TheoremStatement: AI-adaptedProof: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

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Fixed-level SIR has an absolute O(L/N) mean-square bound

Statement

Assume AC. Use the full-resampling fixed-level SIR model of Full-resampling fixed-level SIR splitting with invariant mutations on nested levels $A_{0} \supseteq \dots \supseteq A_{L}$ , with $N \geq 1$ particles and positive finite prefixes $Z_{p} = P (A_{p}) > 0$ . Its event potentials satisfy $0 \leq G_{p} \leq 1$ . Then $E [\hat{p}] = P (A_{L}), E [(\hat{p} - P (A_{L}))^{2}] = Var (\hat{p}) \leq \frac{L + 1}{4 N} .$ The bound is an absolute mean-square error bound for fixed $L$ ; it is not a relative-efficiency claim.

Facts & Assumptions

Given: The nested event model, $L \geq 1$ , $N \geq 1$ , full multinomial resampling, transitions $M_{p}$ , and indicator potentials $G_{p} = 1_{A_{p}}$ in Full-resampling fixed-level SIR splitting with invariant mutations.

[A1]

The Axiom of Choice: AC is inherited through the canonical Feynman–Kac, SIR, and conditional-variance suppliers.

[F1]

Nested rare-event levels and conditional laws: $Z_{p} = P (A_{p}) \in (0, 1]$ for each prefix, and the target is $P (A_{L})$ .

[F2]

Full-resampling fixed-level SIR splitting with invariant mutations: The particle normalizer is multiplied at step $p$ by the empirical mean of $G_{p}$ ; full multinomial resampling followed by $M_{p}$ supplies conditionally independent proposals from the current weighted empirical law followed by $M_{p}$ .

[F3]

A bounded-potential SIR update has a uniform unnormalized variance bound: Given $z \in [0, 1]$ , a probability law $ν$ , transition $M_{p}$ , potential $G_{p} \in [0, 1]$ , and test $f \in [0, 1]$ , the unnormalized update has conditional mean $z ν M_{p} (G_{p} f)$ and conditional variance at most $z^{2} / (4 N)$ .

[F4]

Fixed-level full-resampling SIR splitting is unbiased: Under positive prefixes and full resampling, $E [\hat{p}] = Z_{L} = P (A_{L})$ .

[F5]

Feynman–Kac path targets and normalizers: The finite-horizon normalizer is the integral of the ordered product of potentials against the initial law and transition kernels.

[F6]

Unbiasedness and exact variance of the IID Monte Carlo mean: The mean of $N$ IID square-integrable variables has variance equal to the common variance divided by $N$ .

[F7]

Conditional variance decomposition: For square-integrable $X$ , $Var (X) = E [Var (X ∣ G)] + Var (E [X ∣ G])$ .

[F8]

Moments, variance, and covariance on a probability space: Variance is the expectation of the centered square.

Proof

1.1A1F1F4given

By [F4], the estimate is unbiased for $Z_{L} = P (A_{L})$ , so its mean-square error equals its variance.

1.2F1F2F5giveninduction

Set $Q_{L} \equiv 1$ , and recursively $Q_{p - 1} = M_{p} (G_{p} Q_{p})$ for $p = L, \dots, 1$ . Since $M_{p}$ is a probability kernel and $0 \leq G_{p} \leq 1$ , backward induction gives $0 \leq Q_{p} \leq 1$ . By [F5], $Z_{L} = P (A_{L}) = \int Q_{0} d P$ .

2.1A1F2F3step 1.2givenalgebra

Let $Γ_{p}^{N} = {\hat{Z}}_{p} η_{p}^{N}$ be the unnormalized weighted empirical measure after level $p$ , and put $U_{p} = Γ_{p}^{N} (Q_{p})$ . Conditional on the particle history through level $p - 1$ , [F2] gives IID proposals from $η_{p - 1}^{N} M_{p}$ ; applying [F3] with $z = {\hat{Z}}_{p - 1}$ , $ν = η_{p - 1}^{N}$ , and $f = Q_{p}$ yields $E [U_{p} ∣ F_{p - 1}] = {\hat{Z}}_{p - 1} η_{p - 1}^{N} (M_{p} (G_{p} Q_{p})) = U_{p - 1}, Var (U_{p} ∣ F_{p - 1}) \leq \frac{{\hat{Z}}_{p - 1}^{2}}{4 N} \leq \frac{1}{4 N} .$ Here the final inequality holds because ${\hat{Z}}_{p - 1}$ is a product of averages of potentials in $[0, 1]$ .

3.1A1F2F6F7F8step 2.1giveninduction

The initial particles are IID from $P$ , so $U_{0} = N^{- 1} \sum_{i = 1}^{N} Q_{0} (X_{0}^{i})$ . Since $Q_{0}^{2} \leq Q_{0}$ , [F8] gives $Var (Q_{0} (X_{0})) \leq E Q_{0} - (E Q_{0})^{2} \leq 1 / 4$ ; [F6] then gives $Var (U_{0}) \leq 1 / (4 N)$ . Iterating [F7] with the conditional mean identity from step 2.1 gives $Var (U_{L}) = Var (U_{0}) + \sum_{p = 1}^{L} E [Var (U_{p} ∣ F_{p - 1})] \leq \frac{L + 1}{4 N} .$

4.1F4step 1.1step 3.1algebra∎

Since $Q_{L} \equiv 1$ , $U_{L} = Γ_{L}^{N} (1) = {\hat{Z}}_{L} = \hat{p}$ . Step 1.1 and step 3.1 prove both assertions. The constant does not involve $P (A_{L})$ , so no lower bound on the event probability is needed for this absolute error bound.

Boundary notes

The bound uses indicator potentials in $[0, 1]$ , full multinomial resampling, and positive finite prefixes. It does not bound relative error or a family with unbounded numbers of levels.

Source notes

The cited SMC paper treats related nonasymptotic variance bounds with additional domination assumptions. The absolute $(L + 1) / (4 N)$ estimate here is proved locally from the one-step lemma and bounded-potential backward recursion; the citation is context, not a claim that the source states this exact bound.

Depends on

Used by

Finite-particle SIR absolute and relative error scope Remark

Dependency tree · two levels

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Sources

Adam M. Johansen, Pierre Del Moral, and Arnaud Doucet, Sequential Monte Carlo Samplers for Rare Events, §§2–3 (standard reference, not scraped)
Frédéric Cérou, Pierre Del Moral, and Arnaud Guyader, A nonasymptotic theorem for unnormalized Feynman–Kac particle models, Theorem 1.5 and §5 (standard reference, not scraped)