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.

An invariant particle Gibbs chain can be sticky

Example

Take the one-time model $T = 0$ with state space ${0, 1}$ , target $π (0) = 99 / 100$ , $π (1) = 1 / 100$ , and proposal $q (0) = 1 / 100$ , $q (1) = 99 / 100$ . Both target states are in proposal support. Use $N = 2$ particles: one is the retained reference state, the other is sampled from $q$ , and the next state is drawn from the two importance weights $w (x) = π (x) / q (x)$ . This is the one-time conditional particle Gibbs update.

Facts & Assumptions

Given: The conditional particle filter and invariance results in Conditional particle filter for particle Gibbs, The particle Gibbs extended target has the exact smoothing marginal, and The conditional particle Gibbs sweep is a Gibbs update.

[F1]

Conditional particle filter for particle Gibbs: with one free particle at $T = 0$ , draw it from $q$ and select between it and the fixed reference particle in proportion to their target-to-proposal weights.

[F2]

The conditional particle Gibbs sweep is a Gibbs update: the finite particle Gibbs update preserves the extended target and has the exact posterior marginal.

Verification

Given: The finite two-state model in the Example.

1.1F1givenalgebra

The importance weights are $w (0) = (99 / 100) / (1 / 100) = 99$ and $w (1) = (1 / 100) / (99 / 100) = 1 / 99$ . From reference state $0$ , the free particle is $1$ with probability $99 / 100$ and is selected with probability $(1 / 99) / (99 + 1 / 99) = 1 / 9802$ . Therefore $P (0, 1) = 99 / 980200$ .

1.2F1givenalgebra

From reference state $1$ , the free particle is $0$ with probability $1 / 100$ and is selected with probability $99 / (99 + 1 / 99) = 9801 / 9802$ . Therefore $P (1, 0) = 9801 / 980200$ .

2.1F2step 1.1step 1.2algebra∎

The detailed-balance flows agree: $(99 / 100) P (0, 1) = 9801 / 98020000 = (1 / 100) P (1, 0)$ , consistent with the exact invariance in [F2]. Yet the probability of leaving state $0$ is only $99 / 980200 < 1 / 9000$ , so exact invariance alone gives no useful lower bound on movement.

Depends on

Used by

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Dependency tree · two levels

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Sources

Andrieu, Doucet, and Holenstein, Particle Markov chain Monte Carlo methods, §4.5 and discussion of path degeneracy (standard reference, not scraped)
Lindsten and Schön, On the use of backward simulation in particle Markov chain Monte Carlo methods (arXiv 1110.2873v2), §§1 and 4 (standard reference, not scraped)