Large Signal ModelReading edition

20. Truncating memory without losing the prediction bound

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20. Truncating memory without losing the prediction bound

Section 8 bounded the omitted tail of the exact linear memory expression using the same resolved history. We now close the separate recursion argument.

Write the exact relation as

\[ z_{t+1}=Az_t+\sum_{k=0}^{t-1}K_kz_{t-1-k}+f_t, \qquad K_k=BD^kC,\quad f_t=BD^tq_0. \]

For negative indices set \(z_s=0\), solely as a convention for missing prehistory in this initialized formula. A length-\(H\) approximate model, with \(H\geq0\), is

\[ \widehat z_{t+1}=\widehat A\widehat z_t+ \sum_{k=0}^{H-1}\widehat K_k\widehat z_{t-1-k}. \]

An empty sum is zero. Let \(d_t\) be the exact right-hand side minus this truncated right-hand side evaluated on the true history.

Proposition 19: Uniform prediction bound for a stable finite-memory truncation

Suppose

\[ a=\|\widehat A\|+\sum_{k=0}^{H-1}\|\widehat K_k\|<1, \qquad\sup_{t\geq0}\|d_t\|\leq\Delta. \]

Let \(E_0\) bound the initialization errors at indices \(-H,\ldots,0\). Then the recursively generated predictions satisfy

\[ \boxed{ \sup_{t\geq0}\|z_t-\widehat z_t\| \leq\max\left(E_0,\frac{\Delta}{1-a}\right). } \]

If \(\|z_t\|\leq Z\) for \(t\geq0\) and \(\|D\|\leq\rho<1\), a valid choice of defect bound is

\[ \begin{aligned} \Delta={}&Z\left(\|A-\widehat A\| +\sum_{k=0}^{H-1}\|K_k-\widehat K_k\|\right)\\ &+\|B\|\,\|C\|Z\frac{\rho^H}{1-\rho} +\|B\|\,\|q_0\|. \end{aligned} \]

The last term can be removed if the exact initialized forcing \(f_t\) is retained in the approximate recurrence.

Proof

Subtract the two recurrences and bound every retained history error by the largest of the errors at indices \(t-H,\ldots,t\). This gives \(e_{t+1}\leq a\max_{t-H\leq s\leq t}e_s+\Delta\). The ball of radius \(E=\max(E_0,\Delta/(1-a))\) is invariant because \(aE+\Delta\leq E\). Induction proves the bound. The estimate for \(\Delta\) separates coefficient errors, the geometric kernel tail and the initialized forcing, using Proposition 10. \(\square\)

This norm-sum condition is sufficient, not necessary. A stable recurrence with cancellation can fail this test. Failure to find this certificate is not proof of instability.

For the scalar example with \(q_0=0\), take exact retained coefficients and \(z_0=1\). All outputs lie in \([0,1]\): the full nonnegative memory coefficients sum to at most \(7/8\), so induction bounds every new output by the previous maximum. Thus \(Z=1\) is valid. With

\[ A=1/2,\quad K_k=\frac14\left(\frac13\right)^k, \]

one obtains

\[ a_H=\frac78-\frac38\,3^{-H},\qquad \Delta_H=\frac38\,3^{-H},\qquad \boxed{\sup_t|z_t-\widehat z_t|\leq \frac{3\,3^{-H}}{1+3\,3^{-H}}.} \]

The right-hand side is a theorem for every time in this initialized example. The figure compares it with a separately computed finite-horizon error. The computed maximum is not substituted for the uniform bound.

A memory-tail estimate becomes a recursive prediction certificate only after a stability margin is supplied. The measured curve is the maximum over 80 calculated steps. The upper curve is the proved bound for all steps under the stated scalar initialization.
A memory-tail estimate becomes a recursive prediction certificate only after a stability margin is supplied. The measured curve is the maximum over 80 calculated steps. The upper curve is the proved bound for all steps under the stated scalar initialization.

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