7)$.","Updating first leaves $e_0$ unchanged. Predicting afterward gives $(3/4,1/4)$ instead. The order of operations expresses which time the observation concerns.\n\\vfig{filter_order}\n\\platecaption{Both routes produce valid probability vectors. Their difference is therefore not a normalization bug. It is a different conditioning problem.}\n\\head{Keep an unnormalized path alongside the filter}\nFor a sequence of observations, let $D_r=\\diag(\\lambda^{(r)})$. The forward recursion is\n\\[\n\\widehat p_r=Kp_{r-1},\\qquad p_r=D_r\\widehat p_r/Z_r.\n\\]\nIts unnormalized companion is\n\\[\nv_r=D_rK\\cdots D_1Kp_0,\n\\qquad p_r=\\frac{v_r}{\\ones^{\\mathsf T}v_r},\n\\qquad \\ones^{\\mathsf T}v_r=\\prod_{j=1}^r Z_j.\n\\]\nThe product of evidences records the","Updating first leaves $e_0$ unchanged. Predicting afterward gives $(3/4,1/4)$ instead. The order of operations expresses which time the observation concerns.\n\\vfig{filter_order}\n\\platecaption{Both routes produce valid probability vectors. Their difference is therefore not a normalization bug. It is a different conditioning problem.}\n\\head{Keep an unnormalized path alongside the filter}\nFor a sequence of observations, let $D_r=\\diag(\\lambda^{(r)})$. The forward recursion is\n\\[\n\\widehat p_r=Kp_{r-1},\\qquad p_r=D_r\\widehat p_r/Z_r.\n\\]\nIts unnormalized companion is\n\\[\nv_r=D_rK\\cdots D_1Kp_0,\n\\qquad p_r=\\frac{v_r}{\\ones^{\\mathsf T}v_r},\n\\qquad \\ones^{\\mathsf T}v_r=\\prod_{j=1}^r Z_j.\n\\]\nThe product of evidences records the normalization lost at each individual step. The f","$ unchanged. Predicting afterward gives $(3/4,1/4)$ instead. The order of operations expresses which time the observation concerns.\n\\vfig{filter_order}\n\\platecaption{Both routes produce valid probability vectors. Their difference is therefore not a normalization bug. It is a different conditioning problem.}\n\\head{Keep an unnormalized path alongside the filter}\nFor a sequence of observations, let $D_r=\\diag(\\lambda^{(r)})$. The forward recursion is\n\\[\n\\widehat p_r=Kp_{r-1},\\qquad p_r=D_r\\widehat p_r/Z_r.\n\\]\nIts unnormalized companion is\n\\[\nv_r=D_rK\\cdots D_1Kp_0,\n\\qquad p_r=\\frac{v_r}{\\ones^{\\mathsf T}v_r},\n\\qquad \\ones^{\\mathsf T}v_r=\\prod_{j=1}^r Z_j.\n\\]\nThe product of evidences records the normalization lost at each individual step. The four-state computation on \\pagerefV{26} checks both paths independently, including their coarse versions.","stead. The order of operations expresses which time the observation concerns.\n\\vfig{filter_order}\n\\platecaption{Both routes produce valid probability vectors. Their difference is therefore not a normalization bug. It is a different conditioning problem.}\n\\head{Keep an unnormalized path alongside the filter}\nFor a sequence of observations, let $D_r=\\diag(\\lambda^{(r)})$. The forward recursion is\n\\[\n\\widehat p_r=Kp_{r-1},\\qquad p_r=D_r\\widehat p_r/Z_r.\n\\]\nIts unnormalized companion is\n\\[\nv_r=D_rK\\cdots D_1Kp_0,\n\\qquad p_r=\\frac{v_r}{\\ones^{\\mathsf T}v_r},\n\\qquad \\ones^{\\mathsf T}v_r=\\prod_{j=1}^r Z_j.\n\\]\nThe product of evidences records the normalization lost at each individual step. The four-state computation on \\pagerefV{26} checks both paths independently, including their coarse versions.","_order}\n\\platecaption{Both routes produce valid probability vectors. Their difference is therefore not a normalization bug. It is a different conditioning problem.}\n\\head{Keep an unnormalized path alongside the filter}\nFor a sequence of observations, let $D_r=\\diag(\\lambda^{(r)})$. The forward recursion is\n\\[\n\\widehat p_r=Kp_{r-1},\\qquad p_r=D_r\\widehat p_r/Z_r.\n\\]\nIts unnormalized companion is\n\\[\nv_r=D_rK\\cdots D_1Kp_0,\n\\qquad p_r=\\frac{v_r}{\\ones^{\\mathsf T}v_r},\n\\qquad \\ones^{\\mathsf T}v_r=\\prod_{j=1}^r Z_j.\n\\]\nThe product of evidences records the normalization lost at each individual step. The four-state computation on \\pagerefV{26} checks both paths independently, including their coarse versions.","Here $D_rK$ means an observation after prediction. In the Feynman--Kac example on \\pagerefV{20}, departure-state weighting has forward operator $K D_w$. Both conventions are legitimate once their meanings are fixed, but they cannot be substituted for one another.\n\\takeaway{Write the chronological sequence before multiplying matrices. A product can be normalized correctly and still answer the wrong time-ordered question.}","\\vpage{14}{Conjugacy tracks exponents}{BETA MODEL / POSTERIOR AND EVIDENCE}{Traditional Part I: Example 9.2. A finite-parameter conjugate family, not a universal posterior model.}\nFor a small exact example, take the $\\mathrm{Beta}(2,3)$ density\n\\[\n\\rho(\\theta)=12\\theta(1-\\theta)^2,\\qquad0\\le\\theta\\le1.\n\\]\nObserve a specified ordered sequence with two successes and one failure. Its likelihood is $\\theta^2(1-\\theta)$. Multiplication adds exponents, so the posterior belongs to $\\mathrm{Beta}(4,4)$ with density $140\\theta^3(1-\\theta)^3$.\n\\vfig{beta}\n\\platecaption{The dashed prior is asymmetric. The solid posterior is symmetric with mean $1/2$. Polynomial integration verifies the normalizers exactly; the curves are sampled drawings of those polynomials.}\n\\head{The observation convention changes the evidence}\nThe integral of the unnormalized posterior is\n\\[\n\\int_0^1 12\\theta^3(1-\\theta)^3\\,d\\theta=\\frac3{35}.\n\\]\nDividing by $3/","\\vpage{14}{Conjugacy tracks exponents}{BETA MODEL / POSTERIOR AND EVIDENCE}{Traditional Part I: Example 9.2. A finite-parameter conjugate family, not a universal posterior model.}\nFor a small exact example, take the $\\mathrm{Beta}(2,3)$ density\n\\[\n\\rho(\\theta)=12\\theta(1-\\theta)^2,\\qquad0\\le\\theta\\le1.\n\\]\nObserve a specified ordered sequence with two successes and one failure. Its likelihood is $\\theta^2(1-\\theta)$. Multiplication adds exponents, so the posterior belongs to $\\mathrm{Beta}(4,4)$ with density $140\\theta^3(1-\\theta)^3$.\n\\vfig{beta}\n\\platecaption{The dashed prior is asymmetric. The solid posterior is symmetric with mean $1/2$. Polynomial integration verifies the normalizers exactly; the curves are sampled drawings of those polynomials.}\n\\head{The observation convention changes the evidence}\nThe integral of the unnormalized posterior is\n\\[\n\\int_0^1 12\\theta^3(1-\\theta)^3\\,d\\theta=\\frac3{35}.\n\\]\nDividing by $3/35$ gives the displayed normalized density. If the observation is ins","\\vpage{14}{Conjugacy tracks exponents}{BETA MODEL / POSTERIOR AND EVIDENCE}{Traditional Part I: Example 9.2. A finite-parameter conjugate family, not a universal posterior model.}\nFor a small exact example, take the $\\mathrm{Beta}(2,3)$ density\n\\[\n\\rho(\\theta)=12\\theta(1-\\theta)^2,\\qquad0\\le\\theta\\le1.\n\\]\nObserve a specified ordered sequence with two successes and one failure. Its likelihood is $\\theta^2(1-\\theta)$. Multiplication adds exponents, so the posterior belongs to $\\mathrm{Beta}(4,4)$ with density $140\\theta^3(1-\\theta)^3$.\n\\vfig{beta}\n\\platecaption{The dashed prior is asymmetric. The solid posterior is symmetric with mean $1/2$. Polynomial integration verifies the normalizers exactly; the curves are sampled drawings of those polynomials.}\n\\head{The observation convention changes the evidence}\nThe integral of the unnormalized posterior is\n\\[\n\\int_0^1 12\\theta^3(1-\\theta)^3\\,d\\theta=\\frac3{35}.\n\\]\nDividing by $3/35$ gives the displayed normalized density. If the observation is instead the \\emph{count} of two successes in three trials, there are $\\binom32=3$ compatible ordere…","itional Part I: Example 9.2. A finite-parameter conjugate family, not a universal posterior model.}\nFor a small exact example, take the $\\mathrm{Beta}(2,3)$ density\n\\[\n\\rho(\\theta)=12\\theta(1-\\theta)^2,\\qquad0\\le\\theta\\le1.\n\\]\nObserve a specified ordered sequence with two successes and one failure. Its likelihood is $\\theta^2(1-\\theta)$. Multiplication adds exponents, so the posterior belongs to $\\mathrm{Beta}(4,4)$ with density $140\\theta^3(1-\\theta)^3$.\n\\vfig{beta}\n\\platecaption{The dashed prior is asymmetric. The solid posterior is symmetric with mean $1/2$. Polynomial integration verifies the normalizers exactly; the curves are sampled drawings of those polynomials.}\n\\head{The observation convention changes the evidence}\nThe integral of the unnormalized posterior is\n\\[\n\\int_0^1 12\\theta^3(1-\\theta)^3\\,d\\theta=\\frac3{35}.\n\\]\nDividing by $3/35$ gives the displayed normalized density. If the observation is instead the \\emph{count} of two successes in three trials, there are $\\binom32=3$ compatible ordered sequences. The count likelihood is therefore $3\\theta^2(1-\\theta)$ and its evi…","ite-parameter conjugate family, not a universal posterior model.}\nFor a small exact example, take the $\\mathrm{Beta}(2,3)$ density\n\\[\n\\rho(\\theta)=12\\theta(1-\\theta)^2,\\qquad0\\le\\theta\\le1.\n\\]\nObserve a specified ordered sequence with two successes and one failure. Its likelihood is $\\theta^2(1-\\theta)$. Multiplication adds exponents, so the posterior belongs to $\\mathrm{Beta}(4,4)$ with density $140\\theta^3(1-\\theta)^3$.\n\\vfig{beta}\n\\platecaption{The dashed prior is asymmetric. The solid posterior is symmetric with mean $1/2$. Polynomial integration verifies the normalizers exactly; the curves are sampled drawings of those polynomials.}\n\\head{The observation convention changes the evidence}\nThe integral of the unnormalized posterior is\n\\[\n\\int_0^1 12\\theta^3(1-\\theta)^3\\,d\\theta=\\frac3{35}.\n\\]\nDividing by $3/35$ gives the displayed normalized density. If the observation is instead the \\emph{count} of two successes in three trials, there are $\\binom32=3$ compatible ordered sequences. The count likelihood is therefore $3\\theta^2(1-\\theta)$ and its evidence is $9/35$.","ty\n\\[\n\\rho(\\theta)=12\\theta(1-\\theta)^2,\\qquad0\\le\\theta\\le1.\n\\]\nObserve a specified ordered sequence with two successes and one failure. Its likelihood is $\\theta^2(1-\\theta)$. Multiplication adds exponents, so the posterior belongs to $\\mathrm{Beta}(4,4)$ with density $140\\theta^3(1-\\theta)^3$.\n\\vfig{beta}\n\\platecaption{The dashed prior is asymmetric. The solid posterior is symmetric with mean $1/2$. Polynomial integration verifies the normalizers exactly; the curves are sampled drawings of those polynomials.}\n\\head{The observation convention changes the evidence}\nThe integral of the unnormalized posterior is\n\\[\n\\int_0^1 12\\theta^3(1-\\theta)^3\\,d\\theta=\\frac3{35}.\n\\]\nDividing by $3/35$ gives the displayed normalized density. If the observation is instead the \\emph{count} of two successes in three trials, there are $\\binom32=3$ compatible ordered sequences. The count likelihood is therefore $3\\theta^2(1-\\theta)$ and its evidence is $9/35$.","rior belongs to $\\mathrm{Beta}(4,4)$ with density $140\\theta^3(1-\\theta)^3$.\n\\vfig{beta}\n\\platecaption{The dashed prior is asymmetric. The solid posterior is symmetric with mean $1/2$. Polynomial integration verifies the normalizers exactly; the curves are sampled drawings of those polynomials.}\n\\head{The observation convention changes the evidence}\nThe integral of the unnormalized posterior is\n\\[\n\\int_0^1 12\\theta^3(1-\\theta)^3\\,d\\theta=\\frac3{35}.\n\\]\nDividing by $3/35$ gives the displayed normalized density. If the observation is instead the \\emph{count} of two successes in three trials, there are $\\binom32=3$ compatible ordered sequences. The count likelihood is therefore $3\\theta^2(1-\\theta)$ and its evidence is $9/35$.","a)^3$.\n\\vfig{beta}\n\\platecaption{The dashed prior is asymmetric. The solid posterior is symmetric with mean $1/2$. Polynomial integration verifies the normalizers exactly; the curves are sampled drawings of those polynomials.}\n\\head{The observation convention changes the evidence}\nThe integral of the unnormalized posterior is\n\\[\n\\int_0^1 12\\theta^3(1-\\theta)^3\\,d\\theta=\\frac3{35}.\n\\]\nDividing by $3/35$ gives the displayed normalized density. If the observation is instead the \\emph{count} of two successes in three trials, there are $\\binom32=3$ compatible ordered sequences. The count likelihood is therefore $3\\theta^2(1-\\theta)$ and its evidence is $9/35$.","verifies the normalizers exactly; the curves are sampled drawings of those polynomials.}\n\\head{The observation convention changes the evidence}\nThe integral of the unnormalized posterior is\n\\[\n\\int_0^1 12\\theta^3(1-\\theta)^3\\,d\\theta=\\frac3{35}.\n\\]\nDividing by $3/35$ gives the displayed normalized density. If the observation is instead the \\emph{count} of two successes in three trials, there are $\\binom32=3$ compatible ordered sequences. The count likelihood is therefore $3\\theta^2(1-\\theta)$ and its evidence is $9/35$.","lynomials.}\n\\head{The observation convention changes the evidence}\nThe integral of the unnormalized posterior is\n\\[\n\\int_0^1 12\\theta^3(1-\\theta)^3\\,d\\theta=\\frac3{35}.\n\\]\nDividing by $3/35$ gives the displayed normalized density. If the observation is instead the \\emph{count} of two successes in three trials, there are $\\binom32=3$ compatible ordered sequences. The count likelihood is therefore $3\\theta^2(1-\\theta)$ and its evidence is $9/35$.","on changes the evidence}\nThe integral of the unnormalized posterior is\n\\[\n\\int_0^1 12\\theta^3(1-\\theta)^3\\,d\\theta=\\frac3{35}.\n\\]\nDividing by $3/35$ gives the displayed normalized density. If the observation is instead the \\emph{count} of two successes in three trials, there are $\\binom32=3$ compatible ordered sequences. The count likelihood is therefore $3\\theta^2(1-\\theta)$ and its evidence is $9/35$.","The extra factor does not depend on $\\theta$, so it cancels from the posterior. The two experiments thus have the same posterior distribution but different data probabilities. Evidence is not determined by the final normalized density alone.","The exact check also integrates both densities to one and integrates $\\theta$ against the posterior to obtain $1/2$. This demonstrates one conjugate family in elementary terms. It does not show that arbitrary likelihoods preserve that family or that all posterior information can be compressed into two exponents.\n\\takeaway{Conjugacy gives a closed parameter update when a specified likelihood respects a specified family. Preserve the evidence as well as the normalized posterior.}","\\vpage{15}{The modes stay; their clocks change}{FRACTIONAL DIFFUSION / NEW COUPLINGS}{Traditional Part I: Theorem 10.1. A positive self-adjoint finite model and its spectral powers.}\nReturn to the four-state generator on \\pagerefV{4} and put $L=-Q$. It is real symmetric and nonnegative, with orthogonal projectors $P_0,P_1,P_2,P_3$ and eigenvalues $0,2,4,6$. Its square root is\n\\[\nL^{1/2}=\\sqrt2P_1+2P_2+\\sqrt6P_3,\n\\qquad (L^{1/2})^2=L.\n\\]\nBoth heat evolutions retain these modes. Ordinary diffusion multiplies mode $j$ by $e^{-2jt}$; square-root diffusion multiplies it by $e^{-\\sqrt{2j}\\,t}$. The constant mode is unchanged in both.\n\\vfig{fractional}\n\\platecaption{The highest mode decays at rates $6$ and $\\sqrt6$. The graph beside the curves shows the additional couplings in the fractional operator. Curves display the explicit formulas, not an empirical fit.}\n\\head{Fractional does not mean entrywise}\nAlthough $L_{03}=0$, its spectral squa","\\vpage{15}{The modes stay; their clocks change}{FRACTIONAL DIFFUSION / NEW COUPLINGS}{Traditional Part I: Theorem 10.1. A positive self-adjoint finite model and its spectral powers.}\nReturn to the four-state generator on \\pagerefV{4} and put $L=-Q$. It is real symmetric and nonnegative, with orthogonal projectors $P_0,P_1,P_2,P_3$ and eigenvalues $0,2,4,6$. Its square root is\n\\[\nL^{1/2}=\\sqrt2P_1+2P_2+\\sqrt6P_3,\n\\qquad (L^{1/2})^2=L.\n\\]\nBoth heat evolutions retain these modes. Ordinary diffusion multiplies mode $j$ by $e^{-2jt}$; square-root diffusion multiplies it by $e^{-\\sqrt{2j}\\,t}$. The constant mode is unchanged in both.\n\\vfig{fractional}\n\\platecaption{The highest mode decays at rates $6$ and $\\sqrt6$. The graph beside the curves shows the additional couplings in the fractional operator. Curves display the explicit formulas, not an empirical fit.}\n\\head{Fractional does not mean entrywise}\nAlthough $L_{03}=0$, its spectral square root satisfies\n\\[\n(L^{1/2})_{03}=\\frac{\\sqrt6-\\sqrt2-2}{4}<0.\n\\]\nThus $-L^{1/2}$","\\vpage{15}{The modes stay; their clocks change}{FRACTIONAL DIFFUSION / NEW COUPLINGS}{Traditional Part I: Theorem 10.1. A positive self-adjoint finite model and its spectral powers.}\nReturn to the four-state generator on \\pagerefV{4} and put $L=-Q$. It is real symmetric and nonnegative, with orthogonal projectors $P_0,P_1,P_2,P_3$ and eigenvalues $0,2,4,6$. Its square root is\n\\[\nL^{1/2}=\\sqrt2P_1+2P_2+\\sqrt6P_3,\n\\qquad (L^{1/2})^2=L.\n\\]\nBoth heat evolutions retain these modes. Ordinary diffusion multiplies mode $j$ by $e^{-2jt}$; square-root diffusion multiplies it by $e^{-\\sqrt{2j}\\,t}$. The constant mode is unchanged in both.\n\\vfig{fractional}\n\\platecaption{The highest mode decays at rates $6$ and $\\sqrt6$. The graph beside the curves shows the additional couplings in the fractional operator. Curves display the explicit formulas, not an empirical fit.}\n\\head{Fractional does not mean entrywise}\nAlthough $L_{03}=0$, its spectral square root satisfies\n\\[\n(L^{1/2})_{03}=\\frac{\\sqrt6-\\sqrt2-2}{4}<0.\n\\]\nThus $-L^{1/2}$ permits a direct transitio","\\vpage{15}{The modes stay; their clocks change}{FRACTIONAL DIFFUSION / NEW COUPLINGS}{Traditional Part I: Theorem 10.1. A positive self-adjoint finite model and its spectral powers.}\nReturn to the four-state generator on \\pagerefV{4} and put $L=-Q$. It is real symmetric and nonnegative, with orthogonal projectors $P_0,P_1,P_2,P_3$ and eigenvalues $0,2,4,6$. Its square root is\n\\[\nL^{1/2}=\\sqrt2P_1+2P_2+\\sqrt6P_3,\n\\qquad (L^{1/2})^2=L.\n\\]\nBoth heat evolutions retain these modes. Ordinary diffusion multiplies mode $j$ by $e^{-2jt}$; square-root diffusion multiplies it by $e^{-\\sqrt{2j}\\,t}$. The constant mode is unchanged in both.\n\\vfig{fractional}\n\\platecaption{The highest mode decays at rates $6$ and $\\sqrt6$. The graph beside the curves shows the additional couplings in the fractional operator. Curves display the explicit formulas, not an empirical fit.}\n\\head{Fractional does not mean entrywise}\nAlthough $L_{03}=0$, its spectral square root satisfies\n\\[\n(L^{1/2})_{03}=\\frac{\\sqrt6-\\sqrt2-2}{4}<0.\n\\]\nThus $-L^{1/2}$ permits a direct transition between states $0$ and $3$, which were no…",". A positive self-adjoint finite model and its spectral powers.}\nReturn to the four-state generator on \\pagerefV{4} and put $L=-Q$. It is real symmetric and nonnegative, with orthogonal projectors $P_0,P_1,P_2,P_3$ and eigenvalues $0,2,4,6$. Its square root is\n\\[\nL^{1/2}=\\sqrt2P_1+2P_2+\\sqrt6P_3,\n\\qquad (L^{1/2})^2=L.\n\\]\nBoth heat evolutions retain these modes. Ordinary diffusion multiplies mode $j$ by $e^{-2jt}$; square-root diffusion multiplies it by $e^{-\\sqrt{2j}\\,t}$. The constant mode is unchanged in both.\n\\vfig{fractional}\n\\platecaption{The highest mode decays at rates $6$ and $\\sqrt6$. The graph beside the curves shows the additional couplings in the fractional operator. Curves display the explicit formulas, not an empirical fit.}\n\\head{Fractional does not mean entrywise}\nAlthough $L_{03}=0$, its spectral square root satisfies\n\\[\n(L^{1/2})_{03}=\\frac{\\sqrt6-\\sqrt2-2}{4}<0.\n\\]\nThus $-L^{1/2}$ permits a direct transition between states $0$ and $3$, which were not neighbors for $Q$. Taking the square root of each matrix entry would neither produce this value nor square back to $…","itive self-adjoint finite model and its spectral powers.}\nReturn to the four-state generator on \\pagerefV{4} and put $L=-Q$. It is real symmetric and nonnegative, with orthogonal projectors $P_0,P_1,P_2,P_3$ and eigenvalues $0,2,4,6$. Its square root is\n\\[\nL^{1/2}=\\sqrt2P_1+2P_2+\\sqrt6P_3,\n\\qquad (L^{1/2})^2=L.\n\\]\nBoth heat evolutions retain these modes. Ordinary diffusion multiplies mode $j$ by $e^{-2jt}$; square-root diffusion multiplies it by $e^{-\\sqrt{2j}\\,t}$. The constant mode is unchanged in both.\n\\vfig{fractional}\n\\platecaption{The highest mode decays at rates $6$ and $\\sqrt6$. The graph beside the curves shows the additional couplings in the fractional operator. Curves display the explicit formulas, not an empirical fit.}\n\\head{Fractional does not mean entrywise}\nAlthough $L_{03}=0$, its spectral square root satisfies\n\\[\n(L^{1/2})_{03}=\\frac{\\sqrt6-\\sqrt2-2}{4}<0.\n\\]\nThus $-L^{1/2}$ permits a direct transition between states $0$ and $3$, which were not neighbors for $Q$. Taking the square root of each matrix entry would neither produce this value nor square back to $L$.","wers.}\nReturn to the four-state generator on \\pagerefV{4} and put $L=-Q$. It is real symmetric and nonnegative, with orthogonal projectors $P_0,P_1,P_2,P_3$ and eigenvalues $0,2,4,6$. Its square root is\n\\[\nL^{1/2}=\\sqrt2P_1+2P_2+\\sqrt6P_3,\n\\qquad (L^{1/2})^2=L.\n\\]\nBoth heat evolutions retain these modes. Ordinary diffusion multiplies mode $j$ by $e^{-2jt}$; square-root diffusion multiplies it by $e^{-\\sqrt{2j}\\,t}$. The constant mode is unchanged in both.\n\\vfig{fractional}\n\\platecaption{The highest mode decays at rates $6$ and $\\sqrt6$. The graph beside the curves shows the additional couplings in the fractional operator. Curves display the explicit formulas, not an empirical fit.}\n\\head{Fractional does not mean entrywise}\nAlthough $L_{03}=0$, its spectral square root satisfies\n\\[\n(L^{1/2})_{03}=\\frac{\\sqrt6-\\sqrt2-2}{4}<0.\n\\]\nThus $-L^{1/2}$ permits a direct transition between states $0$ and $3$, which were not neighbors for $Q$. Taking the square root of each matrix entry would neither produce this value nor square back to $L$.","l projectors $P_0,P_1,P_2,P_3$ and eigenvalues $0,2,4,6$. Its square root is\n\\[\nL^{1/2}=\\sqrt2P_1+2P_2+\\sqrt6P_3,\n\\qquad (L^{1/2})^2=L.\n\\]\nBoth heat evolutions retain these modes. Ordinary diffusion multiplies mode $j$ by $e^{-2jt}$; square-root diffusion multiplies it by $e^{-\\sqrt{2j}\\,t}$. The constant mode is unchanged in both.\n\\vfig{fractional}\n\\platecaption{The highest mode decays at rates $6$ and $\\sqrt6$. The graph beside the curves shows the additional couplings in the fractional operator. Curves display the explicit formulas, not an empirical fit.}\n\\head{Fractional does not mean entrywise}\nAlthough $L_{03}=0$, its spectral square root satisfies\n\\[\n(L^{1/2})_{03}=\\frac{\\sqrt6-\\sqrt2-2}{4}<0.\n\\]\nThus $-L^{1/2}$ permits a direct transition between states $0$ and $3$, which were not neighbors for $Q$. Taking the square root of each matrix entry would neither produce this value nor square back to $L$.","tors $P_0,P_1,P_2,P_3$ and eigenvalues $0,2,4,6$. Its square root is\n\\[\nL^{1/2}=\\sqrt2P_1+2P_2+\\sqrt6P_3,\n\\qquad (L^{1/2})^2=L.\n\\]\nBoth heat evolutions retain these modes. Ordinary diffusion multiplies mode $j$ by $e^{-2jt}$; square-root diffusion multiplies it by $e^{-\\sqrt{2j}\\,t}$. The constant mode is unchanged in both.\n\\vfig{fractional}\n\\platecaption{The highest mode decays at rates $6$ and $\\sqrt6$. The graph beside the curves shows the additional couplings in the fractional operator. Curves display the explicit formulas, not an empirical fit.}\n\\head{Fractional does not mean entrywise}\nAlthough $L_{03}=0$, its spectral square root satisfies\n\\[\n(L^{1/2})_{03}=\\frac{\\sqrt6-\\sqrt2-2}{4}<0.\n\\]\nThus $-L^{1/2}$ permits a direct transition between states $0$ and $3$, which were not neighbors for $Q$. Taking the square root of each matrix entry would neither produce this value nor square back to $L$.","$ by $e^{-2jt}$; square-root diffusion multiplies it by $e^{-\\sqrt{2j}\\,t}$. The constant mode is unchanged in both.\n\\vfig{fractional}\n\\platecaption{The highest mode decays at rates $6$ and $\\sqrt6$. The graph beside the curves shows the additional couplings in the fractional operator. Curves display the explicit formulas, not an empirical fit.}\n\\head{Fractional does not mean entrywise}\nAlthough $L_{03}=0$, its spectral square root satisfies\n\\[\n(L^{1/2})_{03}=\\frac{\\sqrt6-\\sqrt2-2}{4}<0.\n\\]\nThus $-L^{1/2}$ permits a direct transition between states $0$ and $3$, which were not neighbors for $Q$. Taking the square root of each matrix entry would neither produce this value nor square back to $L$.","s it by $e^{-\\sqrt{2j}\\,t}$. The constant mode is unchanged in both.\n\\vfig{fractional}\n\\platecaption{The highest mode decays at rates $6$ and $\\sqrt6$. The graph beside the curves shows the additional couplings in the fractional operator. Curves display the explicit formulas, not an empirical fit.}\n\\head{Fractional does not mean entrywise}\nAlthough $L_{03}=0$, its spectral square root satisfies\n\\[\n(L^{1/2})_{03}=\\frac{\\sqrt6-\\sqrt2-2}{4}<0.\n\\]\nThus $-L^{1/2}$ permits a direct transition between states $0$ and $3$, which were not neighbors for $Q$. Taking the square root of each matrix entry would neither produce this value nor square back to $L$.","anged in both.\n\\vfig{fractional}\n\\platecaption{The highest mode decays at rates $6$ and $\\sqrt6$. The graph beside the curves shows the additional couplings in the fractional operator. Curves display the explicit formulas, not an empirical fit.}\n\\head{Fractional does not mean entrywise}\nAlthough $L_{03}=0$, its spectral square root satisfies\n\\[\n(L^{1/2})_{03}=\\frac{\\sqrt6-\\sqrt2-2}{4}<0.\n\\]\nThus $-L^{1/2}$ permits a direct transition between states $0$ and $3$, which were not neighbors for $Q$. Taking the square root of each matrix entry would neither produce this value nor square back to $L$.","hest mode decays at rates $6$ and $\\sqrt6$. The graph beside the curves shows the additional couplings in the fractional operator. Curves display the explicit formulas, not an empirical fit.}\n\\head{Fractional does not mean entrywise}\nAlthough $L_{03}=0$, its spectral square root satisfies\n\\[\n(L^{1/2})_{03}=\\frac{\\sqrt6-\\sqrt2-2}{4}<0.\n\\]\nThus $-L^{1/2}$ permits a direct transition between states $0$ and $3$, which were not neighbors for $Q$. Taking the square root of each matrix entry would neither produce this value nor square back to $L$.","e decays at rates $6$ and $\\sqrt6$. The graph beside the curves shows the additional couplings in the fractional operator. Curves display the explicit formulas, not an empirical fit.}\n\\head{Fractional does not mean entrywise}\nAlthough $L_{03}=0$, its spectral square root satisfies\n\\[\n(L^{1/2})_{03}=\\frac{\\sqrt6-\\sqrt2-2}{4}<0.\n\\]\nThus $-L^{1/2}$ permits a direct transition between states $0$ and $3$, which were not neighbors for $Q$. Taking the square root of each matrix entry would neither produce this value nor square back to $L$.",". The graph beside the curves shows the additional couplings in the fractional operator. Curves display the explicit formulas, not an empirical fit.}\n\\head{Fractional does not mean entrywise}\nAlthough $L_{03}=0$, its spectral square root satisfies\n\\[\n(L^{1/2})_{03}=\\frac{\\sqrt6-\\sqrt2-2}{4}<0.\n\\]\nThus $-L^{1/2}$ permits a direct transition between states $0$ and $3$, which were not neighbors for $Q$. Taking the square root of each matrix entry would neither produce this value nor square back to $L$.","ay the explicit formulas, not an empirical fit.}\n\\head{Fractional does not mean entrywise}\nAlthough $L_{03}=0$, its spectral square root satisfies\n\\[\n(L^{1/2})_{03}=\\frac{\\sqrt6-\\sqrt2-2}{4}<0.\n\\]\nThus $-L^{1/2}$ permits a direct transition between states $0$ and $3$, which were not neighbors for $Q$. Taking the square root of each matrix entry would neither produce this value nor square back to $L$.","The exact block aggregation remains compatible in this example:\n\\[\nC L^{1/2}=\\mat{1&-1\\\\-1&1}C.\n\\]\nThe coarse matrix is the positive square root of $-Q_c$. This is a concrete naturality identity, checked on the full matrices. It is not a claim that an arbitrary partition commutes with every fractional construction.","-1&1}C.\n\\]\nThe coarse matrix is the positive square root of $-Q_c$. This is a concrete naturality identity, checked on the full matrices. It is not a claim that an arbitrary partition commutes with every fractional construction.","\\vpage{16}{A gap changes the error scale}{FRACTIONAL DEFECTS / NEAR ZERO VERSUS AWAY FROM ZERO}{Traditional Part I: Theorem 10.3 and Corollary 10.4. Separate generator error from fractional error.}\nThe function $a\\mapsto\\sqrt a$ changes most rapidly near zero. A scalar model already shows why a spectral gap appears in fractional-defect estimates:\n\\[\n|\\sqrt\\varepsilon-\\sqrt0|=\\sqrt\\varepsilon,\n\\qquad\n\\sqrt{1+\\varepsilon}-1\n=\\frac{\\varepsilon}{\\sqrt{1+\\varepsilon}+1}\\le\\frac\\varepsilon2.\n\\]\nThe same input difference has a square-root scale at zero and a linear bound above a fixed positive lower bound.\n\\vfig{gap}\n\\platecaption{The upper curve compares $0$ with $\\varepsilon$. The lower curve compares $1$ with $1+\\varepsilon$. The dotted line $\\varepsilon/2$ bounds the latter. All plotted values come from the displayed scalar formulas.}\n\\head{Read an operator bound with its hypotheses}\nFor positive self-adjoint finite","\\vpage{16}{A gap changes the error scale}{FRACTIONAL DEFECTS / NEAR ZERO VERSUS AWAY FROM ZERO}{Traditional Part I: Theorem 10.3 and Corollary 10.4. Separate generator error from fractional error.}\nThe function $a\\mapsto\\sqrt a$ changes most rapidly near zero. A scalar model already shows why a spectral gap appears in fractional-defect estimates:\n\\[\n|\\sqrt\\varepsilon-\\sqrt0|=\\sqrt\\varepsilon,\n\\qquad\n\\sqrt{1+\\varepsilon}-1\n=\\frac{\\varepsilon}{\\sqrt{1+\\varepsilon}+1}\\le\\frac\\varepsilon2.\n\\]\nThe same input difference has a square-root scale at zero and a linear bound above a fixed positive lower bound.\n\\vfig{gap}\n\\platecaption{The upper curve compares $0$ with $\\varepsilon$. The lower curve compares $1$ with $1+\\varepsilon$. The dotted line $\\varepsilon/2$ bounds the latter. All plotted values come from the displayed scalar formulas.}\n\\head{Read an operator bound with its hypotheses}\nFor positive self-adjoint finite matrices $A,B\\ge mI$ and $0<\\alpha<1$, the manuscript's gapped estimate has the form\n\\[\n\\norm{CA^\\alpha-B^\\alpha C}\n\\le \\alpha m^{\\alpha-1}\\norm{CA-BC}.\n\\]\nIt concerns an in…","o. A scalar model already shows why a spectral gap appears in fractional-defect estimates:\n\\[\n|\\sqrt\\varepsilon-\\sqrt0|=\\sqrt\\varepsilon,\n\\qquad\n\\sqrt{1+\\varepsilon}-1\n=\\frac{\\varepsilon}{\\sqrt{1+\\varepsilon}+1}\\le\\frac\\varepsilon2.\n\\]\nThe same input difference has a square-root scale at zero and a linear bound above a fixed positive lower bound.\n\\vfig{gap}\n\\platecaption{The upper curve compares $0$ with $\\varepsilon$. The lower curve compares $1$ with $1+\\varepsilon$. The dotted line $\\varepsilon/2$ bounds the latter. All plotted values come from the displayed scalar formulas.}\n\\head{Read an operator bound with its hypotheses}\nFor positive self-adjoint finite matrices $A,B\\ge mI$ and $0<\\alpha<1$, the manuscript's gapped estimate has the form\n\\[\n\\norm{CA^\\alpha-B^\\alpha C}\n\\le \\alpha m^{\\alpha-1}\\norm{CA-BC}.\n\\]\nIt concerns an intertwining defect, not just two eigenvalue lists. With $C=1$, $A=4$, $B=9$, $m=4$, and $\\alpha=1/2$, the generator defect has size $5$, the fractional defect has size $1$, and the bound is $5/4$.","ar model already shows why a spectral gap appears in fractional-defect estimates:\n\\[\n|\\sqrt\\varepsilon-\\sqrt0|=\\sqrt\\varepsilon,\n\\qquad\n\\sqrt{1+\\varepsilon}-1\n=\\frac{\\varepsilon}{\\sqrt{1+\\varepsilon}+1}\\le\\frac\\varepsilon2.\n\\]\nThe same input difference has a square-root scale at zero and a linear bound above a fixed positive lower bound.\n\\vfig{gap}\n\\platecaption{The upper curve compares $0$ with $\\varepsilon$. The lower curve compares $1$ with $1+\\varepsilon$. The dotted line $\\varepsilon/2$ bounds the latter. All plotted values come from the displayed scalar formulas.}\n\\head{Read an operator bound with its hypotheses}\nFor positive self-adjoint finite matrices $A,B\\ge mI$ and $0<\\alpha<1$, the manuscript's gapped estimate has the form\n\\[\n\\norm{CA^\\alpha-B^\\alpha C}\n\\le \\alpha m^{\\alpha-1}\\norm{CA-BC}.\n\\]\nIt concerns an intertwining defect, not just two eigenvalue lists. With $C=1$, $A=4$, $B=9$, $m=4$, and $\\alpha=1/2$, the generator defect has size $5$, the fractional defect has size $1$, and the bound is $5/4$.","p appears in fractional-defect estimates:\n\\[\n|\\sqrt\\varepsilon-\\sqrt0|=\\sqrt\\varepsilon,\n\\qquad\n\\sqrt{1+\\varepsilon}-1\n=\\frac{\\varepsilon}{\\sqrt{1+\\varepsilon}+1}\\le\\frac\\varepsilon2.\n\\]\nThe same input difference has a square-root scale at zero and a linear bound above a fixed positive lower bound.\n\\vfig{gap}\n\\platecaption{The upper curve compares $0$ with $\\varepsilon$. The lower curve compares $1$ with $1+\\varepsilon$. The dotted line $\\varepsilon/2$ bounds the latter. All plotted values come from the displayed scalar formulas.}\n\\head{Read an operator bound with its hypotheses}\nFor positive self-adjoint finite matrices $A,B\\ge mI$ and $0<\\alpha<1$, the manuscript's gapped estimate has the form\n\\[\n\\norm{CA^\\alpha-B^\\alpha C}\n\\le \\alpha m^{\\alpha-1}\\norm{CA-BC}.\n\\]\nIt concerns an intertwining defect, not just two eigenvalue lists. With $C=1$, $A=4$, $B=9$, $m=4$, and $\\alpha=1/2$, the generator defect has size $5$, the fractional defect has size $1$, and the bound is $5/4$.","in fractional-defect estimates:\n\\[\n|\\sqrt\\varepsilon-\\sqrt0|=\\sqrt\\varepsilon,\n\\qquad\n\\sqrt{1+\\varepsilon}-1\n=\\frac{\\varepsilon}{\\sqrt{1+\\varepsilon}+1}\\le\\frac\\varepsilon2.\n\\]\nThe same input difference has a square-root scale at zero and a linear bound above a fixed positive lower bound.\n\\vfig{gap}\n\\platecaption{The upper curve compares $0$ with $\\varepsilon$. The lower curve compares $1$ with $1+\\varepsilon$. The dotted line $\\varepsilon/2$ bounds the latter. All plotted values come from the displayed scalar formulas.}\n\\head{Read an operator bound with its hypotheses}\nFor positive self-adjoint finite matrices $A,B\\ge mI$ and $0<\\alpha<1$, the manuscript's gapped estimate has the form\n\\[\n\\norm{CA^\\alpha-B^\\alpha C}\n\\le \\alpha m^{\\alpha-1}\\norm{CA-BC}.\n\\]\nIt concerns an intertwining defect, not just two eigenvalue lists. With $C=1$, $A=4$, $B=9$, $m=4$, and $\\alpha=1/2$, the generator defect has size $5$, the fractional defect has size $1$, and the bound is $5/4$.","\\[\n|\\sqrt\\varepsilon-\\sqrt0|=\\sqrt\\varepsilon,\n\\qquad\n\\sqrt{1+\\varepsilon}-1\n=\\frac{\\varepsilon}{\\sqrt{1+\\varepsilon}+1}\\le\\frac\\varepsilon2.\n\\]\nThe same input difference has a square-root scale at zero and a linear bound above a fixed positive lower bound.\n\\vfig{gap}\n\\platecaption{The upper curve compares $0$ with $\\varepsilon$. The lower curve compares $1$ with $1+\\varepsilon$. The dotted line $\\varepsilon/2$ bounds the latter. All plotted values come from the displayed scalar formulas.}\n\\head{Read an operator bound with its hypotheses}\nFor positive self-adjoint finite matrices $A,B\\ge mI$ and $0<\\alpha<1$, the manuscript's gapped estimate has the form\n\\[\n\\norm{CA^\\alpha-B^\\alpha C}\n\\le \\alpha m^{\\alpha-1}\\norm{CA-BC}.\n\\]\nIt concerns an intertwining defect, not just two eigenvalue lists. With $C=1$, $A=4$, $B=9$, $m=4$, and $\\alpha=1/2$, the generator defect has size $5$, the fractional defect has size $1$, and the bound is $5/4$.","scale at zero and a linear bound above a fixed positive lower bound.\n\\vfig{gap}\n\\platecaption{The upper curve compares $0$ with $\\varepsilon$. The lower curve compares $1$ with $1+\\varepsilon$. The dotted line $\\varepsilon/2$ bounds the latter. All plotted values come from the displayed scalar formulas.}\n\\head{Read an operator bound with its hypotheses}\nFor positive self-adjoint finite matrices $A,B\\ge mI$ and $0<\\alpha<1$, the manuscript's gapped estimate has the form\n\\[\n\\norm{CA^\\alpha-B^\\alpha C}\n\\le \\alpha m^{\\alpha-1}\\norm{CA-BC}.\n\\]\nIt concerns an intertwining defect, not just two eigenvalue lists. With $C=1$, $A=4$, $B=9$, $m=4$, and $\\alpha=1/2$, the generator defect has size $5$, the fractional defect has size $1$, and the bound is $5/4$.","nd a linear bound above a fixed positive lower bound.\n\\vfig{gap}\n\\platecaption{The upper curve compares $0$ with $\\varepsilon$. The lower curve compares $1$ with $1+\\varepsilon$. The dotted line $\\varepsilon/2$ bounds the latter. All plotted values come from the displayed scalar formulas.}\n\\head{Read an operator bound with its hypotheses}\nFor positive self-adjoint finite matrices $A,B\\ge mI$ and $0<\\alpha<1$, the manuscript's gapped estimate has the form\n\\[\n\\norm{CA^\\alpha-B^\\alpha C}\n\\le \\alpha m^{\\alpha-1}\\norm{CA-BC}.\n\\]\nIt concerns an intertwining defect, not just two eigenvalue lists. With $C=1$, $A=4$, $B=9$, $m=4$, and $\\alpha=1/2$, the generator defect has size $5$, the fractional defect has size $1$, and the bound is $5/4$.","ap}\n\\platecaption{The upper curve compares $0$ with $\\varepsilon$. The lower curve compares $1$ with $1+\\varepsilon$. The dotted line $\\varepsilon/2$ bounds the latter. All plotted values come from the displayed scalar formulas.}\n\\head{Read an operator bound with its hypotheses}\nFor positive self-adjoint finite matrices $A,B\\ge mI$ and $0<\\alpha<1$, the manuscript's gapped estimate has the form\n\\[\n\\norm{CA^\\alpha-B^\\alpha C}\n\\le \\alpha m^{\\alpha-1}\\norm{CA-BC}.\n\\]\nIt concerns an intertwining defect, not just two eigenvalue lists. With $C=1$, $A=4$, $B=9$, $m=4$, and $\\alpha=1/2$, the generator defect has size $5$, the fractional defect has size $1$, and the bound is $5/4$.","ilon/2$ bounds the latter. All plotted values come from the displayed scalar formulas.}\n\\head{Read an operator bound with its hypotheses}\nFor positive self-adjoint finite matrices $A,B\\ge mI$ and $0<\\alpha<1$, the manuscript's gapped estimate has the form\n\\[\n\\norm{CA^\\alpha-B^\\alpha C}\n\\le \\alpha m^{\\alpha-1}\\norm{CA-BC}.\n\\]\nIt concerns an intertwining defect, not just two eigenvalue lists. With $C=1$, $A=4$, $B=9$, $m=4$, and $\\alpha=1/2$, the generator defect has size $5$, the fractional defect has size $1$, and the bound is $5/4$.","bounds the latter. All plotted values come from the displayed scalar formulas.}\n\\head{Read an operator bound with its hypotheses}\nFor positive self-adjoint finite matrices $A,B\\ge mI$ and $0<\\alpha<1$, the manuscript's gapped estimate has the form\n\\[\n\\norm{CA^\\alpha-B^\\alpha C}\n\\le \\alpha m^{\\alpha-1}\\norm{CA-BC}.\n\\]\nIt concerns an intertwining defect, not just two eigenvalue lists. With $C=1$, $A=4$, $B=9$, $m=4$, and $\\alpha=1/2$, the generator defect has size $5$, the fractional defect has size $1$, and the bound is $5/4$.","the latter. All plotted values come from the displayed scalar formulas.}\n\\head{Read an operator bound with its hypotheses}\nFor positive self-adjoint finite matrices $A,B\\ge mI$ and $0<\\alpha<1$, the manuscript's gapped estimate has the form\n\\[\n\\norm{CA^\\alpha-B^\\alpha C}\n\\le \\alpha m^{\\alpha-1}\\norm{CA-BC}.\n\\]\nIt concerns an intertwining defect, not just two eigenvalue lists. With $C=1$, $A=4$, $B=9$, $m=4$, and $\\alpha=1/2$, the generator defect has size $5$, the fractional defect has size $1$, and the bound is $5/4$.","tter. All plotted values come from the displayed scalar formulas.}\n\\head{Read an operator bound with its hypotheses}\nFor positive self-adjoint finite matrices $A,B\\ge mI$ and $0<\\alpha<1$, the manuscript's gapped estimate has the form\n\\[\n\\norm{CA^\\alpha-B^\\alpha C}\n\\le \\alpha m^{\\alpha-1}\\norm{CA-BC}.\n\\]\nIt concerns an intertwining defect, not just two eigenvalue lists. With $C=1$, $A=4$, $B=9$, $m=4$, and $\\alpha=1/2$, the generator defect has size $5$, the fractional defect has size $1$, and the bound is $5/4$.","lotted values come from the displayed scalar formulas.}\n\\head{Read an operator bound with its hypotheses}\nFor positive self-adjoint finite matrices $A,B\\ge mI$ and $0<\\alpha<1$, the manuscript's gapped estimate has the form\n\\[\n\\norm{CA^\\alpha-B^\\alpha C}\n\\le \\alpha m^{\\alpha-1}\\norm{CA-BC}.\n\\]\nIt concerns an intertwining defect, not just two eigenvalue lists. With $C=1$, $A=4$, $B=9$, $m=4$, and $\\alpha=1/2$, the generator defect has size $5$, the fractional defect has size $1$, and the bound is $5/4$.","formulas.}\n\\head{Read an operator bound with its hypotheses}\nFor positive self-adjoint finite matrices $A,B\\ge mI$ and $0<\\alpha<1$, the manuscript's gapped estimate has the form\n\\[\n\\norm{CA^\\alpha-B^\\alpha C}\n\\le \\alpha m^{\\alpha-1}\\norm{CA-BC}.\n\\]\nIt concerns an intertwining defect, not just two eigenvalue lists. With $C=1$, $A=4$, $B=9$, $m=4$, and $\\alpha=1/2$, the generator defect has size $5$, the fractional defect has size $1$, and the bound is $5/4$.","ound with its hypotheses}\nFor positive self-adjoint finite matrices $A,B\\ge mI$ and $0<\\alpha<1$, the manuscript's gapped estimate has the form\n\\[\n\\norm{CA^\\alpha-B^\\alpha C}\n\\le \\alpha m^{\\alpha-1}\\norm{CA-BC}.\n\\]\nIt concerns an intertwining defect, not just two eigenvalue lists. With $C=1$, $A=4$, $B=9$, $m=4$, and $\\alpha=1/2$, the generator defect has size $5$, the fractional defect has size $1$, and the bound is $5/4$.","es}\nFor positive self-adjoint finite matrices $A,B\\ge mI$ and $0<\\alpha<1$, the manuscript's gapped estimate has the form\n\\[\n\\norm{CA^\\alpha-B^\\alpha C}\n\\le \\alpha m^{\\alpha-1}\\norm{CA-BC}.\n\\]\nIt concerns an intertwining defect, not just two eigenvalue lists. With $C=1$, $A=4$, $B=9$, $m=4$, and $\\alpha=1/2$, the generator defect has size $5$, the fractional defect has size $1$, and the bound is $5/4$.","On \\pagerefV{27}, the sequences $1/n^2$ and $1/n$ turn this change in error scale into a concrete distinction between summable and nonsummable defects.\n\\takeaway{A fixed gap is quantitative information. It prevents the derivative of the fractional function from becoming singular along the comparison.}","\\vpage{17}{A random diffusion clock}{SUBORDINATION / A DISTRIBUTION OF ORDINARY TIMES}{Traditional Part I: Theorem 10.2. The Poisson subordination formula for the finite positive model.}\nSquare-root evolution can be written as an average of ordinary diffusion over a positive clock variable $s$:\n\\[\ne^{-t\\sqrt L}=\\int_0^\\infty \\eta_t(s)e^{-sL}\\,ds,\n\\qquad\n\\eta_t(s)=\\frac{t}{2\\sqrt\\pi}s^{-3/2}e^{-t^2/(4s)},\\quad t>0.\n\\]\nHere $t$ is the external time; $s$ is the ordinary diffusion time being mixed. They are not interchangeable variables.\n\\vfig{random_clock}\n\\platecaption{The density $\\eta_1$ peaks at $s=1/6$ but has a long right tail. The drawing ends at $s=2$; it does not claim that the missing tail has zero mass.}\n\\head{Check the normalizer before interpreting a density}\nUnder $s=t^2/(4u^2)$, reversing the integration limits gives\n\\[\n\\eta_t(s)\\,|ds|=\\frac2{\\sqrt\\pi}e^{-u^2}\\,du.\n\\]\nIts integral over $u>0$ is one, so $\\eta_t$ is a probability density. On each eigenmode of $L$, the","\\vpage{17}{A random diffusion clock}{SUBORDINATION / A DISTRIBUTION OF ORDINARY TIMES}{Traditional Part I: Theorem 10.2. The Poisson subordination formula for the finite positive model.}\nSquare-root evolution can be written as an average of ordinary diffusion over a positive clock variable $s$:\n\\[\ne^{-t\\sqrt L}=\\int_0^\\infty \\eta_t(s)e^{-sL}\\,ds,\n\\qquad\n\\eta_t(s)=\\frac{t}{2\\sqrt\\pi}s^{-3/2}e^{-t^2/(4s)},\\quad t>0.\n\\]\nHere $t$ is the external time; $s$ is the ordinary diffusion time being mixed. They are not interchangeable variables.\n\\vfig{random_clock}\n\\platecaption{The density $\\eta_1$ peaks at $s=1/6$ but has a long right tail. The drawing ends at $s=2$; it does not claim that the missing tail has zero mass.}\n\\head{Check the normalizer before interpreting a density}\nUnder $s=t^2/(4u^2)$, reversing the integration limits gives\n\\[\n\\eta_t(s)\\,|ds|=\\frac2{\\sqrt\\pi}e^{-u^2}\\,du.\n\\]\nIts integral over $u>0$ is one, so $\\eta_t$ is a probability density. On each eigenmode of $L$, the operator formula reduces to the scalar Laplace identity\n\\[\n\\int_0^\\infty\\eta_t(s)e^{-\\lambda s}\\,ds=e^{-t\\s…","on clock}{SUBORDINATION / A DISTRIBUTION OF ORDINARY TIMES}{Traditional Part I: Theorem 10.2. The Poisson subordination formula for the finite positive model.}\nSquare-root evolution can be written as an average of ordinary diffusion over a positive clock variable $s$:\n\\[\ne^{-t\\sqrt L}=\\int_0^\\infty \\eta_t(s)e^{-sL}\\,ds,\n\\qquad\n\\eta_t(s)=\\frac{t}{2\\sqrt\\pi}s^{-3/2}e^{-t^2/(4s)},\\quad t>0.\n\\]\nHere $t$ is the external time; $s$ is the ordinary diffusion time being mixed. They are not interchangeable variables.\n\\vfig{random_clock}\n\\platecaption{The density $\\eta_1$ peaks at $s=1/6$ but has a long right tail. The drawing ends at $s=2$; it does not claim that the missing tail has zero mass.}\n\\head{Check the normalizer before interpreting a density}\nUnder $s=t^2/(4u^2)$, reversing the integration limits gives\n\\[\n\\eta_t(s)\\,|ds|=\\frac2{\\sqrt\\pi}e^{-u^2}\\,du.\n\\]\nIts integral over $u>0$ is one, so $\\eta_t$ is a probability density. On each eigenmode of $L$, the operator formula reduces to the scalar Laplace identity\n\\[\n\\int_0^\\infty\\eta_t(s)e^{-\\lambda s}\\,ds=e^{-t\\sqrt\\lambda}.\n\\]\nThe manuscr…","A DISTRIBUTION OF ORDINARY TIMES}{Traditional Part I: Theorem 10.2. The Poisson subordination formula for the finite positive model.}\nSquare-root evolution can be written as an average of ordinary diffusion over a positive clock variable $s$:\n\\[\ne^{-t\\sqrt L}=\\int_0^\\infty \\eta_t(s)e^{-sL}\\,ds,\n\\qquad\n\\eta_t(s)=\\frac{t}{2\\sqrt\\pi}s^{-3/2}e^{-t^2/(4s)},\\quad t>0.\n\\]\nHere $t$ is the external time; $s$ is the ordinary diffusion time being mixed. They are not interchangeable variables.\n\\vfig{random_clock}\n\\platecaption{The density $\\eta_1$ peaks at $s=1/6$ but has a long right tail. The drawing ends at $s=2$; it does not claim that the missing tail has zero mass.}\n\\head{Check the normalizer before interpreting a density}\nUnder $s=t^2/(4u^2)$, reversing the integration limits gives\n\\[\n\\eta_t(s)\\,|ds|=\\frac2{\\sqrt\\pi}e^{-u^2}\\,du.\n\\]\nIts integral over $u>0$ is one, so $\\eta_t$ is a probability density. On each eigenmode of $L$, the operator formula reduces to the scalar Laplace identity\n\\[\n\\int_0^\\infty\\eta_t(s)e^{-\\lambda s}\\,ds=e^{-t\\sqrt\\lambda}.\n\\]\nThe manuscript supplies that scalar i…","Square-root evolution can be written as an average of ordinary diffusion over a positive clock variable $s$:\n\\[\ne^{-t\\sqrt L}=\\int_0^\\infty \\eta_t(s)e^{-sL}\\,ds,\n\\qquad\n\\eta_t(s)=\\frac{t}{2\\sqrt\\pi}s^{-3/2}e^{-t^2/(4s)},\\quad t>0.\n\\]\nHere $t$ is the external time; $s$ is the ordinary diffusion time being mixed. They are not interchangeable variables.\n\\vfig{random_clock}\n\\platecaption{The density $\\eta_1$ peaks at $s=1/6$ but has a long right tail. The drawing ends at $s=2$; it does not claim that the missing tail has zero mass.}\n\\head{Check the normalizer before interpreting a density}\nUnder $s=t^2/(4u^2)$, reversing the integration limits gives\n\\[\n\\eta_t(s)\\,|ds|=\\frac2{\\sqrt\\pi}e^{-u^2}\\,du.\n\\]\nIts integral over $u>0$ is one, so $\\eta_t$ is a probability density. On each eigenmode of $L$, the operator formula reduces to the scalar Laplace identity\n\\[\n\\int_0^\\infty\\eta_t(s)e^{-\\lambda s}\\,ds=e^{-t\\sqrt\\lambda}.\n\\]\nThe manuscript supplies that scalar identity and its spectral assembly. The appendix separately checks the change of variables and normalization; sampled curves are not us…","ion can be written as an average of ordinary diffusion over a positive clock variable $s$:\n\\[\ne^{-t\\sqrt L}=\\int_0^\\infty \\eta_t(s)e^{-sL}\\,ds,\n\\qquad\n\\eta_t(s)=\\frac{t}{2\\sqrt\\pi}s^{-3/2}e^{-t^2/(4s)},\\quad t>0.\n\\]\nHere $t$ is the external time; $s$ is the ordinary diffusion time being mixed. They are not interchangeable variables.\n\\vfig{random_clock}\n\\platecaption{The density $\\eta_1$ peaks at $s=1/6$ but has a long right tail. The drawing ends at $s=2$; it does not claim that the missing tail has zero mass.}\n\\head{Check the normalizer before interpreting a density}\nUnder $s=t^2/(4u^2)$, reversing the integration limits gives\n\\[\n\\eta_t(s)\\,|ds|=\\frac2{\\sqrt\\pi}e^{-u^2}\\,du.\n\\]\nIts integral over $u>0$ is one, so $\\eta_t$ is a probability density. On each eigenmode of $L$, the operator formula reduces to the scalar Laplace identity\n\\[\n\\int_0^\\infty\\eta_t(s)e^{-\\lambda s}\\,ds=e^{-t\\sqrt\\lambda}.\n\\]\nThe manuscript supplies that scalar identity and its spectral assembly. The appendix separately checks the change of variables and normalization; sampled curves are not used to certify an i…","over a positive clock variable $s$:\n\\[\ne^{-t\\sqrt L}=\\int_0^\\infty \\eta_t(s)e^{-sL}\\,ds,\n\\qquad\n\\eta_t(s)=\\frac{t}{2\\sqrt\\pi}s^{-3/2}e^{-t^2/(4s)},\\quad t>0.\n\\]\nHere $t$ is the external time; $s$ is the ordinary diffusion time being mixed. They are not interchangeable variables.\n\\vfig{random_clock}\n\\platecaption{The density $\\eta_1$ peaks at $s=1/6$ but has a long right tail. The drawing ends at $s=2$; it does not claim that the missing tail has zero mass.}\n\\head{Check the normalizer before interpreting a density}\nUnder $s=t^2/(4u^2)$, reversing the integration limits gives\n\\[\n\\eta_t(s)\\,|ds|=\\frac2{\\sqrt\\pi}e^{-u^2}\\,du.\n\\]\nIts integral over $u>0$ is one, so $\\eta_t$ is a probability density. On each eigenmode of $L$, the operator formula reduces to the scalar Laplace identity\n\\[\n\\int_0^\\infty\\eta_t(s)e^{-\\lambda s}\\,ds=e^{-t\\sqrt\\lambda}.\n\\]\nThe manuscript supplies that scalar identity and its spectral assembly. The appendix separately checks the change of variables and normalization; sampled curves are not used to certify an improper integral.","{-3/2}e^{-t^2/(4s)},\\quad t>0.\n\\]\nHere $t$ is the external time; $s$ is the ordinary diffusion time being mixed. They are not interchangeable variables.\n\\vfig{random_clock}\n\\platecaption{The density $\\eta_1$ peaks at $s=1/6$ but has a long right tail. The drawing ends at $s=2$; it does not claim that the missing tail has zero mass.}\n\\head{Check the normalizer before interpreting a density}\nUnder $s=t^2/(4u^2)$, reversing the integration limits gives\n\\[\n\\eta_t(s)\\,|ds|=\\frac2{\\sqrt\\pi}e^{-u^2}\\,du.\n\\]\nIts integral over $u>0$ is one, so $\\eta_t$ is a probability density. On each eigenmode of $L$, the operator formula reduces to the scalar Laplace identity\n\\[\n\\int_0^\\infty\\eta_t(s)e^{-\\lambda s}\\,ds=e^{-t\\sqrt\\lambda}.\n\\]\nThe manuscript supplies that scalar identity and its spectral assembly. The appendix separately checks the change of variables and normalization; sampled curves are not used to certify an improper integral.","al time; $s$ is the ordinary diffusion time being mixed. They are not interchangeable variables.\n\\vfig{random_clock}\n\\platecaption{The density $\\eta_1$ peaks at $s=1/6$ but has a long right tail. The drawing ends at $s=2$; it does not claim that the missing tail has zero mass.}\n\\head{Check the normalizer before interpreting a density}\nUnder $s=t^2/(4u^2)$, reversing the integration limits gives\n\\[\n\\eta_t(s)\\,|ds|=\\frac2{\\sqrt\\pi}e^{-u^2}\\,du.\n\\]\nIts integral over $u>0$ is one, so $\\eta_t$ is a probability density. On each eigenmode of $L$, the operator formula reduces to the scalar Laplace identity\n\\[\n\\int_0^\\infty\\eta_t(s)e^{-\\lambda s}\\,ds=e^{-t\\sqrt\\lambda}.\n\\]\nThe manuscript supplies that scalar identity and its spectral assembly. The appendix separately checks the change of variables and normalization; sampled curves are not used to certify an improper integral.","interchangeable variables.\n\\vfig{random_clock}\n\\platecaption{The density $\\eta_1$ peaks at $s=1/6$ but has a long right tail. The drawing ends at $s=2$; it does not claim that the missing tail has zero mass.}\n\\head{Check the normalizer before interpreting a density}\nUnder $s=t^2/(4u^2)$, reversing the integration limits gives\n\\[\n\\eta_t(s)\\,|ds|=\\frac2{\\sqrt\\pi}e^{-u^2}\\,du.\n\\]\nIts integral over $u>0$ is one, so $\\eta_t$ is a probability density. On each eigenmode of $L$, the operator formula reduces to the scalar Laplace identity\n\\[\n\\int_0^\\infty\\eta_t(s)e^{-\\lambda s}\\,ds=e^{-t\\sqrt\\lambda}.\n\\]\nThe manuscript supplies that scalar identity and its spectral assembly. The appendix separately checks the change of variables and normalization; sampled curves are not used to certify an improper integral.","variables.\n\\vfig{random_clock}\n\\platecaption{The density $\\eta_1$ peaks at $s=1/6$ but has a long right tail. The drawing ends at $s=2$; it does not claim that the missing tail has zero mass.}\n\\head{Check the normalizer before interpreting a density}\nUnder $s=t^2/(4u^2)$, reversing the integration limits gives\n\\[\n\\eta_t(s)\\,|ds|=\\frac2{\\sqrt\\pi}e^{-u^2}\\,du.\n\\]\nIts integral over $u>0$ is one, so $\\eta_t$ is a probability density. On each eigenmode of $L$, the operator formula reduces to the scalar Laplace identity\n\\[\n\\int_0^\\infty\\eta_t(s)e^{-\\lambda s}\\,ds=e^{-t\\sqrt\\lambda}.\n\\]\nThe manuscript supplies that scalar identity and its spectral assembly. The appendix separately checks the change of variables and normalization; sampled curves are not used to certify an improper integral.","$\\eta_1$ peaks at $s=1/6$ but has a long right tail. The drawing ends at $s=2$; it does not claim that the missing tail has zero mass.}\n\\head{Check the normalizer before interpreting a density}\nUnder $s=t^2/(4u^2)$, reversing the integration limits gives\n\\[\n\\eta_t(s)\\,|ds|=\\frac2{\\sqrt\\pi}e^{-u^2}\\,du.\n\\]\nIts integral over $u>0$ is one, so $\\eta_t$ is a probability density. On each eigenmode of $L$, the operator formula reduces to the scalar Laplace identity\n\\[\n\\int_0^\\infty\\eta_t(s)e^{-\\lambda s}\\,ds=e^{-t\\sqrt\\lambda}.\n\\]\nThe manuscript supplies that scalar identity and its spectral assembly. The appendix separately checks the change of variables and normalization; sampled curves are not used to certify an improper integral.","ends at $s=2$; it does not claim that the missing tail has zero mass.}\n\\head{Check the normalizer before interpreting a density}\nUnder $s=t^2/(4u^2)$, reversing the integration limits gives\n\\[\n\\eta_t(s)\\,|ds|=\\frac2{\\sqrt\\pi}e^{-u^2}\\,du.\n\\]\nIts integral over $u>0$ is one, so $\\eta_t$ is a probability density. On each eigenmode of $L$, the operator formula reduces to the scalar Laplace identity\n\\[\n\\int_0^\\infty\\eta_t(s)e^{-\\lambda s}\\,ds=e^{-t\\sqrt\\lambda}.\n\\]\nThe manuscript supplies that scalar identity and its spectral assembly. The appendix separately checks the change of variables and normalization; sampled curves are not used to certify an improper integral.","A normalized density need not have a finite mean. For large $s$, the exponential factor tends to one and $\\eta_t(s)$ is proportional to $s^{-3/2}$. Consequently $s\\eta_t(s)$ has a nonintegrable $s^{-1/2}$ tail. The clock has infinite mean even though its total mass is one.\n\\takeaway{Subordination is a weighted mixture of diffusion times. Replacing the clock distribution by a single average time is not justified, and here that average is infinite.}","\\vpage{18}{A Green kernel is an inverse with units}{DIRICHLET GRID / THE MISSING FACTOR OF h}{Traditional Part I: Theorem 11.1. Exact nodal Green reconstruction for this one-dimensional stencil.}\nTake three interior nodes $x_i=i/4$ on $[0,1]$, with zero boundary values and $h=1/4$. The discrete negative second derivative is\n\\[\nL_h=\\frac1{h^2}\\mat{2&-1&0","\\vpage{18}{A Green kernel is an inverse with units}{DIRICHLET GRID / THE MISSING FACTOR OF h}{Traditional Part I: Theorem 11.1. Exact nodal Green reconstruction for this one-dimensional stencil.}\nTake three interior nodes $x_i=i/4$ on $[0,1]$, with zero boundary values and $h=1/4$. The discrete negative second derivative is\n\\[\nL_h=\\frac1{h^2}\\mat{2&-1&0\\\\-1&2&-1\\\\0&-1&2}.\n\\]\nFor the continuous Dirichlet problem, the Green function is $G(x,y)=\\min(x,y)-xy$. At these nodes the exact comparison is\n\\[\nL_h^{-1}=h\\,[G(x_i,x_j)]_{i,j=1}^3\n=\\frac1{64}\\mat{3&2&1","0&-1&2}.\n\\]\nFor the continuous Dirichlet problem, the Green function is $G(x,y)=\\min(x,y)-xy$. At these nodes the exact comparison is\n\\[\nL_h^{-1}=h\\,[G(x_i,x_j)]_{i,j=1}^3\n=\\frac1{64}\\mat{3&2&1","0&-1&2}.\n\\]\nFor the continuous Dirichlet problem, the Green function is $G(x,y)=\\min(x,y)-xy$. At these nodes the exact comparison is\n\\[\nL_h^{-1}=h\\,[G(x_i,x_j)]_{i,j=1}^3\n=\\frac1{64}\\mat{3&2&1\\\\2&4&2\\\\1&2&3}.\n\\]\n\\vfig{green}\n\\platecaption{The tent is the central column of the continuous Green kernel. It has nodal values $(1,2,1)/8$; its discrete inverse column is $(1,2,1)/32$, smaller by the factor $h$.}\n\\head{A forcing vector is not a quadrature-weighted measure}\nFor sampled forcing $f_j$, the finite solution is\n\\[\nu_i=\\sum_j(L_h^{-1})_{ij}f_j\n=h\\sum_jG(x_i,x_j)f_j.\n\\]\nThe factor $h$ converts samples into the quadrature weights appropriate to this formula. Dropping it multiplies the computed solution by four on the displayed grid.","1&2&3}.\n\\]\n\\vfig{green}\n\\platecaption{The tent is the central column of the continuous Green kernel. It has nodal values $(1,2,1)/8$; its discrete inverse column is $(1,2,1)/32$, smaller by the factor $h$.}\n\\head{A forcing vector is not a quadrature-weighted measure}\nFor sampled forcing $f_j$, the finite solution is\n\\[\nu_i=\\sum_j(L_h^{-1})_{ij}f_j\n=h\\sum_jG(x_i,x_j)f_j.\n\\]\nThe factor $h$ converts samples into the quadrature weights appropriate to this formula. Dropping it multiplies the computed solution by four on the displayed grid.","A discrete unit coordinate vector and a discretized unit point mass use different normalizations. For $f=e_2$, the result is the second inverse column. For $f=e_2/h$, it is the unscaled tent sampled at the nodes. Either convention can be useful, but it must be specified before comparing answers.","Exact checks multiply $L_h$ by the proposed inverse at $1,3,7,$ and $15$ interior nodes. The general identity is the manuscript theorem. These finite checks do not extend it to arbitrary meshes, dimensions, or boundary conditions.\n\\takeaway{An inverse kernel carries a measure convention. A correct-looking tent can give a wrong solution if the grid weight is omitted.}","\\vpage{19}{Convergence with a nonzero exact error}{FINITE DIFFERENCES / A QUARTIC TEST PROBLEM}{Traditional Part I: Theorems 11.1--11.2. Exact nodal inversion and the scale of low-mode errors.}\nUse $u(x)=x-x^4$, which vanishes at both endpoints and solves $-u''=12x^2$. Its centered second difference is not exact:\n\\[\nL_hu|_{\\rm nodes}=12x_i^2+2h^2.\n\\]\nSince the stencil differentiates the quadratic $x(1-x)$ exactly, the solution $u_h$ to the sampled forcing $12x_i^2$ satisfies\n\\[\n\\boxed{u_h(x_i)-u(x_i)=-h^2x_i(1-x_i).}\n\\]\nThis gives both the sign and size of the error, rather than merely a log--log slope.\n\\vfig{mesh}\n\\platecaption{The three curves show magnitudes of the exact nodal errors for $h=1/4,1/8,1/16$. Lines between grid points are visual guides, not an additional continuous numerical solution. The error is negative on every interior node.}\n\\head{One refinement divides this error by four}\nA","\\vpage{19}{Convergence with a nonzero exact error}{FINITE DIFFERENCES / A QUARTIC TEST PROBLEM}{Traditional Part I: Theorems 11.1--11.2. Exact nodal inversion and the scale of low-mode errors.}\nUse $u(x)=x-x^4$, which vanishes at both endpoints and solves $-u''=12x^2$. Its centered second difference is not exact:\n\\[\nL_hu|_{\\rm nodes}=12x_i^2+2h^2.\n\\]\nSince the stencil differentiates the quadratic $x(1-x)$ exactly, the solution $u_h$ to the sampled forcing $12x_i^2$ satisfies\n\\[\n\\boxed{u_h(x_i)-u(x_i)=-h^2x_i(1-x_i).}\n\\]\nThis gives both the sign and size of the error, rather than merely a log--log slope.\n\\vfig{mesh}\n\\platecaption{The three curves show magnitudes of the exact nodal errors for $h=1/4,1/8,1/16$. Lines between grid points are visual guides, not an additional continuous numerical solution. The error is negative on every interior node.}\n\\head{One refinement divides this error by four}\nAll three grids include $x=1/2$, where the maximum absolute","\\vpage{19}{Convergence with a nonzero exact error}{FINITE DIFFERENCES / A QUARTIC TEST PROBLEM}{Traditional Part I: Theorems 11.1--11.2. Exact nodal inversion and the scale of low-mode errors.}\nUse $u(x)=x-x^4$, which vanishes at both endpoints and solves $-u''=12x^2$. Its centered second difference is not exact:\n\\[\nL_hu|_{\\rm nodes}=12x_i^2+2h^2.\n\\]\nSince the stencil differentiates the quadratic $x(1-x)$ exactly, the solution $u_h$ to the sampled forcing $12x_i^2$ satisfies\n\\[\n\\boxed{u_h(x_i)-u(x_i)=-h^2x_i(1-x_i).}\n\\]\nThis gives both the sign and size of the error, rather than merely a log--log slope.\n\\vfig{mesh}\n\\platecaption{The three curves show magnitudes of the exact nodal errors for $h=1/4,1/8,1/16$. Lines between grid points are visual guides, not an additional continuous numerical solution. The error is negative on every interior node.}\n\\head{One refinement divides this error by four}\nAll three grids include $x=1/2$, where the maximum absolute nodal error is $h^2/4$. It equals $1/64$, $1/256$, and $1/1024$, respectively. Exa","vpage{19}{Convergence with a nonzero exact error}{FINITE DIFFERENCES / A QUARTIC TEST PROBLEM}{Traditional Part I: Theorems 11.1--11.2. Exact nodal inversion and the scale of low-mode errors.}\nUse $u(x)=x-x^4$, which vanishes at both endpoints and solves $-u''=12x^2$. Its centered second difference is not exact:\n\\[\nL_hu|_{\\rm nodes}=12x_i^2+2h^2.\n\\]\nSince the stencil differentiates the quadratic $x(1-x)$ exactly, the solution $u_h$ to the sampled forcing $12x_i^2$ satisfies\n\\[\n\\boxed{u_h(x_i)-u(x_i)=-h^2x_i(1-x_i).}\n\\]\nThis gives both the sign and size of the error, rather than merely a log--log slope.\n\\vfig{mesh}\n\\platecaption{The three curves show magnitudes of the exact nodal errors for $h=1/4,1/8,1/16$. Lines between grid points are visual guides, not an additional continuous numerical solution. The error is negative on every interior node.}\n\\head{One refinement divides this error by four}\nAll three grids include $x=1/2$, where the maximum absolute nodal error is $h^2/4$. It equals $1/64$, $1/256$, and $1/1024$, respectively. Exact rational matrix inversion verifies these values …","nzero exact error}{FINITE DIFFERENCES / A QUARTIC TEST PROBLEM}{Traditional Part I: Theorems 11.1--11.2. Exact nodal inversion and the scale of low-mode errors.}\nUse $u(x)=x-x^4$, which vanishes at both endpoints and solves $-u''=12x^2$. Its centered second difference is not exact:\n\\[\nL_hu|_{\\rm nodes}=12x_i^2+2h^2.\n\\]\nSince the stencil differentiates the quadratic $x(1-x)$ exactly, the solution $u_h$ to the sampled forcing $12x_i^2$ satisfies\n\\[\n\\boxed{u_h(x_i)-u(x_i)=-h^2x_i(1-x_i).}\n\\]\nThis gives both the sign and size of the error, rather than merely a log--log slope.\n\\vfig{mesh}\n\\platecaption{The three curves show magnitudes of the exact nodal errors for $h=1/4,1/8,1/16$. Lines between grid points are visual guides, not an additional continuous numerical solution. The error is negative on every interior node.}\n\\head{One refinement divides this error by four}\nAll three grids include $x=1/2$, where the maximum absolute nodal error is $h^2/4$. It equals $1/64$, $1/256$, and $1/1024$, respectively. Exact rational matrix inversion verifies these values independently of the closed err…","FERENCES / A QUARTIC TEST PROBLEM}{Traditional Part I: Theorems 11.1--11.2. Exact nodal inversion and the scale of low-mode errors.}\nUse $u(x)=x-x^4$, which vanishes at both endpoints and solves $-u''=12x^2$. Its centered second difference is not exact:\n\\[\nL_hu|_{\\rm nodes}=12x_i^2+2h^2.\n\\]\nSince the stencil differentiates the quadratic $x(1-x)$ exactly, the solution $u_h$ to the sampled forcing $12x_i^2$ satisfies\n\\[\n\\boxed{u_h(x_i)-u(x_i)=-h^2x_i(1-x_i).}\n\\]\nThis gives both the sign and size of the error, rather than merely a log--log slope.\n\\vfig{mesh}\n\\platecaption{The three curves show magnitudes of the exact nodal errors for $h=1/4,1/8,1/16$. Lines between grid points are visual guides, not an additional continuous numerical solution. The error is negative on every interior node.}\n\\head{One refinement divides this error by four}\nAll three grids include $x=1/2$, where the maximum absolute nodal error is $h^2/4$. It equals $1/64$, $1/256$, and $1/1024$, respectively. Exact rational matrix inversion verifies these values independently of the closed error formula.","TEST PROBLEM}{Traditional Part I: Theorems 11.1--11.2. Exact nodal inversion and the scale of low-mode errors.}\nUse $u(x)=x-x^4$, which vanishes at both endpoints and solves $-u''=12x^2$. Its centered second difference is not exact:\n\\[\nL_hu|_{\\rm nodes}=12x_i^2+2h^2.\n\\]\nSince the stencil differentiates the quadratic $x(1-x)$ exactly, the solution $u_h$ to the sampled forcing $12x_i^2$ satisfies\n\\[\n\\boxed{u_h(x_i)-u(x_i)=-h^2x_i(1-x_i).}\n\\]\nThis gives both the sign and size of the error, rather than merely a log--log slope.\n\\vfig{mesh}\n\\platecaption{The three curves show magnitudes of the exact nodal errors for $h=1/4,1/8,1/16$. Lines between grid points are visual guides, not an additional continuous numerical solution. The error is negative on every interior node.}\n\\head{One refinement divides this error by four}\nAll three grids include $x=1/2$, where the maximum absolute nodal error is $h^2/4$. It equals $1/64$, $1/256$, and $1/1024$, respectively. Exact rational matrix inversion verifies these values independently of the closed error formula.","is not exact:\n\\[\nL_hu|_{\\rm nodes}=12x_i^2+2h^2.\n\\]\nSince the stencil differentiates the quadratic $x(1-x)$ exactly, the solution $u_h$ to the sampled forcing $12x_i^2$ satisfies\n\\[\n\\boxed{u_h(x_i)-u(x_i)=-h^2x_i(1-x_i).}\n\\]\nThis gives both the sign and size of the error, rather than merely a log--log slope.\n\\vfig{mesh}\n\\platecaption{The three curves show magnitudes of the exact nodal errors for $h=1/4,1/8,1/16$. Lines between grid points are visual guides, not an additional continuous numerical solution. The error is negative on every interior node.}\n\\head{One refinement divides this error by four}\nAll three grids include $x=1/2$, where the maximum absolute nodal error is $h^2/4$. It equals $1/64$, $1/256$, and $1/1024$, respectively. Exact rational matrix inversion verifies these values independently of the closed error formula.","ves both the sign and size of the error, rather than merely a log--log slope.\n\\vfig{mesh}\n\\platecaption{The three curves show magnitudes of the exact nodal errors for $h=1/4,1/8,1/16$. Lines between grid points are visual guides, not an additional continuous numerical solution. The error is negative on every interior node.}\n\\head{One refinement divides this error by four}\nAll three grids include $x=1/2$, where the maximum absolute nodal error is $h^2/4$. It equals $1/64$, $1/256$, and $1/1024$, respectively. Exact rational matrix inversion verifies these values independently of the closed error formula.","n merely a log--log slope.\n\\vfig{mesh}\n\\platecaption{The three curves show magnitudes of the exact nodal errors for $h=1/4,1/8,1/16$. Lines between grid points are visual guides, not an additional continuous numerical solution. The error is negative on every interior node.}\n\\head{One refinement divides this error by four}\nAll three grids include $x=1/2$, where the maximum absolute nodal error is $h^2/4$. It equals $1/64$, $1/256$, and $1/1024$, respectively. Exact rational matrix inversion verifies these values independently of the closed error formula.","slope.\n\\vfig{mesh}\n\\platecaption{The three curves show magnitudes of the exact nodal errors for $h=1/4,1/8,1/16$. Lines between grid points are visual guides, not an additional continuous numerical solution. The error is negative on every interior node.}\n\\head{One refinement divides this error by four}\nAll three grids include $x=1/2$, where the maximum absolute nodal error is $h^2/4$. It equals $1/64$, $1/256$, and $1/1024$, respectively. Exact rational matrix inversion verifies these values independently of the closed error formula.","\\vfig{mesh}\n\\platecaption{The three curves show magnitudes of the exact nodal errors for $h=1/4,1/8,1/16$. Lines between grid points are visual guides, not an additional continuous numerical solution. The error is negative on every interior node.}\n\\head{One refinement divides this error by four}\nAll three grids include $x=1/2$, where the maximum absolute nodal error is $h^2/4$. It equals $1/64$, $1/256$, and $1/1024$, respectively. Exact rational matrix inversion verifies these values independently of the closed error formula.","platecaption{The three curves show magnitudes of the exact nodal errors for $h=1/4,1/8,1/16$. Lines between grid points are visual guides, not an additional continuous numerical solution. The error is negative on every interior node.}\n\\head{One refinement divides this error by four}\nAll three grids include $x=1/2$, where the maximum absolute nodal error is $h^2/4$. It equals $1/64$, $1/256$, and $1/1024$, respectively. Exact rational matrix inversion verifies these values independently of the closed error formula.","The manuscript also compares the discrete eigenvalues\n\\[\n\\lambda_{k,h}=4h^{-2}\\sin^2(k\\pi h/2)\n\\]\nwith $(k\\pi)^2$, with error at most $(k\\pi)^4h^2/12$. For a fixed mode $k$, the bound tends to zero quadratically. It does not claim a uniform small error for modes whose index grows like $1/h$.","\\vpage{20}{Eight paths, one weighted expectation}{FEYNMAN--KAC / DEPARTURE-STATE WEIGHTS}{Traditional Part I: Theorem 12.1. A finite path sum with its time convention displayed.}\nLet $K_2=\\frac14\\mat{3&1\\\\1&3}$, start at state $0$, use weights $w=(1,1/2)$, and terminal observable $f=(2,1)$. For three transitions, there are eight possible state sequences after the fixed initial state. We evaluate\n\\[\n\\EE_0\\left[\\prod_{r=0}^{2}w(X_r)\\,f(X_3)\\right].\n\\]\nEach path contributes its transition probability times its three departure weights times its terminal value.\n\\vfig{paths}\n\\platecaption{Every possible path is included. The probability column sums to one. The contribution column sums to $311/256$, agreeing exactly with the transfer calculation.}\n\\head{Check which side of the transition gets weighted}\nThe backward operator on observables is $D_w K_2^{\\mathsf T}$. Therefore the expectation is the zeroth","1&3}$, start at state $0$, use weights $w=(1,1/2)$, and terminal observable $f=(2,1)$. For three transitions, there are eight possible state sequences after the fixed initial state. We evaluate\n\\[\n\\EE_0\\left[\\prod_{r=0}^{2}w(X_r)\\,f(X_3)\\right].\n\\]\nEach path contributes its transition probability times its three departure weights times its terminal value.\n\\vfig{paths}\n\\platecaption{Every possible path is included. The probability column sums to one. The contribution column sums to $311/256$, agreeing exactly with the transfer calculation.}\n\\head{Check which side of the transition gets weighted}\nThe backward operator on observables is $D_w K_2^{\\mathsf T}$. Therefore the expectation is the zeroth coordinate of\n\\[\n(D","1&3}$, start at state $0$, use weights $w=(1,1/2)$, and terminal observable $f=(2,1)$. For three transitions, there are eight possible state sequences after the fixed initial state. We evaluate\n\\[\n\\EE_0\\left[\\prod_{r=0}^{2}w(X_r)\\,f(X_3)\\right].\n\\]\nEach path contributes its transition probability times its three departure weights times its terminal value.\n\\vfig{paths}\n\\platecaption{Every possible path is included. The probability column sums to one. The contribution column sums to $311/256$, agreeing exactly with the transfer calculation.}\n\\head{Check which side of the transition gets weighted}\nThe backward operator on observables is $D_w K_2^{\\mathsf T}$. Therefore the expectation is the zeroth coordinate of\n\\[\n(D_wK_2^{\\mathsf T})^3f.\n\\]","1&3}$, start at state $0$, use weights $w=(1,1/2)$, and terminal observable $f=(2,1)$. For three transitions, there are eight possible state sequences after the fixed initial state. We evaluate\n\\[\n\\EE_0\\left[\\prod_{r=0}^{2}w(X_r)\\,f(X_3)\\right].\n\\]\nEach path contributes its transition probability times its three departure weights times its terminal value.\n\\vfig{paths}\n\\platecaption{Every possible path is included. The probability column sums to one. The contribution column sums to $311/256$, agreeing exactly with the transfer calculation.}\n\\head{Check which side of the transition gets weighted}\nThe backward operator on observables is $D_w K_2^{\\mathsf T}$. Therefore the expectation is the zeroth coordinate of\n\\[\n(D_wK_2^{\\mathsf T})^3f.\n\\]\nIts forward dual is $K_2D_w$: weig","1&3}$, start at state $0$, use weights $w=(1,1/2)$, and terminal observable $f=(2,1)$. For three transitions, there are eight possible state sequences after the fixed initial state. We evaluate\n\\[\n\\EE_0\\left[\\prod_{r=0}^{2}w(X_r)\\,f(X_3)\\right].\n\\]\nEach path contributes its transition probability times its three departure weights times its terminal value.\n\\vfig{paths}\n\\platecaption{Every possible path is included. The probability column sums to one. The contribution column sums to $311/256$, agreeing exactly with the transfer calculation.}\n\\head{Check which side of the transition gets weighted}\nThe backward operator on observables is $D_w K_2^{\\mathsf T}$. Therefore the expectation is the zeroth coordinate of\n\\[\n(D_wK_2^{\\mathsf T})^3f.\n\\]\nIts forward dual is $K_2D_w$: weight the mass at the departing state, then move it. Weighting arrival states instead gives $D_wK_2$, a different expression. The exact tests deliberately check that","For three transitions, there are eight possible state sequences after the fixed initial state. We evaluate\n\\[\n\\EE_0\\left[\\prod_{r=0}^{2}w(X_r)\\,f(X_3)\\right].\n\\]\nEach path contributes its transition probability times its three departure weights times its terminal value.\n\\vfig{paths}\n\\platecaption{Every possible path is included. The probability column sums to one. The contribution column sums to $311/256$, agreeing exactly with the transfer calculation.}\n\\head{Check which side of the transition gets weighted}\nThe backward operator on observables is $D_w K_2^{\\mathsf T}$. Therefore the expectation is the zeroth coordinate of\n\\[\n(D_wK_2^{\\mathsf T})^3f.\n\\]\nIts forward dual is $K_2D_w$: weight the mass at the departing state, then move it. Weighting arrival states instead gives $D_wK_2$, a different expression. The exact tests deliberately check that the two conventions do not agree in this example.","].\n\\]\nEach path contributes its transition probability times its three departure weights times its terminal value.\n\\vfig{paths}\n\\platecaption{Every possible path is included. The probability column sums to one. The contribution column sums to $311/256$, agreeing exactly with the transfer calculation.}\n\\head{Check which side of the transition gets weighted}\nThe backward operator on observables is $D_w K_2^{\\mathsf T}$. Therefore the expectation is the zeroth coordinate of\n\\[\n(D_wK_2^{\\mathsf T})^3f.\n\\]\nIts forward dual is $K_2D_w$: weight the mass at the departing state, then move it. Weighting arrival states instead gives $D_wK_2$, a different expression. The exact tests deliberately check that the two conventions do not agree in this example.","re weights times its terminal value.\n\\vfig{paths}\n\\platecaption{Every possible path is included. The probability column sums to one. The contribution column sums to $311/256$, agreeing exactly with the transfer calculation.}\n\\head{Check which side of the transition gets weighted}\nThe backward operator on observables is $D_w K_2^{\\mathsf T}$. Therefore the expectation is the zeroth coordinate of\n\\[\n(D_wK_2^{\\mathsf T})^3f.\n\\]\nIts forward dual is $K_2D_w$: weight the mass at the departing state, then move it. Weighting arrival states instead gives $D_wK_2$, a different expression. The exact tests deliberately check that the two conventions do not agree in this example.","\\platecaption{Every possible path is included. The probability column sums to one. The contribution column sums to $311/256$, agreeing exactly with the transfer calculation.}\n\\head{Check which side of the transition gets weighted}\nThe backward operator on observables is $D_w K_2^{\\mathsf T}$. Therefore the expectation is the zeroth coordinate of\n\\[\n(D_wK_2^{\\mathsf T})^3f.\n\\]\nIts forward dual is $K_2D_w$: weight the mass at the departing state, then move it. Weighting arrival states instead gives $D_wK_2$, a different expression. The exact tests deliberately check that the two conventions do not agree in this example.","umn sums to $311/256$, agreeing exactly with the transfer calculation.}\n\\head{Check which side of the transition gets weighted}\nThe backward operator on observables is $D_w K_2^{\\mathsf T}$. Therefore the expectation is the zeroth coordinate of\n\\[\n(D_wK_2^{\\mathsf T})^3f.\n\\]\nIts forward dual is $K_2D_w$: weight the mass at the departing state, then move it. Weighting arrival states instead gives $D_wK_2$, a different expression. The exact tests deliberately check that the two conventions do not agree in this example.","The answer $311/256$ exceeds one, but it is not a probability. The terminal observable can equal two. With $0\\le w_i\\le1$, the weighted expectation is bounded by the maximum terminal value, not necessarily by one.","\\vpage{21}{Elimination makes a frequency-dependent term}{SCHUR--FESHBACH / RETAINED RESOLVENT}{Traditional Part I: Theorem 13.1. A two-state Schur complement with the inverse checked directly.}\nThe two-state evolution from \\pagerefV{3} has generator\n\\[\nA=\\mat{-1&1\\\\1&-1}.\n\\]\nRetain the first coordinate and eliminate the second. For $z$ away from the relevant poles, the retained entry of $(zI-A)^{-1}$ is\n\\[\nR_x(z)=\\left[z+1-\\frac1{z+1}\\right]^{-1}\n=\\frac{z+1}{z(z+2)}.\n\\]\nThe term $\\Sigma(z)=1/(z+1)$ is the hidden coordinate's self-energy.\n\\vfig{resolvent}\n\\platecaption{The full inverse gives a rational response with two poles, at $0$ and $-2$. Its partial fractions are $R_x(z)=\\frac12z^{-1}+\\frac12(z+2)^{-1}$. The hidden intermediate inverse has its own excluded point $z=-1$.}\n\\head{An eliminated variable has not disappeared}\nIf the coupling to the second coordinate were simply deleted, the retained resolvent would be $(z+1)^{-1}$. That has a different pole an","1&-1}.\n\\]\nRetain the first coordinate and eliminate the second. For $z$ away from the relevant poles, the retained entry of $(zI-A)^{-1}$ is\n\\[\nR_x(z)=\\left[z+1-\\frac1{z+1}\\right]^{-1}\n=\\frac{z+1}{z(z+2)}.\n\\]\nThe term $\\Sigma(z)=1/(z+1)$ is the hidden coordinate's self-energy.\n\\vfig{resolvent}\n\\platecaption{The full inverse gives a rational response with two poles, at $0$ and $-2$. Its partial fractions are $R_x(z)=\\frac12z^{-1}+\\frac12(z+2)^{-1}$. The hidden intermediate inverse has its own excluded point $z=-1$.}\n\\head{An eliminated variable has not disappeared}\nIf the coupling to the second coordinate were simply deleted, the retained resolvent would be $(z+1)^{-1}$. That has a different pole and a different long-time limit. The correction is frequency depe","1&-1}.\n\\]\nRetain the first coordinate and eliminate the second. For $z$ away from the relevant poles, the retained entry of $(zI-A)^{-1}$ is\n\\[\nR_x(z)=\\left[z+1-\\frac1{z+1}\\right]^{-1}\n=\\frac{z+1}{z(z+2)}.\n\\]\nThe term $\\Sigma(z)=1/(z+1)$ is the hidden coordinate's self-energy.\n\\vfig{resolvent}\n\\platecaption{The full inverse gives a rational response with two poles, at $0$ and $-2$. Its partial fractions are $R_x(z)=\\frac12z^{-1}+\\frac12(z+2)^{-1}$. The hidden intermediate inverse has its own excluded point $z=-1$.}\n\\head{An eliminated variable has not disappeared}\nIf the coupling to the second coordinate were simply deleted, the retained resolvent would be $(z+1)^{-1}$. That has a different pole and a different long-time limit. The correction is frequency dependent, so it cannot be represented by one constant scalar generato","1&-1}.\n\\]\nRetain the first coordinate and eliminate the second. For $z$ away from the relevant poles, the retained entry of $(zI-A)^{-1}$ is\n\\[\nR_x(z)=\\left[z+1-\\frac1{z+1}\\right]^{-1}\n=\\frac{z+1}{z(z+2)}.\n\\]\nThe term $\\Sigma(z)=1/(z+1)$ is the hidden coordinate's self-energy.\n\\vfig{resolvent}\n\\platecaption{The full inverse gives a rational response with two poles, at $0$ and $-2$. Its partial fractions are $R_x(z)=\\frac12z^{-1}+\\frac12(z+2)^{-1}$. The hidden intermediate inverse has its own excluded point $z=-1$.}\n\\head{An eliminated variable has not disappeared}\nIf the coupling to the second coordinate were simply deleted, the retained resolvent would be $(z+1)^{-1}$. That has a different pole and a different long-time limit. The correction is frequency dependent, so it cannot be represented by one constant scalar generator on the retained coordinate.","tain the first coordinate and eliminate the second. For $z$ away from the relevant poles, the retained entry of $(zI-A)^{-1}$ is\n\\[\nR_x(z)=\\left[z+1-\\frac1{z+1}\\right]^{-1}\n=\\frac{z+1}{z(z+2)}.\n\\]\nThe term $\\Sigma(z)=1/(z+1)$ is the hidden coordinate's self-energy.\n\\vfig{resolvent}\n\\platecaption{The full inverse gives a rational response with two poles, at $0$ and $-2$. Its partial fractions are $R_x(z)=\\frac12z^{-1}+\\frac12(z+2)^{-1}$. The hidden intermediate inverse has its own excluded point $z=-1$.}\n\\head{An eliminated variable has not disappeared}\nIf the coupling to the second coordinate were simply deleted, the retained resolvent would be $(z+1)^{-1}$. That has a different pole and a different long-time limit. The correction is frequency dependent, so it cannot be represented by one constant scalar generator on the retained coordinate.","d entry of $(zI-A)^{-1}$ is\n\\[\nR_x(z)=\\left[z+1-\\frac1{z+1}\\right]^{-1}\n=\\frac{z+1}{z(z+2)}.\n\\]\nThe term $\\Sigma(z)=1/(z+1)$ is the hidden coordinate's self-energy.\n\\vfig{resolvent}\n\\platecaption{The full inverse gives a rational response with two poles, at $0$ and $-2$. Its partial fractions are $R_x(z)=\\frac12z^{-1}+\\frac12(z+2)^{-1}$. The hidden intermediate inverse has its own excluded point $z=-1$.}\n\\head{An eliminated variable has not disappeared}\nIf the coupling to the second coordinate were simply deleted, the retained resolvent would be $(z+1)^{-1}$. That has a different pole and a different long-time limit. The correction is frequency dependent, so it cannot be represented by one constant scalar generator on the retained coordinate.","elf-energy.\n\\vfig{resolvent}\n\\platecaption{The full inverse gives a rational response with two poles, at $0$ and $-2$. Its partial fractions are $R_x(z)=\\frac12z^{-1}+\\frac12(z+2)^{-1}$. The hidden intermediate inverse has its own excluded point $z=-1$.}\n\\head{An eliminated variable has not disappeared}\nIf the coupling to the second coordinate were simply deleted, the retained resolvent would be $(z+1)^{-1}$. That has a different pole and a different long-time limit. The correction is frequency dependent, so it cannot be represented by one constant scalar generator on the retained coordinate.","The Schur expression first requires $z+1\\ne0$, as well as invertibility of the full matrix. Its simplified rational form has a removable intermediate singularity at $z=-1$, where the retained entry of the full inverse is well-defined and equals zero. Simplification does not retroactively make the eliminated block invertible there.","For $z>0$, the Laplace transform of $(1+e^{-2t})/2$ is exactly $R_x(z)$. The next page derives the time-domain equation producing that response.\n\\takeaway{The Schur complement is exact elimination, not independent dynamics on a smaller state space. Hidden evolution returns as a $z$-dependent self-energy.}","\\vpage{22}{Hidden state returns as memory}{EXACT RETAINED DYNAMICS / NOT A SCALAR SEMIGROUP}{Traditional Part I: Theorem 13.2. The retained-state memory identity, including hidden initial data.}\nWrite the same system as\n\\[\nx'=-x+y,\\qquad y'=x-y.\n\\]\nSolving the second equation first gives\n\\[\ny(t)=e^{-t}y(0)+\\int_0^t e^{-(t-r)}x(r)\\,dr.\n\\]\nSubstitution is exact:\n\\[\nx'(t)=-x(t)+e^{-t}y(0)+\\int_0^t e^{-(t-r)}x(r)\\,dr.\n\\]\nThe hidden initial state produces a separate forcing term. It must be retained even when the memory kernel is known.\n\\vfig{memory}\n\\platecaption{For $(x(0),y(0))=(1,0)$, the exact retained coordinate is $(1+e^{-2t})/2$. The dashed curve $e^{-t}$ solves the equation obtained by deleting the hidden coordinate rather than eliminating it.}\n\\head{Projection can destroy the semigroup law}\nLet $R(t)=(1+e^{-2t})/2$, the response to this fixed initial preparation. At $t=\\log2$,\n\\[\nR(t)=\\frac58,\\qquad R(2t)=\\frac{17}{32},\n\\qquad","\\vpage{22}{Hidden state returns as memory}{EXACT RETAINED DYNAMICS / NOT A SCALAR SEMIGROUP}{Traditional Part I: Theorem 13.2. The retained-state memory identity, including hidden initial data.}\nWrite the same system as\n\\[\nx'=-x+y,\\qquad y'=x-y.\n\\]\nSolving the second equation first gives\n\\[\ny(t)=e^{-t}y(0)+\\int_0^t e^{-(t-r)}x(r)\\,dr.\n\\]\nSubstitution is exact:\n\\[\nx'(t)=-x(t)+e^{-t}y(0)+\\int_0^t e^{-(t-r)}x(r)\\,dr.\n\\]\nThe hidden initial state produces a separate forcing term. It must be retained even when the memory kernel is known.\n\\vfig{memory}\n\\platecaption{For $(x(0),y(0))=(1,0)$, the exact retained coordinate is $(1+e^{-2t})/2$. The dashed curve $e^{-t}$ solves the equation obtained by deleting the hidden coordinate rather than eliminating it.}\n\\head{Projection can destroy the semigroup law}\nLet $R(t)=(1+e^{-2t})/2$, the response to this fixed initial preparation. At $t=\\log2$,\n\\[\nR(t)=\\frac58,\\qquad R(2t)=\\frac{17}{32},\n\\qquad R(2t)-R(t)^2=\\frac9{64}.\n\\]\nThe full two-state propagator is a semigroup, but this retained","\\vpage{22}{Hidden state returns as memory}{EXACT RETAINED DYNAMICS / NOT A SCALAR SEMIGROUP}{Traditional Part I: Theorem 13.2. The retained-state memory identity, including hidden initial data.}\nWrite the same system as\n\\[\nx'=-x+y,\\qquad y'=x-y.\n\\]\nSolving the second equation first gives\n\\[\ny(t)=e^{-t}y(0)+\\int_0^t e^{-(t-r)}x(r)\\,dr.\n\\]\nSubstitution is exact:\n\\[\nx'(t)=-x(t)+e^{-t}y(0)+\\int_0^t e^{-(t-r)}x(r)\\,dr.\n\\]\nThe hidden initial state produces a separate forcing term. It must be retained even when the memory kernel is known.\n\\vfig{memory}\n\\platecaption{For $(x(0),y(0))=(1,0)$, the exact retained coordinate is $(1+e^{-2t})/2$. The dashed curve $e^{-t}$ solves the equation obtained by deleting the hidden coordinate rather than eliminating it.}\n\\head{Projection can destroy the semigroup law}\nLet $R(t)=(1+e^{-2t})/2$, the response to this fixed initial preparation. At $t=\\log2$,\n\\[\nR(t)=\\frac58,\\qquad R(2t)=\\frac{17}{32},\n\\qquad R(2t)-R(t)^2=\\frac9{64}.\n\\]\nThe full two-state propagator is a semigroup, but this retained scalar response is not. Restarting it as a scalar model silent…","g hidden initial data.}\nWrite the same system as\n\\[\nx'=-x+y,\\qquad y'=x-y.\n\\]\nSolving the second equation first gives\n\\[\ny(t)=e^{-t}y(0)+\\int_0^t e^{-(t-r)}x(r)\\,dr.\n\\]\nSubstitution is exact:\n\\[\nx'(t)=-x(t)+e^{-t}y(0)+\\int_0^t e^{-(t-r)}x(r)\\,dr.\n\\]\nThe hidden initial state produces a separate forcing term. It must be retained even when the memory kernel is known.\n\\vfig{memory}\n\\platecaption{For $(x(0),y(0))=(1,0)$, the exact retained coordinate is $(1+e^{-2t})/2$. The dashed curve $e^{-t}$ solves the equation obtained by deleting the hidden coordinate rather than eliminating it.}\n\\head{Projection can destroy the semigroup law}\nLet $R(t)=(1+e^{-2t})/2$, the response to this fixed initial preparation. At $t=\\log2$,\n\\[\nR(t)=\\frac58,\\qquad R(2t)=\\frac{17}{32},\n\\qquad R(2t)-R(t)^2=\\frac9{64}.\n\\]\nThe full two-state propagator is a semigroup, but this retained scalar response is not. Restarting it as a scalar model silently resets the hidden coordinate, which generally carries information from the first interval.","=-x+y,\\qquad y'=x-y.\n\\]\nSolving the second equation first gives\n\\[\ny(t)=e^{-t}y(0)+\\int_0^t e^{-(t-r)}x(r)\\,dr.\n\\]\nSubstitution is exact:\n\\[\nx'(t)=-x(t)+e^{-t}y(0)+\\int_0^t e^{-(t-r)}x(r)\\,dr.\n\\]\nThe hidden initial state produces a separate forcing term. It must be retained even when the memory kernel is known.\n\\vfig{memory}\n\\platecaption{For $(x(0),y(0))=(1,0)$, the exact retained coordinate is $(1+e^{-2t})/2$. The dashed curve $e^{-t}$ solves the equation obtained by deleting the hidden coordinate rather than eliminating it.}\n\\head{Projection can destroy the semigroup law}\nLet $R(t)=(1+e^{-2t})/2$, the response to this fixed initial preparation. At $t=\\log2$,\n\\[\nR(t)=\\frac58,\\qquad R(2t)=\\frac{17}{32},\n\\qquad R(2t)-R(t)^2=\\frac9{64}.\n\\]\nThe full two-state propagator is a semigroup, but this retained scalar response is not. Restarting it as a scalar model silently resets the hidden coordinate, which generally carries information from the first interval.","e second equation first gives\n\\[\ny(t)=e^{-t}y(0)+\\int_0^t e^{-(t-r)}x(r)\\,dr.\n\\]\nSubstitution is exact:\n\\[\nx'(t)=-x(t)+e^{-t}y(0)+\\int_0^t e^{-(t-r)}x(r)\\,dr.\n\\]\nThe hidden initial state produces a separate forcing term. It must be retained even when the memory kernel is known.\n\\vfig{memory}\n\\platecaption{For $(x(0),y(0))=(1,0)$, the exact retained coordinate is $(1+e^{-2t})/2$. The dashed curve $e^{-t}$ solves the equation obtained by deleting the hidden coordinate rather than eliminating it.}\n\\head{Projection can destroy the semigroup law}\nLet $R(t)=(1+e^{-2t})/2$, the response to this fixed initial preparation. At $t=\\log2$,\n\\[\nR(t)=\\frac58,\\qquad R(2t)=\\frac{17}{32},\n\\qquad R(2t)-R(t)^2=\\frac9{64}.\n\\]\nThe full two-state propagator is a semigroup, but this retained scalar response is not. Restarting it as a scalar model silently resets the hidden coordinate, which generally carries information from the first interval.","\\,dr.\n\\]\nThe hidden initial state produces a separate forcing term. It must be retained even when the memory kernel is known.\n\\vfig{memory}\n\\platecaption{For $(x(0),y(0))=(1,0)$, the exact retained coordinate is $(1+e^{-2t})/2$. The dashed curve $e^{-t}$ solves the equation obtained by deleting the hidden coordinate rather than eliminating it.}\n\\head{Projection can destroy the semigroup law}\nLet $R(t)=(1+e^{-2t})/2$, the response to this fixed initial preparation. At $t=\\log2$,\n\\[\nR(t)=\\frac58,\\qquad R(2t)=\\frac{17}{32},\n\\qquad R(2t)-R(t)^2=\\frac9{64}.\n\\]\nThe full two-state propagator is a semigroup, but this retained scalar response is not. Restarting it as a scalar model silently resets the hidden coordinate, which generally carries information from the first interval.","st be retained even when the memory kernel is known.\n\\vfig{memory}\n\\platecaption{For $(x(0),y(0))=(1,0)$, the exact retained coordinate is $(1+e^{-2t})/2$. The dashed curve $e^{-t}$ solves the equation obtained by deleting the hidden coordinate rather than eliminating it.}\n\\head{Projection can destroy the semigroup law}\nLet $R(t)=(1+e^{-2t})/2$, the response to this fixed initial preparation. At $t=\\log2$,\n\\[\nR(t)=\\frac58,\\qquad R(2t)=\\frac{17}{32},\n\\qquad R(2t)-R(t)^2=\\frac9{64}.\n\\]\nThe full two-state propagator is a semigroup, but this retained scalar response is not. Restarting it as a scalar model silently resets the hidden coordinate, which generally carries information from the first interval.","ed even when the memory kernel is known.\n\\vfig{memory}\n\\platecaption{For $(x(0),y(0))=(1,0)$, the exact retained coordinate is $(1+e^{-2t})/2$. The dashed curve $e^{-t}$ solves the equation obtained by deleting the hidden coordinate rather than eliminating it.}\n\\head{Projection can destroy the semigroup law}\nLet $R(t)=(1+e^{-2t})/2$, the response to this fixed initial preparation. At $t=\\log2$,\n\\[\nR(t)=\\frac58,\\qquad R(2t)=\\frac{17}{32},\n\\qquad R(2t)-R(t)^2=\\frac9{64}.\n\\]\nThe full two-state propagator is a semigroup, but this retained scalar response is not. Restarting it as a scalar model silently resets the hidden coordinate, which generally carries information from the first interval.","The appendix verifies the full differential equation, initial condition, integral memory equation, and Laplace resolvent in this model. Its kernel is $e^{-t}$, whose transform is the self-energy on \\pagerefV{21}.\n\\takeaway{Exact reduction retains a record of the eliminated state through convolution and initial forcing. A memoryless approximation must justify the information it removes.}","\\vpage{23}{A repeated pole remembers a Jordan chain}{NONDIAGONAL HIDDEN DYNAMICS / POLYNOMIAL EXPONENTIALS}{Traditional Part I: Theorem 13.2 and Section 15. Functional calculus must retain nilpotent data.}\nNow use a two-dimensional hidden block\n\\[\nJ=-2I+N,\\qquad N=\\mat{0&1\\\\0&0},\\qquad N^2=0.\n\\]\nChoose the entrance column $V=(0,1)^{\\mathsf T}$ and exit row $U=(1,0)$. The hidden exponential is not $e^{-2t}I$:\n\\[\ne^{tJ}=e^{-2t}(I+tN),\n\\qquad K_{\\rm mem}(t)=Ue^{tJ}V=t e^{-2t}.\n\\]\nThe nilpotent link carries an input from the second hidden coordinate to the first.\n\\vfig{jordan}\n\\platecaption{The memory kernel starts at zero, becomes positive, and peaks at $t=1/2$. A rule that retained the eigenvalue $-2$ but discarded the Jordan chain would predict the zero kernel for these entrance and exit maps.}\n\\head{The transfer function sees pole order}\nThe exact transform is\n\\[\nU(zI-J)^{-1}V=\\frac1{(z+2)^2}\n=\\int_0^\\infty t e^{-(z+2)t}\\,dt,\n\\qquad \\operatorname{Re}z>-2.\n\\]\nThe repeated pole and","0&0},\\qquad N^2=0.\n\\]\nChoose the entrance column $V=(0,1)^{\\mathsf T}$ and exit row $U=(1,0)$. The hidden exponential is not $e^{-2t}I$:\n\\[\ne^{tJ}=e^{-2t}(I+tN),\n\\qquad K_{\\rm mem}(t)=Ue^{tJ}V=t e^{-2t}.\n\\]\nThe nilpotent link carries an input from the second hidden coordinate to the first.\n\\vfig{jordan}\n\\platecaption{The memory kernel starts at zero, becomes positive, and peaks at $t=1/2$. A rule that retained the eigenvalue $-2$ but discarded the Jordan chain would predict the zero kernel for these entrance and exit maps.}\n\\head{The transfer function sees pole order}\nThe exact transform is\n\\[\nU(zI-J)^{-1}V=\\frac1{(z+2)^2}\n=\\int_0^\\infty t e^{-(z+2)t}\\,dt,\n\\qquad \\operatorname{Re}z>-2.\n\\]\nThe repeated pole and the polynomial factor $t$ are the same informatio","0&0},\\qquad N^2=0.\n\\]\nChoose the entrance column $V=(0,1)^{\\mathsf T}$ and exit row $U=(1,0)$. The hidden exponential is not $e^{-2t}I$:\n\\[\ne^{tJ}=e^{-2t}(I+tN),\n\\qquad K_{\\rm mem}(t)=Ue^{tJ}V=t e^{-2t}.\n\\]\nThe nilpotent link carries an input from the second hidden coordinate to the first.\n\\vfig{jordan}\n\\platecaption{The memory kernel starts at zero, becomes positive, and peaks at $t=1/2$. A rule that retained the eigenvalue $-2$ but discarded the Jordan chain would predict the zero kernel for these entrance and exit maps.}\n\\head{The transfer function sees pole order}\nThe exact transform is\n\\[\nU(zI-J)^{-1}V=\\frac1{(z+2)^2}\n=\\int_0^\\infty t e^{-(z+2)t}\\,dt,\n\\qquad \\operatorname{Re}z>-2.\n\\]\nThe repeated pole and the polynomial factor $t$ are the same information in two domains. For a","0&0},\\qquad N^2=0.\n\\]\nChoose the entrance column $V=(0,1)^{\\mathsf T}$ and exit row $U=(1,0)$. The hidden exponential is not $e^{-2t}I$:\n\\[\ne^{tJ}=e^{-2t}(I+tN),\n\\qquad K_{\\rm mem}(t)=Ue^{tJ}V=t e^{-2t}.\n\\]\nThe nilpotent link carries an input from the second hidden coordinate to the first.\n\\vfig{jordan}\n\\platecaption{The memory kernel starts at zero, becomes positive, and peaks at $t=1/2$. A rule that retained the eigenvalue $-2$ but discarded the Jordan chain would predict the zero kernel for these entrance and exit maps.}\n\\head{The transfer function sees pole order}\nThe exact transform is\n\\[\nU(zI-J)^{-1}V=\\frac1{(z+2)^2}\n=\\int_0^\\infty t e^{-(z+2)t}\\,dt,\n\\qquad \\operatorname{Re}z>-2.\n\\]\nThe repeated pole and the polynomial factor $t$ are the same information in two domains. For a longer nilpotent chain, higher powers of","0&0},\\qquad N^2=0.\n\\]\nChoose the entrance column $V=(0,1)^{\\mathsf T}$ and exit row $U=(1,0)$. The hidden exponential is not $e^{-2t}I$:\n\\[\ne^{tJ}=e^{-2t}(I+tN),\n\\qquad K_{\\rm mem}(t)=Ue^{tJ}V=t e^{-2t}.\n\\]\nThe nilpotent link carries an input from the second hidden coordinate to the first.\n\\vfig{jordan}\n\\platecaption{The memory kernel starts at zero, becomes positive, and peaks at $t=1/2$. A rule that retained the eigenvalue $-2$ but discarded the Jordan chain would predict the zero kernel for these entrance and exit maps.}\n\\head{The transfer function sees pole order}\nThe exact transform is\n\\[\nU(zI-J)^{-1}V=\\frac1{(z+2)^2}\n=\\int_0^\\infty t e^{-(z+2)t}\\,dt,\n\\qquad \\operatorname{Re}z>-2.\n\\]\nThe repeated pole and the polynomial factor $t$ are the same information in two domains. For a longer nilpotent chain, higher powers of $t$ occur with factorial denominators.","he entrance column $V=(0,1)^{\\mathsf T}$ and exit row $U=(1,0)$. The hidden exponential is not $e^{-2t}I$:\n\\[\ne^{tJ}=e^{-2t}(I+tN),\n\\qquad K_{\\rm mem}(t)=Ue^{tJ}V=t e^{-2t}.\n\\]\nThe nilpotent link carries an input from the second hidden coordinate to the first.\n\\vfig{jordan}\n\\platecaption{The memory kernel starts at zero, becomes positive, and peaks at $t=1/2$. A rule that retained the eigenvalue $-2$ but discarded the Jordan chain would predict the zero kernel for these entrance and exit maps.}\n\\head{The transfer function sees pole order}\nThe exact transform is\n\\[\nU(zI-J)^{-1}V=\\frac1{(z+2)^2}\n=\\int_0^\\infty t e^{-(z+2)t}\\,dt,\n\\qquad \\operatorname{Re}z>-2.\n\\]\nThe repeated pole and the polynomial factor $t$ are the same information in two domains. For a longer nilpotent chain, higher powers of $t$ occur with factorial denominators.","t}.\n\\]\nThe nilpotent link carries an input from the second hidden coordinate to the first.\n\\vfig{jordan}\n\\platecaption{The memory kernel starts at zero, becomes positive, and peaks at $t=1/2$. A rule that retained the eigenvalue $-2$ but discarded the Jordan chain would predict the zero kernel for these entrance and exit maps.}\n\\head{The transfer function sees pole order}\nThe exact transform is\n\\[\nU(zI-J)^{-1}V=\\frac1{(z+2)^2}\n=\\int_0^\\infty t e^{-(z+2)t}\\,dt,\n\\qquad \\operatorname{Re}z>-2.\n\\]\nThe repeated pole and the polynomial factor $t$ are the same information in two domains. For a longer nilpotent chain, higher powers of $t$ occur with factorial denominators.","at zero, becomes positive, and peaks at $t=1/2$. A rule that retained the eigenvalue $-2$ but discarded the Jordan chain would predict the zero kernel for these entrance and exit maps.}\n\\head{The transfer function sees pole order}\nThe exact transform is\n\\[\nU(zI-J)^{-1}V=\\frac1{(z+2)^2}\n=\\int_0^\\infty t e^{-(z+2)t}\\,dt,\n\\qquad \\operatorname{Re}z>-2.\n\\]\nThe repeated pole and the polynomial factor $t$ are the same information in two domains. For a longer nilpotent chain, higher powers of $t$ occur with factorial denominators.","ut discarded the Jordan chain would predict the zero kernel for these entrance and exit maps.}\n\\head{The transfer function sees pole order}\nThe exact transform is\n\\[\nU(zI-J)^{-1}V=\\frac1{(z+2)^2}\n=\\int_0^\\infty t e^{-(z+2)t}\\,dt,\n\\qquad \\operatorname{Re}z>-2.\n\\]\nThe repeated pole and the polynomial factor $t$ are the same information in two domains. For a longer nilpotent chain, higher powers of $t$ occur with factorial denominators.","A zero value $K_{\\rm mem}(0)$ does not make the entire kernel zero. Nor do stable eigenvalues alone specify every transient contribution: entrance maps, exit maps, and nilpotent structure matter. Here the full kernel is nevertheless elementary, so its peak and transform can be checked without numerical fitting.","\\vpage{24}{A zero initial hazard can still lead to certain exit}{PHASE-TYPE SURVIVAL / TWO SUCCESSIVE STAGES}{Traditional Part I: Theorem 14.1. Survival, density, hazard, and moments have distinct meanings.}\nLet a transient process move through two successive rate-one stages, then leave the system. With column probabilities, the transient generator and starting vector are\n\\[\nT=\\mat{-1&0\\\\1&-1},\\qquad p(0)=e_0.\n\\]\nThus $p(t)=e^{-t}(1,t)^{\\mathsf T}$. Its total remaining mass, exit density, and hazard are\n\\[\nS(t)=(1+t)e^{-t},\\qquad f(t)=t e^{-t},\n\\qquad h(t)=\\frac{t}{1+t}.\n\\]\nThe initial hazard is zero because an exit needs two stages, not one.\n\\vfig{survival}\n\\platecaption{Survival falls from one to zero, while absorbed mass $1-S$ rises to one. The dotted hazard is an instantaneous rate, not a probability, even though it shares the displayed numerical scale. Time is measured in inverse stage-rate units.}\n\\head{Missing transient mass is absorbed mass}\nThe column sums of $T$ are $0$ and $-1$. The exit-rate row is therefore $(0,1)$, and $f(t)=(0,1)p(t)$. Adding an absorbing state restore…","1&-1},\\qquad p(0)=e_0.\n\\]\nThus $p(t)=e^{-t}(1,t)^{\\mathsf T}$. Its total remaining mass, exit density, and hazard are\n\\[\nS(t)=(1+t)e^{-t},\\qquad f(t)=t e^{-t},\n\\qquad h(t)=\\frac{t}{1+t}.\n\\]\nThe initial hazard is zero because an exit needs two stages, not one.\n\\vfig{survival}\n\\platecaption{Survival falls from one to zero, while absorbed mass $1-S$ rises to one. The dotted hazard is an instantaneous rate, not a probability, even though it shares the displayed numerical scale. Time is measured in inverse stage-rate units.}\n\\head{Missing transient mass is absorbed mass}\nThe column sums of $T$ are $0$ and $-1$. The exit-rate row is therefore $(0,1)$, and $f(t)=(0,1)p(t)$. Adding an absorbing state restores a three-state probability vector with total mass","1&-1},\\qquad p(0)=e_0.\n\\]\nThus $p(t)=e^{-t}(1,t)^{\\mathsf T}$. Its total remaining mass, exit density, and hazard are\n\\[\nS(t)=(1+t)e^{-t},\\qquad f(t)=t e^{-t},\n\\qquad h(t)=\\frac{t}{1+t}.\n\\]\nThe initial hazard is zero because an exit needs two stages, not one.\n\\vfig{survival}\n\\platecaption{Survival falls from one to zero, while absorbed mass $1-S$ rises to one. The dotted hazard is an instantaneous rate, not a probability, even though it shares the displayed numerical scale. Time is measured in inverse stage-rate units.}\n\\head{Missing transient mass is absorbed mass}\nThe column sums of $T$ are $0$ and $-1$. The exit-rate row is therefore $(0,1)$, and $f(t)=(0,1)p(t)$. Adding an absorbing state restores a three-state probability vector with total mass one.","initial hazard is zero because an exit needs two stages, not one.\n\\vfig{survival}\n\\platecaption{Survival falls from one to zero, while absorbed mass $1-S$ rises to one. The dotted hazard is an instantaneous rate, not a probability, even though it shares the displayed numerical scale. Time is measured in inverse stage-rate units.}\n\\head{Missing transient mass is absorbed mass}\nThe column sums of $T$ are $0$ and $-1$. The exit-rate row is therefore $(0,1)$, and $f(t)=(0,1)p(t)$. Adding an absorbing state restores a three-state probability vector with total mass one.","hazard is zero because an exit needs two stages, not one.\n\\vfig{survival}\n\\platecaption{Survival falls from one to zero, while absorbed mass $1-S$ rises to one. The dotted hazard is an instantaneous rate, not a probability, even though it shares the displayed numerical scale. Time is measured in inverse stage-rate units.}\n\\head{Missing transient mass is absorbed mass}\nThe column sums of $T$ are $0$ and $-1$. The exit-rate row is therefore $(0,1)$, and $f(t)=(0,1)p(t)$. Adding an absorbing state restores a three-state probability vector with total mass one.","is zero because an exit needs two stages, not one.\n\\vfig{survival}\n\\platecaption{Survival falls from one to zero, while absorbed mass $1-S$ rises to one. The dotted hazard is an instantaneous rate, not a probability, even though it shares the displayed numerical scale. Time is measured in inverse stage-rate units.}\n\\head{Missing transient mass is absorbed mass}\nThe column sums of $T$ are $0$ and $-1$. The exit-rate row is therefore $(0,1)$, and $f(t)=(0,1)p(t)$. Adding an absorbing state restores a three-state probability vector with total mass one.","ges, not one.\n\\vfig{survival}\n\\platecaption{Survival falls from one to zero, while absorbed mass $1-S$ rises to one. The dotted hazard is an instantaneous rate, not a probability, even though it shares the displayed numerical scale. Time is measured in inverse stage-rate units.}\n\\head{Missing transient mass is absorbed mass}\nThe column sums of $T$ are $0$ and $-1$. The exit-rate row is therefore $(0,1)$, and $f(t)=(0,1)p(t)$. Adding an absorbing state restores a three-state probability vector with total mass one.","\\vfig{survival}\n\\platecaption{Survival falls from one to zero, while absorbed mass $1-S$ rises to one. The dotted hazard is an instantaneous rate, not a probability, even though it shares the displayed numerical scale. Time is measured in inverse stage-rate units.}\n\\head{Missing transient mass is absorbed mass}\nThe column sums of $T$ are $0$ and $-1$. The exit-rate row is therefore $(0,1)$, and $f(t)=(0,1)p(t)$. Adding an absorbing state restores a three-state probability vector with total mass one.","Exact integration gives\n\\[\n\\int_0^\\infty f(t)\\,dt=1,\\qquad\n\\EE[\\tau]=2,\\qquad \\EE[\\tau^2]=6,\n\\qquad \\Var(\\tau)=2.\n\\]\nEquivalently, the mean is $\\ones^{\\mathsf T}(-T)^{-1}e_0$. A single exponential waiting time with mean two would have constant hazard $1/2$ and variance four, so matching the mean alone would not reproduce this distribution.","The hazard formula is used where $S(t)>0$. The model has that property for every finite $t\\ge0$. No division by an already vanished survival mass is required.\n\\takeaway{Survival tracks mass, density tracks exits per unit time, and hazard conditions on survival. Equal means or equal initial rates do not identify a waiting-time law.}","\\vpage{25}{Naturality also transports derivatives}{HOLOMORPHIC CALCULUS / A RECTANGULAR COMPARISON}{Traditional Part I: Theorem 15.1. Exact intertwining passes to the specified functional calculus.}\nConsider a fine operator, a coarse operator, and a rectangular map\n\\[\nA_f=\\mat{2&1&0\\\\0&2&0\\\\0&0&9},\\qquad\nA_c=\\mat{2&1\\\\0&2},\\qquad\nC_r=\\mat{1&0&0\\\\0&1&0}.\n\\]\nDirect multiplication gives $C_rA_f=A_cC_r$. For $F(z)=1/(1+z)$, which is defined at both spectra, compute the matrices by inversion:\n\\[\nF(A_f)=\\mat{1/3&-1/9&0","0&1&0}.\n\\]\nDirect multiplication gives $C_rA_f=A_cC_r$. For $F(z)=1/(1+z)$, which is defined at both spectra, compute the matrices by inversion:\n\\[\nF(A_f)=\\mat{1/3&-1/9&0","0&1&0}.\n\\]\nDirect multiplication gives $C_rA_f=A_cC_r$. For $F(z)=1/(1+z)$, which is defined at both spectra, compute the matrices by inversion:\n\\[\nF(A_f)=\\mat{1/3&-1/9&0\\\\0&1/3&0\\\\0&0&1/10}.\n\\]\n\\vfig{naturality}\n\\platecaption{The discarded eigenvalue $9$ contributes $1/10$ only on the fine side. On the retained Jordan block, the off-diagonal entry is $F'(2)=-1/9$, not zero. The rectangular comparison preserves that entry.}\n\\head{Why spectral values are not enough}\nOn the retained block $A_c=2I+N$, with $N^2=0$, one has\n\\[\nF(A_c)=F(2)I+F'(2)N=\\tfrac13I-\\tfrac19N.\n\\]\nHence $C_rF(A_f)=F(A_c)C_r$. Replacing the block by $F(2)I$ would lose the nonzero derivative term and no longer give the actual inverse of $I+A_c$.","0&0&1/10}.\n\\]\n\\vfig{naturality}\n\\platecaption{The discarded eigenvalue $9$ contributes $1/10$ only on the fine side. On the retained Jordan block, the off-diagonal entry is $F'(2)=-1/9$, not zero. The rectangular comparison preserves that entry.}\n\\head{Why spectral values are not enough}\nOn the retained block $A_c=2I+N$, with $N^2=0$, one has\n\\[\nF(A_c)=F(2)I+F'(2)N=\\tfrac13I-\\tfrac19N.\n\\]\nHence $C_rF(A_f)=F(A_c)C_r$. Replacing the block by $F(2)I$ would lose the nonzero derivative term and no longer give the actual inverse of $I+A_c$.","{The discarded eigenvalue $9$ contributes $1/10$ only on the fine side. On the retained Jordan block, the off-diagonal entry is $F'(2)=-1/9$, not zero. The rectangular comparison preserves that entry.}\n\\head{Why spectral values are not enough}\nOn the retained block $A_c=2I+N$, with $N^2=0$, one has\n\\[\nF(A_c)=F(2)I+F'(2)N=\\tfrac13I-\\tfrac19N.\n\\]\nHence $C_rF(A_f)=F(A_c)C_r$. Replacing the block by $F(2)I$ would lose the nonzero derivative term and no longer give the actual inverse of $I+A_c$.","Jordan block, the off-diagonal entry is $F'(2)=-1/9$, not zero. The rectangular comparison preserves that entry.}\n\\head{Why spectral values are not enough}\nOn the retained block $A_c=2I+N$, with $N^2=0$, one has\n\\[\nF(A_c)=F(2)I+F'(2)N=\\tfrac13I-\\tfrac19N.\n\\]\nHence $C_rF(A_f)=F(A_c)C_r$. Replacing the block by $F(2)I$ would lose the nonzero derivative term and no longer give the actual inverse of $I+A_c$.","The same mechanism explains why the Jordan memory kernel on \\pagerefV{23} contains $t e^{-2t}$. Exponentiation also has nonzero derivatives at repeated spectral values.\n\\takeaway{A functional calculus acts on operators, including their nilpotent parts. Exact intertwining transports that full action, not merely a list of scalar eigenvalue images.}","\\vpage{26}{One complete coarse filter}{END-TO-END MODEL / TWO OBSERVATIONS}{Traditional Part I: Proposition 9.3 with Theorems 5.1 and 8.2. Markov closure and likelihood sufficiency are separate.}\nUse the four-state kernel $K$ and aggregation $C$ from \\pagerefV{5}, starting at $p_0=e_0$. Observe two likelihood vectors constant inside both packets:\n\\[\n\\lambda^{(1)}=(3/4,3/4,1/4,1/4),\\qquad\n\\lambda^{(2)}=(1/2,1/2,1,1).\n\\]\nPredict, then update at each observation. The fine calculation gives\n\\[\np_1=\\tfrac1{20}(15,3,2,0)^{\\mathsf T},\\qquad\np_2=\\tfrac1{104}(41,15,40,8)^{\\mathsf T}.\n\\]\n\\vfig{integrated_filter}\n\\platecaption{Within-packet probabilities change during the two observations, but the packet totals agree exactly with an independent two-state filter. The second coarse posterior is $(7/13,6/13)$.}\n\\head{Check the normalizers, not only the final vector}\nThe first and second predictive evidences are\n\\[\nZ_1=\\frac5","\\vpage{26}{One complete coarse filter}{END-TO-END MODEL / TWO OBSERVATIONS}{Traditional Part I: Proposition 9.3 with Theorems 5.1 and 8.2. Markov closure and likelihood sufficiency are separate.}\nUse the four-state kernel $K$ and aggregation $C$ from \\pagerefV{5}, starting at $p_0=e_0$. Observe two likelihood vectors constant inside both packets:\n\\[\n\\lambda^{(1)}=(3/4,3/4,1/4,1/4),\\qquad\n\\lambda^{(2)}=(1/2,1/2,1,1).\n\\]\nPredict, then update at each observation. The fine calculation gives\n\\[\np_1=\\tfrac1{20}(15,3,2,0)^{\\mathsf T},\\qquad\np_2=\\tfrac1{104}(41,15,40,8)^{\\mathsf T}.\n\\]\n\\vfig{integrated_filter}\n\\platecaption{Within-packet probabilities change during the two observations, but the packet totals agree exactly with an independent two-state filter. The second coarse posterior is $(7/13,6/13)$.}\n\\head{Check the normalizers, not only the final vector}\nThe first and second predictive evidences are\n\\[\nZ_1=\\frac58,\\qquad Z_2=\\frac{1","\\vpage{26}{One complete coarse filter}{END-TO-END MODEL / TWO OBSERVATIONS}{Traditional Part I: Proposition 9.3 with Theorems 5.1 and 8.2. Markov closure and likelihood sufficiency are separate.}\nUse the four-state kernel $K$ and aggregation $C$ from \\pagerefV{5}, starting at $p_0=e_0$. Observe two likelihood vectors constant inside both packets:\n\\[\n\\lambda^{(1)}=(3/4,3/4,1/4,1/4),\\qquad\n\\lambda^{(2)}=(1/2,1/2,1,1).\n\\]\nPredict, then update at each observation. The fine calculation gives\n\\[\np_1=\\tfrac1{20}(15,3,2,0)^{\\mathsf T},\\qquad\np_2=\\tfrac1{104}(41,15,40,8)^{\\mathsf T}.\n\\]\n\\vfig{integrated_filter}\n\\platecaption{Within-packet probabilities change during the two observations, but the packet totals agree exactly with an independent two-state filter. The second coarse posterior is $(7/13,6/13)$.}\n\\head{Check the normalizers, not only the final vector}\nThe first and second predictive evidences are\n\\[\nZ_1=\\frac58,\\qquad Z_2=\\frac{13}{20},\n\\qquad Z_1Z_2=\\frac{13}{32}.\n\\]\nT","\\vpage{26}{One complete coarse filter}{END-TO-END MODEL / TWO OBSERVATIONS}{Traditional Part I: Proposition 9.3 with Theorems 5.1 and 8.2. Markov closure and likelihood sufficiency are separate.}\nUse the four-state kernel $K$ and aggregation $C$ from \\pagerefV{5}, starting at $p_0=e_0$. Observe two likelihood vectors constant inside both packets:\n\\[\n\\lambda^{(1)}=(3/4,3/4,1/4,1/4),\\qquad\n\\lambda^{(2)}=(1/2,1/2,1,1).\n\\]\nPredict, then update at each observation. The fine calculation gives\n\\[\np_1=\\tfrac1{20}(15,3,2,0)^{\\mathsf T},\\qquad\np_2=\\tfrac1{104}(41,15,40,8)^{\\mathsf T}.\n\\]\n\\vfig{integrated_filter}\n\\platecaption{Within-packet probabilities change during the two observations, but the packet totals agree exactly with an independent two-state filter. The second coarse posterior is $(7/13,6/13)$.}\n\\head{Check the normalizers, not only the final vector}\nThe first and second predictive evidences are\n\\[\nZ_1=\\frac58,\\qquad Z_2=\\frac{13}{20},\n\\qquad Z_1Z_2=\\frac{13}{32}.\n\\]\nThe unnormalized product $D_2KD_1Kp_0$ has mass $13/32$ and normalizes to $p_2$. Because the likelihood entries here…",": Proposition 9.3 with Theorems 5.1 and 8.2. Markov closure and likelihood sufficiency are separate.}\nUse the four-state kernel $K$ and aggregation $C$ from \\pagerefV{5}, starting at $p_0=e_0$. Observe two likelihood vectors constant inside both packets:\n\\[\n\\lambda^{(1)}=(3/4,3/4,1/4,1/4),\\qquad\n\\lambda^{(2)}=(1/2,1/2,1,1).\n\\]\nPredict, then update at each observation. The fine calculation gives\n\\[\np_1=\\tfrac1{20}(15,3,2,0)^{\\mathsf T},\\qquad\np_2=\\tfrac1{104}(41,15,40,8)^{\\mathsf T}.\n\\]\n\\vfig{integrated_filter}\n\\platecaption{Within-packet probabilities change during the two observations, but the packet totals agree exactly with an independent two-state filter. The second coarse posterior is $(7/13,6/13)$.}\n\\head{Check the normalizers, not only the final vector}\nThe first and second predictive evidences are\n\\[\nZ_1=\\frac58,\\qquad Z_2=\\frac{13}{20},\n\\qquad Z_1Z_2=\\frac{13}{32}.\n\\]\nThe unnormalized product $D_2KD_1Kp_0$ has mass $13/32$ and normalizes to $p_2$. Because the likelihood entries here lie in $[0,1]$, they can serve as probabilities of the specified observations. This makes the…","mbda^{(2)}=(1/2,1/2,1,1).\n\\]\nPredict, then update at each observation. The fine calculation gives\n\\[\np_1=\\tfrac1{20}(15,3,2,0)^{\\mathsf T},\\qquad\np_2=\\tfrac1{104}(41,15,40,8)^{\\mathsf T}.\n\\]\n\\vfig{integrated_filter}\n\\platecaption{Within-packet probabilities change during the two observations, but the packet totals agree exactly with an independent two-state filter. The second coarse posterior is $(7/13,6/13)$.}\n\\head{Check the normalizers, not only the final vector}\nThe first and second predictive evidences are\n\\[\nZ_1=\\frac58,\\qquad Z_2=\\frac{13}{20},\n\\qquad Z_1Z_2=\\frac{13}{32}.\n\\]\nThe unnormalized product $D_2KD_1Kp_0$ has mass $13/32$ and normalizes to $p_2$. Because the likelihood entries here lie in $[0,1]$, they can serve as probabilities of the specified observations. This makes the product their joint evidence in the stated model.",",3,2,0)^{\\mathsf T},\\qquad\np_2=\\tfrac1{104}(41,15,40,8)^{\\mathsf T}.\n\\]\n\\vfig{integrated_filter}\n\\platecaption{Within-packet probabilities change during the two observations, but the packet totals agree exactly with an independent two-state filter. The second coarse posterior is $(7/13,6/13)$.}\n\\head{Check the normalizers, not only the final vector}\nThe first and second predictive evidences are\n\\[\nZ_1=\\frac58,\\qquad Z_2=\\frac{13}{20},\n\\qquad Z_1Z_2=\\frac{13}{32}.\n\\]\nThe unnormalized product $D_2KD_1Kp_0$ has mass $13/32$ and normalizes to $p_2$. Because the likelihood entries here lie in $[0,1]$, they can serve as probabilities of the specified observations. This makes the product their joint evidence in the stated model.","\\platecaption{Within-packet probabilities change during the two observations, but the packet totals agree exactly with an independent two-state filter. The second coarse posterior is $(7/13,6/13)$.}\n\\head{Check the normalizers, not only the final vector}\nThe first and second predictive evidences are\n\\[\nZ_1=\\frac58,\\qquad Z_2=\\frac{13}{20},\n\\qquad Z_1Z_2=\\frac{13}{32}.\n\\]\nThe unnormalized product $D_2KD_1Kp_0$ has mass $13/32$ and normalizes to $p_2$. Because the likelihood entries here lie in $[0,1]$, they can serve as probabilities of the specified observations. This makes the product their joint evidence in the stated model.","cket probabilities change during the two observations, but the packet totals agree exactly with an independent two-state filter. The second coarse posterior is $(7/13,6/13)$.}\n\\head{Check the normalizers, not only the final vector}\nThe first and second predictive evidences are\n\\[\nZ_1=\\frac58,\\qquad Z_2=\\frac{13}{20},\n\\qquad Z_1Z_2=\\frac{13}{32}.\n\\]\nThe unnormalized product $D_2KD_1Kp_0$ has mass $13/32$ and normalizes to $p_2$. Because the likelihood entries here lie in $[0,1]$, they can serve as probabilities of the specified observations. This makes the product their joint evidence in the stated model.","during the two observations, but the packet totals agree exactly with an independent two-state filter. The second coarse posterior is $(7/13,6/13)$.}\n\\head{Check the normalizers, not only the final vector}\nThe first and second predictive evidences are\n\\[\nZ_1=\\frac58,\\qquad Z_2=\\frac{13}{20},\n\\qquad Z_1Z_2=\\frac{13}{32}.\n\\]\nThe unnormalized product $D_2KD_1Kp_0$ has mass $13/32$ and normalizes to $p_2$. Because the likelihood entries here lie in $[0,1]$, they can serve as probabilities of the specified observations. This makes the product their joint evidence in the stated model.","agree exactly with an independent two-state filter. The second coarse posterior is $(7/13,6/13)$.}\n\\head{Check the normalizers, not only the final vector}\nThe first and second predictive evidences are\n\\[\nZ_1=\\frac58,\\qquad Z_2=\\frac{13}{20},\n\\qquad Z_1Z_2=\\frac{13}{32}.\n\\]\nThe unnormalized product $D_2KD_1Kp_0$ has mass $13/32$ and normalizes to $p_2$. Because the likelihood entries here lie in $[0,1]$, they can serve as probabilities of the specified observations. This makes the product their joint evidence in the stated model.","Two distinct equalities justify the coarse computation. First, $CK=K_cC$ transports prediction. Second, the likelihood is constant on each packet, so $CD_r=\\overline D_rC$ transports observation weighting. Their composition preserves both the posterior packet totals and the evidence factors.","\\vpage{27}{Small at each level is not summable across levels}{REFINEMENT LIMITS / WHAT THE FINITE MODELS DO NOT PROVE}{Traditional Part I: Corollary 10.5 and Section 17. Summability and uniform control are additional data.}\nFor each $n\\ge1$, compare the two scalar operators $A_n=0$ and $B_n=1/n^2$ through the identity. Their ordinary defects have sizes $1/n^2$, while their square-root defects have sizes $1/n$:\n\\[\n\\sum_n|A_n-B_n|<\\infty,\n\\qquad\n\\sum_n|A_n^{1/2}-B_n^{1/2}|=\\infty.\n\\]\nThese are independent scalar comparisons, not a constructed inverse system. They isolate an error-scale obstruction without claiming a coherent refinement tower.\n\\vfig{summability}\n\\platecaption{Partial sums through $n=16$ illustrate the contrast. The convergence and divergence statements follow from the displayed series, not from the finite plot. A common positive gap gives the third, summable upper bound.}\n\\head{A uniform hypothesis can change th","\\vpage{27}{Small at each level is not summable across levels}{REFINEMENT LIMITS / WHAT THE FINITE MODELS DO NOT PROVE}{Traditional Part I: Corollary 10.5 and Section 17. Summability and uniform control are additional data.}\nFor each $n\\ge1$, compare the two scalar operators $A_n=0$ and $B_n=1/n^2$ through the identity. Their ordinary defects have sizes $1/n^2$, while their square-root defects have sizes $1/n$:\n\\[\n\\sum_n|A_n-B_n|<\\infty,\n\\qquad\n\\sum_n|A_n^{1/2}-B_n^{1/2}|=\\infty.\n\\]\nThese are independent scalar comparisons, not a constructed inverse system. They isolate an error-scale obstruction without claiming a coherent refinement tower.\n\\vfig{summability}\n\\platecaption{Partial sums through $n=16$ illustrate the contrast. The convergence and divergence statements follow from the displayed series, not from the finite plot. A common positive gap gives the third, summable upper bound.}\n\\head{A uniform hypothesis can change the conclusion}\nReplace the pair by $A_n=1$","\\vpage{27}{Small at each level is not summable across levels}{REFINEMENT LIMITS / WHAT THE FINITE MODELS DO NOT PROVE}{Traditional Part I: Corollary 10.5 and Section 17. Summability and uniform control are additional data.}\nFor each $n\\ge1$, compare the two scalar operators $A_n=0$ and $B_n=1/n^2$ through the identity. Their ordinary defects have sizes $1/n^2$, while their square-root defects have sizes $1/n$:\n\\[\n\\sum_n|A_n-B_n|<\\infty,\n\\qquad\n\\sum_n|A_n^{1/2}-B_n^{1/2}|=\\infty.\n\\]\nThese are independent scalar comparisons, not a constructed inverse system. They isolate an error-scale obstruction without claiming a coherent refinement tower.\n\\vfig{summability}\n\\platecaption{Partial sums through $n=16$ illustrate the contrast. The convergence and divergence statements follow from the displayed series, not from the finite plot. A common positive gap gives the third, summable upper bound.}\n\\head{A uniform hypothesis can change the conclusion}\nReplace the pair by $A_n=1$ and $B_n=1+1/n^2","\\vpage{27}{Small at each level is not summable across levels}{REFINEMENT LIMITS / WHAT THE FINITE MODELS DO NOT PROVE}{Traditional Part I: Corollary 10.5 and Section 17. Summability and uniform control are additional data.}\nFor each $n\\ge1$, compare the two scalar operators $A_n=0$ and $B_n=1/n^2$ through the identity. Their ordinary defects have sizes $1/n^2$, while their square-root defects have sizes $1/n$:\n\\[\n\\sum_n|A_n-B_n|<\\infty,\n\\qquad\n\\sum_n|A_n^{1/2}-B_n^{1/2}|=\\infty.\n\\]\nThese are independent scalar comparisons, not a constructed inverse system. They isolate an error-scale obstruction without claiming a coherent refinement tower.\n\\vfig{summability}\n\\platecaption{Partial sums through $n=16$ illustrate the contrast. The convergence and divergence statements follow from the displayed series, not from the finite plot. A common positive gap gives the third, summable upper bound.}\n\\head{A uniform hypothesis can change the conclusion}\nReplace the pair by $A_n=1$ and $B_n=1+1/n^2$. Then\n\\[\n\\left|B_n^{1/2}-A_n^{1/2}\\right|\n=\\frac{1/n^2}{\\sqrt{","7}{Small at each level is not summable across levels}{REFINEMENT LIMITS / WHAT THE FINITE MODELS DO NOT PROVE}{Traditional Part I: Corollary 10.5 and Section 17. Summability and uniform control are additional data.}\nFor each $n\\ge1$, compare the two scalar operators $A_n=0$ and $B_n=1/n^2$ through the identity. Their ordinary defects have sizes $1/n^2$, while their square-root defects have sizes $1/n$:\n\\[\n\\sum_n|A_n-B_n|<\\infty,\n\\qquad\n\\sum_n|A_n^{1/2}-B_n^{1/2}|=\\infty.\n\\]\nThese are independent scalar comparisons, not a constructed inverse system. They isolate an error-scale obstruction without claiming a coherent refinement tower.\n\\vfig{summability}\n\\platecaption{Partial sums through $n=16$ illustrate the contrast. The convergence and divergence statements follow from the displayed series, not from the finite plot. A common positive gap gives the third, summable upper bound.}\n\\head{A uniform hypothesis can change the conclusion}\nReplace the pair by $A_n=1$ and $B_n=1+1/n^2$. Then\n\\[\n\\left|B_n^{1/2}-A_n^{1/2}\\right|\n=\\frac{1/n^2}{\\sqrt{1+1/n^2}+1}\n\\le\\frac1{2n^2}.\n\\]\nThis is the ele…","at each level is not summable across levels}{REFINEMENT LIMITS / WHAT THE FINITE MODELS DO NOT PROVE}{Traditional Part I: Corollary 10.5 and Section 17. Summability and uniform control are additional data.}\nFor each $n\\ge1$, compare the two scalar operators $A_n=0$ and $B_n=1/n^2$ through the identity. Their ordinary defects have sizes $1/n^2$, while their square-root defects have sizes $1/n$:\n\\[\n\\sum_n|A_n-B_n|<\\infty,\n\\qquad\n\\sum_n|A_n^{1/2}-B_n^{1/2}|=\\infty.\n\\]\nThese are independent scalar comparisons, not a constructed inverse system. They isolate an error-scale obstruction without claiming a coherent refinement tower.\n\\vfig{summability}\n\\platecaption{Partial sums through $n=16$ illustrate the contrast. The convergence and divergence statements follow from the displayed series, not from the finite plot. A common positive gap gives the third, summable upper bound.}\n\\head{A uniform hypothesis can change the conclusion}\nReplace the pair by $A_n=1$ and $B_n=1+1/n^2$. Then\n\\[\n\\left|B_n^{1/2}-A_n^{1/2}\\right|\n=\\frac{1/n^2}{\\sqrt{1+1/n^2}+1}\n\\le\\frac1{2n^2}.\n\\]\nThis is the elementary f…","gh the identity. Their ordinary defects have sizes $1/n^2$, while their square-root defects have sizes $1/n$:\n\\[\n\\sum_n|A_n-B_n|<\\infty,\n\\qquad\n\\sum_n|A_n^{1/2}-B_n^{1/2}|=\\infty.\n\\]\nThese are independent scalar comparisons, not a constructed inverse system. They isolate an error-scale obstruction without claiming a coherent refinement tower.\n\\vfig{summability}\n\\platecaption{Partial sums through $n=16$ illustrate the contrast. The convergence and divergence statements follow from the displayed series, not from the finite plot. A common positive gap gives the third, summable upper bound.}\n\\head{A uniform hypothesis can change the conclusion}\nReplace the pair by $A_n=1$ and $B_n=1+1/n^2$. Then\n\\[\n\\left|B_n^{1/2}-A_n^{1/2}\\right|\n=\\frac{1/n^2}{\\sqrt{1+1/n^2}+1}\n\\le\\frac1{2n^2}.\n\\]\nThis is the elementary form of the uniform-gap mechanism on \\pagerefV{16}. Without such information, transferring a summable generator-defect budget through a fractional function can fail.","e an error-scale obstruction without claiming a coherent refinement tower.\n\\vfig{summability}\n\\platecaption{Partial sums through $n=16$ illustrate the contrast. The convergence and divergence statements follow from the displayed series, not from the finite plot. A common positive gap gives the third, summable upper bound.}\n\\head{A uniform hypothesis can change the conclusion}\nReplace the pair by $A_n=1$ and $B_n=1+1/n^2$. Then\n\\[\n\\left|B_n^{1/2}-A_n^{1/2}\\right|\n=\\frac{1/n^2}{\\sqrt{1+1/n^2}+1}\n\\le\\frac1{2n^2}.\n\\]\nThis is the elementary form of the uniform-gap mechanism on \\pagerefV{16}. Without such information, transferring a summable generator-defect budget through a fractional function can fail.","cale obstruction without claiming a coherent refinement tower.\n\\vfig{summability}\n\\platecaption{Partial sums through $n=16$ illustrate the contrast. The convergence and divergence statements follow from the displayed series, not from the finite plot. A common positive gap gives the third, summable upper bound.}\n\\head{A uniform hypothesis can change the conclusion}\nReplace the pair by $A_n=1$ and $B_n=1+1/n^2$. Then\n\\[\n\\left|B_n^{1/2}-A_n^{1/2}\\right|\n=\\frac{1/n^2}{\\sqrt{1+1/n^2}+1}\n\\le\\frac1{2n^2}.\n\\]\nThis is the elementary form of the uniform-gap mechanism on \\pagerefV{16}. Without such information, transferring a summable generator-defect budget through a fractional function can fail.","out claiming a coherent refinement tower.\n\\vfig{summability}\n\\platecaption{Partial sums through $n=16$ illustrate the contrast. The convergence and divergence statements follow from the displayed series, not from the finite plot. A common positive gap gives the third, summable upper bound.}\n\\head{A uniform hypothesis can change the conclusion}\nReplace the pair by $A_n=1$ and $B_n=1+1/n^2$. Then\n\\[\n\\left|B_n^{1/2}-A_n^{1/2}\\right|\n=\\frac{1/n^2}{\\sqrt{1+1/n^2}+1}\n\\le\\frac1{2n^2}.\n\\]\nThis is the elementary form of the uniform-gap mechanism on \\pagerefV{16}. Without such information, transferring a summable generator-defect budget through a fractional function can fail.","\\vpage{29}{Acyclic paths stop after finitely many transfers}{DIRECTED TRANSFER / A FINITE DYSON EXAMPLE}{Traditional Part II: Theorem E2.8 and Corollary E2.9. Acyclicity concerns the transfer graph.}\nLet three scalar packets form the directed chain $0\\to1\\to2$, with transfer coefficients $2$ and $3$. The generator and its powers are\n\\[\nA=\\mat{0&0&0","\\vpage{29}{Acyclic paths stop after finitely many transfers}{DIRECTED TRANSFER / A FINITE DYSON EXAMPLE}{Traditional Part II: Theorem E2.8 and Corollary E2.9. Acyclicity concerns the transfer graph.}\nLet three scalar packets form the directed chain $0\\to1\\to2$, with transfer coefficients $2$ and $3$. The generator and its powers are\n\\[\nA=\\mat{0&0&0\\\\2&0&0\\\\0&3&0},\\qquad\nA^2=\\mat{0&0&0\\\\0&0&0\\\\6&0&0},\\qquad A^3=0.\n\\]\nTherefore the exponential is the exact polynomial\n\\[\ne^{tA}=I+tA+\\frac{t^2}{2}A^2.\n\\]\n\\vfig{ext_paths}\n\\platecaption{Only the empty path, one-edge paths, and the two-edge path exist. The coefficient $3t^2$ includes both the edge product $2\\cdot3$ and the ordered-time simplex area $t^2/2$.}\n\\head{Path length counts transfers, not elapsed time}\nStarting at $e_0$, the complete state is $(1,2t,3t^2)^{\\mathsf T}$. No fourth packet or repeated cycle can contribute. Direct differentiation gives $x'(t)=Ax(t)$, including the relation $x_2'(t)=3x_1(t)$.","6&0&0},\\qquad A^3=0.\n\\]\nTherefore the exponential is the exact polynomial\n\\[\ne^{tA}=I+tA+\\frac{t^2}{2}A^2.\n\\]\n\\vfig{ext_paths}\n\\platecaption{Only the empty path, one-edge paths, and the two-edge path exist. The coefficient $3t^2$ includes both the edge product $2\\cdot3$ and the ordered-time simplex area $t^2/2$.}\n\\head{Path length counts transfers, not elapsed time}\nStarting at $e_0$, the complete state is $(1,2t,3t^2)^{\\mathsf T}$. No fourth packet or repeated cycle can contribute. Direct differentiation gives $x'(t)=Ax(t)$, including the relation $x_2'(t)=3x_1(t)$.","ad A^3=0.\n\\]\nTherefore the exponential is the exact polynomial\n\\[\ne^{tA}=I+tA+\\frac{t^2}{2}A^2.\n\\]\n\\vfig{ext_paths}\n\\platecaption{Only the empty path, one-edge paths, and the two-edge path exist. The coefficient $3t^2$ includes both the edge product $2\\cdot3$ and the ordered-time simplex area $t^2/2$.}\n\\head{Path length counts transfers, not elapsed time}\nStarting at $e_0$, the complete state is $(1,2t,3t^2)^{\\mathsf T}$. No fourth packet or repeated cycle can contribute. Direct differentiation gives $x'(t)=Ax(t)$, including the relation $x_2'(t)=3x_1(t)$.","t_paths}\n\\platecaption{Only the empty path, one-edge paths, and the two-edge path exist. The coefficient $3t^2$ includes both the edge product $2\\cdot3$ and the ordered-time simplex area $t^2/2$.}\n\\head{Path length counts transfers, not elapsed time}\nStarting at $e_0$, the complete state is $(1,2t,3t^2)^{\\mathsf T}$. No fourth packet or repeated cycle can contribute. Direct differentiation gives $x'(t)=Ax(t)$, including the relation $x_2'(t)=3x_1(t)$.","path, one-edge paths, and the two-edge path exist. The coefficient $3t^2$ includes both the edge product $2\\cdot3$ and the ordered-time simplex area $t^2/2$.}\n\\head{Path length counts transfers, not elapsed time}\nStarting at $e_0$, the complete state is $(1,2t,3t^2)^{\\mathsf T}$. No fourth packet or repeated cycle can contribute. Direct differentiation gives $x'(t)=Ax(t)$, including the relation $x_2'(t)=3x_1(t)$.","This is a linear transfer example. The columns of $A$ do not sum to zero; the total of the displayed state is $1+2t+3t^2$. It is consequently not a Markov probability model. Positivity of off-diagonal coefficients alone does not imply conservation of probability.\n\\takeaway{An acyclic packet graph terminates the transfer expansion. Nilpotence makes this particular full exponential polynomial; the two statements have different hypotheses.}","\\vpage{30}{Convergence does not close the dynamics}{REFINEMENT / A PERSISTENT RANGE EXCURSION}{Traditional Part II: Theorem E4.8 and Example E4.10. Summability and closure are different tests.}\nLet the coarse space be $\\RR$ and every finer space be $\\RR^2$. The first connecting maps are $C(x,y)=x$ and $Rx=(x,0)$; all later maps are identities. Let the coarse evolution be the identity and all fine evolutions be rotations:\n\\[\nT(t)=\\mat{\\cos t&-\\sin t","not close the dynamics}{REFINEMENT / A PERSISTENT RANGE EXCURSION}{Traditional Part II: Theorem E4.8 and Example E4.10. Summability and closure are different tests.}\nLet the coarse space be $\\RR$ and every finer space be $\\RR^2$. The first connecting maps are $C(x,y)=x$ and $Rx=(x,0)$; all later maps are identities. Let the coarse evolution be the identity and all fine evolutions be rotations:\n\\[\nT(t)=\\mat{\\cos t&-\\sin t\\\\\\sin t&\\cos t},\\qquad\nU_n(t)=CT(t)R=\\cos t\\quad(n\\ge1).\n\\]\nThe reconstructed sequence is already constant above its first level. Only one adjacent defect is nonzero, so its defect sum is finite on every bounded time interval.\n\\vfig{ext_closure}\n\\platecaption{The full state returns to the coarse line with negative sign after two quarter turns. Resetting the hidden coordinate after the first quarter turn loses the returning contribution.}\n\\head{A limit can fail the semigroup identity}\nAt $s=t=\\pi/2$,\n\\[\nU(s)U(t)=0,\\qquad U(s+t)=-1.\n\\]\nWrite $P=RC=\\diag(1,0)$. The exact algebraic difference is\n\\[\nU(s)U(t)-U(s+t)=CT(s)(P-I)T(t)R.\n\\]\nIts value is $1$ at these times, matc…","eady constant above its first level. Only one adjacent defect is nonzero, so its defect sum is finite on every bounded time interval.\n\\vfig{ext_closure}\n\\platecaption{The full state returns to the coarse line with negative sign after two quarter turns. Resetting the hidden coordinate after the first quarter turn loses the returning contribution.}\n\\head{A limit can fail the semigroup identity}\nAt $s=t=\\pi/2$,\n\\[\nU(s)U(t)=0,\\qquad U(s+t)=-1.\n\\]\nWrite $P=RC=\\diag(1,0)$. The exact algebraic difference is\n\\[\nU(s)U(t)-U(s+t)=CT(s)(P-I)T(t)R.\n\\]\nIts value is $1$ at these times, matching the missing hidden excursion. This is a failure of closure, not a failure of convergence.","above its first level. Only one adjacent defect is nonzero, so its defect sum is finite on every bounded time interval.\n\\vfig{ext_closure}\n\\platecaption{The full state returns to the coarse line with negative sign after two quarter turns. Resetting the hidden coordinate after the first quarter turn loses the returning contribution.}\n\\head{A limit can fail the semigroup identity}\nAt $s=t=\\pi/2$,\n\\[\nU(s)U(t)=0,\\qquad U(s+t)=-1.\n\\]\nWrite $P=RC=\\diag(1,0)$. The exact algebraic difference is\n\\[\nU(s)U(t)-U(s+t)=CT(s)(P-I)T(t)R.\n\\]\nIts value is $1$ at these times, matching the missing hidden excursion. This is a failure of closure, not a failure of convergence.","defect is nonzero, so its defect sum is finite on every bounded time interval.\n\\vfig{ext_closure}\n\\platecaption{The full state returns to the coarse line with negative sign after two quarter turns. Resetting the hidden coordinate after the first quarter turn loses the returning contribution.}\n\\head{A limit can fail the semigroup identity}\nAt $s=t=\\pi/2$,\n\\[\nU(s)U(t)=0,\\qquad U(s+t)=-1.\n\\]\nWrite $P=RC=\\diag(1,0)$. The exact algebraic difference is\n\\[\nU(s)U(t)-U(s+t)=CT(s)(P-I)T(t)R.\n\\]\nIts value is $1$ at these times, matching the missing hidden excursion. This is a failure of closure, not a failure of convergence.","ry bounded time interval.\n\\vfig{ext_closure}\n\\platecaption{The full state returns to the coarse line with negative sign after two quarter turns. Resetting the hidden coordinate after the first quarter turn loses the returning contribution.}\n\\head{A limit can fail the semigroup identity}\nAt $s=t=\\pi/2$,\n\\[\nU(s)U(t)=0,\\qquad U(s+t)=-1.\n\\]\nWrite $P=RC=\\diag(1,0)$. The exact algebraic difference is\n\\[\nU(s)U(t)-U(s+t)=CT(s)(P-I)T(t)R.\n\\]\nIts value is $1$ at these times, matching the missing hidden excursion. This is a failure of closure, not a failure of convergence.","caption{The full state returns to the coarse line with negative sign after two quarter turns. Resetting the hidden coordinate after the first quarter turn loses the returning contribution.}\n\\head{A limit can fail the semigroup identity}\nAt $s=t=\\pi/2$,\n\\[\nU(s)U(t)=0,\\qquad U(s+t)=-1.\n\\]\nWrite $P=RC=\\diag(1,0)$. The exact algebraic difference is\n\\[\nU(s)U(t)-U(s+t)=CT(s)(P-I)T(t)R.\n\\]\nIts value is $1$ at these times, matching the missing hidden excursion. This is a failure of closure, not a failure of convergence.","\\vpage{31}{A tail bound states what refinement buys}{SUMMABLE DEFECTS / A SCALAR COMPLETION}{Traditional Part II: Theorem E4.5 and Corollary E4.7. This example has identity connecting maps.}\nTake $X_n=\\RR$, $C_n=R_n=1$, and\n\\[\na_n=1-2^{-n},\\qquad A_n=-a_n,\\qquad T_n(t)=e^{-a_nt}\\quad(n\\ge0).\n\\]\nEach semigroup is contractive for $t\\ge0$. The exact adjacent generator-defect magnitude is\n\\[\n|A_{n+1}-A_n|=2^{-(n+1)},\\qquad\n\\sum_{j=n}^{\\infty}|A_{j+1}-A_j|=2^{-n}.\n\\]\n\\vfig{ext_tails}\n\\platecaption{The bars show the certified tail $2^{-n}$ at horizon $T=1$. They are bounds on a whole time interval, not errors sampled at the right endpoint.}\n\\head{The estimate follows before any graph is drawn}\nFor $0\\le t\\le T$, the derivative of $a\\mapsto e^{-at}$ has absolute value at most $T$ when $a\\ge0$. Thus\n\\[\n\\sup_{0\\le t\\le T}|T_n(t)-e^{-t}|\\le T2^{-n}.\n\\]\nThe limit is $e^{-t}$, which does have the semigr","\\vpage{31}{A tail bound states what refinement buys}{SUMMABLE DEFECTS / A SCALAR COMPLETION}{Traditional Part II: Theorem E4.5 and Corollary E4.7. This example has identity connecting maps.}\nTake $X_n=\\RR$, $C_n=R_n=1$, and\n\\[\na_n=1-2^{-n},\\qquad A_n=-a_n,\\qquad T_n(t)=e^{-a_nt}\\quad(n\\ge0).\n\\]\nEach semigroup is contractive for $t\\ge0$. The exact adjacent generator-defect magnitude is\n\\[\n|A_{n+1}-A_n|=2^{-(n+1)},\\qquad\n\\sum_{j=n}^{\\infty}|A_{j+1}-A_j|=2^{-n}.\n\\]\n\\vfig{ext_tails}\n\\platecaption{The bars show the certified tail $2^{-n}$ at horizon $T=1$. They are bounds on a whole time interval, not errors sampled at the right endpoint.}\n\\head{The estimate follows before any graph is drawn}\nFor $0\\le t\\le T$, the derivative of $a\\mapsto e^{-at}$ has absolute value at most $T$ when $a\\ge0$. Thus\n\\[\n\\sup_{0\\le t\\le T}|T_n(t)-e^{-t}|\\le T2^{-n}.\n\\]\nThe limit is $e^{-t}$, which does have the semigroup law. Here","\\vpage{31}{A tail bound states what refinement buys}{SUMMABLE DEFECTS / A SCALAR COMPLETION}{Traditional Part II: Theorem E4.5 and Corollary E4.7. This example has identity connecting maps.}\nTake $X_n=\\RR$, $C_n=R_n=1$, and\n\\[\na_n=1-2^{-n},\\qquad A_n=-a_n,\\qquad T_n(t)=e^{-a_nt}\\quad(n\\ge0).\n\\]\nEach semigroup is contractive for $t\\ge0$. The exact adjacent generator-defect magnitude is\n\\[\n|A_{n+1}-A_n|=2^{-(n+1)},\\qquad\n\\sum_{j=n}^{\\infty}|A_{j+1}-A_j|=2^{-n}.\n\\]\n\\vfig{ext_tails}\n\\platecaption{The bars show the certified tail $2^{-n}$ at horizon $T=1$. They are bounds on a whole time interval, not errors sampled at the right endpoint.}\n\\head{The estimate follows before any graph is drawn}\nFor $0\\le t\\le T$, the derivative of $a\\mapsto e^{-at}$ has absolute value at most $T$ when $a\\ge0$. Thus\n\\[\n\\sup_{0\\le t\\le T}|T_n(t)-e^{-t}|\\le T2^{-n}.\n\\]\nThe limit is $e^{-t}$, which does have the semigroup law. Here every reconstruction is the identity, so the hidden-range excursion is iden","\\vpage{31}{A tail bound states what refinement buys}{SUMMABLE DEFECTS / A SCALAR COMPLETION}{Traditional Part II: Theorem E4.5 and Corollary E4.7. This example has identity connecting maps.}\nTake $X_n=\\RR$, $C_n=R_n=1$, and\n\\[\na_n=1-2^{-n},\\qquad A_n=-a_n,\\qquad T_n(t)=e^{-a_nt}\\quad(n\\ge0).\n\\]\nEach semigroup is contractive for $t\\ge0$. The exact adjacent generator-defect magnitude is\n\\[\n|A_{n+1}-A_n|=2^{-(n+1)},\\qquad\n\\sum_{j=n}^{\\infty}|A_{j+1}-A_j|=2^{-n}.\n\\]\n\\vfig{ext_tails}\n\\platecaption{The bars show the certified tail $2^{-n}$ at horizon $T=1$. They are bounds on a whole time interval, not errors sampled at the right endpoint.}\n\\head{The estimate follows before any graph is drawn}\nFor $0\\le t\\le T$, the derivative of $a\\mapsto e^{-at}$ has absolute value at most $T$ when $a\\ge0$. Thus\n\\[\n\\sup_{0\\le t\\le T}|T_n(t)-e^{-t}|\\le T2^{-n}.\n\\]\nThe limit is $e^{-t}$, which does have the semigroup law. Here every reconstruction is the identity, so the hidden-range excursion is identically zero. That closure fact is addition","\\vpage{31}{A tail bound states what refinement buys}{SUMMABLE DEFECTS / A SCALAR COMPLETION}{Traditional Part II: Theorem E4.5 and Corollary E4.7. This example has identity connecting maps.}\nTake $X_n=\\RR$, $C_n=R_n=1$, and\n\\[\na_n=1-2^{-n},\\qquad A_n=-a_n,\\qquad T_n(t)=e^{-a_nt}\\quad(n\\ge0).\n\\]\nEach semigroup is contractive for $t\\ge0$. The exact adjacent generator-defect magnitude is\n\\[\n|A_{n+1}-A_n|=2^{-(n+1)},\\qquad\n\\sum_{j=n}^{\\infty}|A_{j+1}-A_j|=2^{-n}.\n\\]\n\\vfig{ext_tails}\n\\platecaption{The bars show the certified tail $2^{-n}$ at horizon $T=1$. They are bounds on a whole time interval, not errors sampled at the right endpoint.}\n\\head{The estimate follows before any graph is drawn}\nFor $0\\le t\\le T$, the derivative of $a\\mapsto e^{-at}$ has absolute value at most $T$ when $a\\ge0$. Thus\n\\[\n\\sup_{0\\le t\\le T}|T_n(t)-e^{-t}|\\le T2^{-n}.\n\\]\nThe limit is $e^{-t}$, which does have the semigroup law. Here every reconstruction is the identity, so the hidden-range excursion is identically zero. That closure fact is additional to the tail estimate.","orollary E4.7. This example has identity connecting maps.}\nTake $X_n=\\RR$, $C_n=R_n=1$, and\n\\[\na_n=1-2^{-n},\\qquad A_n=-a_n,\\qquad T_n(t)=e^{-a_nt}\\quad(n\\ge0).\n\\]\nEach semigroup is contractive for $t\\ge0$. The exact adjacent generator-defect magnitude is\n\\[\n|A_{n+1}-A_n|=2^{-(n+1)},\\qquad\n\\sum_{j=n}^{\\infty}|A_{j+1}-A_j|=2^{-n}.\n\\]\n\\vfig{ext_tails}\n\\platecaption{The bars show the certified tail $2^{-n}$ at horizon $T=1$. They are bounds on a whole time interval, not errors sampled at the right endpoint.}\n\\head{The estimate follows before any graph is drawn}\nFor $0\\le t\\le T$, the derivative of $a\\mapsto e^{-at}$ has absolute value at most $T$ when $a\\ge0$. Thus\n\\[\n\\sup_{0\\le t\\le T}|T_n(t)-e^{-t}|\\le T2^{-n}.\n\\]\nThe limit is $e^{-t}$, which does have the semigroup law. Here every reconstruction is the identity, so the hidden-range excursion is identically zero. That closure fact is additional to the tail estimate.","example has identity connecting maps.}\nTake $X_n=\\RR$, $C_n=R_n=1$, and\n\\[\na_n=1-2^{-n},\\qquad A_n=-a_n,\\qquad T_n(t)=e^{-a_nt}\\quad(n\\ge0).\n\\]\nEach semigroup is contractive for $t\\ge0$. The exact adjacent generator-defect magnitude is\n\\[\n|A_{n+1}-A_n|=2^{-(n+1)},\\qquad\n\\sum_{j=n}^{\\infty}|A_{j+1}-A_j|=2^{-n}.\n\\]\n\\vfig{ext_tails}\n\\platecaption{The bars show the certified tail $2^{-n}$ at horizon $T=1$. They are bounds on a whole time interval, not errors sampled at the right endpoint.}\n\\head{The estimate follows before any graph is drawn}\nFor $0\\le t\\le T$, the derivative of $a\\mapsto e^{-at}$ has absolute value at most $T$ when $a\\ge0$. Thus\n\\[\n\\sup_{0\\le t\\le T}|T_n(t)-e^{-t}|\\le T2^{-n}.\n\\]\nThe limit is $e^{-t}$, which does have the semigroup law. Here every reconstruction is the identity, so the hidden-range excursion is identically zero. That closure fact is additional to the tail estimate.","emigroup is contractive for $t\\ge0$. The exact adjacent generator-defect magnitude is\n\\[\n|A_{n+1}-A_n|=2^{-(n+1)},\\qquad\n\\sum_{j=n}^{\\infty}|A_{j+1}-A_j|=2^{-n}.\n\\]\n\\vfig{ext_tails}\n\\platecaption{The bars show the certified tail $2^{-n}$ at horizon $T=1$. They are bounds on a whole time interval, not errors sampled at the right endpoint.}\n\\head{The estimate follows before any graph is drawn}\nFor $0\\le t\\le T$, the derivative of $a\\mapsto e^{-at}$ has absolute value at most $T$ when $a\\ge0$. Thus\n\\[\n\\sup_{0\\le t\\le T}|T_n(t)-e^{-t}|\\le T2^{-n}.\n\\]\nThe limit is $e^{-t}$, which does have the semigroup law. Here every reconstruction is the identity, so the hidden-range excursion is identically zero. That closure fact is additional to the tail estimate.","0$. The exact adjacent generator-defect magnitude is\n\\[\n|A_{n+1}-A_n|=2^{-(n+1)},\\qquad\n\\sum_{j=n}^{\\infty}|A_{j+1}-A_j|=2^{-n}.\n\\]\n\\vfig{ext_tails}\n\\platecaption{The bars show the certified tail $2^{-n}$ at horizon $T=1$. They are bounds on a whole time interval, not errors sampled at the right endpoint.}\n\\head{The estimate follows before any graph is drawn}\nFor $0\\le t\\le T$, the derivative of $a\\mapsto e^{-at}$ has absolute value at most $T$ when $a\\ge0$. Thus\n\\[\n\\sup_{0\\le t\\le T}|T_n(t)-e^{-t}|\\le T2^{-n}.\n\\]\nThe limit is $e^{-t}$, which does have the semigroup law. Here every reconstruction is the identity, so the hidden-range excursion is identically zero. That closure fact is additional to the tail estimate.","ude is\n\\[\n|A_{n+1}-A_n|=2^{-(n+1)},\\qquad\n\\sum_{j=n}^{\\infty}|A_{j+1}-A_j|=2^{-n}.\n\\]\n\\vfig{ext_tails}\n\\platecaption{The bars show the certified tail $2^{-n}$ at horizon $T=1$. They are bounds on a whole time interval, not errors sampled at the right endpoint.}\n\\head{The estimate follows before any graph is drawn}\nFor $0\\le t\\le T$, the derivative of $a\\mapsto e^{-at}$ has absolute value at most $T$ when $a\\ge0$. Thus\n\\[\n\\sup_{0\\le t\\le T}|T_n(t)-e^{-t}|\\le T2^{-n}.\n\\]\nThe limit is $e^{-t}$, which does have the semigroup law. Here every reconstruction is the identity, so the hidden-range excursion is identically zero. That closure fact is additional to the tail estimate.","|A_{n+1}-A_n|=2^{-(n+1)},\\qquad\n\\sum_{j=n}^{\\infty}|A_{j+1}-A_j|=2^{-n}.\n\\]\n\\vfig{ext_tails}\n\\platecaption{The bars show the certified tail $2^{-n}$ at horizon $T=1$. They are bounds on a whole time interval, not errors sampled at the right endpoint.}\n\\head{The estimate follows before any graph is drawn}\nFor $0\\le t\\le T$, the derivative of $a\\mapsto e^{-at}$ has absolute value at most $T$ when $a\\ge0$. Thus\n\\[\n\\sup_{0\\le t\\le T}|T_n(t)-e^{-t}|\\le T2^{-n}.\n\\]\nThe limit is $e^{-t}$, which does have the semigroup law. Here every reconstruction is the identity, so the hidden-range excursion is identically zero. That closure fact is additional to the tail estimate.","2^{-(n+1)},\\qquad\n\\sum_{j=n}^{\\infty}|A_{j+1}-A_j|=2^{-n}.\n\\]\n\\vfig{ext_tails}\n\\platecaption{The bars show the certified tail $2^{-n}$ at horizon $T=1$. They are bounds on a whole time interval, not errors sampled at the right endpoint.}\n\\head{The estimate follows before any graph is drawn}\nFor $0\\le t\\le T$, the derivative of $a\\mapsto e^{-at}$ has absolute value at most $T$ when $a\\ge0$. Thus\n\\[\n\\sup_{0\\le t\\le T}|T_n(t)-e^{-t}|\\le T2^{-n}.\n\\]\nThe limit is $e^{-t}$, which does have the semigroup law. Here every reconstruction is the identity, so the hidden-range excursion is identically zero. That closure fact is additional to the tail estimate.","fig{ext_tails}\n\\platecaption{The bars show the certified tail $2^{-n}$ at horizon $T=1$. They are bounds on a whole time interval, not errors sampled at the right endpoint.}\n\\head{The estimate follows before any graph is drawn}\nFor $0\\le t\\le T$, the derivative of $a\\mapsto e^{-at}$ has absolute value at most $T$ when $a\\ge0$. Thus\n\\[\n\\sup_{0\\le t\\le T}|T_n(t)-e^{-t}|\\le T2^{-n}.\n\\]\nThe limit is $e^{-t}$, which does have the semigroup law. Here every reconstruction is the identity, so the hidden-range excursion is identically zero. That closure fact is additional to the tail estimate.","\\vpage{32}{The obstruction lives in the forgotten directions}{MARKOV DESCENT / TEST THE AGGREGATION KERNEL}{Traditional Part II: Theorem E5.3 and Corollary E5.4. A quotient criterion with a concrete witness.}\nReturn to the four-state model of V--4 and the perturbed stochastic matrix $K_\\varepsilon$ of V--6. Let\n\\[\nC=\\mat{1&1&0&0","\\vpage{32}{The obstruction lives in the forgotten directions}{MARKOV DESCENT / TEST THE AGGREGATION KERNEL}{Traditional Part II: Theorem E5.3 and Corollary E5.4. A quotient criterion with a concrete witness.}\nReturn to the four-state model of V--4 and the perturbed stochastic matrix $K_\\varepsilon$ of V--6. Let\n\\[\nC=\\mat{1&1&0&0\\\\0&0&1&1},\\qquad w=e_0-e_1.\n\\]\nThe vector $w$ changes the fine distribution inside the first packet without changing its mass: $Cw=0$. Yet\n\\[\nCK_\\varepsilon w=\\mat{-1/16","0&0&1&1},\\qquad w=e_0-e_1.\n\\]\nThe vector $w$ changes the fine distribution inside the first packet without changing its mass: $Cw=0$. Yet\n\\[\nCK_\\varepsilon w=\\mat{-1/16","0&0&1&1},\\qquad w=e_0-e_1.\n\\]\nThe vector $w$ changes the fine distribution inside the first packet without changing its mass: $Cw=0$. Yet\n\\[\nCK_\\varepsilon w=\\mat{-1/16\\\\1/16}\\ne0.\n\\]\n\\vfig{ext_obstruction}\n\\platecaption{The coarse input cannot see $w$. The next coarse output can. Therefore no linear coarse operator can reproduce this fine step for every input.}\n\\head{Why one witness rules out every coarse matrix}\nIf some $\\bar K$ satisfied $CK_\\varepsilon=\\bar KC$, then $CK_\\varepsilon w=\\bar KCw=0$, contradicting the displayed vector. The failure is not repaired by fitting a different coarse transition matrix to selected priors.","1/16}\\ne0.\n\\]\n\\vfig{ext_obstruction}\n\\platecaption{The coarse input cannot see $w$. The next coarse output can. Therefore no linear coarse operator can reproduce this fine step for every input.}\n\\head{Why one witness rules out every coarse matrix}\nIf some $\\bar K$ satisfied $CK_\\varepsilon=\\bar KC$, then $CK_\\varepsilon w=\\bar KCw=0$, contradicting the displayed vector. The failure is not repaired by fitting a different coarse transition matrix to selected priors.","More generally, for a surjective aggregation map $C$, a linear map $M$ descends exactly if $CM$ vanishes on $\\ker C$. The descended map is defined by $\\bar M(Cx)=CMx$; the kernel condition makes this independent of the representative $x$.","For a deterministic partition of a finite stochastic system, differences $e_i-e_j$ within the same source block span the forgotten directions. Testing them gives the familiar equality of aggregate columns. Nonnegativity and column sums of the descended stochastic matrix then follow from the fine stochastic matrix and the partition structure.\n\\takeaway{The canonical obstruction is the action on information that aggregation discards. A nonzero witness in that kernel rules out exact descent at once.}","\\vpage{33}{Gaussian packets have separate clocks}{ORNSTEIN--UHLENBECK / HERMITE ENERGY}{Traditional Part II: Theorem E7.1 and Corollaries E7.2, E7.4, and E7.5. Standard Gaussian measure.}\nIn $L^2(\\gamma)$ for the standard Gaussian law on $\\RR$, use the probabilists' Hermite polynomials\n\\[\nH_1(x)=x,\\qquad H_2(x)=x^2-1,\\qquad\nH_4(x)=x^4-6x^2+3.\n\\]\nThe Ornstein--Uhlenbeck generator is $\\mathcal Lf=f''-xf'$. Direct polynomial differentiation gives $\\mathcal LH_n=-nH_n$ for these three polynomials. Their squared norms are $n!$ and different Hermite degrees are orthogonal.\n\\vfig{ext_hermite}\n\\platecaption{For $f=H_1+H_2$, the two bars are its separate squared-energy contributions. At $t=\\log2$, their sum is $3/8$, below the spectral-gap bound $3/4$.}\n\\head{Reconstruction preserves the modes; time changes their weights}\nThe semigroup sends $f=H_1+H_2$ to\n\\[\nT(t)f=e^{-t}H_1+e^{-2t}H_2,\\qquad\n\\|T(t)","\\vpage{33}{Gaussian packets have separate clocks}{ORNSTEIN--UHLENBECK / HERMITE ENERGY}{Traditional Part II: Theorem E7.1 and Corollaries E7.2, E7.4, and E7.5. Standard Gaussian measure.}\nIn $L^2(\\gamma)$ for the standard Gaussian law on $\\RR$, use the probabilists' Hermite polynomials\n\\[\nH_1(x)=x,\\qquad H_2(x)=x^2-1,\\qquad\nH_4(x)=x^4-6x^2+3.\n\\]\nThe Ornstein--Uhlenbeck generator is $\\mathcal Lf=f''-xf'$. Direct polynomial differentiation gives $\\mathcal LH_n=-nH_n$ for these three polynomials. Their squared norms are $n!$ and different Hermite degrees are orthogonal.\n\\vfig{ext_hermite}\n\\platecaption{For $f=H_1+H_2$, the two bars are its separate squared-energy contributions. At $t=\\log2$, their sum is $3/8$, below the spectral-gap bound $3/4$.}\n\\head{Reconstruction preserves the modes; time changes their weights}\nThe semigroup sends $f=H_1+H_2$ to\n\\[\nT(t)f=e^{-t}H_1+e^{-2t}H_2,\\qquad\n\\|T(t)f\\|_2^2=e^{-2t}+2e^{-4t}.\n\\]\nIts mean i","\\vpage{33}{Gaussian packets have separate clocks}{ORNSTEIN--UHLENBECK / HERMITE ENERGY}{Traditional Part II: Theorem E7.1 and Corollaries E7.2, E7.4, and E7.5. Standard Gaussian measure.}\nIn $L^2(\\gamma)$ for the standard Gaussian law on $\\RR$, use the probabilists' Hermite polynomials\n\\[\nH_1(x)=x,\\qquad H_2(x)=x^2-1,\\qquad\nH_4(x)=x^4-6x^2+3.\n\\]\nThe Ornstein--Uhlenbeck generator is $\\mathcal Lf=f''-xf'$. Direct polynomial differentiation gives $\\mathcal LH_n=-nH_n$ for these three polynomials. Their squared norms are $n!$ and different Hermite degrees are orthogonal.\n\\vfig{ext_hermite}\n\\platecaption{For $f=H_1+H_2$, the two bars are its separate squared-energy contributions. At $t=\\log2$, their sum is $3/8$, below the spectral-gap bound $3/4$.}\n\\head{Reconstruction preserves the modes; time changes their weights}\nThe semigroup sends $f=H_1+H_2$ to\n\\[\nT(t)f=e^{-t}H_1+e^{-2t}H_2,\\qquad\n\\|T(t)f\\|_2^2=e^{-2t}+2e^{-4t}.\n\\]\nIts mean is zero. The Gaussian gap bounds this by $e^{-2t}\\|f\\|_2^2=3e^{-2t}$. It need not be an equality when mu","\\vpage{33}{Gaussian packets have separate clocks}{ORNSTEIN--UHLENBECK / HERMITE ENERGY}{Traditional Part II: Theorem E7.1 and Corollaries E7.2, E7.4, and E7.5. Standard Gaussian measure.}\nIn $L^2(\\gamma)$ for the standard Gaussian law on $\\RR$, use the probabilists' Hermite polynomials\n\\[\nH_1(x)=x,\\qquad H_2(x)=x^2-1,\\qquad\nH_4(x)=x^4-6x^2+3.\n\\]\nThe Ornstein--Uhlenbeck generator is $\\mathcal Lf=f''-xf'$. Direct polynomial differentiation gives $\\mathcal LH_n=-nH_n$ for these three polynomials. Their squared norms are $n!$ and different Hermite degrees are orthogonal.\n\\vfig{ext_hermite}\n\\platecaption{For $f=H_1+H_2$, the two bars are its separate squared-energy contributions. At $t=\\log2$, their sum is $3/8$, below the spectral-gap bound $3/4$.}\n\\head{Reconstruction preserves the modes; time changes their weights}\nThe semigroup sends $f=H_1+H_2$ to\n\\[\nT(t)f=e^{-t}H_1+e^{-2t}H_2,\\qquad\n\\|T(t)f\\|_2^2=e^{-2t}+2e^{-4t}.\n\\]\nIts mean is zero. The Gaussian gap bounds this by $e^{-2t}\\|f\\|_2^2=3e^{-2t}$. It need not be an equality when multiple degrees are present.","{ORNSTEIN--UHLENBECK / HERMITE ENERGY}{Traditional Part II: Theorem E7.1 and Corollaries E7.2, E7.4, and E7.5. Standard Gaussian measure.}\nIn $L^2(\\gamma)$ for the standard Gaussian law on $\\RR$, use the probabilists' Hermite polynomials\n\\[\nH_1(x)=x,\\qquad H_2(x)=x^2-1,\\qquad\nH_4(x)=x^4-6x^2+3.\n\\]\nThe Ornstein--Uhlenbeck generator is $\\mathcal Lf=f''-xf'$. Direct polynomial differentiation gives $\\mathcal LH_n=-nH_n$ for these three polynomials. Their squared norms are $n!$ and different Hermite degrees are orthogonal.\n\\vfig{ext_hermite}\n\\platecaption{For $f=H_1+H_2$, the two bars are its separate squared-energy contributions. At $t=\\log2$, their sum is $3/8$, below the spectral-gap bound $3/4$.}\n\\head{Reconstruction preserves the modes; time changes their weights}\nThe semigroup sends $f=H_1+H_2$ to\n\\[\nT(t)f=e^{-t}H_1+e^{-2t}H_2,\\qquad\n\\|T(t)f\\|_2^2=e^{-2t}+2e^{-4t}.\n\\]\nIts mean is zero. The Gaussian gap bounds this by $e^{-2t}\\|f\\|_2^2=3e^{-2t}$. It need not be an equality when multiple degrees are present.","d Corollaries E7.2, E7.4, and E7.5. Standard Gaussian measure.}\nIn $L^2(\\gamma)$ for the standard Gaussian law on $\\RR$, use the probabilists' Hermite polynomials\n\\[\nH_1(x)=x,\\qquad H_2(x)=x^2-1,\\qquad\nH_4(x)=x^4-6x^2+3.\n\\]\nThe Ornstein--Uhlenbeck generator is $\\mathcal Lf=f''-xf'$. Direct polynomial differentiation gives $\\mathcal LH_n=-nH_n$ for these three polynomials. Their squared norms are $n!$ and different Hermite degrees are orthogonal.\n\\vfig{ext_hermite}\n\\platecaption{For $f=H_1+H_2$, the two bars are its separate squared-energy contributions. At $t=\\log2$, their sum is $3/8$, below the spectral-gap bound $3/4$.}\n\\head{Reconstruction preserves the modes; time changes their weights}\nThe semigroup sends $f=H_1+H_2$ to\n\\[\nT(t)f=e^{-t}H_1+e^{-2t}H_2,\\qquad\n\\|T(t)f\\|_2^2=e^{-2t}+2e^{-4t}.\n\\]\nIts mean is zero. The Gaussian gap bounds this by $e^{-2t}\\|f\\|_2^2=3e^{-2t}$. It need not be an equality when multiple degrees are present.","standard Gaussian law on $\\RR$, use the probabilists' Hermite polynomials\n\\[\nH_1(x)=x,\\qquad H_2(x)=x^2-1,\\qquad\nH_4(x)=x^4-6x^2+3.\n\\]\nThe Ornstein--Uhlenbeck generator is $\\mathcal Lf=f''-xf'$. Direct polynomial differentiation gives $\\mathcal LH_n=-nH_n$ for these three polynomials. Their squared norms are $n!$ and different Hermite degrees are orthogonal.\n\\vfig{ext_hermite}\n\\platecaption{For $f=H_1+H_2$, the two bars are its separate squared-energy contributions. At $t=\\log2$, their sum is $3/8$, below the spectral-gap bound $3/4$.}\n\\head{Reconstruction preserves the modes; time changes their weights}\nThe semigroup sends $f=H_1+H_2$ to\n\\[\nT(t)f=e^{-t}H_1+e^{-2t}H_2,\\qquad\n\\|T(t)f\\|_2^2=e^{-2t}+2e^{-4t}.\n\\]\nIts mean is zero. The Gaussian gap bounds this by $e^{-2t}\\|f\\|_2^2=3e^{-2t}$. It need not be an equality when multiple degrees are present.","[\nH_1(x)=x,\\qquad H_2(x)=x^2-1,\\qquad\nH_4(x)=x^4-6x^2+3.\n\\]\nThe Ornstein--Uhlenbeck generator is $\\mathcal Lf=f''-xf'$. Direct polynomial differentiation gives $\\mathcal LH_n=-nH_n$ for these three polynomials. Their squared norms are $n!$ and different Hermite degrees are orthogonal.\n\\vfig{ext_hermite}\n\\platecaption{For $f=H_1+H_2$, the two bars are its separate squared-energy contributions. At $t=\\log2$, their sum is $3/8$, below the spectral-gap bound $3/4$.}\n\\head{Reconstruction preserves the modes; time changes their weights}\nThe semigroup sends $f=H_1+H_2$ to\n\\[\nT(t)f=e^{-t}H_1+e^{-2t}H_2,\\qquad\n\\|T(t)f\\|_2^2=e^{-2t}+2e^{-4t}.\n\\]\nIts mean is zero. The Gaussian gap bounds this by $e^{-2t}\\|f\\|_2^2=3e^{-2t}$. It need not be an equality when multiple degrees are present.","=x^2-1,\\qquad\nH_4(x)=x^4-6x^2+3.\n\\]\nThe Ornstein--Uhlenbeck generator is $\\mathcal Lf=f''-xf'$. Direct polynomial differentiation gives $\\mathcal LH_n=-nH_n$ for these three polynomials. Their squared norms are $n!$ and different Hermite degrees are orthogonal.\n\\vfig{ext_hermite}\n\\platecaption{For $f=H_1+H_2$, the two bars are its separate squared-energy contributions. At $t=\\log2$, their sum is $3/8$, below the spectral-gap bound $3/4$.}\n\\head{Reconstruction preserves the modes; time changes their weights}\nThe semigroup sends $f=H_1+H_2$ to\n\\[\nT(t)f=e^{-t}H_1+e^{-2t}H_2,\\qquad\n\\|T(t)f\\|_2^2=e^{-2t}+2e^{-4t}.\n\\]\nIts mean is zero. The Gaussian gap bounds this by $e^{-2t}\\|f\\|_2^2=3e^{-2t}$. It need not be an equality when multiple degrees are present.","The Ornstein--Uhlenbeck generator is $\\mathcal Lf=f''-xf'$. Direct polynomial differentiation gives $\\mathcal LH_n=-nH_n$ for these three polynomials. Their squared norms are $n!$ and different Hermite degrees are orthogonal.\n\\vfig{ext_hermite}\n\\platecaption{For $f=H_1+H_2$, the two bars are its separate squared-energy contributions. At $t=\\log2$, their sum is $3/8$, below the spectral-gap bound $3/4$.}\n\\head{Reconstruction preserves the modes; time changes their weights}\nThe semigroup sends $f=H_1+H_2$ to\n\\[\nT(t)f=e^{-t}H_1+e^{-2t}H_2,\\qquad\n\\|T(t)f\\|_2^2=e^{-2t}+2e^{-4t}.\n\\]\nIts mean is zero. The Gaussian gap bounds this by $e^{-2t}\\|f\\|_2^2=3e^{-2t}$. It need not be an equality when multiple degrees are present.","s $\\mathcal LH_n=-nH_n$ for these three polynomials. Their squared norms are $n!$ and different Hermite degrees are orthogonal.\n\\vfig{ext_hermite}\n\\platecaption{For $f=H_1+H_2$, the two bars are its separate squared-energy contributions. At $t=\\log2$, their sum is $3/8$, below the spectral-gap bound $3/4$.}\n\\head{Reconstruction preserves the modes; time changes their weights}\nThe semigroup sends $f=H_1+H_2$ to\n\\[\nT(t)f=e^{-t}H_1+e^{-2t}H_2,\\qquad\n\\|T(t)f\\|_2^2=e^{-2t}+2e^{-4t}.\n\\]\nIts mean is zero. The Gaussian gap bounds this by $e^{-2t}\\|f\\|_2^2=3e^{-2t}$. It need not be an equality when multiple degrees are present.","=-nH_n$ for these three polynomials. Their squared norms are $n!$ and different Hermite degrees are orthogonal.\n\\vfig{ext_hermite}\n\\platecaption{For $f=H_1+H_2$, the two bars are its separate squared-energy contributions. At $t=\\log2$, their sum is $3/8$, below the spectral-gap bound $3/4$.}\n\\head{Reconstruction preserves the modes; time changes their weights}\nThe semigroup sends $f=H_1+H_2$ to\n\\[\nT(t)f=e^{-t}H_1+e^{-2t}H_2,\\qquad\n\\|T(t)f\\|_2^2=e^{-2t}+2e^{-4t}.\n\\]\nIts mean is zero. The Gaussian gap bounds this by $e^{-2t}\\|f\\|_2^2=3e^{-2t}$. It need not be an equality when multiple degrees are present.","_hermite}\n\\platecaption{For $f=H_1+H_2$, the two bars are its separate squared-energy contributions. At $t=\\log2$, their sum is $3/8$, below the spectral-gap bound $3/4$.}\n\\head{Reconstruction preserves the modes; time changes their weights}\nThe semigroup sends $f=H_1+H_2$ to\n\\[\nT(t)f=e^{-t}H_1+e^{-2t}H_2,\\qquad\n\\|T(t)f\\|_2^2=e^{-2t}+2e^{-4t}.\n\\]\nIts mean is zero. The Gaussian gap bounds this by $e^{-2t}\\|f\\|_2^2=3e^{-2t}$. It need not be an equality when multiple degrees are present.","For the nonnegative operator $-\\mathcal L$, a square-root evolution replaces rate $n$ by $\\sqrt n$. On $g=H_1+H_4$, the coefficients at $t=\\log2$ are $1/2$ and $1/4$, whereas ordinary OU evolution gives $1/2$ and $1/16$. The eigenfunctions agree; the rates differ.","These calculations concern polynomial elements of Gaussian $L^2$. The general expansion and generator-domain statements belong to the traditional proof; a finite list of Hermite checks does not establish completeness of the basis.\n\\takeaway{Orthogonal packets make the energy accounting exact. A fractional generator alters the spectral clock while keeping the same Gaussian modes.}","\\vpage{34}{Several observations determine one sufficient partition}{FIXED LIKELIHOODS / JOINT PROFILES}{Traditional Part II: Theorem E9.3. A finite family of strictly positive likelihoods.}\nOn four states, consider two likelihoods\n\\[\n\\lambda^{(1)}=(2,2,1,1),\\qquad\n\\lambda^{(2)}=(1,1,3,1).\n\\]\nTwo states may share a packet in the coarsest partition on which every listed likelihood is constant exactly when their entire likelihood profiles agree. The profiles are\n\\[\n(2,1),\\ (2,1),\\ (1,3),\\ (1,1),\n\\]\nso the partition is $\\{0,1\\},\\{2\\},\\{3\\}$.\n\\vfig{ext_profiles}\n\\platecaption{The first observation alone allows the last two states to be merged. The second observation distinguishes them. The shared partition must respect both likelihoods.}\n\\head{A complete two-stage calculation}\nStart from the uniform prior. The coarse prior is $(1/2,1/4,1/4)$. After the first likelihood, it becomes $(2/3,1/6,1/6)$; after the second, it is\n\\[\n(1/2,3/8,1/8).\n\\]\nThe direct fine update multiplies the l","}{FIXED LIKELIHOODS / JOINT PROFILES}{Traditional Part II: Theorem E9.3. A finite family of strictly positive likelihoods.}\nOn four states, consider two likelihoods\n\\[\n\\lambda^{(1)}=(2,2,1,1),\\qquad\n\\lambda^{(2)}=(1,1,3,1).\n\\]\nTwo states may share a packet in the coarsest partition on which every listed likelihood is constant exactly when their entire likelihood profiles agree. The profiles are\n\\[\n(2,1),\\ (2,1),\\ (1,3),\\ (1,1),\n\\]\nso the partition is $\\{0,1\\},\\{2\\},\\{3\\}$.\n\\vfig{ext_profiles}\n\\platecaption{The first observation alone allows the last two states to be merged. The second observation distinguishes them. The shared partition must respect both likelihoods.}\n\\head{A complete two-stage calculation}\nStart from the uniform prior. The coarse prior is $(1/2,1/4,1/4)$. After the first likelihood, it becomes $(2/3,1/6,1/6)$; after the second, it is\n\\[\n(1/2,3/8,1/8).\n\\]\nThe direct fine update multiplies the likelihoods componentwise, giving weights $(2,2,3,1)$ and posterior $(2,2,3,1)/8$. Its packet masses are precisely the displayed coarse posterior.","I: Theorem E9.3. A finite family of strictly positive likelihoods.}\nOn four states, consider two likelihoods\n\\[\n\\lambda^{(1)}=(2,2,1,1),\\qquad\n\\lambda^{(2)}=(1,1,3,1).\n\\]\nTwo states may share a packet in the coarsest partition on which every listed likelihood is constant exactly when their entire likelihood profiles agree. The profiles are\n\\[\n(2,1),\\ (2,1),\\ (1,3),\\ (1,1),\n\\]\nso the partition is $\\{0,1\\},\\{2\\},\\{3\\}$.\n\\vfig{ext_profiles}\n\\platecaption{The first observation alone allows the last two states to be merged. The second observation distinguishes them. The shared partition must respect both likelihoods.}\n\\head{A complete two-stage calculation}\nStart from the uniform prior. The coarse prior is $(1/2,1/4,1/4)$. After the first likelihood, it becomes $(2/3,1/6,1/6)$; after the second, it is\n\\[\n(1/2,3/8,1/8).\n\\]\nThe direct fine update multiplies the likelihoods componentwise, giving weights $(2,2,3,1)$ and posterior $(2,2,3,1)/8$. Its packet masses are precisely the displayed coarse posterior.","files agree. The profiles are\n\\[\n(2,1),\\ (2,1),\\ (1,3),\\ (1,1),\n\\]\nso the partition is $\\{0,1\\},\\{2\\},\\{3\\}$.\n\\vfig{ext_profiles}\n\\platecaption{The first observation alone allows the last two states to be merged. The second observation distinguishes them. The shared partition must respect both likelihoods.}\n\\head{A complete two-stage calculation}\nStart from the uniform prior. The coarse prior is $(1/2,1/4,1/4)$. After the first likelihood, it becomes $(2/3,1/6,1/6)$; after the second, it is\n\\[\n(1/2,3/8,1/8).\n\\]\nThe direct fine update multiplies the likelihoods componentwise, giving weights $(2,2,3,1)$ and posterior $(2,2,3,1)/8$. Its packet masses are precisely the displayed coarse posterior.","(1,1),\n\\]\nso the partition is $\\{0,1\\},\\{2\\},\\{3\\}$.\n\\vfig{ext_profiles}\n\\platecaption{The first observation alone allows the last two states to be merged. The second observation distinguishes them. The shared partition must respect both likelihoods.}\n\\head{A complete two-stage calculation}\nStart from the uniform prior. The coarse prior is $(1/2,1/4,1/4)$. After the first likelihood, it becomes $(2/3,1/6,1/6)$; after the second, it is\n\\[\n(1/2,3/8,1/8).\n\\]\nThe direct fine update multiplies the likelihoods componentwise, giving weights $(2,2,3,1)$ and posterior $(2,2,3,1)/8$. Its packet masses are precisely the displayed coarse posterior.","2\\},\\{3\\}$.\n\\vfig{ext_profiles}\n\\platecaption{The first observation alone allows the last two states to be merged. The second observation distinguishes them. The shared partition must respect both likelihoods.}\n\\head{A complete two-stage calculation}\nStart from the uniform prior. The coarse prior is $(1/2,1/4,1/4)$. After the first likelihood, it becomes $(2/3,1/6,1/6)$; after the second, it is\n\\[\n(1/2,3/8,1/8).\n\\]\nThe direct fine update multiplies the likelihoods componentwise, giving weights $(2,2,3,1)$ and posterior $(2,2,3,1)/8$. Its packet masses are precisely the displayed coarse posterior.","to be merged. The second observation distinguishes them. The shared partition must respect both likelihoods.}\n\\head{A complete two-stage calculation}\nStart from the uniform prior. The coarse prior is $(1/2,1/4,1/4)$. After the first likelihood, it becomes $(2/3,1/6,1/6)$; after the second, it is\n\\[\n(1/2,3/8,1/8).\n\\]\nThe direct fine update multiplies the likelihoods componentwise, giving weights $(2,2,3,1)$ and posterior $(2,2,3,1)/8$. Its packet masses are precisely the displayed coarse posterior.","observation distinguishes them. The shared partition must respect both likelihoods.}\n\\head{A complete two-stage calculation}\nStart from the uniform prior. The coarse prior is $(1/2,1/4,1/4)$. After the first likelihood, it becomes $(2/3,1/6,1/6)$; after the second, it is\n\\[\n(1/2,3/8,1/8).\n\\]\nThe direct fine update multiplies the likelihoods componentwise, giving weights $(2,2,3,1)$ and posterior $(2,2,3,1)/8$. Its packet masses are precisely the displayed coarse posterior.","The individual evidences are $Z_1=3/2$ and $Z_2=4/3$, and their product is $2$, the direct evidence of the product likelihood. These are relative likelihood weights, so evidence larger than one is allowed.","This partition supplies fixed coarse likelihoods that work for every prior. A single-packet aggregation always gives posterior $(1)$; that mass-only commutation does not identify likelihood profiles. A prior-dependent conditional average can summarize one update on a coarser partition, but requires the within-packet information that a fixed sufficient rule avoids.\n\\takeaway{Equal full likelihood profiles determine the coarsest partition on which every listed likelihood is constant. This partition admits fixed sufficient update rules.}","\\vpage{35}{Information loss has an exact location}{RELATIVE ENTROPY / COARSE AND CONDITIONAL TERMS}{Traditional Part II: Theorem E9.8 and Corollary E9.9. Positive reference masses in this example.}\nTake\n\\[\np=(1/2,1/4,1/8,1/8),\\qquad q=(1/4,1/4,1/4,1/4)\n\\]\nand group the first two and last two states. Their coarse laws are $P=(3/4,1/4)$ and $Q=(1/2,1/2)$. Define $D(p\\Vert q)=\\sum_i p_i\\log(p_i/q_i)$ with natural logarithms.\n\\vfig{ext_information}\n\\platecaption{The bar lengths use the displayed exact formulas evaluated for illustration. The split is an identity: coarse information plus posterior-weighted conditional information. It is not a fitted allocation.}\n\\head{The chain rule identifies exactly what aggregation removes}\nInside packet A, the conditional laws are $(2/3,1/3)$ and $(1/2,1/2)$; inside B they agree. Hence\n\\[\n\\begin{aligned}\nD(p\\Vert q)&=\\tfrac14\\log2,","=\\sum_i p_i\\log(p_i/q_i)$ with natural logarithms.\n\\vfig{ext_information}\n\\platecaption{The bar lengths use the displayed exact formulas evaluated for illustration. The split is an identity: coarse information plus posterior-weighted conditional information. It is not a fitted allocation.}\n\\head{The chain rule identifies exactly what aggregation removes}\nInside packet A, the conditional laws are $(2/3,1/3)$ and $(1/2,1/2)$; inside B they agree. Hence\n\\[\n\\begin{aligned}\nD(p\\Vert q)&=\\tfrac14\\log2,","p_i/q_i)$ with natural logarithms.\n\\vfig{ext_information}\n\\platecaption{The bar lengths use the displayed exact formulas evaluated for illustration. The split is an identity: coarse information plus posterior-weighted conditional information. It is not a fitted allocation.}\n\\head{The chain rule identifies exactly what aggregation removes}\nInside packet A, the conditional laws are $(2/3,1/3)$ and $(1/2,1/2)$; inside B they agree. Hence\n\\[\n\\begin{aligned}\nD(p\\Vert q)&=\\tfrac14\\log2,","ext_information}\n\\platecaption{The bar lengths use the displayed exact formulas evaluated for illustration. The split is an identity: coarse information plus posterior-weighted conditional information. It is not a fitted allocation.}\n\\head{The chain rule identifies exactly what aggregation removes}\nInside packet A, the conditional laws are $(2/3,1/3)$ and $(1/2,1/2)$; inside B they agree. Hence\n\\[\n\\begin{aligned}\nD(p\\Vert q)&=\\tfrac14\\log2,\\\\\nD(P\\Vert Q)&=\\tfrac34\\log3-\\log2,\\\\\n\\sum_aP_aD(p(\\cdot\\mid a)\\Vert q(\\cdot\\mid a))\n&=\\tfrac54\\log2-\\tfrac34\\log3.\n\\end{aligned}\n\\]\nAdding the last two lines recovers the first. Every conditional term is nonnegative, so aggregation cannot increase relative entropy. Equality holds precisely when the occupied packets have the same conditional laws, under the stated support convention.","The weighting is by the packet mass $P_a$ from the first argument of $D(p\\Vert q)$. Replacing it by $Q_a$ generally breaks the identity. Likewise, forgetting zero-mass conventions or allowing a positive $p_i$ over a zero $q_i$ changes which expressions are finite.\n\\takeaway{Coarse relative entropy measures distinguishability that survives aggregation. The conditional term records exactly the distinguishability removed inside the packets.}","\\vpage{36}{The score rises because it has variance}{BAYESIAN FLOW / EVIDENCE CUMULANTS}{Traditional Part II: Theorems E10.1 and E10.2. Exponential tilting of a fixed finite prior.}\nStart with equal prior masses on scores $u=(0,1,2)$ and likelihood $e^{tu}$. Put $q=e^t$. The evidence and tilted distribution are\n\\[\nZ(t)=\\frac{1+q+q^2}{3},\\qquad\np(t)=\\frac{(1,q,q^2)}{1+q+q^2}.\n\\]\nAt $t=\\log2$, they are $Z=7/3$ and $p=(1,2,4)/7$.\n\\vfig{ext_cumulants}\n\\platecaption{The score values stay fixed. Their probabilities change. The rising mean score is a property of this tilted family, not an increase in the individual scores.}\n\\head{Two derivatives have different statistical meanings}\nThe first derivative of log evidence is the current mean score, and the second is its variance:\n\\[\n\\frac{d}{dt}\\log Z(t)=\\EE_{p(t)}u,\n\\qquad\n\\frac{d^2}{dt^2}\\log Z(t)=\\Var_{p(t)}(u)\\ge0.\n\\]\nAt $q=2$ the exact values are\n\\[\n\\EE u=\\frac{10}{7},\\qquad","\\vpage{36}{The score rises because it has variance}{BAYESIAN FLOW / EVIDENCE CUMULANTS}{Traditional Part II: Theorems E10.1 and E10.2. Exponential tilting of a fixed finite prior.}\nStart with equal prior masses on scores $u=(0,1,2)$ and likelihood $e^{tu}$. Put $q=e^t$. The evidence and tilted distribution are\n\\[\nZ(t)=\\frac{1+q+q^2}{3},\\qquad\np(t)=\\frac{(1,q,q^2)}{1+q+q^2}.\n\\]\nAt $t=\\log2$, they are $Z=7/3$ and $p=(1,2,4)/7$.\n\\vfig{ext_cumulants}\n\\platecaption{The score values stay fixed. Their probabilities change. The rising mean score is a property of this tilted family, not an increase in the individual scores.}\n\\head{Two derivatives have different statistical meanings}\nThe first derivative of log evidence is the current mean score, and the second is its variance:\n\\[\n\\frac{d}{dt}\\log Z(t)=\\EE_{p(t)}u,\n\\qquad\n\\frac{d^2}{dt^2}\\log Z(t)=\\Var_{p(t)}(u)\\ge0.\n\\]\nAt $q=2$ the exact values are\n\\[\n\\EE u=\\frac{10}{7},\\qquad\n\\EE u^2=\\frac{18}{7},\\q","\\vpage{36}{The score rises because it has variance}{BAYESIAN FLOW / EVIDENCE CUMULANTS}{Traditional Part II: Theorems E10.1 and E10.2. Exponential tilting of a fixed finite prior.}\nStart with equal prior masses on scores $u=(0,1,2)$ and likelihood $e^{tu}$. Put $q=e^t$. The evidence and tilted distribution are\n\\[\nZ(t)=\\frac{1+q+q^2}{3},\\qquad\np(t)=\\frac{(1,q,q^2)}{1+q+q^2}.\n\\]\nAt $t=\\log2$, they are $Z=7/3$ and $p=(1,2,4)/7$.\n\\vfig{ext_cumulants}\n\\platecaption{The score values stay fixed. Their probabilities change. The rising mean score is a property of this tilted family, not an increase in the individual scores.}\n\\head{Two derivatives have different statistical meanings}\nThe first derivative of log evidence is the current mean score, and the second is its variance:\n\\[\n\\frac{d}{dt}\\log Z(t)=\\EE_{p(t)}u,\n\\qquad\n\\frac{d^2}{dt^2}\\log Z(t)=\\Var_{p(t)}(u)\\ge0.\n\\]\nAt $q=2$ the exact values are\n\\[\n\\EE u=\\frac{10}{7},\\qquad\n\\EE u^2=\\frac{18}{7},\\qquad\n\\Var(u)=","\\vpage{36}{The score rises because it has variance}{BAYESIAN FLOW / EVIDENCE CUMULANTS}{Traditional Part II: Theorems E10.1 and E10.2. Exponential tilting of a fixed finite prior.}\nStart with equal prior masses on scores $u=(0,1,2)$ and likelihood $e^{tu}$. Put $q=e^t$. The evidence and tilted distribution are\n\\[\nZ(t)=\\frac{1+q+q^2}{3},\\qquad\np(t)=\\frac{(1,q,q^2)}{1+q+q^2}.\n\\]\nAt $t=\\log2$, they are $Z=7/3$ and $p=(1,2,4)/7$.\n\\vfig{ext_cumulants}\n\\platecaption{The score values stay fixed. Their probabilities change. The rising mean score is a property of this tilted family, not an increase in the individual scores.}\n\\head{Two derivatives have different statistical meanings}\nThe first derivative of log evidence is the current mean score, and the second is its variance:\n\\[\n\\frac{d}{dt}\\log Z(t)=\\EE_{p(t)}u,\n\\qquad\n\\frac{d^2}{dt^2}\\log Z(t)=\\Var_{p(t)}(u)\\ge0.\n\\]\nAt $q=2$ the exact values are\n\\[\n\\EE u=\\frac{10}{7},\\qquad\n\\EE u^2=\\frac{18}{7},\\qquad\n\\Var(u)=\\frac{26}{49}.\n\\]\nThus the derivative of the mean is $26/49$ at this time. The flow equation $p_i'=(u_i-\\EE u","\\vpage{36}{The score rises because it has variance}{BAYESIAN FLOW / EVIDENCE CUMULANTS}{Traditional Part II: Theorems E10.1 and E10.2. Exponential tilting of a fixed finite prior.}\nStart with equal prior masses on scores $u=(0,1,2)$ and likelihood $e^{tu}$. Put $q=e^t$. The evidence and tilted distribution are\n\\[\nZ(t)=\\frac{1+q+q^2}{3},\\qquad\np(t)=\\frac{(1,q,q^2)}{1+q+q^2}.\n\\]\nAt $t=\\log2$, they are $Z=7/3$ and $p=(1,2,4)/7$.\n\\vfig{ext_cumulants}\n\\platecaption{The score values stay fixed. Their probabilities change. The rising mean score is a property of this tilted family, not an increase in the individual scores.}\n\\head{Two derivatives have different statistical meanings}\nThe first derivative of log evidence is the current mean score, and the second is its variance:\n\\[\n\\frac{d}{dt}\\log Z(t)=\\EE_{p(t)}u,\n\\qquad\n\\frac{d^2}{dt^2}\\log Z(t)=\\Var_{p(t)}(u)\\ge0.\n\\]\nAt $q=2$ the exact values are\n\\[\n\\EE u=\\frac{10}{7},\\qquad\n\\EE u^2=\\frac{18}{7},\\qquad\n\\Var(u)=\\frac{26}{49}.\n\\]\nThus the derivative of the mean is $26/49$ at this time. The flow equation $p_i'=(u_i-\\EE u)p_i$ gives de","ge{36}{The score rises because it has variance}{BAYESIAN FLOW / EVIDENCE CUMULANTS}{Traditional Part II: Theorems E10.1 and E10.2. Exponential tilting of a fixed finite prior.}\nStart with equal prior masses on scores $u=(0,1,2)$ and likelihood $e^{tu}$. Put $q=e^t$. The evidence and tilted distribution are\n\\[\nZ(t)=\\frac{1+q+q^2}{3},\\qquad\np(t)=\\frac{(1,q,q^2)}{1+q+q^2}.\n\\]\nAt $t=\\log2$, they are $Z=7/3$ and $p=(1,2,4)/7$.\n\\vfig{ext_cumulants}\n\\platecaption{The score values stay fixed. Their probabilities change. The rising mean score is a property of this tilted family, not an increase in the individual scores.}\n\\head{Two derivatives have different statistical meanings}\nThe first derivative of log evidence is the current mean score, and the second is its variance:\n\\[\n\\frac{d}{dt}\\log Z(t)=\\EE_{p(t)}u,\n\\qquad\n\\frac{d^2}{dt^2}\\log Z(t)=\\Var_{p(t)}(u)\\ge0.\n\\]\nAt $q=2$ the exact values are\n\\[\n\\EE u=\\frac{10}{7},\\qquad\n\\EE u^2=\\frac{18}{7},\\qquad\n\\Var(u)=\\frac{26}{49}.\n\\]\nThus the derivative of the mean is $26/49$ at this time. The flow equation $p_i'=(u_i-\\EE u)p_i$ gives derivatives $(-…","core rises because it has variance}{BAYESIAN FLOW / EVIDENCE CUMULANTS}{Traditional Part II: Theorems E10.1 and E10.2. Exponential tilting of a fixed finite prior.}\nStart with equal prior masses on scores $u=(0,1,2)$ and likelihood $e^{tu}$. Put $q=e^t$. The evidence and tilted distribution are\n\\[\nZ(t)=\\frac{1+q+q^2}{3},\\qquad\np(t)=\\frac{(1,q,q^2)}{1+q+q^2}.\n\\]\nAt $t=\\log2$, they are $Z=7/3$ and $p=(1,2,4)/7$.\n\\vfig{ext_cumulants}\n\\platecaption{The score values stay fixed. Their probabilities change. The rising mean score is a property of this tilted family, not an increase in the individual scores.}\n\\head{Two derivatives have different statistical meanings}\nThe first derivative of log evidence is the current mean score, and the second is its variance:\n\\[\n\\frac{d}{dt}\\log Z(t)=\\EE_{p(t)}u,\n\\qquad\n\\frac{d^2}{dt^2}\\log Z(t)=\\Var_{p(t)}(u)\\ge0.\n\\]\nAt $q=2$ the exact values are\n\\[\n\\EE u=\\frac{10}{7},\\qquad\n\\EE u^2=\\frac{18}{7},\\qquad\n\\Var(u)=\\frac{26}{49}.\n\\]\nThus the derivative of the mean is $26/49$ at this time. The flow equation $p_i'=(u_i-\\EE u)p_i$ gives derivatives $(-10,-6,16)/49…","t $t=\\log2$, they are $Z=7/3$ and $p=(1,2,4)/7$.\n\\vfig{ext_cumulants}\n\\platecaption{The score values stay fixed. Their probabilities change. The rising mean score is a property of this tilted family, not an increase in the individual scores.}\n\\head{Two derivatives have different statistical meanings}\nThe first derivative of log evidence is the current mean score, and the second is its variance:\n\\[\n\\frac{d}{dt}\\log Z(t)=\\EE_{p(t)}u,\n\\qquad\n\\frac{d^2}{dt^2}\\log Z(t)=\\Var_{p(t)}(u)\\ge0.\n\\]\nAt $q=2$ the exact values are\n\\[\n\\EE u=\\frac{10}{7},\\qquad\n\\EE u^2=\\frac{18}{7},\\qquad\n\\Var(u)=\\frac{26}{49}.\n\\]\nThus the derivative of the mean is $26/49$ at this time. The flow equation $p_i'=(u_i-\\EE u)p_i$ gives derivatives $(-10,-6,16)/49$, which sum to zero.","lues stay fixed. Their probabilities change. The rising mean score is a property of this tilted family, not an increase in the individual scores.}\n\\head{Two derivatives have different statistical meanings}\nThe first derivative of log evidence is the current mean score, and the second is its variance:\n\\[\n\\frac{d}{dt}\\log Z(t)=\\EE_{p(t)}u,\n\\qquad\n\\frac{d^2}{dt^2}\\log Z(t)=\\Var_{p(t)}(u)\\ge0.\n\\]\nAt $q=2$ the exact values are\n\\[\n\\EE u=\\frac{10}{7},\\qquad\n\\EE u^2=\\frac{18}{7},\\qquad\n\\Var(u)=\\frac{26}{49}.\n\\]\nThus the derivative of the mean is $26/49$ at this time. The flow equation $p_i'=(u_i-\\EE u)p_i$ gives derivatives $(-10,-6,16)/49$, which sum to zero.","bilities change. The rising mean score is a property of this tilted family, not an increase in the individual scores.}\n\\head{Two derivatives have different statistical meanings}\nThe first derivative of log evidence is the current mean score, and the second is its variance:\n\\[\n\\frac{d}{dt}\\log Z(t)=\\EE_{p(t)}u,\n\\qquad\n\\frac{d^2}{dt^2}\\log Z(t)=\\Var_{p(t)}(u)\\ge0.\n\\]\nAt $q=2$ the exact values are\n\\[\n\\EE u=\\frac{10}{7},\\qquad\n\\EE u^2=\\frac{18}{7},\\qquad\n\\Var(u)=\\frac{26}{49}.\n\\]\nThus the derivative of the mean is $26/49$ at this time. The flow equation $p_i'=(u_i-\\EE u)p_i$ gives derivatives $(-10,-6,16)/49$, which sum to zero.","}\n\\head{Two derivatives have different statistical meanings}\nThe first derivative of log evidence is the current mean score, and the second is its variance:\n\\[\n\\frac{d}{dt}\\log Z(t)=\\EE_{p(t)}u,\n\\qquad\n\\frac{d^2}{dt^2}\\log Z(t)=\\Var_{p(t)}(u)\\ge0.\n\\]\nAt $q=2$ the exact values are\n\\[\n\\EE u=\\frac{10}{7},\\qquad\n\\EE u^2=\\frac{18}{7},\\qquad\n\\Var(u)=\\frac{26}{49}.\n\\]\nThus the derivative of the mean is $26/49$ at this time. The flow equation $p_i'=(u_i-\\EE u)p_i$ gives derivatives $(-10,-6,16)/49$, which sum to zero.","tatistical meanings}\nThe first derivative of log evidence is the current mean score, and the second is its variance:\n\\[\n\\frac{d}{dt}\\log Z(t)=\\EE_{p(t)}u,\n\\qquad\n\\frac{d^2}{dt^2}\\log Z(t)=\\Var_{p(t)}(u)\\ge0.\n\\]\nAt $q=2$ the exact values are\n\\[\n\\EE u=\\frac{10}{7},\\qquad\n\\EE u^2=\\frac{18}{7},\\qquad\n\\Var(u)=\\frac{26}{49}.\n\\]\nThus the derivative of the mean is $26/49$ at this time. The flow equation $p_i'=(u_i-\\EE u)p_i$ gives derivatives $(-10,-6,16)/49$, which sum to zero.","e of log evidence is the current mean score, and the second is its variance:\n\\[\n\\frac{d}{dt}\\log Z(t)=\\EE_{p(t)}u,\n\\qquad\n\\frac{d^2}{dt^2}\\log Z(t)=\\Var_{p(t)}(u)\\ge0.\n\\]\nAt $q=2$ the exact values are\n\\[\n\\EE u=\\frac{10}{7},\\qquad\n\\EE u^2=\\frac{18}{7},\\qquad\n\\Var(u)=\\frac{26}{49}.\n\\]\nThus the derivative of the mean is $26/49$ at this time. The flow equation $p_i'=(u_i-\\EE u)p_i$ gives derivatives $(-10,-6,16)/49$, which sum to zero.","As $t\\to\\infty$, mass concentrates on the largest score present in the initial support. A score with zero initial prior mass is never created by multiplication. If several occupied states tie for the maximum, their relative prior proportions remain unchanged within that maximal packet.\n\\takeaway{Log evidence is a cumulant generator for the tilted law. Its curvature is variance, so the mean score is monotone even though the flow is nonlinear.}","\\vpage{37}{Keep blocks by their certified contribution}{MEMORY COMPRESSION / AN OPTIMIZATION WITH A DEFINED COST}{Traditional Part II: Theorem E15.1 and Corollary E15.2. Stable primary blocks; block count is the budget.}\nConsider the scalar memory kernel\n\\[\n\\mathcal K(t)=\\tfrac14e^{-t}+(3+2t)e^{-2t}+\\tfrac13e^{-3t}.\n\\]\nIt has three primary blocks. The middle block carries a Jordan term and must be retained or omitted as a whole for this selection rule. Their certified integrated weights are\n\\[\nw_1=\\frac14,\\qquad w_2=\\frac32+\\frac{2}{2^2}=2,\n\\qquad w_3=\\frac{1/3}{3}=\\frac19.\n\\]\n\\vfig{ext_selection}\n\\platecaption{With one primary block retained, choose the middle block and certify omitted weight $13/36$. With two retained, add the rate-one block and the remaining certificate is $1/9$.}\n\\head{The slowest eigenvalue is not necessarily the largest effect}\nThe rate-two block has the greatest weight because its couplings and nilpotent contribution dominate. Retaining the slowest rate alone would omit weight $2","A DEFINED COST}{Traditional Part II: Theorem E15.1 and Corollary E15.2. Stable primary blocks; block count is the budget.}\nConsider the scalar memory kernel\n\\[\n\\mathcal K(t)=\\tfrac14e^{-t}+(3+2t)e^{-2t}+\\tfrac13e^{-3t}.\n\\]\nIt has three primary blocks. The middle block carries a Jordan term and must be retained or omitted as a whole for this selection rule. Their certified integrated weights are\n\\[\nw_1=\\frac14,\\qquad w_2=\\frac32+\\frac{2}{2^2}=2,\n\\qquad w_3=\\frac{1/3}{3}=\\frac19.\n\\]\n\\vfig{ext_selection}\n\\platecaption{With one primary block retained, choose the middle block and certify omitted weight $13/36$. With two retained, add the rate-one block and the remaining certificate is $1/9$.}\n\\head{The slowest eigenvalue is not necessarily the largest effect}\nThe rate-two block has the greatest weight because its couplings and nilpotent contribution dominate. Retaining the slowest rate alone would omit weight $2+1/9$, a much larger certificate.","rac13e^{-3t}.\n\\]\nIt has three primary blocks. The middle block carries a Jordan term and must be retained or omitted as a whole for this selection rule. Their certified integrated weights are\n\\[\nw_1=\\frac14,\\qquad w_2=\\frac32+\\frac{2}{2^2}=2,\n\\qquad w_3=\\frac{1/3}{3}=\\frac19.\n\\]\n\\vfig{ext_selection}\n\\platecaption{With one primary block retained, choose the middle block and certify omitted weight $13/36$. With two retained, add the rate-one block and the remaining certificate is $1/9$.}\n\\head{The slowest eigenvalue is not necessarily the largest effect}\nThe rate-two block has the greatest weight because its couplings and nilpotent contribution dominate. Retaining the slowest rate alone would omit weight $2+1/9$, a much larger certificate.","and must be retained or omitted as a whole for this selection rule. Their certified integrated weights are\n\\[\nw_1=\\frac14,\\qquad w_2=\\frac32+\\frac{2}{2^2}=2,\n\\qquad w_3=\\frac{1/3}{3}=\\frac19.\n\\]\n\\vfig{ext_selection}\n\\platecaption{With one primary block retained, choose the middle block and certify omitted weight $13/36$. With two retained, add the rate-one block and the remaining certificate is $1/9$.}\n\\head{The slowest eigenvalue is not necessarily the largest effect}\nThe rate-two block has the greatest weight because its couplings and nilpotent contribution dominate. Retaining the slowest rate alone would omit weight $2+1/9$, a much larger certificate.","n{With one primary block retained, choose the middle block and certify omitted weight $13/36$. With two retained, add the rate-one block and the remaining certificate is $1/9$.}\n\\head{The slowest eigenvalue is not necessarily the largest effect}\nThe rate-two block has the greatest weight because its couplings and nilpotent contribution dominate. Retaining the slowest rate alone would omit weight $2+1/9$, a much larger certificate.","The descending order is $2,1/4,1/9$. A target tolerance $1/8$ needs two blocks according to this bound: one block leaves $13/36>1/8$, while two leave $1/9\\le1/8$. Every term in this particular scalar kernel is nonnegative, so its omitted integral equals the certificate.","\\vpage{38}{A kernel certificate becomes a trajectory certificate}{VOLTERRA STABILITY / SAME INITIAL DATA AND FORCING}{Traditional Part II: Theorem E15.6. Norms and a finite time horizon are explicit.}\nLet $N=\\mat{0&1\\\\0&0}$, so $N^2=0$ and $\\|N\\|_2=1$. On $[0,T]$, compare the memory kernels $K(t)=N$ and $\\widetilde K(t)=0$, with identity free semigroup, no forcing, and initial state $x_0=e_1=(0,1)^{\\mathsf T}$. The equations are\n\\[\nx'(t)=\\int_0^t N x(r)\\,dr,\\qquad\n\\widetilde x'(t)=0.\n\\]\nTheir exact solutions are $x(t)=(t^2/2,1)^{\\mathsf T}$ and $\\widetilde x(t)=(0,1)^{\\mathsf T}$.\n\\vfig{ext_trajectory}\n\\platecaption{The first-coordinate error is exactly $t^2/2$. The certificate bounds its maximum on the entire interval. The bound is deliberately shown separately from the trajectory.}\n\\head{Track every constant before applying the estimate}\nHere $M_T=1$, $B=\\sup\\|\\widetilde x\\|=1$, and\n\\[\nL=\\|K\\|_{L^1(0,T)}=T,","0&0}$, so $N^2=0$ and $\\|N\\|_2=1$. On $[0,T]$, compare the memory kernels $K(t)=N$ and $\\widetilde K(t)=0$, with identity free semigroup, no forcing, and initial state $x_0=e_1=(0,1)^{\\mathsf T}$. The equations are\n\\[\nx'(t)=\\int_0^t N x(r)\\,dr,\\qquad\n\\widetilde x'(t)=0.\n\\]\nTheir exact solutions are $x(t)=(t^2/2,1)^{\\mathsf T}$ and $\\widetilde x(t)=(0,1)^{\\mathsf T}$.\n\\vfig{ext_trajectory}\n\\platecaption{The first-coordinate error is exactly $t^2/2$. The certificate bounds its maximum on the entire interval. The bound is deliberately shown separately from the trajectory.}\n\\head{Track every constant before applying the estimate}\nHere $M_T=1$, $B=\\sup\\|\\widetilde x\\|=1$, and\n\\[\nL=\\|K\\|_{L^1(0,T)}=T,\\qquad\n\\vare","0&0}$, so $N^2=0$ and $\\|N\\|_2=1$. On $[0,T]$, compare the memory kernels $K(t)=N$ and $\\widetilde K(t)=0$, with identity free semigroup, no forcing, and initial state $x_0=e_1=(0,1)^{\\mathsf T}$. The equations are\n\\[\nx'(t)=\\int_0^t N x(r)\\,dr,\\qquad\n\\widetilde x'(t)=0.\n\\]\nTheir exact solutions are $x(t)=(t^2/2,1)^{\\mathsf T}$ and $\\widetilde x(t)=(0,1)^{\\mathsf T}$.\n\\vfig{ext_trajectory}\n\\platecaption{The first-coordinate error is exactly $t^2/2$. The certificate bounds its maximum on the entire interval. The bound is deliberately shown separately from the trajectory.}\n\\head{Track every constant before applying the estimate}\nHere $M_T=1$, $B=\\sup\\|\\widetilde x\\|=1$, and\n\\[\nL=\\|K\\|_{L^1(0,T)}=T,\\qquad\n\\varepsilon=\\|K-\\wide","0&0}$, so $N^2=0$ and $\\|N\\|_2=1$. On $[0,T]$, compare the memory kernels $K(t)=N$ and $\\widetilde K(t)=0$, with identity free semigroup, no forcing, and initial state $x_0=e_1=(0,1)^{\\mathsf T}$. The equations are\n\\[\nx'(t)=\\int_0^t N x(r)\\,dr,\\qquad\n\\widetilde x'(t)=0.\n\\]\nTheir exact solutions are $x(t)=(t^2/2,1)^{\\mathsf T}$ and $\\widetilde x(t)=(0,1)^{\\mathsf T}$.\n\\vfig{ext_trajectory}\n\\platecaption{The first-coordinate error is exactly $t^2/2$. The certificate bounds its maximum on the entire interval. The bound is deliberately shown separately from the trajectory.}\n\\head{Track every constant before applying the estimate}\nHere $M_T=1$, $B=\\sup\\|\\widetilde x\\|=1$, and\n\\[\nL=\\|K\\|_{L^1(0,T)}=T,\\qquad\n\\varepsilon=\\|K-\\widetilde K\\|_{L","0&0}$, so $N^2=0$ and $\\|N\\|_2=1$. On $[0,T]$, compare the memory kernels $K(t)=N$ and $\\widetilde K(t)=0$, with identity free semigroup, no forcing, and initial state $x_0=e_1=(0,1)^{\\mathsf T}$. The equations are\n\\[\nx'(t)=\\int_0^t N x(r)\\,dr,\\qquad\n\\widetilde x'(t)=0.\n\\]\nTheir exact solutions are $x(t)=(t^2/2,1)^{\\mathsf T}$ and $\\widetilde x(t)=(0,1)^{\\mathsf T}$.\n\\vfig{ext_trajectory}\n\\platecaption{The first-coordinate error is exactly $t^2/2$. The certificate bounds its maximum on the entire interval. The bound is deliberately shown separately from the trajectory.}\n\\head{Track every constant before applying the estimate}\nHere $M_T=1$, $B=\\sup\\|\\widetilde x\\|=1$, and\n\\[\nL=\\|K\\|_{L^1(0,T)}=T,\\qquad\n\\varepsilon=\\|K-\\widetilde K\\|_{L^1(0,T)}=T.\n\\]\nThe theorem yields\n\\[","0&0}$, so $N^2=0$ and $\\|N\\|_2=1$. On $[0,T]$, compare the memory kernels $K(t)=N$ and $\\widetilde K(t)=0$, with identity free semigroup, no forcing, and initial state $x_0=e_1=(0,1)^{\\mathsf T}$. The equations are\n\\[\nx'(t)=\\int_0^t N x(r)\\,dr,\\qquad\n\\widetilde x'(t)=0.\n\\]\nTheir exact solutions are $x(t)=(t^2/2,1)^{\\mathsf T}$ and $\\widetilde x(t)=(0,1)^{\\mathsf T}$.\n\\vfig{ext_trajectory}\n\\platecaption{The first-coordinate error is exactly $t^2/2$. The certificate bounds its maximum on the entire interval. The bound is deliberately shown separately from the trajectory.}\n\\head{Track every constant before applying the estimate}\nHere $M_T=1$, $B=\\sup\\|\\widetilde x\\|=1$, and\n\\[\nL=\\|K\\|_{L^1(0,T)}=T,\\qquad\n\\varepsilon=\\|K-\\widetilde K\\|_{L^1(0,T)}=T.\n\\]\nThe theorem yields\n\\[\n\\sup_{0\\le t\\le T}\\|x(t)","0&0}$, so $N^2=0$ and $\\|N\\|_2=1$. On $[0,T]$, compare the memory kernels $K(t)=N$ and $\\widetilde K(t)=0$, with identity free semigroup, no forcing, and initial state $x_0=e_1=(0,1)^{\\mathsf T}$. The equations are\n\\[\nx'(t)=\\int_0^t N x(r)\\,dr,\\qquad\n\\widetilde x'(t)=0.\n\\]\nTheir exact solutions are $x(t)=(t^2/2,1)^{\\mathsf T}$ and $\\widetilde x(t)=(0,1)^{\\mathsf T}$.\n\\vfig{ext_trajectory}\n\\platecaption{The first-coordinate error is exactly $t^2/2$. The certificate bounds its maximum on the entire interval. The bound is deliberately shown separately from the trajectory.}\n\\head{Track every constant before applying the estimate}\nHere $M_T=1$, $B=\\sup\\|\\widetilde x\\|=1$, and\n\\[\nL=\\|K\\|_{L^1(0,T)}=T,\\qquad\n\\varepsilon=\\|K-\\widetilde K\\|_{L^1(0,T)}=T.\n\\]\nThe theorem yields\n\\[\n\\sup_{0\\le t\\le T}\\|x(t)-\\widetilde x(t)\\|\n=\\frac{T^2}{2}\\le T^2e^{T^2}.\n\\]\nAt $T=1/2$, the exact error is $1/8$","0&0}$, so $N^2=0$ and $\\|N\\|_2=1$. On $[0,T]$, compare the memory kernels $K(t)=N$ and $\\widetilde K(t)=0$, with identity free semigroup, no forcing, and initial state $x_0=e_1=(0,1)^{\\mathsf T}$. The equations are\n\\[\nx'(t)=\\int_0^t N x(r)\\,dr,\\qquad\n\\widetilde x'(t)=0.\n\\]\nTheir exact solutions are $x(t)=(t^2/2,1)^{\\mathsf T}$ and $\\widetilde x(t)=(0,1)^{\\mathsf T}$.\n\\vfig{ext_trajectory}\n\\platecaption{The first-coordinate error is exactly $t^2/2$. The certificate bounds its maximum on the entire interval. The bound is deliberately shown separately from the trajectory.}\n\\head{Track every constant before applying the estimate}\nHere $M_T=1$, $B=\\sup\\|\\widetilde x\\|=1$, and\n\\[\nL=\\|K\\|_{L^1(0,T)}=T,\\qquad\n\\varepsilon=\\|K-\\widetilde K\\|_{L^1(0,T)}=T.\n\\]\nThe theorem yields\n\\[\n\\sup_{0\\le t\\le T}\\|x(t)-\\widetilde x(t)\\|\n=\\frac{T^2}{2}\\le T^2e^{T^2}.\n\\]\nAt $T=1/2$, the exact error is $1/8$ and the certificate is $e^{1/4}/4$. Nilpotence makes this example polynomial;","0&0}$, so $N^2=0$ and $\\|N\\|_2=1$. On $[0,T]$, compare the memory kernels $K(t)=N$ and $\\widetilde K(t)=0$, with identity free semigroup, no forcing, and initial state $x_0=e_1=(0,1)^{\\mathsf T}$. The equations are\n\\[\nx'(t)=\\int_0^t N x(r)\\,dr,\\qquad\n\\widetilde x'(t)=0.\n\\]\nTheir exact solutions are $x(t)=(t^2/2,1)^{\\mathsf T}$ and $\\widetilde x(t)=(0,1)^{\\mathsf T}$.\n\\vfig{ext_trajectory}\n\\platecaption{The first-coordinate error is exactly $t^2/2$. The certificate bounds its maximum on the entire interval. The bound is deliberately shown separately from the trajectory.}\n\\head{Track every constant before applying the estimate}\nHere $M_T=1$, $B=\\sup\\|\\widetilde x\\|=1$, and\n\\[\nL=\\|K\\|_{L^1(0,T)}=T,\\qquad\n\\varepsilon=\\|K-\\widetilde K\\|_{L^1(0,T)}=T.\n\\]\nThe theorem yields\n\\[\n\\sup_{0\\le t\\le T}\\|x(t)-\\widetilde x(t)\\|\n=\\frac{T^2}{2}\\le T^2e^{T^2}.\n\\]\nAt $T=1/2$, the exact error is $1/8$ and the certificate is $e^{1/4}/4$. Nilpotence makes this example polynomial; it does not remove the time horizon from the general th","0&0}$, so $N^2=0$ and $\\|N\\|_2=1$. On $[0,T]$, compare the memory kernels $K(t)=N$ and $\\widetilde K(t)=0$, with identity free semigroup, no forcing, and initial state $x_0=e_1=(0,1)^{\\mathsf T}$. The equations are\n\\[\nx'(t)=\\int_0^t N x(r)\\,dr,\\qquad\n\\widetilde x'(t)=0.\n\\]\nTheir exact solutions are $x(t)=(t^2/2,1)^{\\mathsf T}$ and $\\widetilde x(t)=(0,1)^{\\mathsf T}$.\n\\vfig{ext_trajectory}\n\\platecaption{The first-coordinate error is exactly $t^2/2$. The certificate bounds its maximum on the entire interval. The bound is deliberately shown separately from the trajectory.}\n\\head{Track every constant before applying the estimate}\nHere $M_T=1$, $B=\\sup\\|\\widetilde x\\|=1$, and\n\\[\nL=\\|K\\|_{L^1(0,T)}=T,\\qquad\n\\varepsilon=\\|K-\\widetilde K\\|_{L^1(0,T)}=T.\n\\]\nThe theorem yields\n\\[\n\\sup_{0\\le t\\le T}\\|x(t)-\\widetilde x(t)\\|\n=\\frac{T^2}{2}\\le T^2e^{T^2}.\n\\]\nAt $T=1/2$, the exact error is $1/8$ and the certificate is $e^{1/4}/4$. Nilpotence makes this example polynomial; it does not remove the time horizon from the general theorem.",", compare the memory kernels $K(t)=N$ and $\\widetilde K(t)=0$, with identity free semigroup, no forcing, and initial state $x_0=e_1=(0,1)^{\\mathsf T}$. The equations are\n\\[\nx'(t)=\\int_0^t N x(r)\\,dr,\\qquad\n\\widetilde x'(t)=0.\n\\]\nTheir exact solutions are $x(t)=(t^2/2,1)^{\\mathsf T}$ and $\\widetilde x(t)=(0,1)^{\\mathsf T}$.\n\\vfig{ext_trajectory}\n\\platecaption{The first-coordinate error is exactly $t^2/2$. The certificate bounds its maximum on the entire interval. The bound is deliberately shown separately from the trajectory.}\n\\head{Track every constant before applying the estimate}\nHere $M_T=1$, $B=\\sup\\|\\widetilde x\\|=1$, and\n\\[\nL=\\|K\\|_{L^1(0,T)}=T,\\qquad\n\\varepsilon=\\|K-\\widetilde K\\|_{L^1(0,T)}=T.\n\\]\nThe theorem yields\n\\[\n\\sup_{0\\le t\\le T}\\|x(t)-\\widetilde x(t)\\|\n=\\frac{T^2}{2}\\le T^2e^{T^2}.\n\\]\nAt $T=1/2$, the exact error is $1/8$ and the certificate is $e^{1/4}/4$. Nilpotence makes this example polynomial; it does not remove the time horizon from the general theorem.",",dr,\\qquad\n\\widetilde x'(t)=0.\n\\]\nTheir exact solutions are $x(t)=(t^2/2,1)^{\\mathsf T}$ and $\\widetilde x(t)=(0,1)^{\\mathsf T}$.\n\\vfig{ext_trajectory}\n\\platecaption{The first-coordinate error is exactly $t^2/2$. The certificate bounds its maximum on the entire interval. The bound is deliberately shown separately from the trajectory.}\n\\head{Track every constant before applying the estimate}\nHere $M_T=1$, $B=\\sup\\|\\widetilde x\\|=1$, and\n\\[\nL=\\|K\\|_{L^1(0,T)}=T,\\qquad\n\\varepsilon=\\|K-\\widetilde K\\|_{L^1(0,T)}=T.\n\\]\nThe theorem yields\n\\[\n\\sup_{0\\le t\\le T}\\|x(t)-\\widetilde x(t)\\|\n=\\frac{T^2}{2}\\le T^2e^{T^2}.\n\\]\nAt $T=1/2$, the exact error is $1/8$ and the certificate is $e^{1/4}/4$. Nilpotence makes this example polynomial; it does not remove the time horizon from the general theorem.","d\n\\widetilde x'(t)=0.\n\\]\nTheir exact solutions are $x(t)=(t^2/2,1)^{\\mathsf T}$ and $\\widetilde x(t)=(0,1)^{\\mathsf T}$.\n\\vfig{ext_trajectory}\n\\platecaption{The first-coordinate error is exactly $t^2/2$. The certificate bounds its maximum on the entire interval. The bound is deliberately shown separately from the trajectory.}\n\\head{Track every constant before applying the estimate}\nHere $M_T=1$, $B=\\sup\\|\\widetilde x\\|=1$, and\n\\[\nL=\\|K\\|_{L^1(0,T)}=T,\\qquad\n\\varepsilon=\\|K-\\widetilde K\\|_{L^1(0,T)}=T.\n\\]\nThe theorem yields\n\\[\n\\sup_{0\\le t\\le T}\\|x(t)-\\widetilde x(t)\\|\n=\\frac{T^2}{2}\\le T^2e^{T^2}.\n\\]\nAt $T=1/2$, the exact error is $1/8$ and the certificate is $e^{1/4}/4$. Nilpotence makes this example polynomial; it does not remove the time horizon from the general theorem.","act solutions are $x(t)=(t^2/2,1)^{\\mathsf T}$ and $\\widetilde x(t)=(0,1)^{\\mathsf T}$.\n\\vfig{ext_trajectory}\n\\platecaption{The first-coordinate error is exactly $t^2/2$. The certificate bounds its maximum on the entire interval. The bound is deliberately shown separately from the trajectory.}\n\\head{Track every constant before applying the estimate}\nHere $M_T=1$, $B=\\sup\\|\\widetilde x\\|=1$, and\n\\[\nL=\\|K\\|_{L^1(0,T)}=T,\\qquad\n\\varepsilon=\\|K-\\widetilde K\\|_{L^1(0,T)}=T.\n\\]\nThe theorem yields\n\\[\n\\sup_{0\\le t\\le T}\\|x(t)-\\widetilde x(t)\\|\n=\\frac{T^2}{2}\\le T^2e^{T^2}.\n\\]\nAt $T=1/2$, the exact error is $1/8$ and the certificate is $e^{1/4}/4$. Nilpotence makes this example polynomial; it does not remove the time horizon from the general theorem.","trajectory}\n\\platecaption{The first-coordinate error is exactly $t^2/2$. The certificate bounds its maximum on the entire interval. The bound is deliberately shown separately from the trajectory.}\n\\head{Track every constant before applying the estimate}\nHere $M_T=1$, $B=\\sup\\|\\widetilde x\\|=1$, and\n\\[\nL=\\|K\\|_{L^1(0,T)}=T,\\qquad\n\\varepsilon=\\|K-\\widetilde K\\|_{L^1(0,T)}=T.\n\\]\nThe theorem yields\n\\[\n\\sup_{0\\le t\\le T}\\|x(t)-\\widetilde x(t)\\|\n=\\frac{T^2}{2}\\le T^2e^{T^2}.\n\\]\nAt $T=1/2$, the exact error is $1/8$ and the certificate is $e^{1/4}/4$. Nilpotence makes this example polynomial; it does not remove the time horizon from the general theorem.","ificate bounds its maximum on the entire interval. The bound is deliberately shown separately from the trajectory.}\n\\head{Track every constant before applying the estimate}\nHere $M_T=1$, $B=\\sup\\|\\widetilde x\\|=1$, and\n\\[\nL=\\|K\\|_{L^1(0,T)}=T,\\qquad\n\\varepsilon=\\|K-\\widetilde K\\|_{L^1(0,T)}=T.\n\\]\nThe theorem yields\n\\[\n\\sup_{0\\le t\\le T}\\|x(t)-\\widetilde x(t)\\|\n=\\frac{T^2}{2}\\le T^2e^{T^2}.\n\\]\nAt $T=1/2$, the exact error is $1/8$ and the certificate is $e^{1/4}/4$. Nilpotence makes this example polynomial; it does not remove the time horizon from the general theorem.","n the entire interval. The bound is deliberately shown separately from the trajectory.}\n\\head{Track every constant before applying the estimate}\nHere $M_T=1$, $B=\\sup\\|\\widetilde x\\|=1$, and\n\\[\nL=\\|K\\|_{L^1(0,T)}=T,\\qquad\n\\varepsilon=\\|K-\\widetilde K\\|_{L^1(0,T)}=T.\n\\]\nThe theorem yields\n\\[\n\\sup_{0\\le t\\le T}\\|x(t)-\\widetilde x(t)\\|\n=\\frac{T^2}{2}\\le T^2e^{T^2}.\n\\]\nAt $T=1/2$, the exact error is $1/8$ and the certificate is $e^{1/4}/4$. Nilpotence makes this example polynomial; it does not remove the time horizon from the general theorem.","und is deliberately shown separately from the trajectory.}\n\\head{Track every constant before applying the estimate}\nHere $M_T=1$, $B=\\sup\\|\\widetilde x\\|=1$, and\n\\[\nL=\\|K\\|_{L^1(0,T)}=T,\\qquad\n\\varepsilon=\\|K-\\widetilde K\\|_{L^1(0,T)}=T.\n\\]\nThe theorem yields\n\\[\n\\sup_{0\\le t\\le T}\\|x(t)-\\widetilde x(t)\\|\n=\\frac{T^2}{2}\\le T^2e^{T^2}.\n\\]\nAt $T=1/2$, the exact error is $1/8$ and the certificate is $e^{1/4}/4$. Nilpotence makes this example polynomial; it does not remove the time horizon from the general theorem.","The two solutions share initial state, free semigroup, and forcing. If those are changed as well, their errors need additional terms. An integrated kernel error alone does not automatically control a trajectory on an infinite horizon.\n\\takeaway{The compression estimate transports an $L^1$ kernel budget into a finite-horizon state bound, using an explicit bound on the comparison trajectory.}","\\vpage{39}{Two auxiliary states realize a repeated pole}{FINITE REALIZATION / AN EXACT DIFFERENTIAL MODEL}{Traditional Part II: Corollaries E15.4 and E15.5. This is a supplied rational realization.}\nThe kernel $K(t)=te^{-2t}$ from V--23 has transform $1/(z+2)^2$. It is realized by two auxiliary states:\n\\[\n\\begin{aligned}\n\\dot y_1&=-2y_1+y_2,& y_1(0)&=0,","\\vpage{39}{Two auxiliary states realize a repeated pole}{FINITE REALIZATION / AN EXACT DIFFERENTIAL MODEL}{Traditional Part II: Corollaries E15.4 and E15.5. This is a supplied rational realization.}\nThe kernel $K(t)=te^{-2t}$ from V--23 has transform $1/(z+2)^2$. It is realized by two auxiliary states:\n\\[\n\\begin{aligned}\n\\dot y_1&=-2y_1+y_2,& y_1(0)&=0,\\\\\n\\dot y_2&=-2y_2+x,& y_2(0)&=0,\\\\\n\\dot x&=a x+y_1.&\n\\end{aligned}\n\\]\nWith $D=\\mat{-2&1","\\dot x&=a x+y_1.&\n\\end{aligned}\n\\]\nWith $D=\\mat{-2&1\\\\0&-2}$, $B=(1,0)$, and $C=(0,1)^{\\mathsf T}$, the feedback kernel is $Be^{tD}C=te^{-2t}$.\n\\vfig{ext_auxiliary}\n\\platecaption{An input first enters $y_2$, transfers to $y_1$, and then returns to $x$. The two equal decay rates produce a repeated pole through the intervening link.}\n\\head{Elimination recovers the convolution, including its preparation}\nVariation of constants gives\n\\[\ny(t)=e^{tD}y(0)+\\int_0^t e^{(t-r)D}Cx(r)\\,dr,\n\\]\nso the retained equation is\n\\[\n\\dot x(t)=a x(t)+Be^{tD}y(0)\n+\\int_0^t (t-r)e^{-2(t-r)}x(r)\\,dr.\n\\]\nThe convolution-only version is exact for $y(0)=0$. For a nonzero preparation, $Be^{tD}y(0)=e^{-2t}(y_1(0)+t y_2(0))$ is an additional forcing.","0&-2}$, $B=(1,0)$, and $C=(0,1)^{\\mathsf T}$, the feedback kernel is $Be^{tD}C=te^{-2t}$.\n\\vfig{ext_auxiliary}\n\\platecaption{An input first enters $y_2$, transfers to $y_1$, and then returns to $x$. The two equal decay rates produce a repeated pole through the intervening link.}\n\\head{Elimination recovers the convolution, including its preparation}\nVariation of constants gives\n\\[\ny(t)=e^{tD}y(0)+\\int_0^t e^{(t-r)D}Cx(r)\\,dr,\n\\]\nso the retained equation is\n\\[\n\\dot x(t)=a x(t)+Be^{tD}y(0)\n+\\int_0^t (t-r)e^{-2(t-r)}x(r)\\,dr.\n\\]\nThe convolution-only version is exact for $y(0)=0$. For a nonzero preparation, $Be^{tD}y(0)=e^{-2t}(y_1(0)+t y_2(0))$ is an additional forcing.","nel is $Be^{tD}C=te^{-2t}$.\n\\vfig{ext_auxiliary}\n\\platecaption{An input first enters $y_2$, transfers to $y_1$, and then returns to $x$. The two equal decay rates produce a repeated pole through the intervening link.}\n\\head{Elimination recovers the convolution, including its preparation}\nVariation of constants gives\n\\[\ny(t)=e^{tD}y(0)+\\int_0^t e^{(t-r)D}Cx(r)\\,dr,\n\\]\nso the retained equation is\n\\[\n\\dot x(t)=a x(t)+Be^{tD}y(0)\n+\\int_0^t (t-r)e^{-2(t-r)}x(r)\\,dr.\n\\]\nThe convolution-only version is exact for $y(0)=0$. For a nonzero preparation, $Be^{tD}y(0)=e^{-2t}(y_1(0)+t y_2(0))$ is an additional forcing.","and then returns to $x$. The two equal decay rates produce a repeated pole through the intervening link.}\n\\head{Elimination recovers the convolution, including its preparation}\nVariation of constants gives\n\\[\ny(t)=e^{tD}y(0)+\\int_0^t e^{(t-r)D}Cx(r)\\,dr,\n\\]\nso the retained equation is\n\\[\n\\dot x(t)=a x(t)+Be^{tD}y(0)\n+\\int_0^t (t-r)e^{-2(t-r)}x(r)\\,dr.\n\\]\nThe convolution-only version is exact for $y(0)=0$. For a nonzero preparation, $Be^{tD}y(0)=e^{-2t}(y_1(0)+t y_2(0))$ is an additional forcing.","l decay rates produce a repeated pole through the intervening link.}\n\\head{Elimination recovers the convolution, including its preparation}\nVariation of constants gives\n\\[\ny(t)=e^{tD}y(0)+\\int_0^t e^{(t-r)D}Cx(r)\\,dr,\n\\]\nso the retained equation is\n\\[\n\\dot x(t)=a x(t)+Be^{tD}y(0)\n+\\int_0^t (t-r)e^{-2(t-r)}x(r)\\,dr.\n\\]\nThe convolution-only version is exact for $y(0)=0$. For a nonzero preparation, $Be^{tD}y(0)=e^{-2t}(y_1(0)+t y_2(0))$ is an additional forcing.","\\vpage{40}{Survivors have a different mixture of causes}{COMPETING RISKS / RESIDUAL LIFE}{Traditional Part II: Proposition E16.4 and Theorem E16.5. This example is stable and absorbs surely.}\nTwo transient states have no internal transitions. Their exit rates are $1$ and $2$, their initial masses are both $1/2$, and the two states exit by distinct causes. Thus\n\\[\nA=\\diag(-1,-2),\\qquad\nS(t)=\\tfrac12e^{-t}+\\tfrac12e^{-2t},\n\\]\nwith cause densities $f_1(t)=e^{-t}/2$ and $f_2(t)=e^{-2t}$.\n\\vfig{ext_risks}\n\\platecaption{At $t=\\log2$, surviving mass is $3/8$. Cause 1 has accumulated probability $1/4$ and cause 2 has accumulated probability $3/8$; the three pieces sum to one.}\n\\head{Condition on survival before restarting the clock}\nAt this time the transient masses are $(1/4,1/8)$, so the conditional distribution of survivors is $(2/3,1/3)$. Consequently\n\\[\n\\frac{S(t+s)}{S(t)}=\\frac23e^{-s}+\\frac13e^{-2s},\n\\qquad\n\\EE[\\tau-t\\mid\\tau>t]=\\frac23+\\frac16=\\frac56.","\\vpage{40}{Survivors have a different mixture of causes}{COMPETING RISKS / RESIDUAL LIFE}{Traditional Part II: Proposition E16.4 and Theorem E16.5. This example is stable and absorbs surely.}\nTwo transient states have no internal transitions. Their exit rates are $1$ and $2$, their initial masses are both $1/2$, and the two states exit by distinct causes. Thus\n\\[\nA=\\diag(-1,-2),\\qquad\nS(t)=\\tfrac12e^{-t}+\\tfrac12e^{-2t},\n\\]\nwith cause densities $f_1(t)=e^{-t}/2$ and $f_2(t)=e^{-2t}$.\n\\vfig{ext_risks}\n\\platecaption{At $t=\\log2$, surviving mass is $3/8$. Cause 1 has accumulated probability $1/4$ and cause 2 has accumulated probability $3/8$; the three pieces sum to one.}\n\\head{Condition on survival before restarting the clock}\nAt this time the transient masses are $(1/4,1/8)$, so the conditional distribution of survivors is $(2/3,1/3)$. Consequently\n\\[\n\\frac{S(t+s)}{S(t)}=\\frac23e^{-s}+\\frac13e^{-2s},\n\\qquad\n\\EE[\\tau-t\\mid\\tau>t]=\\frac23+\\frac16=\\frac56.\n\\]\nThe","\\vpage{40}{Survivors have a different mixture of causes}{COMPETING RISKS / RESIDUAL LIFE}{Traditional Part II: Proposition E16.4 and Theorem E16.5. This example is stable and absorbs surely.}\nTwo transient states have no internal transitions. Their exit rates are $1$ and $2$, their initial masses are both $1/2$, and the two states exit by distinct causes. Thus\n\\[\nA=\\diag(-1,-2),\\qquad\nS(t)=\\tfrac12e^{-t}+\\tfrac12e^{-2t},\n\\]\nwith cause densities $f_1(t)=e^{-t}/2$ and $f_2(t)=e^{-2t}$.\n\\vfig{ext_risks}\n\\platecaption{At $t=\\log2$, surviving mass is $3/8$. Cause 1 has accumulated probability $1/4$ and cause 2 has accumulated probability $3/8$; the three pieces sum to one.}\n\\head{Condition on survival before restarting the clock}\nAt this time the transient masses are $(1/4,1/8)$, so the conditional distribution of survivors is $(2/3,1/3)$. Consequently\n\\[\n\\frac{S(t+s)}{S(t)}=\\frac23e^{-s}+\\frac13e^{-2s},\n\\qquad\n\\EE[\\tau-t\\mid\\tau>t]=\\frac23+\\frac16=\\frac56.\n\\]\nThe original mean lifetime was $\\tfrac12\\","\\vpage{40}{Survivors have a different mixture of causes}{COMPETING RISKS / RESIDUAL LIFE}{Traditional Part II: Proposition E16.4 and Theorem E16.5. This example is stable and absorbs surely.}\nTwo transient states have no internal transitions. Their exit rates are $1$ and $2$, their initial masses are both $1/2$, and the two states exit by distinct causes. Thus\n\\[\nA=\\diag(-1,-2),\\qquad\nS(t)=\\tfrac12e^{-t}+\\tfrac12e^{-2t},\n\\]\nwith cause densities $f_1(t)=e^{-t}/2$ and $f_2(t)=e^{-2t}$.\n\\vfig{ext_risks}\n\\platecaption{At $t=\\log2$, surviving mass is $3/8$. Cause 1 has accumulated probability $1/4$ and cause 2 has accumulated probability $3/8$; the three pieces sum to one.}\n\\head{Condition on survival before restarting the clock}\nAt this time the transient masses are $(1/4,1/8)$, so the conditional distribution of survivors is $(2/3,1/3)$. Consequently\n\\[\n\\frac{S(t+s)}{S(t)}=\\frac23e^{-s}+\\frac13e^{-2s},\n\\qquad\n\\EE[\\tau-t\\mid\\tau>t]=\\frac23+\\frac16=\\frac56.\n\\]\nThe original mean lifetime was $\\tfrac12\\cdot1+\\tfrac12\\cdot\\tfrac12=3/4$. Surviving longer changes the mixture toward the slower …","auses}{COMPETING RISKS / RESIDUAL LIFE}{Traditional Part II: Proposition E16.4 and Theorem E16.5. This example is stable and absorbs surely.}\nTwo transient states have no internal transitions. Their exit rates are $1$ and $2$, their initial masses are both $1/2$, and the two states exit by distinct causes. Thus\n\\[\nA=\\diag(-1,-2),\\qquad\nS(t)=\\tfrac12e^{-t}+\\tfrac12e^{-2t},\n\\]\nwith cause densities $f_1(t)=e^{-t}/2$ and $f_2(t)=e^{-2t}$.\n\\vfig{ext_risks}\n\\platecaption{At $t=\\log2$, surviving mass is $3/8$. Cause 1 has accumulated probability $1/4$ and cause 2 has accumulated probability $3/8$; the three pieces sum to one.}\n\\head{Condition on survival before restarting the clock}\nAt this time the transient masses are $(1/4,1/8)$, so the conditional distribution of survivors is $(2/3,1/3)$. Consequently\n\\[\n\\frac{S(t+s)}{S(t)}=\\frac23e^{-s}+\\frac13e^{-2s},\n\\qquad\n\\EE[\\tau-t\\mid\\tau>t]=\\frac23+\\frac16=\\frac56.\n\\]\nThe original mean lifetime was $\\tfrac12\\cdot1+\\tfrac12\\cdot\\tfrac12=3/4$. Surviving longer changes the mixture toward the slower state; it does not restart the original mixture.","/ RESIDUAL LIFE}{Traditional Part II: Proposition E16.4 and Theorem E16.5. This example is stable and absorbs surely.}\nTwo transient states have no internal transitions. Their exit rates are $1$ and $2$, their initial masses are both $1/2$, and the two states exit by distinct causes. Thus\n\\[\nA=\\diag(-1,-2),\\qquad\nS(t)=\\tfrac12e^{-t}+\\tfrac12e^{-2t},\n\\]\nwith cause densities $f_1(t)=e^{-t}/2$ and $f_2(t)=e^{-2t}$.\n\\vfig{ext_risks}\n\\platecaption{At $t=\\log2$, surviving mass is $3/8$. Cause 1 has accumulated probability $1/4$ and cause 2 has accumulated probability $3/8$; the three pieces sum to one.}\n\\head{Condition on survival before restarting the clock}\nAt this time the transient masses are $(1/4,1/8)$, so the conditional distribution of survivors is $(2/3,1/3)$. Consequently\n\\[\n\\frac{S(t+s)}{S(t)}=\\frac23e^{-s}+\\frac13e^{-2s},\n\\qquad\n\\EE[\\tau-t\\mid\\tau>t]=\\frac23+\\frac16=\\frac56.\n\\]\nThe original mean lifetime was $\\tfrac12\\cdot1+\\tfrac12\\cdot\\tfrac12=3/4$. Surviving longer changes the mixture toward the slower state; it does not restart the original mixture.","16.4 and Theorem E16.5. This example is stable and absorbs surely.}\nTwo transient states have no internal transitions. Their exit rates are $1$ and $2$, their initial masses are both $1/2$, and the two states exit by distinct causes. Thus\n\\[\nA=\\diag(-1,-2),\\qquad\nS(t)=\\tfrac12e^{-t}+\\tfrac12e^{-2t},\n\\]\nwith cause densities $f_1(t)=e^{-t}/2$ and $f_2(t)=e^{-2t}$.\n\\vfig{ext_risks}\n\\platecaption{At $t=\\log2$, surviving mass is $3/8$. Cause 1 has accumulated probability $1/4$ and cause 2 has accumulated probability $3/8$; the three pieces sum to one.}\n\\head{Condition on survival before restarting the clock}\nAt this time the transient masses are $(1/4,1/8)$, so the conditional distribution of survivors is $(2/3,1/3)$. Consequently\n\\[\n\\frac{S(t+s)}{S(t)}=\\frac23e^{-s}+\\frac13e^{-2s},\n\\qquad\n\\EE[\\tau-t\\mid\\tau>t]=\\frac23+\\frac16=\\frac56.\n\\]\nThe original mean lifetime was $\\tfrac12\\cdot1+\\tfrac12\\cdot\\tfrac12=3/4$. Surviving longer changes the mixture toward the slower state; it does not restart the original mixture.","example is stable and absorbs surely.}\nTwo transient states have no internal transitions. Their exit rates are $1$ and $2$, their initial masses are both $1/2$, and the two states exit by distinct causes. Thus\n\\[\nA=\\diag(-1,-2),\\qquad\nS(t)=\\tfrac12e^{-t}+\\tfrac12e^{-2t},\n\\]\nwith cause densities $f_1(t)=e^{-t}/2$ and $f_2(t)=e^{-2t}$.\n\\vfig{ext_risks}\n\\platecaption{At $t=\\log2$, surviving mass is $3/8$. Cause 1 has accumulated probability $1/4$ and cause 2 has accumulated probability $3/8$; the three pieces sum to one.}\n\\head{Condition on survival before restarting the clock}\nAt this time the transient masses are $(1/4,1/8)$, so the conditional distribution of survivors is $(2/3,1/3)$. Consequently\n\\[\n\\frac{S(t+s)}{S(t)}=\\frac23e^{-s}+\\frac13e^{-2s},\n\\qquad\n\\EE[\\tau-t\\mid\\tau>t]=\\frac23+\\frac16=\\frac56.\n\\]\nThe original mean lifetime was $\\tfrac12\\cdot1+\\tfrac12\\cdot\\tfrac12=3/4$. Surviving longer changes the mixture toward the slower state; it does not restart the original mixture.","transient states have no internal transitions. Their exit rates are $1$ and $2$, their initial masses are both $1/2$, and the two states exit by distinct causes. Thus\n\\[\nA=\\diag(-1,-2),\\qquad\nS(t)=\\tfrac12e^{-t}+\\tfrac12e^{-2t},\n\\]\nwith cause densities $f_1(t)=e^{-t}/2$ and $f_2(t)=e^{-2t}$.\n\\vfig{ext_risks}\n\\platecaption{At $t=\\log2$, surviving mass is $3/8$. Cause 1 has accumulated probability $1/4$ and cause 2 has accumulated probability $3/8$; the three pieces sum to one.}\n\\head{Condition on survival before restarting the clock}\nAt this time the transient masses are $(1/4,1/8)$, so the conditional distribution of survivors is $(2/3,1/3)$. Consequently\n\\[\n\\frac{S(t+s)}{S(t)}=\\frac23e^{-s}+\\frac13e^{-2s},\n\\qquad\n\\EE[\\tau-t\\mid\\tau>t]=\\frac23+\\frac16=\\frac56.\n\\]\nThe original mean lifetime was $\\tfrac12\\cdot1+\\tfrac12\\cdot\\tfrac12=3/4$. Surviving longer changes the mixture toward the slower state; it does not restart the original mixture.","Their exit rates are $1$ and $2$, their initial masses are both $1/2$, and the two states exit by distinct causes. Thus\n\\[\nA=\\diag(-1,-2),\\qquad\nS(t)=\\tfrac12e^{-t}+\\tfrac12e^{-2t},\n\\]\nwith cause densities $f_1(t)=e^{-t}/2$ and $f_2(t)=e^{-2t}$.\n\\vfig{ext_risks}\n\\platecaption{At $t=\\log2$, surviving mass is $3/8$. Cause 1 has accumulated probability $1/4$ and cause 2 has accumulated probability $3/8$; the three pieces sum to one.}\n\\head{Condition on survival before restarting the clock}\nAt this time the transient masses are $(1/4,1/8)$, so the conditional distribution of survivors is $(2/3,1/3)$. Consequently\n\\[\n\\frac{S(t+s)}{S(t)}=\\frac23e^{-s}+\\frac13e^{-2s},\n\\qquad\n\\EE[\\tau-t\\mid\\tau>t]=\\frac23+\\frac16=\\frac56.\n\\]\nThe original mean lifetime was $\\tfrac12\\cdot1+\\tfrac12\\cdot\\tfrac12=3/4$. Surviving longer changes the mixture toward the slower state; it does not restart the original mixture.","-1,-2),\\qquad\nS(t)=\\tfrac12e^{-t}+\\tfrac12e^{-2t},\n\\]\nwith cause densities $f_1(t)=e^{-t}/2$ and $f_2(t)=e^{-2t}$.\n\\vfig{ext_risks}\n\\platecaption{At $t=\\log2$, surviving mass is $3/8$. Cause 1 has accumulated probability $1/4$ and cause 2 has accumulated probability $3/8$; the three pieces sum to one.}\n\\head{Condition on survival before restarting the clock}\nAt this time the transient masses are $(1/4,1/8)$, so the conditional distribution of survivors is $(2/3,1/3)$. Consequently\n\\[\n\\frac{S(t+s)}{S(t)}=\\frac23e^{-s}+\\frac13e^{-2s},\n\\qquad\n\\EE[\\tau-t\\mid\\tau>t]=\\frac23+\\frac16=\\frac56.\n\\]\nThe original mean lifetime was $\\tfrac12\\cdot1+\\tfrac12\\cdot\\tfrac12=3/4$. Surviving longer changes the mixture toward the slower state; it does not restart the original mixture.","use densities $f_1(t)=e^{-t}/2$ and $f_2(t)=e^{-2t}$.\n\\vfig{ext_risks}\n\\platecaption{At $t=\\log2$, surviving mass is $3/8$. Cause 1 has accumulated probability $1/4$ and cause 2 has accumulated probability $3/8$; the three pieces sum to one.}\n\\head{Condition on survival before restarting the clock}\nAt this time the transient masses are $(1/4,1/8)$, so the conditional distribution of survivors is $(2/3,1/3)$. Consequently\n\\[\n\\frac{S(t+s)}{S(t)}=\\frac23e^{-s}+\\frac13e^{-2s},\n\\qquad\n\\EE[\\tau-t\\mid\\tau>t]=\\frac23+\\frac16=\\frac56.\n\\]\nThe original mean lifetime was $\\tfrac12\\cdot1+\\tfrac12\\cdot\\tfrac12=3/4$. Surviving longer changes the mixture toward the slower state; it does not restart the original mixture.","}/2$ and $f_2(t)=e^{-2t}$.\n\\vfig{ext_risks}\n\\platecaption{At $t=\\log2$, surviving mass is $3/8$. Cause 1 has accumulated probability $1/4$ and cause 2 has accumulated probability $3/8$; the three pieces sum to one.}\n\\head{Condition on survival before restarting the clock}\nAt this time the transient masses are $(1/4,1/8)$, so the conditional distribution of survivors is $(2/3,1/3)$. Consequently\n\\[\n\\frac{S(t+s)}{S(t)}=\\frac23e^{-s}+\\frac13e^{-2s},\n\\qquad\n\\EE[\\tau-t\\mid\\tau>t]=\\frac23+\\frac16=\\frac56.\n\\]\nThe original mean lifetime was $\\tfrac12\\cdot1+\\tfrac12\\cdot\\tfrac12=3/4$. Surviving longer changes the mixture toward the slower state; it does not restart the original mixture.","nd cause 2 has accumulated probability $3/8$; the three pieces sum to one.}\n\\head{Condition on survival before restarting the clock}\nAt this time the transient masses are $(1/4,1/8)$, so the conditional distribution of survivors is $(2/3,1/3)$. Consequently\n\\[\n\\frac{S(t+s)}{S(t)}=\\frac23e^{-s}+\\frac13e^{-2s},\n\\qquad\n\\EE[\\tau-t\\mid\\tau>t]=\\frac23+\\frac16=\\frac56.\n\\]\nThe original mean lifetime was $\\tfrac12\\cdot1+\\tfrac12\\cdot\\tfrac12=3/4$. Surviving longer changes the mixture toward the slower state; it does not restart the original mixture.","The cause densities at $t=\\log2$ both equal $1/4$, so the cause-specific hazards are both $2/3$ and total hazard is $4/3$. Hazards are rates and need not lie in $[0,1]$. Each eventual cause probability is $1/2$, obtained by integrating its density. Stability of the subgenerator is part of that eventual-absorption statement.\n\\takeaway{Residual life restarts the same transient semigroup from the conditional surviving distribution. Cause probabilities, densities, and hazards answer different questions.}","\\vpage{41}{An operator substitution transports the defect too}{FUNCTIONAL CALCULUS / AN EXACT RATIONAL TEST}{Traditional Part II: Theorem E3.4; compare Traditional Part I, Theorem 15.1, and V--25.}\nUse the operators from V--25, but perturb the rectangular comparison map:\n\\[\nA=\\mat{2&1&0\\\\0&2&0\\\\0&0&9},\\quad\nB=\\mat{2&1\\\\0&2},\\quad\nC_\\eta=\\mat{1&0&\\eta\\\\0&1&0}.\n\\]\nThe generator defect is $\\Delta=C_\\eta A-BC_\\eta$, with only one nonzero entry, $\\Delta_{13}=7\\eta$ (matrix indices begin at one).\n\\vfig{ext_substitution}\n\\platecaption{The perturbation lets the discarded eigenvalue enter the comparison. Exact naturality holds at $\\eta=0$; a nonzero defect is transformed, not erased, by the rational calculus.}\n\\head{A resolvent identity supplies the sign and both inverse factors}\nFor $F(z)=1/(1+z)$, all required inverses exist, and\n\\[\nC_\\eta F(A)-F(B)C_\\eta\n=-(I+B)^{-1}\\Delta(I+A)^{-1}\n=\\mat{0&0&-7\\eta/30","0&1&0}.\n\\]\nThe generator defect is $\\Delta=C_\\eta A-BC_\\eta$, with only one nonzero entry, $\\Delta_{13}=7\\eta$ (matrix indices begin at one).\n\\vfig{ext_substitution}\n\\platecaption{The perturbation lets the discarded eigenvalue enter the comparison. Exact naturality holds at $\\eta=0$; a nonzero defect is transformed, not erased, by the rational calculus.}\n\\head{A resolvent identity supplies the sign and both inverse factors}\nFor $F(z)=1/(1+z)$, all required inverses exist, and\n\\[\nC_\\eta F(A)-F(B)C_\\eta\n=-(I+B)^{-1}\\Delta(I+A)^{-1}\n=\\mat{0&0&-7\\eta/30","s $\\Delta=C_\\eta A-BC_\\eta$, with only one nonzero entry, $\\Delta_{13}=7\\eta$ (matrix indices begin at one).\n\\vfig{ext_substitution}\n\\platecaption{The perturbation lets the discarded eigenvalue enter the comparison. Exact naturality holds at $\\eta=0$; a nonzero defect is transformed, not erased, by the rational calculus.}\n\\head{A resolvent identity supplies the sign and both inverse factors}\nFor $F(z)=1/(1+z)$, all required inverses exist, and\n\\[\nC_\\eta F(A)-F(B)C_\\eta\n=-(I+B)^{-1}\\Delta(I+A)^{-1}\n=\\mat{0&0&-7\\eta/30","entry, $\\Delta_{13}=7\\eta$ (matrix indices begin at one).\n\\vfig{ext_substitution}\n\\platecaption{The perturbation lets the discarded eigenvalue enter the comparison. Exact naturality holds at $\\eta=0$; a nonzero defect is transformed, not erased, by the rational calculus.}\n\\head{A resolvent identity supplies the sign and both inverse factors}\nFor $F(z)=1/(1+z)$, all required inverses exist, and\n\\[\nC_\\eta F(A)-F(B)C_\\eta\n=-(I+B)^{-1}\\Delta(I+A)^{-1}\n=\\mat{0&0&-7\\eta/30\\\\0&0&0}.\n\\]\nThis follows by multiplying the difference by $I+B$ on the left and $I+A$ on the right. The middle numerator is $(I+B)C_\\eta-C_\\eta(I+A)=-\\Delta$.","0&0&0}.\n\\]\nThis follows by multiplying the difference by $I+B$ on the left and $I+A$ on the right. The middle numerator is $(I+B)C_\\eta-C_\\eta(I+A)=-\\Delta$.","The retained Jordan block still contributes $F'(2)=-1/9$, as on V--25. The new third-column discrepancy compares $F(9)=1/10$ with $F(2)=1/3$, multiplied by $\\eta$.","font=\\small,align=center,anchor=east,font=\\scriptsize] at (0.64,2.004167) {1};\n\\draw[black!13,line width=.25pt] (0.8,3.358333) -- (9.6,3.358333);\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (0.64,3.358333) {2};\n\\draw[->,line width=.6pt] (0.8,0.65) -- (9.72,0.65);\n\\draw[->,line width=.6pt] (0.8,0.65) -- (0.8,4.02);\n\\node[font=\\small,align=center,font=\\scriptsize] at (5.2,0.1) {$\\theta$};\n\\node[font=\\small,align=center,anchor=west,font=\\scriptsize] at (0.8,4.3) {density};\n\\draw[dashed,line width=.8pt] plot coordinates {(0.8,0.65) (0.888,0.809266) (0.976,0.96213) (1.064,1.108689) (1.152,1.24904) (1.24,1.383281) (1.328,1.51151) (1.416,1.633824) (1.504,1.75032) (1.592,1.861096) (1.68,1.96625) (1.768,2.065879) (1.856,2.16008) (1.944,2.248951) (2.032,2.33259) (2.12,2.411094) (2.208,2.48456) (2.296,2.553086) (2.384,2.61677) (2.472,2.675709) (2.56,2.73) (2.648,2.779741) (2.736,2.82503) (2.824,2.865964) (2.912,2.90264) (3,2.935156) (3.088,2.96361) (3.176,2.988099) (3.264,3.00872) (3.352,3.025571) (3.44,3.03875) (3.528,3.048354) (3.616,3.05448) (3.704,3.057226) (3.792,3.0566…","(7.928,1.34106) (8.016,1.25962) (8.104,1.182576) (8.192,1.110254) (8.28,1.042944) (8.368,0.980887) (8.456,0.924276) (8.544,0.87325) (8.632,0.827889) (8.72,0.788206) (8.808,0.754148) (8.896,0.725585) (8.984,0.702305) (9.072,0.684012) (9.16,0.670318) (9.248,0.660735) (9.336,0.654672) (9.424,0.651427) (9.512,0.650184) (9.6,0.65)};\n\\node[font=\\small,align=center,font=\\small] at (11.65,3.75) {Prior: $\\mathrm{Beta}(2,3)$};\n\\node[font=\\small,align=center,font=\\small] at (11.65,2.95) {Posterior: $\\mathrm{Beta}(4,4)$};\n\\node[font=\\small,align=center,font=\\small] at (11.65,1.8) {ordered data: $3/35$","887) (8.456,0.924276) (8.544,0.87325) (8.632,0.827889) (8.72,0.788206) (8.808,0.754148) (8.896,0.725585) (8.984,0.702305) (9.072,0.684012) (9.16,0.670318) (9.248,0.660735) (9.336,0.654672) (9.424,0.651427) (9.512,0.650184) (9.6,0.65)};\n\\node[font=\\small,align=center,font=\\small] at (11.65,3.75) {Prior: $\\mathrm{Beta}(2,3)$};\n\\node[font=\\small,align=center,font=\\small] at (11.65,2.95) {Posterior: $\\mathrm{Beta}(4,4)$};\n\\node[font=\\small,align=center,font=\\small] at (11.65,1.8) {ordered data: $3/35$","725585) (8.984,0.702305) (9.072,0.684012) (9.16,0.670318) (9.248,0.660735) (9.336,0.654672) (9.424,0.651427) (9.512,0.650184) (9.6,0.65)};\n\\node[font=\\small,align=center,font=\\small] at (11.65,3.75) {Prior: $\\mathrm{Beta}(2,3)$};\n\\node[font=\\small,align=center,font=\\small] at (11.65,2.95) {Posterior: $\\mathrm{Beta}(4,4)$};\n\\node[font=\\small,align=center,font=\\small] at (11.65,1.8) {ordered data: $3/35$","count data: $9/35$};\n\\node[font=\\small,align=center,font=\\small] at (11.65,0.6) {Same posterior,","\\begin{tikzpicture}[x=1cm,y=1cm,>=Stealth,line cap=round,line join=round]\n\\path[use as bounding box] (0,-.6) rectangle (14,4.8);\n\\draw[line width=.6pt] (0.5,0.8) -- (4,0.8);\n\\node[font=\\small,align=center] at (2.25,4.15) {$t=0$};\n\\path[draw=black,fill=black!12,line width=.6pt] (0.92,0.8) rectangle (1.83,3.4);\n\\node[font=\\small,align=center,font=\\scriptsize] at (1.375,0.5) {state 0};\n\\node[font=\\small,align=center,font=\\scriptsize] at (1.375,3.62) {$1$};\n\\path[draw=black,fill=black!12,line width=.6pt] (2.67,0.8) rectangle (3.58,0.8);\n\\node[font=\\small,align=center,font=\\scriptsize] at (3.125,0.5) {state 1};\n\\node[font=\\small,align=center,font=\\scriptsize] at (3.125,1.02) {$0$};\n\\draw[line width=.6pt] (5,0.8) -- (8.5,0","und,line join=round]\n\\path[use as bounding box] (0,-.6) rectangle (14,4.8);\n\\draw[line width=.6pt] (0.5,0.8) -- (4,0.8);\n\\node[font=\\small,align=center] at (2.25,4.15) {$t=0$};\n\\path[draw=black,fill=black!12,line width=.6pt] (0.92,0.8) rectangle (1.83,3.4);\n\\node[font=\\small,align=center,font=\\scriptsize] at (1.375,0.5) {state 0};\n\\node[font=\\small,align=center,font=\\scriptsize] at (1.375,3.62) {$1$};\n\\path[draw=black,fill=black!12,line width=.6pt] (2.67,0.8) rectangle (3.58,0.8);\n\\node[font=\\small,align=center,font=\\scriptsize] at (3.125,0.5) {state 1};\n\\node[font=\\small,align=center,font=\\scriptsize] at (3.125,1.02) {$0$};\n\\draw[line width=.6pt] (5,0.8) -- (8.5,0.8);\n\\node[font=\\small,align=center] at (6.75,4.15) {$t=\\log(2)/2$};\n\\path[draw=black,fill=black!12,line width=.6pt] (5.42,0.8) rectangle (6.33,2.75);\n\\node[font=\\small,align=center,font=\\scriptsize] at (5.875,0.5) {state 0};\n\\n","92,0.8) rectangle (1.83,3.4);\n\\node[font=\\small,align=center,font=\\scriptsize] at (1.375,0.5) {state 0};\n\\node[font=\\small,align=center,font=\\scriptsize] at (1.375,3.62) {$1$};\n\\path[draw=black,fill=black!12,line width=.6pt] (2.67,0.8) rectangle (3.58,0.8);\n\\node[font=\\small,align=center,font=\\scriptsize] at (3.125,0.5) {state 1};\n\\node[font=\\small,align=center,font=\\scriptsize] at (3.125,1.02) {$0$};\n\\draw[line width=.6pt] (5,0.8) -- (8.5,0.8);\n\\node[font=\\small,align=center] at (6.75,4.15) {$t=\\log(2)/2$};\n\\path[draw=black,fill=black!12,line width=.6pt] (5.42,0.8) rectangle (6.33,2.75);\n\\node[font=\\small,align=center,font=\\scriptsize] at (5.875,0.5) {state 0};\n\\node[font=\\small,align=center,font=\\scriptsize] at (5.875,2.97) {$\\frac{3}{4}$};\n\\path[draw=black,fill=black!12,line width=.6pt] (7.17,0.8) rectangle (8.08,1.45);\n\\node[font=\\small,align=center,font=\\scriptsize] at (7.625,0.5) {s","e 0};\n\\node[font=\\small,align=center,font=\\scriptsize] at (1.375,3.62) {$1$};\n\\path[draw=black,fill=black!12,line width=.6pt] (2.67,0.8) rectangle (3.58,0.8);\n\\node[font=\\small,align=center,font=\\scriptsize] at (3.125,0.5) {state 1};\n\\node[font=\\small,align=center,font=\\scriptsize] at (3.125,1.02) {$0$};\n\\draw[line width=.6pt] (5,0.8) -- (8.5,0.8);\n\\node[font=\\small,align=center] at (6.75,4.15) {$t=\\log(2)/2$};\n\\path[draw=black,fill=black!12,line width=.6pt] (5.42,0.8) rectangle (6.33,2.75);\n\\node[font=\\small,align=center,font=\\scriptsize] at (5.875,0.5) {state 0};\n\\node[font=\\small,align=center,font=\\scriptsize] at (5.875,2.97) {$\\frac{3}{4}$};\n\\path[draw=black,fill=black!12,line width=.6pt] (7.17,0.8) rectangle (8.08,1.45);\n\\node[font=\\small,align=center,font=\\scriptsize] at (7.625,0.5) {state 1};\n\\node[font=\\small,align=center,font=\\scriptsize] at (7.625,1.67) {$\\frac{1}{4}$};\n\\draw[line width=.","[font=\\small,align=center,font=\\scriptsize] at (3.125,1.02) {$0$};\n\\draw[line width=.6pt] (5,0.8) -- (8.5,0.8);\n\\node[font=\\small,align=center] at (6.75,4.15) {$t=\\log(2)/2$};\n\\path[draw=black,fill=black!12,line width=.6pt] (5.42,0.8) rectangle (6.33,2.75);\n\\node[font=\\small,align=center,font=\\scriptsize] at (5.875,0.5) {state 0};\n\\node[font=\\small,align=center,font=\\scriptsize] at (5.875,2.97) {$\\frac{3}{4}$};\n\\path[draw=black,fill=black!12,line width=.6pt] (7.17,0.8) rectangle (8.08,1.45);\n\\node[font=\\small,align=center,font=\\scriptsize] at (7.625,0.5) {state 1};\n\\node[font=\\small,align=center,font=\\scriptsize] at (7.625,1.67) {$\\frac{1}{4}$};\n\\draw[line width=.6pt] (9.5,0.8) -- (13,0.8);\n\\node[font=\\small,align=center] at (11.25,4.15) {$t=\\log(2)$};\n\\path[draw=black,fill=black!12,line width=.6pt] (9.92,0.8) rectangle (10.83,2.425);\n\\node[font=\\small,align=center,font=\\scriptsize] at (10.375,0.5)","angle (6.33,2.75);\n\\node[font=\\small,align=center,font=\\scriptsize] at (5.875,0.5) {state 0};\n\\node[font=\\small,align=center,font=\\scriptsize] at (5.875,2.97) {$\\frac{3}{4}$};\n\\path[draw=black,fill=black!12,line width=.6pt] (7.17,0.8) rectangle (8.08,1.45);\n\\node[font=\\small,align=center,font=\\scriptsize] at (7.625,0.5) {state 1};\n\\node[font=\\small,align=center,font=\\scriptsize] at (7.625,1.67) {$\\frac{1}{4}$};\n\\draw[line width=.6pt] (9.5,0.8) -- (13,0.8);\n\\node[font=\\small,align=center] at (11.25,4.15) {$t=\\log(2)$};\n\\path[draw=black,fill=black!12,line width=.6pt] (9.92,0.8) rectangle (10.83,2.425);\n\\node[font=\\small,align=center,font=\\scriptsize] at (10.375,0.5) {state 0};\n\\node[font=\\small,align=center,font=\\scriptsize] at (10.375,2.645) {$\\frac{5}{8}$};\n\\path[draw=black,fill=black!12,line width=.6pt] (11.67,0.8) rectangle (12.58,1.775);\n\\node[font=\\small,align=center,font=\\scriptsize] at (12.12",",align=center,font=\\scriptsize] at (5.875,2.97) {$\\frac{3}{4}$};\n\\path[draw=black,fill=black!12,line width=.6pt] (7.17,0.8) rectangle (8.08,1.45);\n\\node[font=\\small,align=center,font=\\scriptsize] at (7.625,0.5) {state 1};\n\\node[font=\\small,align=center,font=\\scriptsize] at (7.625,1.67) {$\\frac{1}{4}$};\n\\draw[line width=.6pt] (9.5,0.8) -- 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{3};\n\\node[font=\\small,align=center,font=\\scriptsize] at (2.5,4.5) {Solid: $L$; dashed: new in $\\sqrt L$};\n\\draw[black!13,line width=.25pt] (6.6,0.65) -- (6.6,3.45);\n\\draw[] (6.6,0.59) -- (6.6,0.71);\n\\node[font=\\small,align=center,font=\\scriptsize] at (6.6,0.39) {0};\n\\draw[black!13,line width=.25pt] (7.92,0.65) -- (7.92,3.45);\n\\draw[] (7.92,0.59) -- (7.92,0.71);\n\\node[font=\\small,align=center,font=\\scriptsize] at (7.92,0.39) {0.5};\n\\draw[black!13,line width=.25pt] (9.24,0.65) -- (9.24,3.45);\n\\draw[] (9.24,0.59) -- (9.24,0.71);\n\\node[font=\\small,align=center,font=\\scripts","white] (4,3.6) circle (0.21);\n\\node[font=\\small,align=center,font=\\scriptsize] at (4,3.6) {1};\n\\path[draw,fill=white] (1,1) circle (0.21);\n\\node[font=\\small,align=center,font=\\scriptsize] at (1,1) {2};\n\\path[draw,fill=white] (4,1) circle (0.21);\n\\node[font=\\small,align=center,font=\\scriptsize] at (4,1) {3};\n\\node[font=\\small,align=center,font=\\scriptsize] at (2.5,4.5) 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$t$};\n\\node[font=\\small,align=center,anchor=west,font=\\scriptsize] at (6.6,3.85) {fastest mode amplitude};\n\\draw[line width=1.1pt] plot coordinates {(6.6,3.45) (6.666,3.28368) (6.732,3.12724) (6.798,2.980092) (6.864,2.841685) (6.93,2.711499) (6.996,2.589046) (7.062,2.473867) (7.128,2.365529) (7.194,2.263627) (7.26,2.167777) (7.326,2.077621) (7.392,1.992821) (7.458,1.913057) (7.524,1.838032) (7.59,1.767463) (7.656,1.701086) (7.722,1.638651) (7.788,1.579925) (7.854,1.524688) (7.92,1.472731) (7.986,1","650013) (12.078,0.650011) (12.144,0.650009) (12.21,0.650008) (12.276,0.650007) (12.342,0.650006) (12.408,0.650005) (12.474,0.650004) (12.54,0.650004) (12.606,0.650003) (12.672,0.650003) (12.738,0.650002) (12.804,0.650002) (12.87,0.650002) (12.936,0.650002) (13.002,0.650001) (13.068,0.650001) (13.134,0.650001) (13.2,0.650001)};\n\\node[font=\\small,align=center,font=\\scriptsize] at (10,4.45) {solid: $e^{-t\\sqrt6}$; dashed: $e^{-6t}$};\n\\end{tikzpicture}",") (12.144,0.650009) 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rectangle (12.58,1.63);\n\\node[font=\\small,align=center,font=\\small] at (12.15,1.2) {$3$};\n\\node[font=\\small,align=center,font=\\small] at (11.3,0.4) {$L_h^{-1}=hG_{\\rm nodes}$};\n\\end{tikzpicture}","round]\n\\path[use as bounding box] (0,-.6) rectangle (14,4.8);\n\\draw[line width=.6pt] (0.5,0.65) -- (4,0.65);\n\\node[font=\\small,align=center] at (2.25,3.85) {Both coarse priors};\n\\path[draw=black,fill=black!12,line width=.6pt] (0.92,0.65) rectangle (1.83,1.875);\n\\node[font=\\small,align=center,font=\\scriptsize] at (1.375,0.35) {A};\n\\node[font=\\small,align=center,font=\\scriptsize] at (1.375,2.095) {$\\frac{1}{2}$};\n\\path[draw=black,fill=black!12,line width=.6pt] (2.67,0.65) rectangle (3.58,1.875);\n\\node[font=\\small,align=center,font=\\scriptsize] at (3.125,0.35) {B};\n\\node[font=\\small,align=center,font=\\scriptsize] at (3.125,2.095) {$\\frac{1}{2}$};\n\\draw[line width=.6pt] (5.15,0.65) -- (8.65,0.65);\n\\node[font=\\small,align=center] at (6.9,3.85) {From $p^{(a)}$};\n\\path[draw=black,fill=black!12,line width=.6pt] (5.57,0.65) rectangle (6.48,2.61);\n\\node[font=\\small,align=center,font=\\scriptsize] at (6.025,0.","rectangle (1.83,1.875);\n\\node[font=\\small,align=center,font=\\scriptsize] at (1.375,0.35) {A};\n\\node[font=\\small,align=center,font=\\scriptsize] at (1.375,2.095) {$\\frac{1}{2}$};\n\\path[draw=black,fill=black!12,line width=.6pt] (2.67,0.65) rectangle (3.58,1.875);\n\\node[font=\\small,align=center,font=\\scriptsize] at (3.125,0.35) {B};\n\\node[font=\\small,align=center,font=\\scriptsize] at (3.125,2.095) {$\\frac{1}{2}$};\n\\draw[line width=.6pt] (5.15,0.65) -- (8.65,0.65);\n\\node[font=\\small,align=center] at (6.9,3.85) {From $p^{(a)}$};\n\\path[draw=black,fill=black!12,line width=.6pt] (5.57,0.65) rectangle (6.48,2.61);\n\\node[font=\\small,align=center,font=\\scriptsize] at (6.025,0.35) {A};\n\\node[font=\\small,align=center,font=\\scriptsize] at (6.025,2.83) 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(6.775,0.3) {1};\n\\node[font=\\small,align=center,font=\\scriptsize] at (6.775,1.291154) {$\\frac{15}{104}$};\n\\path[draw=black,fill=black!12,line width=.6pt] (7.552,0.6) rectangle (8.098,1.85641);\n\\node[font=\\small,align=center,font=\\scriptsize] at (7.825,0.3) {2};\n\\node[font=\\small,align=center,font=\\scriptsize] at (7.825,2.07641) {$\\frac{5}{13}$};\n\\path[dr","98,1.887821);\n\\node[font=\\small,align=center,font=\\scriptsize] at (5.725,0.3) {0};\n\\node[font=\\small,align=center,font=\\scriptsize] at (5.725,2.107821) {$\\frac{41}{104}$};\n\\path[draw=black,fill=black!12,line width=.6pt] (6.502,0.6) rectangle (7.048,1.071154);\n\\node[font=\\small,align=center,font=\\scriptsize] at (6.775,0.3) {1};\n\\node[font=\\small,align=center,font=\\scriptsize] at (6.775,1.291154) {$\\frac{15}{104}$};\n\\path[draw=black,fill=black!12,line width=.6pt] (7.552,0.6) rectangle (8.098,1.85641);\n\\node[font=\\small,align=center,font=\\scriptsize] at (7.825,0.3) {2};\n\\node[font=\\small,align=center,font=\\scriptsize] at (7.825,2.07641) {$\\frac{5}{13}$};\n\\path[draw=black,fill=black!12,line width=.6pt] (8.602,0.6) rectangle (9.148,0.851282);\n\\node[font=\\small,align=center,font=\\scriptsize] at (8.875,0.3) {3};\n\\node[font=\\small,align=center,font=\\scriptsize] at (8.875,1.071282) {$\\frac{1}{13}$};\n\\draw[line",".048,1.071154);\n\\node[font=\\small,align=center,font=\\scriptsize] at (6.775,0.3) {1};\n\\node[font=\\small,align=center,font=\\scriptsize] at (6.775,1.291154) {$\\frac{15}{104}$};\n\\path[draw=black,fill=black!12,line width=.6pt] (7.552,0.6) rectangle (8.098,1.85641);\n\\node[font=\\small,align=center,font=\\scriptsize] at (7.825,0.3) {2};\n\\node[font=\\small,align=center,font=\\scriptsize] at (7.825,2.07641) {$\\frac{5}{13}$};\n\\path[draw=black,fill=black!12,line width=.6pt] (8.602,0.6) rectangle (9.148,0.851282);\n\\node[font=\\small,align=center,font=\\scriptsize] at (8.875,0.3) {3};\n\\node[font=\\small,align=center,font=\\scriptsize] at (8.875,1.071282) {$\\frac{1}{13}$};\n\\draw[line width=.6pt] (10.3,0.6) -- (13.5,0.6);\n\\node[font=\\small,align=center] at (11.9,3.8) {Coarse posterior};\n\\path[draw=black,fill=black!12,line width=.6pt] (10.684,0.6) rectangle (11.516,2.358974);\n\\node[font=\\small,align=center,font=\\scriptsize","(8.098,1.85641);\n\\node[font=\\small,align=center,font=\\scriptsize] at (7.825,0.3) {2};\n\\node[font=\\small,align=center,font=\\scriptsize] at (7.825,2.07641) {$\\frac{5}{13}$};\n\\path[draw=black,fill=black!12,line width=.6pt] (8.602,0.6) rectangle (9.148,0.851282);\n\\node[font=\\small,align=center,font=\\scriptsize] at (8.875,0.3) {3};\n\\node[font=\\small,align=center,font=\\scriptsize] at (8.875,1.071282) {$\\frac{1}{13}$};\n\\draw[line width=.6pt] (10.3,0.6) -- (13.5,0.6);\n\\node[font=\\small,align=center] at (11.9,3.8) {Coarse posterior};\n\\path[draw=black,fill=black!12,line width=.6pt] (10.684,0.6) rectangle (11.516,2.358974);\n\\node[font=\\small,align=center,font=\\scriptsize] at (11.1,0.3) {A};\n\\node[font=\\small,align=center,font=\\scriptsize] at (11.1,2.578974) {$\\frac{7}{13}$};\n\\path[draw=black,fill=black!12,line width=.6pt] (12.284,0.6) rectangle (13.116,2.107692);\n\\node[font=\\small,align=center,font=\\scriptsize",",font=\\scriptsize] at (8.875,1.071282) {$\\frac{1}{13}$};\n\\draw[line width=.6pt] (10.3,0.6) -- (13.5,0.6);\n\\node[font=\\small,align=center] at (11.9,3.8) {Coarse posterior};\n\\path[draw=black,fill=black!12,line width=.6pt] (10.684,0.6) rectangle (11.516,2.358974);\n\\node[font=\\small,align=center,font=\\scriptsize] at (11.1,0.3) {A};\n\\node[font=\\small,align=center,font=\\scriptsize] at (11.1,2.578974) {$\\frac{7}{13}$};\n\\path[draw=black,fill=black!12,line width=.6pt] (12.284,0.6) rectangle (13.116,2.107692);\n\\node[font=\\small,align=center,font=\\scriptsize] at (12.7,0.3) {B};\n\\node[font=\\small,align=center,font=\\scriptsize] at (12.7,2.327692) {$\\frac{6}{13}$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (7,4.45) {Evidence $Z_1=5/8$, $Z_2=13/20$, and $Z_1Z_2=13/32$};\n\\end{tikzpicture}","11.516,2.358974);\n\\node[font=\\small,align=center,font=\\scriptsize] at (11.1,0.3) {A};\n\\node[font=\\small,align=center,font=\\scriptsize] at (11.1,2.578974) {$\\frac{7}{13}$};\n\\path[draw=black,fill=black!12,line width=.6pt] (12.284,0.6) rectangle (13.116,2.107692);\n\\node[font=\\small,align=center,font=\\scriptsize] at (12.7,0.3) {B};\n\\node[font=\\small,align=center,font=\\scriptsize] at (12.7,2.327692) {$\\frac{6}{13}$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (7,4.45) {Evidence $Z_1=5/8$, $Z_2=13/20$, and $Z_1Z_2=13/32$};\n\\end{tikzpicture}","ode[font=\\small,align=center,font=\\scriptsize] at (11.1,2.578974) {$\\frac{7}{13}$};\n\\path[draw=black,fill=black!12,line width=.6pt] (12.284,0.6) rectangle (13.116,2.107692);\n\\node[font=\\small,align=center,font=\\scriptsize] at (12.7,0.3) {B};\n\\node[font=\\small,align=center,font=\\scriptsize] at (12.7,2.327692) {$\\frac{6}{13}$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (7,4.45) {Evidence $Z_1=5/8$, $Z_2=13/20$, and $Z_1Z_2=13/32$};\n\\end{tikzpicture}","mall,align=center,font=\\scriptsize] at (11.1,2.578974) {$\\frac{7}{13}$};\n\\path[draw=black,fill=black!12,line width=.6pt] (12.284,0.6) rectangle (13.116,2.107692);\n\\node[font=\\small,align=center,font=\\scriptsize] at (12.7,0.3) {B};\n\\node[font=\\small,align=center,font=\\scriptsize] at (12.7,2.327692) {$\\frac{6}{13}$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (7,4.45) {Evidence $Z_1=5/8$, $Z_2=13/20$, and $Z_1Z_2=13/32$};\n\\end{tikzpicture}",",font=\\scriptsize] at (11.1,2.578974) {$\\frac{7}{13}$};\n\\path[draw=black,fill=black!12,line width=.6pt] (12.284,0.6) rectangle (13.116,2.107692);\n\\node[font=\\small,align=center,font=\\scriptsize] at (12.7,0.3) {B};\n\\node[font=\\small,align=center,font=\\scriptsize] at (12.7,2.327692) {$\\frac{6}{13}$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (7,4.45) {Evidence $Z_1=5/8$, $Z_2=13/20$, and $Z_1Z_2=13/32$};\n\\end{tikzpicture}","\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (0.64,2.15) {0.1};\n\\draw[black!13,line width=.25pt] (0.8,3.65) -- (9.3,3.65);\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (0.64,3.65) {0.2};\n\\draw[->,line width=.6pt] (0.8,0.65) -- (9.42,0.65);\n\\draw[->,line width=.6pt] (0.8,0.65) -- (0.8,3.92);\n\\node[font=\\small,align=center,font=\\scriptsize] at (5.05,0.1) {time $t$};\n\\node[font=\\small,align=center,anchor=west,font=\\scriptsize] at (0.8,4.2) {memory coefficient $t e^{-2t}$};\n\\draw[line width=1.1pt] plot coordinates {(0.8,0.65) (0.885,1.073794) (0.97,1.448228) (1.055,1.777615) (1.14,2.06593) (1.225,2.316841) (1.31,2.533726) (1.395,2.719697) (1.48,2.87762) (1.565,3.01013) (1.65,3.119652) (1.735,3.208414) (1.82,3.278462) (1.905,3.331675) (1.99,3.369776) 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{$\\frac{1}{4}$};\n\\path[draw=black!30,fill=black!8] (5.4,1.49) rectangle (6.48,2.57);\n\\node[font=\\small,align=center,font=\\small] at (5.94,2.03) {$\\frac{1}{4}$};\n\\path[draw=black!30,fill=black!8] (6.48,1.49) rectangle (7.56,2.57);\n\\node[font=\\small,align=center,font=\\small] at (7.02,2.03) {$\\frac{1}{4}$};\n\\path[draw=black!30,fill=black!8] (7.56,1.49) rectangle (8.64,2.57);\n\\node[font=\\small,align=center,font=\\small] at (8.1,2.03) {$\\frac{3}{4}$};\n\\path[draw=black!30,fill=black!8] (8.64,1.49) rectangle (9.72,2.57);\n\\node[font=\\small,align=center,font=\\small] at (9.18,2.03) {$\\frac{3}{4}$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (5.25,3.11) {A};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (5.25,2.03) {B};\n\\node[font=\\small,align=center,font=\\scriptsize] at (5.94,3.88) {0};\n\\node[font=\\small,align=center,font=\\scriptsize] at (7.02,3.88) {1};\n\\node[font=\\small,align=center,font=\\scriptsize] at (8.1,3.88) {2};\n\\node[font=\\small,align=cent…","0,fill=black!8] (5.4,1.49) rectangle (6.48,2.57);\n\\node[font=\\small,align=center,font=\\small] at (5.94,2.03) {$\\frac{1}{4}$};\n\\path[draw=black!30,fill=black!8] (6.48,1.49) rectangle (7.56,2.57);\n\\node[font=\\small,align=center,font=\\small] at (7.02,2.03) {$\\frac{1}{4}$};\n\\path[draw=black!30,fill=black!8] (7.56,1.49) rectangle (8.64,2.57);\n\\node[font=\\small,align=center,font=\\small] at (8.1,2.03) {$\\frac{3}{4}$};\n\\path[draw=black!30,fill=black!8] (8.64,1.49) rectangle (9.72,2.57);\n\\node[font=\\small,align=center,font=\\small] at (9.18,2.03) {$\\frac{3}{4}$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (5.25,3.11) {A};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (5.25,2.03) {B};\n\\node[font=\\small,align=center,font=\\scriptsize] at (5.94,3.88) {0};\n\\node[font=\\small,align=center,font=\\scriptsize] at (7.02,3.88) {1};\n\\node[font=\\small,align=center,font=\\scriptsize] at (8.1,3.88) {2};\n\\node[font=\\small,align=center,font=\\scriptsize] at (9.18,3.88) {3};\n\\node[font=\\small,align=center,font=\\small] at (7.55,0.6) {Columns 0 and 1 agree; columns 2 and 3 agree…",",fill=black!8] (6.48,1.49) rectangle (7.56,2.57);\n\\node[font=\\small,align=center,font=\\small] at (7.02,2.03) {$\\frac{1}{4}$};\n\\path[draw=black!30,fill=black!8] (7.56,1.49) rectangle (8.64,2.57);\n\\node[font=\\small,align=center,font=\\small] at (8.1,2.03) {$\\frac{3}{4}$};\n\\path[draw=black!30,fill=black!8] (8.64,1.49) rectangle (9.72,2.57);\n\\node[font=\\small,align=center,font=\\small] at (9.18,2.03) {$\\frac{3}{4}$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (5.25,3.11) {A};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (5.25,2.03) {B};\n\\node[font=\\small,align=center,font=\\scriptsize] at (5.94,3.88) {0};\n\\node[font=\\small,align=center,font=\\scriptsize] at (7.02,3.88) {1};\n\\node[font=\\small,align=center,font=\\scriptsize] at (8.1,3.88) {2};\n\\node[font=\\small,align=center,font=\\scriptsize] at (9.18,3.88) {3};\n\\node[font=\\small,align=center,font=\\small] at (7.55,0.6) {Columns 0 and 1 agree; columns 2 and 3 agree.};\n\\node[font=\\small,align=center] at (12,3.55) {$\\bar K=\\begin{pmatrix}3/4&1/4","r,font=\\scriptsize] at (5.94,3.88) {0};\n\\node[font=\\small,align=center,font=\\scriptsize] at (7.02,3.88) {1};\n\\node[font=\\small,align=center,font=\\scriptsize] at (8.1,3.88) {2};\n\\node[font=\\small,align=center,font=\\scriptsize] at (9.18,3.88) {3};\n\\node[font=\\small,align=center,font=\\small] at (7.55,0.6) {Columns 0 and 1 agree; columns 2 and 3 agree.};\n\\node[font=\\small,align=center] at (12,3.55) {$\\bar K=\\begin{pmatrix}3/4&1/4\\\\1/4&3/4\\end{pmatrix}$};\n\\node[font=\\small,align=center,text width=2.5cm,font=\\small] at (12,1.6) {Fine trajectories differ.",";\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (0.64,2.25) {0.5};\n\\draw[black!13,line width=.25pt] (0.8,3.85) -- (9.3,3.85);\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (0.64,3.85) {1};\n\\draw[->,line width=.6pt] (0.8,0.65) -- (9.42,0.65);\n\\draw[->,line width=.6pt] (0.8,0.65) -- (0.8,3.97);\n\\node[font=\\small,align=center,font=\\scriptsize] at (5.05,0.1) {time $t$};\n\\node[font=\\small,align=center,anchor=west,font=\\scriptsize] at (0.8,4.25) {retained state $x(t)$};\n\\draw[line width=1.1pt] plot coordinates {(0.8,3.85) (0.885,3.726986) (0.97,3.61343) (1.055,3.508605) (1.14,3.411838) (1.225,3.322512) (1.31,3.240053) (1.395,3.163935) (1.48,3.093668) (1.565,3.028804) (1.65,2.968926) (1.735,2.913653) (1.82,2.862629) (1.905,2.815527) (1.99,2.772048) (2.075,2.731911) (2.16,2.69486) (2.245,2.660657) (2.33,2.629084) (2.415,2.599939) (2.5,2.573034) (2.585,2.548198) (2.67,2.525272) (2.755,2.504108) (2.84,2.484571) (2.925,2.466536) (3.01,2.449888) (3.095,2.43452) (3.18,2.420334) (3.265,2.407238) (3.35,2.395149) (3.435,2.383989) (3.52,2.373688) (3.605,2.364178) (…","13,line width=.25pt] (0.8,3.85) -- (9.3,3.85);\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (0.64,3.85) {1};\n\\draw[->,line width=.6pt] (0.8,0.65) -- (9.42,0.65);\n\\draw[->,line width=.6pt] (0.8,0.65) -- (0.8,3.97);\n\\node[font=\\small,align=center,font=\\scriptsize] at (5.05,0.1) {time $t$};\n\\node[font=\\small,align=center,anchor=west,font=\\scriptsize] at (0.8,4.25) {retained state $x(t)$};\n\\draw[line width=1.1pt] plot coordinates {(0.8,3.85) (0.885,3.726986) (0.97,3.61343) (1.055,3.508605) (1.14,3.411838) (1.225,3.322512) (1.31,3.240053) (1.395,3.163935) (1.48,3.093668) (1.565,3.028804) (1.65,2.968926) (1.735,2.913653) (1.82,2.862629) (1.905,2.815527) (1.99,2.772048) (2.075,2.731911) (2.16,2.69486) (2.245,2.660657) (2.33,2.629084) (2.415,2.599939) (2.5,2.573034) (2.585,2.548198) (2.67,2.525272) (2.755,2.504108) (2.84,2.484571) (2.925,2.466536) (3.01,2.449888) (3.095,2.43452) (3.18,2.420334) (3.265,2.407238) (3.35,2.395149) (3.435,2.383989) (3.52,2.373688) (3.605,2.364178) (3.69,2.3554) (3.775,2.347296) (3.86,2.339816) (3.945,2.33291) (4.03,2.326536) (4.115,2.320651) (4…","(7.6,0.780439) (7.685,0.775324) (7.77,0.77041) (7.855,0.765689) (7.94,0.761153) (8.025,0.756794) (8.11,0.752607) (8.195,0.748584) (8.28,0.744718) (8.365,0.741004) (8.45,0.737436) (8.535,0.734008) (8.62,0.730714) (8.705,0.727549) (8.79,0.724508) (8.875,0.721586) (8.96,0.71878) (9.045,0.716083) (9.13,0.713492) (9.215,0.711002) (9.3,0.70861)};\n\\node[font=\\small,align=center] at (11.7,3.75) {exact: $\\dfrac{1+e^{-2t}}2$};\n\\node[font=\\small,align=center] at (11.7,2.7) {naive deletion: $e^{-t}$};\n\\node[font=\\small,align=center,font=\\small] at (11.7,1.25) {$\\displaystyle K(t)=e^{-t}$","5,0.756794) (8.11,0.752607) (8.195,0.748584) (8.28,0.744718) (8.365,0.741004) (8.45,0.737436) (8.535,0.734008) (8.62,0.730714) (8.705,0.727549) (8.79,0.724508) (8.875,0.721586) (8.96,0.71878) (9.045,0.716083) (9.13,0.713492) (9.215,0.711002) (9.3,0.70861)};\n\\node[font=\\small,align=center] at (11.7,3.75) {exact: $\\dfrac{1+e^{-2t}}2$};\n\\node[font=\\small,align=center] at (11.7,2.7) {naive deletion: $e^{-t}$};\n\\node[font=\\small,align=center,font=\\small] at (11.7,1.25) {$\\displaystyle K(t)=e^{-t}$","41004) (8.45,0.737436) (8.535,0.734008) (8.62,0.730714) (8.705,0.727549) (8.79,0.724508) (8.875,0.721586) (8.96,0.71878) (9.045,0.716083) (9.13,0.713492) (9.215,0.711002) (9.3,0.70861)};\n\\node[font=\\small,align=center] at (11.7,3.75) {exact: $\\dfrac{1+e^{-2t}}2$};\n\\node[font=\\small,align=center] at (11.7,2.7) {naive deletion: $e^{-t}$};\n\\node[font=\\small,align=center,font=\\small] at (11.7,1.25) {$\\displaystyle K(t)=e^{-t}$","4,0.59) -- (8.4,0.71);\n\\node[font=\\small,align=center,font=\\scriptsize] at (8.4,0.39) {1};\n\\draw[black!13,line width=.25pt] (0.8,0.65) -- (8.4,0.65);\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (0.64,0.65) {0};\n\\draw[->,line width=.6pt] (0.8,0.65) -- (8.52,0.65);\n\\draw[->,line width=.6pt] (0.8,0.65) -- (0.8,3.87);\n\\node[font=\\small,align=center,font=\\scriptsize] at (4.6,0.1) {$x$};\n\\node[font=\\small,align=center,anchor=west,font=\\scriptsize] at (0.8,4.15) {$|u_h-u|=h^2x(1-x)$};\n\\draw[line width=1.1pt] plot coordinates {(0.8,0.65) (0.876,0.765088) (0.952,0.87785) (1.028,0.988287) (1.104,1.0964) (1.18,1.202187) (1.256,1.30565) (1.332,1.406788) (1.408,1.5056) (1.484,1.602088) (1.56,1.69625) (1.636,1.788088) (1.712,1.8776) (1.788,1.964788) (1.864,2.04965) (1.94,2.132188) (2.016,2.2124) (2.092,2.290288) (2.168,2.36585) (2.244,2.439087) (2.32,2.51) (2.396,2.578588) (2.472,2.","39) {1};\n\\draw[black!13,line width=.25pt] (0.8,0.65) -- (8.4,0.65);\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (0.64,0.65) {0};\n\\draw[->,line width=.6pt] (0.8,0.65) -- (8.52,0.65);\n\\draw[->,line width=.6pt] (0.8,0.65) -- (0.8,3.87);\n\\node[font=\\small,align=center,font=\\scriptsize] at (4.6,0.1) {$x$};\n\\node[font=\\small,align=center,anchor=west,font=\\scriptsize] at (0.8,4.15) {$|u_h-u|=h^2x(1-x)$};\n\\draw[line width=1.1pt] plot coordinates {(0.8,0.65) (0.876,0.765088) (0.952,0.87785) (1.028,0.988287) (1.104,1.0964) (1.18,1.202187) (1.256,1.30565) (1.332,1.406788) (1.408,1.5056) (1.484,1.602088) (1.56,1.69625) (1.636,1.788088) (1.712,1.8776) (1.788,1.964788) (1.864,2.04965) (1.94,2.132188) (2.016,2.2124) (2.092,2.290288) (2.168,2.36585) (2.244,2.439087) (2.32,2.51) (2.396,2.578588) (2.472,2.64485) (2.548,2.708788) (2.624,2.7704) (2.7,2.829687) (2.776,2.88665) (2.852,2.941288) (2.928,2.99","56,0.761818) (7.032,0.757241) (7.108,0.752518) (7.184,0.74765) (7.26,0.742637) (7.336,0.737478) (7.412,0.732174) (7.488,0.726725) (7.564,0.72113) (7.64,0.715391) (7.716,0.709505) (7.792,0.703475) (7.868,0.697299) (7.944,0.690978) (8.02,0.684512) (8.096,0.6779) (8.172,0.671143) (8.248,0.664241) (8.324,0.657193) (8.4,0.65)};\n\\node[font=\\small,align=center,font=\\small] at (11.35,3.95) {grid spacing $h$\\qquad maximum error};\n\\node[font=\\small,align=center] at (11.35,3) {$\\frac{1}{4}$\\qquad $\\frac{1}{64}$};\n\\node[font=\\small,align=center] at (11.35,2) {$\\frac{1}{8}$\\qquad $\\frac{1}{256}$};\n\\node[font=\\small,align=center] at (11.35,1) {$\\frac{1}{16}$\\qquad $\\frac{1}{1024}$};\n\\end{tikzpicture}","42637) (7.336,0.737478) (7.412,0.732174) (7.488,0.726725) (7.564,0.72113) (7.64,0.715391) (7.716,0.709505) (7.792,0.703475) (7.868,0.697299) (7.944,0.690978) (8.02,0.684512) (8.096,0.6779) (8.172,0.671143) (8.248,0.664241) (8.324,0.657193) (8.4,0.65)};\n\\node[font=\\small,align=center,font=\\small] at (11.35,3.95) {grid spacing $h$\\qquad maximum error};\n\\node[font=\\small,align=center] at (11.35,3) {$\\frac{1}{4}$\\qquad $\\frac{1}{64}$};\n\\node[font=\\small,align=center] at (11.35,2) {$\\frac{1}{8}$\\qquad $\\frac{1}{256}$};\n\\node[font=\\small,align=center] at (11.35,1) {$\\frac{1}{16}$\\qquad $\\frac{1}{1024}$};\n\\end{tikzpicture}","78) (7.412,0.732174) (7.488,0.726725) (7.564,0.72113) (7.64,0.715391) (7.716,0.709505) (7.792,0.703475) (7.868,0.697299) (7.944,0.690978) (8.02,0.684512) (8.096,0.6779) (8.172,0.671143) (8.248,0.664241) (8.324,0.657193) (8.4,0.65)};\n\\node[font=\\small,align=center,font=\\small] at (11.35,3.95) {grid spacing $h$\\qquad maximum error};\n\\node[font=\\small,align=center] at (11.35,3) {$\\frac{1}{4}$\\qquad $\\frac{1}{64}$};\n\\node[font=\\small,align=center] at (11.35,2) {$\\frac{1}{8}$\\qquad $\\frac{1}{256}$};\n\\node[font=\\small,align=center] at (11.35,1) {$\\frac{1}{16}$\\qquad $\\frac{1}{1024}$};\n\\end{tikzpicture}","15391) (7.716,0.709505) (7.792,0.703475) (7.868,0.697299) (7.944,0.690978) (8.02,0.684512) (8.096,0.6779) (8.172,0.671143) (8.248,0.664241) (8.324,0.657193) (8.4,0.65)};\n\\node[font=\\small,align=center,font=\\small] at (11.35,3.95) {grid spacing $h$\\qquad maximum error};\n\\node[font=\\small,align=center] at (11.35,3) {$\\frac{1}{4}$\\qquad $\\frac{1}{64}$};\n\\node[font=\\small,align=center] at (11.35,2) {$\\frac{1}{8}$\\qquad $\\frac{1}{256}$};\n\\node[font=\\small,align=center] at (11.35,1) {$\\frac{1}{16}$\\qquad $\\frac{1}{1024}$};\n\\end{tikzpicture}","05) (7.792,0.703475) (7.868,0.697299) (7.944,0.690978) (8.02,0.684512) (8.096,0.6779) (8.172,0.671143) (8.248,0.664241) (8.324,0.657193) (8.4,0.65)};\n\\node[font=\\small,align=center,font=\\small] at (11.35,3.95) {grid spacing $h$\\qquad maximum error};\n\\node[font=\\small,align=center] at (11.35,3) {$\\frac{1}{4}$\\qquad $\\frac{1}{64}$};\n\\node[font=\\small,align=center] at (11.35,2) {$\\frac{1}{8}$\\qquad $\\frac{1}{256}$};\n\\node[font=\\small,align=center] at (11.35,1) {$\\frac{1}{16}$\\qquad $\\frac{1}{1024}$};\n\\end{tikzpicture}","84512) (8.096,0.6779) (8.172,0.671143) (8.248,0.664241) (8.324,0.657193) (8.4,0.65)};\n\\node[font=\\small,align=center,font=\\small] at (11.35,3.95) {grid spacing $h$\\qquad maximum error};\n\\node[font=\\small,align=center] at (11.35,3) {$\\frac{1}{4}$\\qquad $\\frac{1}{64}$};\n\\node[font=\\small,align=center] at (11.35,2) {$\\frac{1}{8}$\\qquad $\\frac{1}{256}$};\n\\node[font=\\small,align=center] at (11.35,1) {$\\frac{1}{16}$\\qquad $\\frac{1}{1024}$};\n\\end{tikzpicture}","(8.172,0.671143) (8.248,0.664241) (8.324,0.657193) (8.4,0.65)};\n\\node[font=\\small,align=center,font=\\small] at (11.35,3.95) {grid spacing $h$\\qquad maximum error};\n\\node[font=\\small,align=center] at (11.35,3) {$\\frac{1}{4}$\\qquad $\\frac{1}{64}$};\n\\node[font=\\small,align=center] at (11.35,2) {$\\frac{1}{8}$\\qquad $\\frac{1}{256}$};\n\\node[font=\\small,align=center] at (11.35,1) {$\\frac{1}{16}$\\qquad $\\frac{1}{1024}$};\n\\end{tikzpicture}","\\begin{tikzpicture}[x=1cm,y=1cm,>=Stealth,line cap=round,line join=round]\n\\path[use as bounding box] (0,-.6) rectangle 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{$0$};\n\\path[draw=black!30,fill=white] (6.32,0.84) rectangle (7.24,1.76);\n\\node[font=\\small,align=center,font=\\small] at (6.78,1.3) {$0$};\n\\path[draw=black!30,fil…","h[draw=black!30,fill=black!8] (6.32,2.68) rectangle (7.24,3.6);\n\\node[font=\\small,align=center,font=\\small] at (6.78,3.14) {$- \\frac{1}{9}$};\n\\path[draw=black!30,fill=white] (7.24,2.68) rectangle (8.16,3.6);\n\\node[font=\\small,align=center,font=\\small] at (7.7,3.14) {$0$};\n\\path[draw=black!30,fill=white] (5.4,1.76) rectangle (6.32,2.68);\n\\node[font=\\small,align=center,font=\\small] at (5.86,2.22) {$0$};\n\\path[draw=black!30,fill=black!8] (6.32,1.76) rectangle (7.24,2.68);\n\\node[font=\\small,align=center,font=\\small] at (6.78,2.22) {$\\frac{1}{3}$};\n\\path[draw=black!30,fill=white] (7.24,1.76) rectangle (8.16,2.68);\n\\node[font=\\small,align=center,font=\\small] at (7.7,2.22) {$0$};\n\\path[draw=black!30,fill=white] (5.4,0.84) rectangle (6.32,1.76);\n\\node[font=\\small,align=center,font=\\small] at (5.86,1.3) {$0$};\n\\path[draw=black!30,fill=white] (6.32,0.84) rectangle (7.24,1.76);\n\\node[font=\\small,align=center,font=\\small] at (6.78,1.3) {$0$};\n\\path[draw=black!30,fill=black!8] (7.24,0.84) rectangle (8.16,1.76);\n\\node[font=\\small,align=center,font=\\small] at (7.7,1.3) {$\\frac{1}{10}$};\n\\path[draw=…","{9}$};\n\\path[draw=black!30,fill=white] (7.24,2.68) rectangle (8.16,3.6);\n\\node[font=\\small,align=center,font=\\small] at (7.7,3.14) {$0$};\n\\path[draw=black!30,fill=white] (5.4,1.76) rectangle (6.32,2.68);\n\\node[font=\\small,align=center,font=\\small] at (5.86,2.22) {$0$};\n\\path[draw=black!30,fill=black!8] (6.32,1.76) rectangle (7.24,2.68);\n\\node[font=\\small,align=center,font=\\small] at (6.78,2.22) {$\\frac{1}{3}$};\n\\path[draw=black!30,fill=white] (7.24,1.76) rectangle (8.16,2.68);\n\\node[font=\\small,align=center,font=\\small] at (7.7,2.22) {$0$};\n\\path[draw=black!30,fill=white] (5.4,0.84) rectangle (6.32,1.76);\n\\node[font=\\small,align=center,font=\\small] at (5.86,1.3) {$0$};\n\\path[draw=black!30,fill=white] (6.32,0.84) rectangle (7.24,1.76);\n\\node[font=\\small,align=center,font=\\small] at (6.78,1.3) {$0$};\n\\path[draw=black!30,fill=black!8] (7.24,0.84) rectangle (8.16,1.76);\n\\node[font=\\small,align=center,font=\\small] at (7.7,1.3) {$\\frac{1}{10}$};\n\\path[draw=black,dashed,line width=1pt] (5.38,1.73) rectangle (7.26,3.63);\n\\node[font=\\small,align=center] at (11.7,3.6) {$F(2)=1/3$};\n\\node[font=…","h[draw=black!30,fill=white] (5.4,1.76) rectangle (6.32,2.68);\n\\node[font=\\small,align=center,font=\\small] at (5.86,2.22) {$0$};\n\\path[draw=black!30,fill=black!8] (6.32,1.76) rectangle (7.24,2.68);\n\\node[font=\\small,align=center,font=\\small] at (6.78,2.22) {$\\frac{1}{3}$};\n\\path[draw=black!30,fill=white] (7.24,1.76) rectangle (8.16,2.68);\n\\node[font=\\small,align=center,font=\\small] at (7.7,2.22) {$0$};\n\\path[draw=black!30,fill=white] (5.4,0.84) rectangle (6.32,1.76);\n\\node[font=\\small,align=center,font=\\small] at (5.86,1.3) {$0$};\n\\path[draw=black!30,fill=white] (6.32,0.84) rectangle (7.24,1.76);\n\\node[font=\\small,align=center,font=\\small] at (6.78,1.3) {$0$};\n\\path[draw=black!30,fill=black!8] (7.24,0.84) rectangle (8.16,1.76);\n\\node[font=\\small,align=center,font=\\small] at (7.7,1.3) {$\\frac{1}{10}$};\n\\path[draw=black,dashed,line width=1pt] (5.38,1.73) rectangle (7.26,3.63);\n\\node[font=\\small,align=center] at (11.7,3.6) {$F(2)=1/3$};\n\\node[font=\\small,align=center] at (11.7,2.65) {$F'(2)=-1/9$};\n\\node[font=\\small,align=center,font=\\small] at (11.7,1.25) {The dashed block is","th[draw=black!30,fill=black!8] (6.32,1.76) rectangle (7.24,2.68);\n\\node[font=\\small,align=center,font=\\small] at (6.78,2.22) {$\\frac{1}{3}$};\n\\path[draw=black!30,fill=white] (7.24,1.76) rectangle (8.16,2.68);\n\\node[font=\\small,align=center,font=\\small] at (7.7,2.22) {$0$};\n\\path[draw=black!30,fill=white] (5.4,0.84) rectangle (6.32,1.76);\n\\node[font=\\small,align=center,font=\\small] at (5.86,1.3) {$0$};\n\\path[draw=black!30,fill=white] (6.32,0.84) rectangle (7.24,1.76);\n\\node[font=\\small,align=center,font=\\small] at (6.78,1.3) {$0$};\n\\path[draw=black!30,fill=black!8] (7.24,0.84) rectangle (8.16,1.76);\n\\node[font=\\small,align=center,font=\\small] at (7.7,1.3) {$\\frac{1}{10}$};\n\\path[draw=black,dashed,line width=1pt] (5.38,1.73) rectangle (7.26,3.63);\n\\node[font=\\small,align=center] at (11.7,3.6) {$F(2)=1/3$};\n\\node[font=\\small,align=center] at (11.7,2.65) {$F'(2)=-1/9$};\n\\node[font=\\small,align=center,font=\\small] at (11.7,1.25) {The dashed block is","{1}{3}$};\n\\path[draw=black!30,fill=white] (7.24,1.76) rectangle (8.16,2.68);\n\\node[font=\\small,align=center,font=\\small] at (7.7,2.22) {$0$};\n\\path[draw=black!30,fill=white] (5.4,0.84) rectangle (6.32,1.76);\n\\node[font=\\small,align=center,font=\\small] at (5.86,1.3) {$0$};\n\\path[draw=black!30,fill=white] (6.32,0.84) rectangle (7.24,1.76);\n\\node[font=\\small,align=center,font=\\small] at (6.78,1.3) {$0$};\n\\path[draw=black!30,fill=black!8] (7.24,0.84) rectangle (8.16,1.76);\n\\node[font=\\small,align=center,font=\\small] at (7.7,1.3) {$\\frac{1}{10}$};\n\\path[draw=black,dashed,line width=1pt] (5.38,1.73) rectangle (7.26,3.63);\n\\node[font=\\small,align=center] at (11.7,3.6) {$F(2)=1/3$};\n\\node[font=\\small,align=center] at (11.7,2.65) {$F'(2)=-1/9$};\n\\node[font=\\small,align=center,font=\\small] at (11.7,1.25) {The dashed block is",") {$0$};\n\\path[draw=black!30,fill=white] (5.4,0.84) rectangle (6.32,1.76);\n\\node[font=\\small,align=center,font=\\small] at (5.86,1.3) {$0$};\n\\path[draw=black!30,fill=white] (6.32,0.84) rectangle (7.24,1.76);\n\\node[font=\\small,align=center,font=\\small] at (6.78,1.3) {$0$};\n\\path[draw=black!30,fill=black!8] (7.24,0.84) rectangle (8.16,1.76);\n\\node[font=\\small,align=center,font=\\small] at (7.7,1.3) {$\\frac{1}{10}$};\n\\path[draw=black,dashed,line width=1pt] (5.38,1.73) rectangle (7.26,3.63);\n\\node[font=\\small,align=center] at (11.7,3.6) {$F(2)=1/3$};\n\\node[font=\\small,align=center] at (11.7,2.65) {$F'(2)=-1/9$};\n\\node[font=\\small,align=center,font=\\small] at (11.7,1.25) {The dashed block is","\\path[draw=black!30,fill=white] (6.32,0.84) rectangle (7.24,1.76);\n\\node[font=\\small,align=center,font=\\small] at (6.78,1.3) {$0$};\n\\path[draw=black!30,fill=black!8] (7.24,0.84) rectangle (8.16,1.76);\n\\node[font=\\small,align=center,font=\\small] at (7.7,1.3) {$\\frac{1}{10}$};\n\\path[draw=black,dashed,line width=1pt] (5.38,1.73) rectangle (7.26,3.63);\n\\node[font=\\small,align=center] at (11.7,3.6) {$F(2)=1/3$};\n\\node[font=\\small,align=center] at (11.7,2.65) {$F'(2)=-1/9$};\n\\node[font=\\small,align=center,font=\\small] at (11.7,1.25) {The dashed block is","1.76);\n\\node[font=\\small,align=center,font=\\small] at (6.78,1.3) {$0$};\n\\path[draw=black!30,fill=black!8] (7.24,0.84) rectangle (8.16,1.76);\n\\node[font=\\small,align=center,font=\\small] at (7.7,1.3) {$\\frac{1}{10}$};\n\\path[draw=black,dashed,line width=1pt] (5.38,1.73) rectangle (7.26,3.63);\n\\node[font=\\small,align=center] at (11.7,3.6) {$F(2)=1/3$};\n\\node[font=\\small,align=center] at (11.7,2.65) {$F'(2)=-1/9$};\n\\node[font=\\small,align=center,font=\\small] at (11.7,1.25) {The dashed block is","$F(A_c)$, not just $F(2)I$.};\n\\end{tikzpicture}","2.25);\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (0.64,2.25) {0.5};\n\\draw[black!13,line width=.25pt] (0.8,3.9) -- (9.5,3.9);\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (0.64,3.9) {1};\n\\draw[->,line width=.6pt] (0.8,0.6) -- (9.62,0.6);\n\\draw[->,line width=.6pt] (0.8,0.6) -- (0.8,4.02);\n\\node[font=\\small,align=center,font=\\scriptsize] at (5.15,0.05) {time $t$};\n\\node[font=\\small,align=center,anchor=west,font=\\scriptsize] at (0.8,4.3) {probability};\n\\draw[line width=1.1pt] plot coordinates {(0.8,2.25) (0.887,2.282996) (0.974,2.315965) (1.061,2.348881) (1.148,2.381719) (1.235,2.414452) (1.322,2.447055) (1.409,2.479503) (1.496,2.51177) (1.583,2.543833) (1.67,2.575669) (1.757,2.607255) (1.844,2.638568) (1.931,2.669588) (2.018,2.700293) (2.105,2.730666) (2.192,2.760686) (2.279,2.790338) (2.366,2.819603) (2.453,2.848467) (2.54,2.876916) (2.627,2.904935) (2.714,2.932513) (2.801,2.959639) (2.888,2.986302) (2.975,3.012493) (3.062,3.038205) (3.149,3.06343) (3.236,3.088163) (3.323,3.112398) (3.41,3.136132) (3.497,3.159361) (3.584,3.182084) (3.671,3.2043)",") (7.76,0.729247) (7.847,0.72437) (7.934,0.71967) (8.021,0.715142) (8.108,0.710778) (8.195,0.706575) (8.282,0.702526) (8.369,0.698626) (8.456,0.69487) (8.543,0.691253) (8.63,0.68777) (8.717,0.684417) (8.804,0.681188) (8.891,0.67808) (8.978,0.675088) (9.065,0.672208) (9.152,0.669436) (9.239,0.666769) (9.326,0.664202) (9.413,0.661731) (9.5,0.659354)};\n\\node[font=\\small,align=center] at (11.6,3.7) {$p_0(t)=\\dfrac{e^t}{1+e^t}$};\n\\node[font=\\small,align=center] at (11.6,2.7) {$p_1(t)=\\dfrac{1}{1+e^t}$};\n\\node[font=\\small,align=center] at (11.6,1.5) {$\\dot p_0=p_0(1-p_0)$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (11.6,0.65) {Mass stays one.","10778) (8.195,0.706575) (8.282,0.702526) (8.369,0.698626) (8.456,0.69487) (8.543,0.691253) (8.63,0.68777) (8.717,0.684417) (8.804,0.681188) (8.891,0.67808) (8.978,0.675088) (9.065,0.672208) (9.152,0.669436) (9.239,0.666769) (9.326,0.664202) (9.413,0.661731) (9.5,0.659354)};\n\\node[font=\\small,align=center] at (11.6,3.7) {$p_0(t)=\\dfrac{e^t}{1+e^t}$};\n\\node[font=\\small,align=center] at (11.6,2.7) {$p_1(t)=\\dfrac{1}{1+e^t}$};\n\\node[font=\\small,align=center] at (11.6,1.5) {$\\dot p_0=p_0(1-p_0)$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (11.6,0.65) {Mass stays one.","8.543,0.691253) (8.63,0.68777) (8.717,0.684417) (8.804,0.681188) (8.891,0.67808) (8.978,0.675088) (9.065,0.672208) (9.152,0.669436) (9.239,0.666769) (9.326,0.664202) (9.413,0.661731) (9.5,0.659354)};\n\\node[font=\\small,align=center] at (11.6,3.7) {$p_0(t)=\\dfrac{e^t}{1+e^t}$};\n\\node[font=\\small,align=center] at (11.6,2.7) {$p_1(t)=\\dfrac{1}{1+e^t}$};\n\\node[font=\\small,align=center] at (11.6,1.5) {$\\dot p_0=p_0(1-p_0)$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (11.6,0.65) {Mass stays one.","\\node[font=\\small,align=center,font=\\small] at (4.4,4.4) {probability};\n\\node[font=\\small,align=center,font=\\small] at (7,4.4) {weight};\n\\node[font=\\small,align=center,font=\\small] at (9.5,4.4) {terminal};\n\\node[font=\\small,align=center,font=\\small] at (12,4.4) {contribution};\n\\node[font=\\small,align=center,font=\\small] at (1.6,3.8) {\\texttt{0000}};\n\\node[font=\\small,align=center] at (4.4,3.8) {$27/64$};\n\\node[font=\\small,align=center] at 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{$1/2$};\n\\node[font=\\small,align=center] at (9.5,1.45) {$1$};\n\\node[font=\\small,align=center] at (12,1.45) {$1/128$};\n\\node[font=\\small,align=center,font=\\s","64$};\n\\node[font=\\small,align=center] at (7,2.39) {$1/2$};\n\\node[font=\\small,align=center] at (9.5,2.39) {$1$};\n\\node[font=\\small,align=center] at (12,2.39) {$9/128$};\n\\node[font=\\small,align=center,font=\\small] at (1.6,1.92) {\\texttt{0100}};\n\\node[font=\\small,align=center] at (4.4,1.92) {$3/64$};\n\\node[font=\\small,align=center] at (7,1.92) {$1/2$};\n\\node[font=\\small,align=center] at (9.5,1.92) {$2$};\n\\node[font=\\small,align=center] at (12,1.92) {$3/64$};\n\\draw[black!18,line width=.3pt] (0.6,1.7) -- (13.4,1.7);\n\\node[font=\\small,align=center,font=\\small] at (1.6,1.45) {\\texttt{0101}};\n\\node[font=\\small,align=center] at (4.4,1.45) {$1/64$};\n\\node[font=\\small,align=center] at (7,1.45) {$1/2$};\n\\node[font=\\small,align=center] at (9.5,1.45) {$1$};\n\\node[font=\\small,align=center] at (12,1.45) {$1/128$};\n\\node[font=\\small,align=center,font=\\small] at (1.6,0.98) {\\texttt{0110}};\n\\node[font=\\smal","1/2$};\n\\node[font=\\small,align=center] at (9.5,2.39) {$1$};\n\\node[font=\\small,align=center] at (12,2.39) {$9/128$};\n\\node[font=\\small,align=center,font=\\small] at (1.6,1.92) {\\texttt{0100}};\n\\node[font=\\small,align=center] at (4.4,1.92) {$3/64$};\n\\node[font=\\small,align=center] at (7,1.92) {$1/2$};\n\\node[font=\\small,align=center] at (9.5,1.92) {$2$};\n\\node[font=\\small,align=center] at (12,1.92) {$3/64$};\n\\draw[black!18,line width=.3pt] (0.6,1.7) -- (13.4,1.7);\n\\node[font=\\small,align=center,font=\\small] at (1.6,1.45) {\\texttt{0101}};\n\\node[font=\\small,align=center] at (4.4,1.45) {$1/64$};\n\\node[font=\\small,align=center] at (7,1.45) {$1/2$};\n\\node[font=\\small,align=center] at (9.5,1.45) {$1$};\n\\node[font=\\small,align=center] at (12,1.45) {$1/128$};\n\\node[font=\\small,align=center,font=\\small] at (1.6,0.98) {\\texttt{0110}};\n\\node[font=\\small,align=center] at (4.4,0.98) {$3/64$};\n\\node[font=\\sma","};\n\\node[font=\\small,align=center] at (4.4,1.92) {$3/64$};\n\\node[font=\\small,align=center] at (7,1.92) {$1/2$};\n\\node[font=\\small,align=center] at (9.5,1.92) {$2$};\n\\node[font=\\small,align=center] at (12,1.92) {$3/64$};\n\\draw[black!18,line width=.3pt] (0.6,1.7) -- (13.4,1.7);\n\\node[font=\\small,align=center,font=\\small] at (1.6,1.45) {\\texttt{0101}};\n\\node[font=\\small,align=center] at (4.4,1.45) {$1/64$};\n\\node[font=\\small,align=center] at (7,1.45) {$1/2$};\n\\node[font=\\small,align=center] at (9.5,1.45) {$1$};\n\\node[font=\\small,align=center] at (12,1.45) {$1/128$};\n\\node[font=\\small,align=center,font=\\small] at (1.6,0.98) {\\texttt{0110}};\n\\node[font=\\small,align=center] at (4.4,0.98) {$3/64$};\n\\node[font=\\small,align=center] at (7,0.98) {$1/4$};\n\\node[font=\\small,align=center] at (9.5,0.98) {$2$};\n\\node[font=\\small,align=center] at (12,0.98) {$3/128$};\n\\draw[black!18,line width=.3pt] (0.6,0.76","4$};\n\\node[font=\\small,align=center] at (7,1.92) {$1/2$};\n\\node[font=\\small,align=center] at (9.5,1.92) {$2$};\n\\node[font=\\small,align=center] at (12,1.92) {$3/64$};\n\\draw[black!18,line width=.3pt] (0.6,1.7) -- (13.4,1.7);\n\\node[font=\\small,align=center,font=\\small] at (1.6,1.45) {\\texttt{0101}};\n\\node[font=\\small,align=center] at (4.4,1.45) {$1/64$};\n\\node[font=\\small,align=center] at (7,1.45) {$1/2$};\n\\node[font=\\small,align=center] at (9.5,1.45) {$1$};\n\\node[font=\\small,align=center] at (12,1.45) {$1/128$};\n\\node[font=\\small,align=center,font=\\small] at (1.6,0.98) {\\texttt{0110}};\n\\node[font=\\small,align=center] at (4.4,0.98) {$3/64$};\n\\node[font=\\small,align=center] at (7,0.98) {$1/4$};\n\\node[font=\\small,align=center] at (9.5,0.98) {$2$};\n\\node[font=\\small,align=center] at (12,0.98) {$3/128$};\n\\draw[black!18,line width=.3pt] (0.6,0.76) -- (13.4,0.76);\n\\node[font=\\small,align=center,font","};\n\\node[font=\\small,align=center] at (9.5,1.92) {$2$};\n\\node[font=\\small,align=center] at (12,1.92) {$3/64$};\n\\draw[black!18,line width=.3pt] (0.6,1.7) -- (13.4,1.7);\n\\node[font=\\small,align=center,font=\\small] at (1.6,1.45) {\\texttt{0101}};\n\\node[font=\\small,align=center] at (4.4,1.45) {$1/64$};\n\\node[font=\\small,align=center] at (7,1.45) {$1/2$};\n\\node[font=\\small,align=center] at (9.5,1.45) {$1$};\n\\node[font=\\small,align=center] at (12,1.45) {$1/128$};\n\\node[font=\\small,align=center,font=\\small] at (1.6,0.98) {\\texttt{0110}};\n\\node[font=\\small,align=center] at (4.4,0.98) {$3/64$};\n\\node[font=\\small,align=center] at (7,0.98) {$1/4$};\n\\node[font=\\small,align=center] at (9.5,0.98) {$2$};\n\\node[font=\\small,align=center] at (12,0.98) {$3/128$};\n\\draw[black!18,line width=.3pt] (0.6,0.76) -- (13.4,0.76);\n\\node[font=\\small,align=center,font=\\small] at (1.6,0.51) {\\texttt{0111}};\n\\node[font=\\s","$};\n\\node[font=\\small,align=center] at (12,1.92) {$3/64$};\n\\draw[black!18,line width=.3pt] (0.6,1.7) -- (13.4,1.7);\n\\node[font=\\small,align=center,font=\\small] at (1.6,1.45) {\\texttt{0101}};\n\\node[font=\\small,align=center] at (4.4,1.45) {$1/64$};\n\\node[font=\\small,align=center] at (7,1.45) {$1/2$};\n\\node[font=\\small,align=center] at (9.5,1.45) {$1$};\n\\node[font=\\small,align=center] at (12,1.45) {$1/128$};\n\\node[font=\\small,align=center,font=\\small] at (1.6,0.98) {\\texttt{0110}};\n\\node[font=\\small,align=center] at (4.4,0.98) {$3/64$};\n\\node[font=\\small,align=center] at (7,0.98) {$1/4$};\n\\node[font=\\small,align=center] at (9.5,0.98) {$2$};\n\\node[font=\\small,align=center] at (12,0.98) {$3/128$};\n\\draw[black!18,line width=.3pt] (0.6,0.76) -- (13.4,0.76);\n\\node[font=\\small,align=center,font=\\small] at (1.6,0.51) {\\texttt{0111}};\n\\node[font=\\small,align=center] at (4.4,0.51) {$9/64$};\n\\node[font=\\s","l,align=center,font=\\small] at (1.6,1.45) {\\texttt{0101}};\n\\node[font=\\small,align=center] at (4.4,1.45) {$1/64$};\n\\node[font=\\small,align=center] at (7,1.45) {$1/2$};\n\\node[font=\\small,align=center] at (9.5,1.45) {$1$};\n\\node[font=\\small,align=center] at (12,1.45) {$1/128$};\n\\node[font=\\small,align=center,font=\\small] at (1.6,0.98) {\\texttt{0110}};\n\\node[font=\\small,align=center] at (4.4,0.98) {$3/64$};\n\\node[font=\\small,align=center] at (7,0.98) {$1/4$};\n\\node[font=\\small,align=center] at (9.5,0.98) {$2$};\n\\node[font=\\small,align=center] at (12,0.98) {$3/128$};\n\\draw[black!18,line width=.3pt] (0.6,0.76) -- (13.4,0.76);\n\\node[font=\\small,align=center,font=\\small] at (1.6,0.51) {\\texttt{0111}};\n\\node[font=\\small,align=center] at (4.4,0.51) {$9/64$};\n\\node[font=\\small,align=center] at (7,0.51) {$1/4$};\n\\node[font=\\small,align=center] at (9.5,0.51) {$1$};\n\\node[font=\\small,align=center] at (12","1}};\n\\node[font=\\small,align=center] at (4.4,1.45) {$1/64$};\n\\node[font=\\small,align=center] at (7,1.45) {$1/2$};\n\\node[font=\\small,align=center] at (9.5,1.45) {$1$};\n\\node[font=\\small,align=center] at (12,1.45) {$1/128$};\n\\node[font=\\small,align=center,font=\\small] at (1.6,0.98) {\\texttt{0110}};\n\\node[font=\\small,align=center] at (4.4,0.98) {$3/64$};\n\\node[font=\\small,align=center] at (7,0.98) {$1/4$};\n\\node[font=\\small,align=center] at (9.5,0.98) {$2$};\n\\node[font=\\small,align=center] at (12,0.98) {$3/128$};\n\\draw[black!18,line width=.3pt] (0.6,0.76) -- (13.4,0.76);\n\\node[font=\\small,align=center,font=\\small] at (1.6,0.51) {\\texttt{0111}};\n\\node[font=\\small,align=center] at (4.4,0.51) {$9/64$};\n\\node[font=\\small,align=center] at (7,0.51) {$1/4$};\n\\node[font=\\small,align=center] at (9.5,0.51) {$1$};\n\\node[font=\\small,align=center] at (12,0.51) {$9/256$};\n\\node[font=\\small,align=center,font","64$};\n\\node[font=\\small,align=center] at (7,1.45) {$1/2$};\n\\node[font=\\small,align=center] at (9.5,1.45) {$1$};\n\\node[font=\\small,align=center] at (12,1.45) {$1/128$};\n\\node[font=\\small,align=center,font=\\small] at (1.6,0.98) {\\texttt{0110}};\n\\node[font=\\small,align=center] at (4.4,0.98) {$3/64$};\n\\node[font=\\small,align=center] at (7,0.98) {$1/4$};\n\\node[font=\\small,align=center] at (9.5,0.98) {$2$};\n\\node[font=\\small,align=center] at (12,0.98) {$3/128$};\n\\draw[black!18,line width=.3pt] (0.6,0.76) -- (13.4,0.76);\n\\node[font=\\small,align=center,font=\\small] at (1.6,0.51) {\\texttt{0111}};\n\\node[font=\\small,align=center] at (4.4,0.51) {$9/64$};\n\\node[font=\\small,align=center] at (7,0.51) {$1/4$};\n\\node[font=\\small,align=center] at (9.5,0.51) {$1$};\n\\node[font=\\small,align=center] at (12,0.51) {$9/256$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (7,-0.22) {Total probability $=1$; w","1/2$};\n\\node[font=\\small,align=center] at (9.5,1.45) {$1$};\n\\node[font=\\small,align=center] at (12,1.45) {$1/128$};\n\\node[font=\\small,align=center,font=\\small] at (1.6,0.98) {\\texttt{0110}};\n\\node[font=\\small,align=center] at (4.4,0.98) {$3/64$};\n\\node[font=\\small,align=center] at (7,0.98) {$1/4$};\n\\node[font=\\small,align=center] at (9.5,0.98) {$2$};\n\\node[font=\\small,align=center] at (12,0.98) {$3/128$};\n\\draw[black!18,line width=.3pt] (0.6,0.76) -- (13.4,0.76);\n\\node[font=\\small,align=center,font=\\small] at (1.6,0.51) {\\texttt{0111}};\n\\node[font=\\small,align=center] at (4.4,0.51) {$9/64$};\n\\node[font=\\small,align=center] at (7,0.51) {$1/4$};\n\\node[font=\\small,align=center] at (9.5,0.51) {$1$};\n\\node[font=\\small,align=center] at (12,0.51) {$9/256$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (7,-0.22) {Total probability $=1$; weighted terminal expectation $=311/256$.};\n\\end{tikzpict","\\node[font=\\small,align=center] at (4.4,0.98) {$3/64$};\n\\node[font=\\small,align=center] at (7,0.98) {$1/4$};\n\\node[font=\\small,align=center] at (9.5,0.98) {$2$};\n\\node[font=\\small,align=center] at (12,0.98) {$3/128$};\n\\draw[black!18,line width=.3pt] (0.6,0.76) -- (13.4,0.76);\n\\node[font=\\small,align=center,font=\\small] at (1.6,0.51) {\\texttt{0111}};\n\\node[font=\\small,align=center] at (4.4,0.51) {$9/64$};\n\\node[font=\\small,align=center] at (7,0.51) {$1/4$};\n\\node[font=\\small,align=center] at (9.5,0.51) {$1$};\n\\node[font=\\small,align=center] at (12,0.51) {$9/256$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (7,-0.22) {Total probability $=1$; weighted terminal expectation $=311/256$.};\n\\end{tikzpicture}",";\n\\node[font=\\small,align=center] at (7,0.98) {$1/4$};\n\\node[font=\\small,align=center] at (9.5,0.98) {$2$};\n\\node[font=\\small,align=center] at (12,0.98) {$3/128$};\n\\draw[black!18,line width=.3pt] (0.6,0.76) -- (13.4,0.76);\n\\node[font=\\small,align=center,font=\\small] at (1.6,0.51) {\\texttt{0111}};\n\\node[font=\\small,align=center] at (4.4,0.51) {$9/64$};\n\\node[font=\\small,align=center] at (7,0.51) {$1/4$};\n\\node[font=\\small,align=center] at (9.5,0.51) {$1$};\n\\node[font=\\small,align=center] at (12,0.51) {$9/256$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (7,-0.22) {Total probability $=1$; weighted terminal expectation $=311/256$.};\n\\end{tikzpicture}","\\node[font=\\small,align=center] at (9.5,0.98) {$2$};\n\\node[font=\\small,align=center] at (12,0.98) {$3/128$};\n\\draw[black!18,line width=.3pt] (0.6,0.76) -- (13.4,0.76);\n\\node[font=\\small,align=center,font=\\small] at (1.6,0.51) {\\texttt{0111}};\n\\node[font=\\small,align=center] at (4.4,0.51) {$9/64$};\n\\node[font=\\small,align=center] at (7,0.51) {$1/4$};\n\\node[font=\\small,align=center] at (9.5,0.51) {$1$};\n\\node[font=\\small,align=center] at (12,0.51) {$9/256$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (7,-0.22) {Total probability $=1$; weighted terminal expectation $=311/256$.};\n\\end{tikzpicture}","\\node[font=\\small,align=center] at (12,0.98) {$3/128$};\n\\draw[black!18,line width=.3pt] (0.6,0.76) -- (13.4,0.76);\n\\node[font=\\small,align=center,font=\\small] at (1.6,0.51) {\\texttt{0111}};\n\\node[font=\\small,align=center] at (4.4,0.51) {$9/64$};\n\\node[font=\\small,align=center] at (7,0.51) {$1/4$};\n\\node[font=\\small,align=center] at (9.5,0.51) {$1$};\n\\node[font=\\small,align=center] at (12,0.51) {$9/256$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (7,-0.22) {Total probability $=1$; weighted terminal expectation $=311/256$.};\n\\end{tikzpicture}",".6,0.76) -- (13.4,0.76);\n\\node[font=\\small,align=center,font=\\small] at (1.6,0.51) {\\texttt{0111}};\n\\node[font=\\small,align=center] at (4.4,0.51) {$9/64$};\n\\node[font=\\small,align=center] at (7,0.51) {$1/4$};\n\\node[font=\\small,align=center] at (9.5,0.51) {$1$};\n\\node[font=\\small,align=center] at (12,0.51) {$9/256$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (7,-0.22) {Total probability $=1$; weighted terminal expectation $=311/256$.};\n\\end{tikzpicture}","\\small,align=center,font=\\small] at (1.6,0.51) {\\texttt{0111}};\n\\node[font=\\small,align=center] at (4.4,0.51) {$9/64$};\n\\node[font=\\small,align=center] at (7,0.51) {$1/4$};\n\\node[font=\\small,align=center] at (9.5,0.51) {$1$};\n\\node[font=\\small,align=center] at (12,0.51) {$9/256$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (7,-0.22) {Total probability $=1$; weighted terminal expectation $=311/256$.};\n\\end{tikzpicture}","gn=center,anchor=east,font=\\scriptsize] at (0.64,2.65) {0.75};\n\\draw[black!13,line width=.25pt] (0.8,3.316667) -- (9.5,3.316667);\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (0.64,3.316667) {1};\n\\draw[->,line width=.6pt] (0.8,0.65) -- (9.62,0.65);\n\\draw[->,line width=.6pt] (0.8,0.65) -- (0.8,3.57);\n\\node[font=\\small,align=center,font=\\scriptsize] at (5.15,0.1) {diffusion time $s$};\n\\node[font=\\small,align=center,anchor=west,font=\\scriptsize] at (0.8,3.85) {clock density $\\eta_1(s)$};\n\\draw[line width=1.1pt] plot coordinates {(0.8522,0.650001) (0.888383,0.651177) (0.924567,0.675089) (0.96075,0.772122) (0.996933,0.962153) (1.033116,1.221126) (1.0693,1.5109) (1.105483,1.799624) (1.141666,2.066984) (1.177849,2.302583) (1.214033,2.502789) (1.250216,2.668017) (1.286399,2.800825) (1.322582,2.904726) (1.358766,2.983489) (1.394949,3.040763) (1.431132,3.079892) (1.467315,3.103843) (1.503499,3.115192) (1.539682,3.116147) (1.575865,3.108582) (1.612049,3.094078) (1.648232,3.073964) (1.684415,3.04935) (1.720598,3.021164) (1.756782,2.99018) (1.792965,2.957042) (1.829148,2.922282…","(0.8,3.316667) -- (9.5,3.316667);\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (0.64,3.316667) {1};\n\\draw[->,line width=.6pt] (0.8,0.65) -- (9.62,0.65);\n\\draw[->,line width=.6pt] (0.8,0.65) -- (0.8,3.57);\n\\node[font=\\small,align=center,font=\\scriptsize] at (5.15,0.1) {diffusion time $s$};\n\\node[font=\\small,align=center,anchor=west,font=\\scriptsize] at (0.8,3.85) {clock density $\\eta_1(s)$};\n\\draw[line width=1.1pt] plot coordinates {(0.8522,0.650001) (0.888383,0.651177) (0.924567,0.675089) (0.96075,0.772122) (0.996933,0.962153) (1.033116,1.221126) (1.0693,1.5109) (1.105483,1.799624) (1.141666,2.066984) (1.177849,2.302583) (1.214033,2.502789) (1.250216,2.668017) (1.286399,2.800825) (1.322582,2.904726) (1.358766,2.983489) (1.394949,3.040763) (1.431132,3.079892) (1.467315,3.103843) (1.503499,3.115192) (1.539682,3.116147) (1.575865,3.108582) (1.612049,3.094078) (1.648232,3.073964) (1.684415,3.04935) (1.720598,3.021164) (1.756782,2.99018) (1.792965,2.957042) (1.829148,2.922282) (1.865331,2.886345) (1.901515,2.849598) (1.937698,2.812344) (1.973881,2.774834) (2.010064,2.73…","9618,0.903302) (9.065801,0.901786) (9.101984,0.900285) (9.138167,0.898799) (9.174351,0.897328) (9.210534,0.895871) (9.246717,0.894428) (9.2829,0.892999) (9.319084,0.891584) (9.355267,0.890183) (9.39145,0.888795) (9.427633,0.88742) (9.463817,0.886059) (9.5,0.88471)};\n\\draw[densely dotted] (1.525,0.65) -- (1.525,3.116885);\n\\node[font=\\small,align=center,font=\\scriptsize] at (3,4.45) {Clock mode at $s=1/6$};\n\\node[font=\\small,align=center] at (11.8,3.8) {$\\displaystyle\\int_0^\\infty\\eta_t(s)\\,ds=1$};\n\\node[font=\\small,align=center] at (11.8,2.7) {$\\displaystyle\\int_0^\\infty s\\eta_t(s)\\,ds=\\infty$};\n\\node[font=\\small,align=center,font=\\small] at (11.8,1.2) {A probability clock",".138167,0.898799) (9.174351,0.897328) (9.210534,0.895871) (9.246717,0.894428) (9.2829,0.892999) (9.319084,0.891584) (9.355267,0.890183) (9.39145,0.888795) (9.427633,0.88742) (9.463817,0.886059) (9.5,0.88471)};\n\\draw[densely dotted] (1.525,0.65) -- (1.525,3.116885);\n\\node[font=\\small,align=center,font=\\scriptsize] at (3,4.45) {Clock mode at $s=1/6$};\n\\node[font=\\small,align=center] at (11.8,3.8) {$\\displaystyle\\int_0^\\infty\\eta_t(s)\\,ds=1$};\n\\node[font=\\small,align=center] at (11.8,2.7) {$\\displaystyle\\int_0^\\infty s\\eta_t(s)\\,ds=\\infty$};\n\\node[font=\\small,align=center,font=\\small] at (11.8,1.2) {A probability clock","9) (9.319084,0.891584) (9.355267,0.890183) (9.39145,0.888795) (9.427633,0.88742) (9.463817,0.886059) (9.5,0.88471)};\n\\draw[densely dotted] (1.525,0.65) -- (1.525,3.116885);\n\\node[font=\\small,align=center,font=\\scriptsize] at (3,4.45) {Clock mode at $s=1/6$};\n\\node[font=\\small,align=center] at (11.8,3.8) {$\\displaystyle\\int_0^\\infty\\eta_t(s)\\,ds=1$};\n\\node[font=\\small,align=center] at (11.8,2.7) {$\\displaystyle\\int_0^\\infty s\\eta_t(s)\\,ds=\\infty$};\n\\node[font=\\small,align=center,font=\\small] at (11.8,1.2) {A probability clock","16667);\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (0.64,2.816667) {2};\n\\draw[black!13,line width=.25pt] (0.8,3.9) -- (9.3,3.9);\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (0.64,3.9) {3};\n\\draw[->,line width=.6pt] (0.8,0.65) -- (9.42,0.65);\n\\draw[->,line width=.6pt] (0.8,0.65) -- (0.8,4.02);\n\\node[font=\\small,align=center,font=\\scriptsize] at (5.05,0.1) {$z>0$};\n\\node[font=\\small,align=center,anchor=west,font=\\scriptsize] at (0.8,4.3) {resolvent coefficient};\n\\draw[line width=1.1pt] plot coordinates {(1.225,3.604545) (1.30575,3.167942) (1.3865,2.850551) (1.46725,2.609135) (1.548,2.419126) (1.62875,2.265528) (1.7095,2.138668) (1.79025,2.032029) (1.871,1.941056) (1.95175,1.862472) (2.0325,1.793856) (2.11325,1.733384) (2.194,1.679652) (2.27475,1.631564) (2.3555,1.588249) (2.43625,1.549011) (2.517,1.513281) (2.59775,1.480594) (2.6785,1.450563) (2.75925,1.422866) (2.84,1.397232) (2.92075,1.373428) (3.0015,1.351259) (3.08225,1.330554) (3.163,1.311168) (3.24375,1.292972) (3.3245,1.275856) (3.40525,1.259723) (3.486,1.244486) (3.56675,1.230069) (3.647…",") (7.8465,0.901004) (7.92725,0.898813) (8.008,0.896661) (8.08875,0.894545) (8.1695,0.892465) (8.25025,0.89042) (8.331,0.88841) (8.41175,0.886432) (8.4925,0.884488) (8.57325,0.882575) (8.654,0.880693) (8.73475,0.878841) (8.8155,0.877019) (8.89625,0.875225) (8.977,0.87346) (9.05775,0.871722) (9.1385,0.870011) (9.21925,0.868326) (9.3,0.866667)};\n\\node[font=\\small,align=center] at (11.7,3.7) {exact: $\\dfrac{z+1}{z(z+2)}$};\n\\node[font=\\small,align=center] at (11.7,2.5) {omit hidden state: $\\dfrac1{z+1}$};\n\\node[font=\\small,align=center,font=\\small] at (11.7,1) {The pole at zero survives","5) (8.25025,0.89042) (8.331,0.88841) (8.41175,0.886432) (8.4925,0.884488) (8.57325,0.882575) (8.654,0.880693) (8.73475,0.878841) (8.8155,0.877019) (8.89625,0.875225) (8.977,0.87346) (9.05775,0.871722) (9.1385,0.870011) (9.21925,0.868326) (9.3,0.866667)};\n\\node[font=\\small,align=center] at (11.7,3.7) {exact: $\\dfrac{z+1}{z(z+2)}$};\n\\node[font=\\small,align=center] at (11.7,2.5) {omit hidden state: $\\dfrac1{z+1}$};\n\\node[font=\\small,align=center,font=\\small] at (11.7,1) {The pole at zero survives","nter,anchor=east,font=\\scriptsize] at (0.64,2.478571) {2};\n\\draw[black!13,line width=.25pt] (0.8,3.392857) -- (9.3,3.392857);\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (0.64,3.392857) {3};\n\\draw[->,line width=.6pt] (0.8,0.65) -- (9.42,0.65);\n\\draw[->,line width=.6pt] (0.8,0.65) -- (0.8,3.97);\n\\node[font=\\small,align=center,font=\\scriptsize] at (5.05,0.1) {partial-sum cutoff $N$};\n\\node[font=\\small,align=center,anchor=west,font=\\scriptsize] at (0.8,4.25) {sum of defect norms};\n\\draw[line width=1.1pt] plot coordinates {(1.33125,1.564286) (1.8625,2.021429) (2.39375,2.32619) (2.925,2.554762) (3.45625,2.737619) (3.9875,2.89) (4.51875,3.020612) (5.05,3.134898) (5.58125,3.236485) (6.1125,3.327914) (6.64375,3.411031) (7.175,3.487221) (7.70625,3.557551) (8.2375,3.622857) (8.76875,3.683809) (9.3,3.740952)};\n\\draw[dashed,line width=.9pt] plot coordinates {(1.33125,1.564286) (1.8625,1.792857) (2.39375,1.894444) (2.925,1.951587) (3.45625,1.988159) (3.9875,2.013556) (4.51875,2.032214) (5.05,2.0465) (5.58125,2.057788) (6.1125,2.06693) (6.64375,2.074487) (7.175,2.080836) (7.706…","dotted,line width=.9pt] plot coordinates {(1.33125,1.107143) (1.8625,1.221429) (2.39375,1.272222) (2.925,1.300794) (3.45625,1.319079) (3.9875,1.331778) (4.51875,1.341107) (5.05,1.34825) (5.58125,1.353894) (6.1125,1.358465) (6.64375,1.362243) (7.175,1.365418) (7.70625,1.368123) (8.2375,1.370455) (8.76875,1.372487) (9.3,1.374273)};\n\\node[font=\\small,align=center,font=\\small] at (11.7,3.95) {solid: $\\sum_{n\\le N}n^{-1}$};\n\\node[font=\\small,align=center,font=\\small] at (11.7,2.95) {dashed: $\\sum_{n\\le N}n^{-2}$};\n\\node[font=\\small,align=center,font=\\small] at (11.7,1.95) {dotted: $\\frac12\\sum_{n\\le N}n^{-2}$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (11.7,0.55) {The infinite-series conclusion","2222) (2.925,1.300794) (3.45625,1.319079) (3.9875,1.331778) (4.51875,1.341107) (5.05,1.34825) (5.58125,1.353894) (6.1125,1.358465) (6.64375,1.362243) (7.175,1.365418) (7.70625,1.368123) (8.2375,1.370455) (8.76875,1.372487) (9.3,1.374273)};\n\\node[font=\\small,align=center,font=\\small] at (11.7,3.95) {solid: $\\sum_{n\\le N}n^{-1}$};\n\\node[font=\\small,align=center,font=\\small] at (11.7,2.95) {dashed: $\\sum_{n\\le N}n^{-2}$};\n\\node[font=\\small,align=center,font=\\small] at (11.7,1.95) {dotted: $\\frac12\\sum_{n\\le N}n^{-2}$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (11.7,0.55) {The infinite-series conclusion",") (5.58125,1.353894) (6.1125,1.358465) (6.64375,1.362243) (7.175,1.365418) (7.70625,1.368123) (8.2375,1.370455) (8.76875,1.372487) (9.3,1.374273)};\n\\node[font=\\small,align=center,font=\\small] at (11.7,3.95) {solid: $\\sum_{n\\le N}n^{-1}$};\n\\node[font=\\small,align=center,font=\\small] at (11.7,2.95) {dashed: $\\sum_{n\\le N}n^{-2}$};\n\\node[font=\\small,align=center,font=\\small] at (11.7,1.95) {dotted: $\\frac12\\sum_{n\\le N}n^{-2}$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (11.7,0.55) {The infinite-series conclusion",";\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (0.64,2.25) {0.5};\n\\draw[black!13,line width=.25pt] (0.8,3.85) -- (9.3,3.85);\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (0.64,3.85) {1};\n\\draw[->,line width=.6pt] (0.8,0.65) -- (9.42,0.65);\n\\draw[->,line width=.6pt] (0.8,0.65) -- (0.8,3.97);\n\\node[font=\\small,align=center,font=\\scriptsize] at (5.05,0.1) {time $t$};\n\\node[font=\\small,align=center,anchor=west,font=\\scriptsize] at (0.8,4.25) {probability or rate};\n\\draw[line width=1.1pt] plot coordinates {(0.8,3.85) (0.885,3.846131) (0.97,3.835028) (1.055,3.817405) (1.14,3.793926) (1.225,3.765203) (1.31,3.731804) (1.395,3.694253) (1.48,3.653034) (1.565,3.608595) (1.65,3.561347) (1.735,3.511671) (1.82,3.459916) (1.905,3.406402) (1.99,3.351424) (2.075,3.295253) (2.16,3.238135) (2.245,3.180296) (2.33,3.121944) (2.415,3.063264) (2.5,3.004428) (2.585,2.945592) (2.67,2.886894) (2.755,2.828461) (2.84,2.770407) (2.925,2.712835) (3.01,2.655834) (3.095,2.599487) (3.18,2.543865) (3.265,2.489031) (3.35,2.435041) (3.435,2.381943) (3.52,2.329779) (3.605,2.278583) …",".21) (7.685,3.216337) (7.77,3.222549) (7.855,3.228641) (7.94,3.234615) (8.025,3.240476) (8.11,3.246226) (8.195,3.251869) (8.28,3.257407) (8.365,3.262844) (8.45,3.268182) (8.535,3.273423) (8.62,3.278571) (8.705,3.283628) (8.79,3.288596) (8.875,3.293478) (8.96,3.298276) (9.045,3.302991) (9.13,3.307627) (9.215,3.312185) (9.3,3.316667)};\n\\node[font=\\small,align=center] at (11.7,3.8) {solid: survival $S(t)$};\n\\node[font=\\small,align=center] at (11.7,2.8) {dashed: absorbed mass};\n\\node[font=\\small,align=center] at (11.7,1.8) {dotted: hazard $h(t)$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (11.7,0.6) {$h(0)=0$, but eventual exit","5,3.262844) (8.45,3.268182) (8.535,3.273423) (8.62,3.278571) (8.705,3.283628) (8.79,3.288596) (8.875,3.293478) (8.96,3.298276) (9.045,3.302991) (9.13,3.307627) (9.215,3.312185) (9.3,3.316667)};\n\\node[font=\\small,align=center] at (11.7,3.8) {solid: survival $S(t)$};\n\\node[font=\\small,align=center] at (11.7,2.8) {dashed: absorbed mass};\n\\node[font=\\small,align=center] at (11.7,1.8) {dotted: hazard $h(t)$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (11.7,0.6) {$h(0)=0$, but eventual exit","628) (8.79,3.288596) (8.875,3.293478) (8.96,3.298276) (9.045,3.302991) (9.13,3.307627) (9.215,3.312185) (9.3,3.316667)};\n\\node[font=\\small,align=center] at (11.7,3.8) {solid: survival $S(t)$};\n\\node[font=\\small,align=center] at (11.7,2.8) {dashed: absorbed mass};\n\\node[font=\\small,align=center] at (11.7,1.8) {dotted: hazard $h(t)$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (11.7,0.6) {$h(0)=0$, but eventual exit","join=round]\n\\path[use as bounding box] (0,-.6) rectangle (14,4.8);\n\\draw[line width=.6pt] (0.2,0.55) -- (4.2,0.55);\n\\node[font=\\small,align=center] at (2.2,3.8) {Keep the full state};\n\\path[draw=black,fill=black!12,line width=.6pt] (0.44,0.55) rectangle (0.96,3.05);\n\\node[font=\\small,align=center,font=\\scriptsize] at (0.7,0.25) {0};\n\\node[font=\\small,align=center,font=\\scriptsize] at (0.7,3.27) {$\\frac{1}{2}$};\n\\path[draw=black,fill=black!12,line width=.6pt] (1.44,0.55) rectangle (1.96,0.55);\n\\node[font=\\small,align=center,font=\\scriptsize] at (1.7,0.25) {1};\n\\node[font=\\small,align=center,font=\\scriptsize] at (1.7,0.77) {$0$};\n\\path[draw=black,fill=black!12,line width=.6pt] (2.44,0.55) rectangle (2.96,1.8);\n\\node[font=\\small,align=center,font=\\scriptsize] at (2.7,0.25) {2};\n\\node[font=\\small,align=center,font=\\scriptsize] at (2.7,2.02) {$\\frac{1}{4}$};\n\\path[draw=black,fill=black!12,line width=.6p","(0.44,0.55) rectangle (0.96,3.05);\n\\node[font=\\small,align=center,font=\\scriptsize] at (0.7,0.25) {0};\n\\node[font=\\small,align=center,font=\\scriptsize] at (0.7,3.27) {$\\frac{1}{2}$};\n\\path[draw=black,fill=black!12,line width=.6pt] (1.44,0.55) rectangle (1.96,0.55);\n\\node[font=\\small,align=center,font=\\scriptsize] at (1.7,0.25) {1};\n\\node[font=\\small,align=center,font=\\scriptsize] at (1.7,0.77) {$0$};\n\\path[draw=black,fill=black!12,line width=.6pt] (2.44,0.55) rectangle (2.96,1.8);\n\\node[font=\\small,align=center,font=\\scriptsize] at (2.7,0.25) {2};\n\\node[font=\\small,align=center,font=\\scriptsize] at (2.7,2.02) {$\\frac{1}{4}$};\n\\path[draw=black,fill=black!12,line width=.6pt] (3.44,0.55) rectangle (3.96,1.8);\n\\node[font=\\small,align=center,font=\\scriptsize] at (3.7,0.25) {3};\n\\node[font=\\small,align=center,font=\\scriptsize] at (3.7,2.02) {$\\frac{1}{4}$};\n\\draw[line width=.6pt] (5.2,0.55) --","width=.6pt] (1.44,0.55) rectangle (1.96,0.55);\n\\node[font=\\small,align=center,font=\\scriptsize] at (1.7,0.25) {1};\n\\node[font=\\small,align=center,font=\\scriptsize] at (1.7,0.77) {$0$};\n\\path[draw=black,fill=black!12,line width=.6pt] (2.44,0.55) rectangle (2.96,1.8);\n\\node[font=\\small,align=center,font=\\scriptsize] at (2.7,0.25) {2};\n\\node[font=\\small,align=center,font=\\scriptsize] at (2.7,2.02) {$\\frac{1}{4}$};\n\\path[draw=black,fill=black!12,line width=.6pt] (3.44,0.55) rectangle (3.96,1.8);\n\\node[font=\\small,align=center,font=\\scriptsize] at (3.7,0.25) {3};\n\\node[font=\\small,align=center,font=\\scriptsize] at (3.7,2.02) {$\\frac{1}{4}$};\n\\draw[line width=.6pt] (5.2,0.55) -- (8.3,0.55);\n\\node[font=\\small,align=center] at (6.75,3.8) {Retain only masses};\n\\path[draw=black,fill=black!12,line width=.6pt] (5.572,0.55) rectangle (6.378,3.05);\n\\node[font=\\small,align=center,font=\\scriptsize] at (5.975,0.25)","t] (2.44,0.55) rectangle 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\\frac{6}{5}","frac{16}{5}\\end{matrix}\\right]$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (6.9,0.7) {$D_1f=\\left[\\begin{matrix}- \\frac{6}{5}\\\\- \\frac{6}{5}\\\\\\frac{18}{35}\\\\\\frac{18}{35}\\end{matrix}\\right]$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (11.5,0.7) {$D_2f=\\left[\\begin{matrix}-2","frac{18}{35}\\end{matrix}\\right]$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (11.5,0.7) {$D_2f=\\left[\\begin{matrix}-2\\\\1\\\\- \\frac{12}{7}\\\\\\frac{9}{7}\\end{matrix}\\right]$};\n\\end{tikzpicture}","\\begin{tikzpicture}[x=1cm,y=1cm,>=Stealth,line cap=round,line join=round]\n\\path[use as bounding box] (0,-.6) rectangle (14,4.8);\n\\draw[line width=.6pt] (0.4,0.65) -- (6.1,0.65);\n\\node[font=\\small,align=center] at (3.25,3.85) {Observable $f$};\n\\path[draw=black,fill=black!12,line width=.6pt] (0.742,0.65) rectangle (1.483,0.65);\n\\node[font=\\small,align=center,font=\\scriptsize] at (1.1125,0.35) {0};\n\\node[font=\\small,align=center,font=\\scriptsize] at (1.1125,0.87) {$0$};\n\\path[draw=black,fill=black!12,line width=.6pt] (2.167,0.65) rectangle (2.908,2.12);\n\\node[font=\\small,align=center,font=\\scriptsize] at (2.5375,0.35) {1};\n\\node[font=\\small,align=center,font=\\scriptsize] at (2.5375,2.34) {$3$};\n\\path[draw=black,fill=black!12,line wi","round]\n\\path[use as bounding box] (0,-.6) rectangle (14,4.8);\n\\draw[line width=.6pt] (0.4,0.65) -- (6.1,0.65);\n\\node[font=\\small,align=center] at (3.25,3.85) {Observable $f$};\n\\path[draw=black,fill=black!12,line width=.6pt] (0.742,0.65) rectangle (1.483,0.65);\n\\node[font=\\small,align=center,font=\\scriptsize] at (1.1125,0.35) {0};\n\\node[font=\\small,align=center,font=\\scriptsize] at (1.1125,0.87) {$0$};\n\\path[draw=black,fill=black!12,line width=.6pt] (2.167,0.65) rectangle (2.908,2.12);\n\\node[font=\\small,align=center,font=\\scriptsize] at (2.5375,0.35) {1};\n\\node[font=\\small,align=center,font=\\scriptsize] at (2.5375,2.34) {$3$};\n\\path[draw=black,fill=black!12,line width=.6pt] (3.592,0.65) rectangle (4.333,1.63);\n\\node[font=\\small,align=center,font=\\scriptsize] at (3.9625,0.35) {2};\n\\node[font=\\small,align=center,font=\\scriptsize] at (3.9625,1.85) {$2$};\n\\path[draw=black,fill=black!12,line wi","2,0.65) rectangle (1.483,0.65);\n\\node[font=\\small,align=center,font=\\scriptsize] at (1.1125,0.35) {0};\n\\node[font=\\small,align=center,font=\\scriptsize] at (1.1125,0.87) {$0$};\n\\path[draw=black,fill=black!12,line width=.6pt] (2.167,0.65) rectangle (2.908,2.12);\n\\node[font=\\small,align=center,font=\\scriptsize] at (2.5375,0.35) {1};\n\\node[font=\\small,align=center,font=\\scriptsize] at (2.5375,2.34) {$3$};\n\\path[draw=black,fill=black!12,line width=.6pt] (3.592,0.65) rectangle (4.333,1.63);\n\\node[font=\\small,align=center,font=\\scriptsize] at (3.9625,0.35) {2};\n\\node[font=\\small,align=center,font=\\scriptsize] at (3.9625,1.85) {$2$};\n\\path[draw=black,fill=black!12,line width=.6pt] (5.017,0.65) rectangle (5.758,3.1);\n\\node[font=\\small,align=center,font=\\scriptsize] at (5.3875,0.35) {3};\n\\node[font=\\small,align=center,font=\\scriptsize] at (5.3875,3.32) {$5$};\n\\draw[line width=.6pt] (7.6,0.65) -- (1","7,0.65) rectangle (2.908,2.12);\n\\node[font=\\small,align=center,font=\\scriptsize] at (2.5375,0.35) {1};\n\\node[font=\\small,align=center,font=\\scriptsize] at (2.5375,2.34) {$3$};\n\\path[draw=black,fill=black!12,line width=.6pt] (3.592,0.65) rectangle (4.333,1.63);\n\\node[font=\\small,align=center,font=\\scriptsize] at (3.9625,0.35) {2};\n\\node[font=\\small,align=center,font=\\scriptsize] at (3.9625,1.85) {$2$};\n\\path[draw=black,fill=black!12,line width=.6pt] (5.017,0.65) rectangle (5.758,3.1);\n\\node[font=\\small,align=center,font=\\scriptsize] at (5.3875,0.35) {3};\n\\node[font=\\small,align=center,font=\\scriptsize] at (5.3875,3.32) {$5$};\n\\draw[line width=.6pt] (7.6,0.65) -- (13.3,0.65);\n\\node[font=\\small,align=center] at (10.45,3.85) {Conditional expectation $Ef$};\n\\path[draw=black,fill=black!12,line width=.6pt] (7.942,0.65) rectangle (8.683,1.63);\n\\node[font=\\small,align=center,font=\\scriptsize] at (","92,0.65) rectangle (4.333,1.63);\n\\node[font=\\small,align=center,font=\\scriptsize] at (3.9625,0.35) {2};\n\\node[font=\\small,align=center,font=\\scriptsize] at (3.9625,1.85) {$2$};\n\\path[draw=black,fill=black!12,line width=.6pt] (5.017,0.65) rectangle (5.758,3.1);\n\\node[font=\\small,align=center,font=\\scriptsize] at (5.3875,0.35) {3};\n\\node[font=\\small,align=center,font=\\scriptsize] at (5.3875,3.32) {$5$};\n\\draw[line width=.6pt] (7.6,0.65) -- (13.3,0.65);\n\\node[font=\\small,align=center] at (10.45,3.85) {Conditional expectation $Ef$};\n\\path[draw=black,fill=black!12,line width=.6pt] (7.942,0.65) rectangle (8.683,1.63);\n\\node[font=\\small,align=center,font=\\scriptsize] at (8.3125,0.35) {0};\n\\node[font=\\small,align=center,font=\\scriptsize] at (8.3125,1.85) {$2$};\n\\path[draw=black,fill=black!12,line width=.6pt] (9.367,0.65) rectangle (10.108,1.63);\n\\node[font=\\small,align=center,font=\\scriptsize] at","enter,font=\\scriptsize] at (3.9625,1.85) {$2$};\n\\path[draw=black,fill=black!12,line width=.6pt] (5.017,0.65) rectangle (5.758,3.1);\n\\node[font=\\small,align=center,font=\\scriptsize] at (5.3875,0.35) {3};\n\\node[font=\\small,align=center,font=\\scriptsize] at (5.3875,3.32) {$5$};\n\\draw[line width=.6pt] (7.6,0.65) -- (13.3,0.65);\n\\node[font=\\small,align=center] at (10.45,3.85) {Conditional expectation $Ef$};\n\\path[draw=black,fill=black!12,line width=.6pt] (7.942,0.65) rectangle (8.683,1.63);\n\\node[font=\\small,align=center,font=\\scriptsize] at (8.3125,0.35) {0};\n\\node[font=\\small,align=center,font=\\scriptsize] at (8.3125,1.85) {$2$};\n\\path[draw=black,fill=black!12,line width=.6pt] (9.367,0.65) rectangle (10.108,1.63);\n\\node[font=\\small,align=center,font=\\scriptsize] at (9.7375,0.35) {1};\n\\node[font=\\small,align=center,font=\\scriptsize] at (9.7375,1.85) {$2$};\n\\path[draw=black,fill=black!12,line w","ter,font=\\scriptsize] at (5.3875,3.32) {$5$};\n\\draw[line width=.6pt] (7.6,0.65) -- (13.3,0.65);\n\\node[font=\\small,align=center] at (10.45,3.85) {Conditional expectation $Ef$};\n\\path[draw=black,fill=black!12,line width=.6pt] (7.942,0.65) rectangle (8.683,1.63);\n\\node[font=\\small,align=center,font=\\scriptsize] at (8.3125,0.35) {0};\n\\node[font=\\small,align=center,font=\\scriptsize] at (8.3125,1.85) {$2$};\n\\path[draw=black,fill=black!12,line width=.6pt] (9.367,0.65) rectangle (10.108,1.63);\n\\node[font=\\small,align=center,font=\\scriptsize] at (9.7375,0.35) {1};\n\\node[font=\\small,align=center,font=\\scriptsize] at (9.7375,1.85) {$2$};\n\\path[draw=black,fill=black!12,line width=.6pt] (10.792,0.65) rectangle (11.533,2.47);\n\\node[font=\\small,align=center,font=\\scriptsize] at (11.1625,0.35) {2};\n\\node[font=\\small,align=center,font=\\scriptsize] at (11.1625,2.69) {$\\frac{26}{7}$};\n\\path[draw=black,fill=",",0.65) rectangle (8.683,1.63);\n\\node[font=\\small,align=center,font=\\scriptsize] at (8.3125,0.35) {0};\n\\node[font=\\small,align=center,font=\\scriptsize] at (8.3125,1.85) {$2$};\n\\path[draw=black,fill=black!12,line width=.6pt] (9.367,0.65) rectangle (10.108,1.63);\n\\node[font=\\small,align=center,font=\\scriptsize] at (9.7375,0.35) {1};\n\\node[font=\\small,align=center,font=\\scriptsize] at (9.7375,1.85) {$2$};\n\\path[draw=black,fill=black!12,line width=.6pt] (10.792,0.65) rectangle (11.533,2.47);\n\\node[font=\\small,align=center,font=\\scriptsize] at (11.1625,0.35) {2};\n\\node[font=\\small,align=center,font=\\scriptsize] at (11.1625,2.69) {$\\frac{26}{7}$};\n\\path[draw=black,fill=black!12,line width=.6pt] (12.217,0.65) rectangle (12.958,2.47);\n\\node[font=\\small,align=center,font=\\scriptsize] at (12.5875,0.35) {3};\n\\node[font=\\small,align=center,font=\\scriptsize] at (12.5875,2.69) {$\\frac{26}{7}$};\n\\node[fo","5) rectangle (10.108,1.63);\n\\node[font=\\small,align=center,font=\\scriptsize] at (9.7375,0.35) {1};\n\\node[font=\\small,align=center,font=\\scriptsize] at (9.7375,1.85) {$2$};\n\\path[draw=black,fill=black!12,line width=.6pt] (10.792,0.65) rectangle (11.533,2.47);\n\\node[font=\\small,align=center,font=\\scriptsize] at (11.1625,0.35) {2};\n\\node[font=\\small,align=center,font=\\scriptsize] at (11.1625,2.69) {$\\frac{26}{7}$};\n\\path[draw=black,fill=black!12,line width=.6pt] (12.217,0.65) rectangle (12.958,2.47);\n\\node[font=\\small,align=center,font=\\scriptsize] at (12.5875,0.35) {3};\n\\node[font=\\small,align=center,font=\\scriptsize] at (12.5875,2.69) {$\\frac{26}{7}$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (7,4.45) {Weights $\\pi=(1,2,3,4)/10$; the block values are 2 and $26/7$.};\n\\end{tikzpicture}","(11.533,2.47);\n\\node[font=\\small,align=center,font=\\scriptsize] at (11.1625,0.35) {2};\n\\node[font=\\small,align=center,font=\\scriptsize] at (11.1625,2.69) {$\\frac{26}{7}$};\n\\path[draw=black,fill=black!12,line width=.6pt] (12.217,0.65) rectangle (12.958,2.47);\n\\node[font=\\small,align=center,font=\\scriptsize] at (12.5875,0.35) {3};\n\\node[font=\\small,align=center,font=\\scriptsize] at (12.5875,2.69) {$\\frac{26}{7}$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (7,4.45) {Weights $\\pi=(1,2,3,4)/10$; the block values are 2 and $26/7$.};\n\\end{tikzpicture}","\\node[font=\\small,align=center,font=\\scriptsize] at (11.1625,2.69) {$\\frac{26}{7}$};\n\\path[draw=black,fill=black!12,line width=.6pt] (12.217,0.65) rectangle (12.958,2.47);\n\\node[font=\\small,align=center,font=\\scriptsize] at (12.5875,0.35) {3};\n\\node[font=\\small,align=center,font=\\scriptsize] at (12.5875,2.69) 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