ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedjudge 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.

A dominant particle weight and the effective sample size

Example

Use proposal probabilities $q (A) = 1 / 4$ , $q (B) = 3 / 4$ and target probabilities $π (A) = 99 / 100$ , $π (B) = 1 / 100$ . Compute the weight-based effective sample size for four proposal draws containing one $A$ and three $B$ values.

Facts & Assumptions

Given: The proposal, target, and observed finite draw described in the Example.

[F1]

Finite-horizon SMC CLT classes and variance recursions: Importance weights are Radon–Nikodym derivatives of target law with respect to proposal law and have proposal mean one.

[F2]

Weight-based effective sample size: For nonnegative weights with positive sum, ${ESS}_{w} = (\sum_{i} w_{i})^{2} / \sum_{i} w_{i}^{2}$ ; it depends only on observed weights.

Verification

Given: $q (A) = 1 / 4$ , $q (B) = 3 / 4$ , $π (A) = 99 / 100$ , and one observed $A$ with three observed $B$ values.

1.1F1givenalgebra

The proposal-to-target ratios are $w (A) = π (A) / q (A) = 99 / 25$ and $w (B) = π (B) / q (B) = 1 / 75$ , and $q (A) w (A) + q (B) w (B) = 1$ . The event of exactly one $A$ in four IID proposal draws has probability $(\binom{4}{1}) (1 / 4) (3 / 4)^{3} = 27 / 64 > 0$ .

2.1F2step 1.1givenalgebra∎

The four unnormalized weights sum to $99 / 25 + 3 / 75 = 4$ , so their normalized values are $99 / 100, 1 / 300, 1 / 300, 1 / 300$ . By [F2], ${ESS}_{w} = {((99 / 100)^{2} + 3 (1 / 300)^{2})}^{- 1} = \frac{7500}{7351} \approx 1.0203.$ The first particle carries $99 %$ of the observed weight, which explains the low ESS; the calculation concerns weight concentration only.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

8 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Simo Särkkä and Lennart Svensson, Bayesian Filtering and Smoothing, 2nd ed., §§11.2, 11.4–11.5 (standard reference, not scraped)