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Conditional future means alone do not bound finite-particle SIR error

Statement refuted

The following proposed implication is false as stated. For every finite-level full-resampling SIR event model with positive finite prefix normalizers, define $Q_{p, L} 1 (x_{p}) = E ⁣ [\prod_{j = p + 1}^{L} G_{j} (X_{j}) ∣ X_{p} = x_{p}], h_{p, L} = \frac{Q_{p, L} 1}{Z_{L} / Z_{p}}, η_{p} = the normalized level- p law .$ If $\int h_{p, L}^{2} d η_{p} \leq B$ for all $p < L$ , the asserted finite-particle bound is $E [({\hat{Z}}_{L} - Z_{L})^{2}] \leq Z_{L}^{2} [{(1 + \frac{B - 1}{N})}^{L} - 1] .$

Facts & Assumptions

Given: The two-state probability space $Ω = {0, 1}$ with uniform law $P$ ; $A_{0} = A_{1} = Ω$ , $A_{2} = {0}$ ; $L = 2$ ; and $N \geq 1$ .

[A1]

The Axiom of Choice: AC is inherited through the general SIR supplier; the finite model and its calculations require no additional choice.

[F1]

Nested rare-event levels and conditional laws: $p_{ℓ} = P (A_{ℓ}) / P (A_{ℓ - 1})$ and $π_{ℓ} = P (\cdot ∣ A_{ℓ})$ .

[F2]

Full-resampling fixed-level SIR splitting with invariant mutations: The SIR potentials are event indicators, the level-one mutation is $K_{1}$ , and the estimator is the product of average incremental weights under full multinomial resampling.

[F3]

Sequential importance resampling for Feynman–Kac path targets: With proposal kernels equal to target transitions, incremental Radon–Nikodym weights are one; the normalizer update multiplies the previous estimate by the sample mean of the next potential.

Counterexample

1.1A1F1F2given

Here $p_{1} = 1$ , $p_{2} = 1 / 2$ , $Z_{0} = Z_{1} = 1$ , $Z_{2} = 1 / 2$ , and $η_{0} = η_{1} = P$ . Use $M_{1}$ equal to the identity and choose the level-one mutation $K_{1} (x, \cdot) = P (\cdot)$ . This refresh kernel is supported on $A_{1} = Ω$ and leaves $π_{1} = P$ invariant, so it satisfies the contract's mutation hypotheses.

1.2F1F2givenalgebra

For $p = 0$ , the future event probability $Q_{0, 2} 1 (x_{0})$ is $1 / 2$ for every state because the refresh is independent of $x_{0}$ ; for $p = 1$ , $Q_{1, 2} 1 (x_{1}) = K_{1} 1_{A_{2}} (x_{1}) = 1 / 2$ for every state. Both normalizers $Z_{2} / Z_{p}$ equal $1 / 2$ . Thus $h_{0, 2} = h_{1, 2} = 1$ , each $η_{p}$ -second moment is $1$ , and $B = 1$ satisfies the stated condition.

2.1F2F3step 1.1algebra

Since $G_{1} \equiv 1$ , the first resampling weights are equal. The refresh mutation ignores the resampled ancestors, so $X_{2}^{1}, \dots, X_{2}^{N}$ are IID with law $P$ . By [F3], ${\hat{Z}}_{2} = N^{- 1} \sum_{i = 1}^{N} 1_{{X_{2}^{i} = 0}}$ , hence $E [({\hat{Z}}_{2} - Z_{2})^{2}] = Var ({\hat{Z}}_{2}) = 1 / (4 N)$ .

3.1step 2.1step 1.2algebra∎

With $B = 1$ , the proposed right-hand side is $Z_{2}^{2} [(1 + (B - 1) / N)^{2} - 1] = 0$ , contradicting the strictly positive error $1 / (4 N)$ in step 2.1. Therefore the proposed implication is false.

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