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.

Pseudo-marginal rejection retains the current likelihood estimate

Example

Let $Θ = {0, 1}$ with uniform prior, exact likelihoods $L_{0} = L_{1} = 1$ , and a symmetric proposal that flips the parameter. At $θ = 0$ let the unbiased likelihood estimator be ${\hat{L}}_{0} = 1$ surely. At $θ = 1$ let ${\hat{L}}_{1} = 1 / 2$ or $3 / 2$ , each with probability $1 / 2$ . The pseudo-marginal target on parameter and estimate has masses $\tilde{π} (0, 1) = 1 / 2, \tilde{π} (1, 1 / 2) = 1 / 8, \tilde{π} (1, 3 / 2) = 3 / 8.$ The parameter marginal is uniform, although the estimate conditional on $θ = 1$ is size-biased. A rejection at $(1, 3 / 2)$ retains $3 / 2$ for the next ratio calculation.

Facts & Assumptions

Given: The PMMH target and update in Particle marginal Metropolis–Hastings.

[F1]

Particle marginal Metropolis–Hastings: the extended target is proportional to prior times auxiliary-law density times the nonnegative unbiased likelihood estimate, and a rejected proposal retains the full current auxiliary state.

[F2]

PMMH preserves the exact parameter posterior: under unbiasedness and finite positive evidence, the parameter marginal of the extended target is the exact posterior.

Verification

Given: The finite two-point example above.

1.1F1algebra

The prior mass is $1 / 2$ at each parameter. Multiplying by the estimator law and its value gives masses $1 / 2$ at $(0, 1)$ , $(1 / 2) (1 / 2) (1 / 2) = 1 / 8$ at $(1, 1 / 2)$ , and $(1 / 2) (1 / 2) (3 / 2) = 3 / 8$ at $(1, 3 / 2)$ . These sum to one, and the two masses over $θ = 1$ sum to $1 / 2$ .

2.1F1F2step 1.1algebra∎

From $(0, 1)$ the proposal $(1, 3 / 2)$ occurs with probability $1 / 2$ and is accepted with probability one, so the stationary flow is $(1 / 2) (1 / 2) = 1 / 4$ . In reverse, from $(1, 3 / 2)$ the proposal $(0, 1)$ is accepted with probability $(1) / (3 / 2) = 2 / 3$ , so its flow is $(3 / 8) (2 / 3) = 1 / 4$ . The extended ratio uses the retained value $3 / 2$ until that move is accepted or rejected.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources

Andrieu, Doucet, and Holenstein, Particle Markov chain Monte Carlo methods, §4.4 (standard reference, not scraped)
Särkkä and Svensson, Bayesian Filtering and Smoothing, 2nd ed., §16.3.5 (standard reference, not scraped)