CounterexampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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Zero limiting remainder does not make finite residual offspring deterministic

Statement refuted

If $\tilde{π} (r (υ)) = 0$ , then standard residual resampling has deterministic offspring counts for every finite particle sample, so its residual covariance may always be set to zero.

Facts & Assumptions

Given: The residual-weight notation in Finite-horizon SMC CLT classes and variance recursions and the standard normalized-weight residual algorithm.

[F1]

Finite-horizon SMC CLT classes and variance recursions: $r (u) = u - ⌊ u ⌋$ and a normalized Radon–Nikodym weight satisfies $\tilde{π} (υ) = 1$ .

[F2]

Multinomial and residual offspring resampling: Residual counts are $⌊ N p_{j} ⌋$ plus multinomial draws from the fractional remainders of the normalized weights $N p_{j}$ .

Counterexample

Given: Let the proposal law on ${A, C, B}$ assign probabilities $(1 / 4, 1 / 4, 1 / 2)$ , and let the importance weights be $(2, 2, 0)$ . Their proposal expectation is one, the target law is $(1 / 2, 1 / 2, 0)$ , and $\tilde{π} (r (υ)) = 0$ .

1.1F1givenalgebra

For $N = 3$ , the event that the proposal sample contains one $A$ , one $C$ , and one $B$ has probability $3! (1 / 4) (1 / 4) (1 / 2) = 3 / 16 > 0$ . On this event the total weight is four and the normalized weights are $(1 / 2, 1 / 2, 0)$ .

2.1F2step 1.1algebra∎

The scaled normalized weights are $(N p_{A}, N p_{C}, N p_{B}) = (3 / 2, 3 / 2, 0)$ , so their floor counts are $(1, 1, 0)$ and their fractional remainders are $(1 / 2, 1 / 2, 0)$ . By [F2] one residual child is drawn, with equal probability from $A$ and $C$ ; hence the offspring-count vector is random although the limiting proposal remainder integral is zero. This contradicts the Statement.

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