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

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Residual resampling is discontinuous at positive integer weights

Statement refuted

Under the recursive weighted-moment class and normalization hypotheses in Finite-horizon SMC CLT classes and variance recursions, the residual offspring covariance is always the displayed $R_{t} (ϕ)$ computed from the limiting fractional parts $r (υ_{t})$ , including when the importance-weight law has an atom at a positive integer.

Facts & Assumptions

Given: The CLT classes and residual covariance formula in the cited definition, and the standard normalized-weight residual procedure.

[F1]

Finite-horizon SMC CLT classes and variance recursions: $υ_{t} = d π_{t} / d {\tilde{π}}_{t}$ has ${\tilde{π}}_{t}$ -mean one; the formal source candidate $R_{t}^{c a n d}$ is computed from $r (υ_{t})$ .

[F2]

Multinomial and residual offspring resampling: With normalized input weights $p_{i}$ , residual resampling uses floors $⌊ N p_{i} ⌋$ and distributes the remaining $N - \sum_{i} ⌊ N p_{i} ⌋$ children according to normalized fractional remainders.

[F3]

Lindeberg-Levy iid central limit theorem: IID real variables with finite positive variance have the usual normalized Gaussian sample-mean limit.

[F4]

Slutsky's theorem for real random variables: Multiplying a distributionally convergent variable by a factor converging in probability to a constant gives the corresponding scaled limit.

[F5]

Continuous mapping theorem: A measurable map continuous outside a null set of the limiting law preserves weak convergence.

Counterexample

Given: At time $0$ , let the proposal law on ${A, B}$ be ${\tilde{π}}_{0} (A) = 1 / 4$ , ${\tilde{π}}_{0} (B) = 3 / 4$ . Let the target be $π_{0} (A) = π_{0} (B) = 1 / 2$ , with Radon–Nikodym weights $υ_{0} (A) = 2$ , $υ_{0} (B) = 2 / 3$ . Use $ϕ = 1_{{A}}$ . Both weights are positive and finite, $ϕ$ is bounded, and ${\tilde{π}}_{0} (r (υ_{0})) = 1 / 2 > 0$ .

1.1F3F4givenalgebra

For $N$ IID proposal particles, let $M_{N}$ be the number at $A$ and $p_{N} = M_{N} / N$ . The weighted empirical estimate is $S_{N} = \frac{2 M_{N}}{2 M_{N} + (2 / 3) (N - M_{N})} = \frac{3 p_{N}}{1 + 2 p_{N}} .$ Since $M_{N}$ is binomial with parameters $(N, 1 / 4)$ , [F3] gives $\sqrt{N} (p_{N} - 1 / 4) \Rightarrow N (0, 3 / 16)$ . The identity $\sqrt{N} (S_{N} - 1 / 2) = 2 \sqrt{N} (p_{N} - 1 / 4) / (1 + 2 p_{N})$ and [F4] yield $Z_{N} : = \sqrt{N} (S_{N} - 1 / 2) \Rightarrow Z \sim N (0, 1 / 3)$ . Moreover, $Z_{N} > 0$ exactly when $p_{N} > 1 / 4$ .

1.2F1givenalgebra

The limiting-weight candidate in [F1] gives $r (2) = 0$ and $r (2 / 3) = 2 / 3$ . Because $ϕ (B) = 0$ , both ${\tilde{π}}_{0} (r (υ_{0}) ϕ^{2})$ and ${\tilde{π}}_{0} (r (υ_{0}) ϕ)$ are zero, so $R_{0}^{c a n d} (ϕ) = 0$ . If used as a universal residual CLT variance, it predicts a post-resampling Gaussian limit with variance $V_{0} (ϕ) = {\tilde{π}}_{0} [υ_{0}^{2} (ϕ - 1 / 2)^{2}] = 1 / 3$ .

1.3F2givenalgebra

Put ${\overset{ˉ}{w}}_{N} = 2 / 3 + (4 / 3) p_{N}$ and $u_{A} = 2 / {\overset{ˉ}{w}}_{N}$ , $u_{B} = (2 / 3) / {\overset{ˉ}{w}}_{N}$ . On $1 / 4 < p_{N} < 1 / 3$ , the normalized residual sizes are $u_{A} \in (1, 2)$ and $u_{B} \in (0, 1)$ ; hence each $A$ particle receives one deterministic child, each $B$ particle receives none, and $R = N - M_{N}$ children are sampled from the remainders. If $X_{N}$ is the number of residual children at $A$ , then conditionally on the proposal sample, $X_{N} \sim B i n (R, a_{N})$ with $a_{N} = M_{N} (u_{A} - 1) / R$ . As $p_{N} ↓ 1 / 4$ , $R / N \to 3 / 4$ and $a_{N} \to 1 / 3$ .

2.1F2step 1.3algebra

On that event, the post-resampling mean is $(M_{N} + X_{N}) / N$ and its conditional mean is $S_{N}$ . A Taylor expansion of the binomial characteristic function, uniform for $0 \leq a_{N} \leq 1$ and $R \leq N$ , gives $E ⁣ [\exp ⁣ (i u \frac{X_{N} - R a_{N}}{\sqrt{N}}) | M_{N}] = \exp ⁣ (- \frac{u^{2}}{2} \frac{R}{N} a_{N} (1 - a_{N}) + O (N^{- 1 / 2})) .$ Thus, on the positive branch $p_{N} > 1 / 4$ , the extra resampling fluctuation has limiting variance $(3 / 4) (1 / 3) (2 / 3) = 1 / 6$ .

2.2F2F3step 1.3algebra

On the negative branch with $p_{N} < 1 / 4$ and $p_{N} \to 1 / 4$ , one has $u_{A} \in (2, 3)$ and $u_{B} \in (0, 1)$ . The floor for each $A$ particle is two, and its total fractional remainder is $M_{N} (u_{A} - 2) = O_{P} (\sqrt{N})$ , because $u_{A} - 2 = 2 (1 / {\overset{ˉ}{w}}_{N} - 1) = O_{P} (∣ p_{N} - 1 / 4 ∣)$ . Conditional variance of the residual $A$ count is at most this total remainder. Consequently the variance of $\sqrt{N}$ times the post-resampling proportion about its weighted conditional mean is $O_{P} (N^{- 1 / 2})$ , so the extra resampling fluctuation tends to zero in probability on the negative branch. The events $p_{N} \notin (1 / 5, 1 / 3)$ have probability tending to zero by [F3].

3.1F5step 1.1step 2.1step 2.2step 1.2algebra∎

Put $U_{N} = \sqrt{N} ({\overset{ˇ}{η}}_{0}^{N} (ϕ) - S_{N})$ , the post-resampling fluctuation centered at the weighted mean, and condition on the proposal count $M_{N}$ . For any real $v, s$ , steps 2.1, 2.2 and the negligible exceptional event show that $E [e^{i s U_{N}} ∣ M_{N}]$ converges in probability to $e^{- s^{2} / 12}$ when $Z_{N} > 0$ and to $1$ when $Z_{N} < 0$ . Since $Z_{N} \Rightarrow Z$ and $P (Z = 0) = 0$ , [F5] gives $E [e^{i v Z_{N} + i s U_{N}}] ⟶ E ⁣ [e^{i v Z} (1_{{Z > 0}} e^{- s^{2} / 12} + 1_{{Z < 0}})] .$ This is the joint characteristic function of $(Z, 1_{{Z > 0}} W)$ for an independent $W \sim N (0, 1 / 6)$ , so the post-resampling limit is $Z + 1_{{Z > 0}} W$ . Its mean is zero and its third moment is $3 E [Z 1_{{Z > 0}}] E [W^{2}] > 0$ , so it is not Gaussian. This contradicts the variance- $1 / 3$ Gaussian predicted in step 1.2.

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Sources

Nicolas Chopin, Central Limit Theorem for Sequential Monte Carlo Methods and its Application to Bayesian Inference, Theorem 2 and Appendix A.4 (standard reference, not scraped)