ExampleConstruction: AI-generatedVerification: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Multinomial and residual resampling on three weights

Example

Resample $N = 5$ particles from normalized weights $p = (31 / 50, 7 / 25, 1 / 10, 0, 0) = (0.62, 0.28, 0.10, 0, 0)$ . The last two particles have zero weight, so compare the three nonzero-weight coordinates; both zero-weight offspring counts remain zero. Compare the expected offspring counts and marginal count variances for the multinomial and residual schemes.

Facts & Assumptions

Given: The normalized weights and the two procedures in the Example.

[F1]

Multinomial and residual offspring resampling: Multinomial counts are sums of conditionally independent categorical indicators; residual counts add integer floors to categorical draws from normalized fractional remainders.

[F2]

Multinomial and residual resampling have unbiased offspring counts: Both schemes have conditional expected count $N p_{j}$ for each particle.

Verification

Given: $N = 5$ and $p = (31 / 50, 7 / 25, 1 / 10, 0, 0)$ .

1.1F2givenalgebra

The vector is a probability vector, and $N p = (31 / 10, 7 / 5, 1 / 2, 0, 0) = (3.1, 1.4, 0.5, 0, 0)$ . By [F2], both schemes therefore have those expected counts.

1.2F1givenalgebra

In multinomial resampling each count is a sum of five independent Bernoulli indicators with success probability $p_{j}$ , so its variance is $5 p_{j} (1 - p_{j})$ . The first three marginal variances are $1.178$ , $1.008$ , and $0.45$ ; the last two are zero.

1.3F1givenalgebra

Residual resampling has deterministic floors $(3, 1, 0, 0, 0)$ and fractional remainders $(0.1, 0.4, 0.5, 0, 0)$ , whose sum is $R = 1$ . It draws one extra index with these probabilities, so the marginal count variances are $0.1 (0.9) = 0.09$ , $0.4 (0.6) = 0.24$ , and $0.5 (0.5) = 0.25$ on the first three coordinates, with variance zero on the last two.

2.1step 1.2step 1.3algebra∎

The expected counts agree componentwise, while for these weights each of the first three residual marginal variances is smaller than its multinomial counterpart, while the last two variances are zero in both schemes. This is a calculation for this input vector, not a general variance-ordering claim.

Remarks

The comparison concerns offspring counts only. It does not compare posterior accuracy or claim that residual resampling uniformly improves every statistic.

Depends on

Used by

Nothing in the library uses this result yet.

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Sources

Arnaud Doucet and Adam M. Johansen, A Tutorial on Particle Filtering and Smoothing: Fifteen Years Later, §3.4 (standard reference, not scraped)
Simo Särkkä and Lennart Svensson, Bayesian Filtering and Smoothing, 2nd ed., §11.4 (standard reference, not scraped)
Nicolas Chopin, Central Limit Theorem for Sequential Monte Carlo Methods and its Application to Bayesian Inference, §2.1 (standard reference, not scraped)