Paper VI: Finite state Feynman Kac transfer

E13 Finite state Feynman Kac transfer

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How to read the result

The potential below is real. Its exponential is a potential weight in general, and a survival weight when the potential is nonnegative. This distinction does not change the finite transfer formula.

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Let \(K\) be column stochastic. Let \(V\) be a real potential on the states and let

\[ D_V={\operatorname{diag}}(V). \]

For functions written as columns, the backward Markov operator is \(K^T\).

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Theorem E13.1: Discrete Feynman Kac formula

Define

\[ M_V={\operatorname{diag}}({\mathrm e}^{-V})K^T. \]

For terminal data \(f\),

\[ (M_V^nf)_i = {\mathbb E}_i\left[ \exp\left( -\sum_{r=0}^{n-1}V(X_r) \right) f(X_n) \right]. \]

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Proof of Theorem E13.1

For one step,

\[ (M_Vf)_i = {\mathrm e}^{-V(i)}\sum_jK_{ji}f_j. \]

Iterating 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.

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Theorem E13.2: Continuous time finite state formula

Let \(Q\) be a finite Markov generator and let \((X_t)_{t\geq0}\) be the corresponding chain. Then

\[ \left( {\mathrm e}^{t(Q^T-D_V)}f \right)_i = {\mathbb E}_i\left[ \exp\left( -\int_0^tV(X_s){\,\mathrm d} s \right) f(X_t) \right]. \]

If \(V\geq v_0\), then

\[ {\left\lVert{\mathrm e}^{t(Q^T-D_V)}\right\rVert}_{\infty\to\infty} \leq {\mathrm e}^{-v_0t}. \]

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Proof of Theorem E13.2

Conditioning over the first small time interval gives the backward equation

\[ \dot u=(Q^T-D_V)u, \qquad u(0)=f. \]

The finite system has a unique solution, namely the displayed matrix exponential. The expectation bound follows from

\[ \exp\left(-\int_0^tV(X_s){\,\mathrm d} s\right) \leq {\mathrm e}^{-v_0t}. \]

This finite argument is the matrix form of the Feynman Kac principle originating in [Kac1949].

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Theorem E13.3: Alternating propagation and killing

For every \(t\geq0\),

\[ {\mathrm e}^{t(Q^T-D_V)} = \lim_{n\to\infty} \left( {\mathrm e}^{-tD_V/n}{\mathrm e}^{tQ^T/n} \right)^n \]

in operator norm. Each finite product alternates one Markov propagation step with one diagonal survival weight.

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Proof of Theorem E13.3

Set \(B=-D_V\) and \(C=Q^T\). For bounded matrices,

\[ {\mathrm e}^{tB/n}{\mathrm e}^{tC/n} = I+\frac{t}{n}(B+C)+O(n^{-2}) \]

uniformly in operator norm. The standard telescoping comparison with \( (I+t(B+C)/n)^n \) shows that the product converges to \({\mathrm e}^{t(B+C)}\). This is the displayed factor order, so the limit follows.

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Corollary E13.4: Positivity and composition

The operators

\[ F_V(t)={\mathrm e}^{t(Q^T-D_V)} \]

form a positive semigroup. If \(f\geq0\), then \(F_V(t)f\geq0\).

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Proof of Corollary E13.4

Both claims follow from the expectation formula.

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