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Particle MCMC: PMMH and particle Gibbs — Examples

1 · Prerequisites

2 · Summary

Finite examples verify the pseudo-marginal size bias and retained-estimate rejection rule, then contrast MCWM's unequal transition flows. A two-state, two-particle particle Gibbs update has the correct posterior as its invariant law but leaves the dominant state with probability below $1 / 9000$ . A final support counterexample shows why a proposal missing a positive-mass path does not provide the Radon–Nikodym weights required by the exactness theorem.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: AI-generatedVerification: AI-generatedjudge pass (gpt-6-luna)Open item page →

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.

CounterexampleConstruction: AI-generatedVerification: AI-generatedjudge pass (gpt-6-luna)Open item page →

Refreshing the current estimate can destroy exactness

Statement refuted

Replacing an exact likelihood by an unbiased estimate freshly drawn at both the current and proposed parameter in every Metropolis step necessarily leaves the exact parameter posterior invariant.

Facts & Assumptions

Given: The MCWM procedure in Monte Carlo within Metropolis.

[F1]

Monte Carlo within Metropolis: MCWM refreshes the current estimate at every iteration instead of retaining it after rejection.

Counterexample

Take $Θ = {0, 1}$ , uniform prior, exact likelihoods $L_{0} = L_{1} = 1$ , and a proposal that always flips the parameter. At $0$ use estimator ${\hat{L}}_{0} = 1$ surely; at $1$ use ${\hat{L}}_{1} \in {1 / 2, 3 / 2}$ with equal probabilities. Both estimators are strictly positive and unbiased.

1.1F1givenalgebra

Under MCWM, a move $0 \to 1$ accepts with conditional probabilities $1 / 2$ and $1$ for the two proposed estimates. Thus $P (0, 1) = (1 / 2) (1 / 2) + (1 / 2) (1) = 3 / 4$ .

1.2F1givenalgebra

For a move $1 \to 0$ , the fresh current estimate is $1 / 2$ or $3 / 2$ with equal probability and the proposed estimate is $1$ . The acceptance probabilities are $1$ and $2 / 3$ , so $P (1, 0) = (1 / 2) (1) + (1 / 2) (2 / 3) = 5 / 6$ .

2.1F1step 1.1step 1.2algebra∎

Starting from the claimed uniform posterior, the next mass at state $1$ is $(1 / 2) (3 / 4) + (1 / 2) (1 - 5 / 6) = 11 / 24 \neq 1 / 2$ . Hence the exact posterior is not invariant. The unequal cross-flows also violate detailed balance.

ExampleConstruction: AI-generatedVerification: AI-generatedjudge pass (gpt-6-luna)Open item page →

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.

CounterexampleConstruction: AI-generatedVerification: AI-generatedjudge pass (gpt-6-luna)Open item page →

A particle Gibbs proposal that misses posterior path support

Statement refuted

A particle Gibbs exactness construction remains well-defined with its usual finite importance weights even if its proposal assigns zero probability to a positive-mass posterior path.

Facts & Assumptions

Given: The support requirements in Conditional particle filter for particle Gibbs and Feynman–Kac path Radon–Nikodym weight.

[F1]

Conditional particle filter for particle Gibbs: its reference path and proposal weights must be defined under the full target-path support condition.

[F2]

Feynman–Kac path Radon–Nikodym weight: the target path measure is absolutely continuous with respect to the proposal path measure and the displayed importance weight is its Radon–Nikodym density.

Counterexample

Use the one-time path space ${0, 1}$ with posterior $π = (δ_{0} + δ_{1}) / 2$ and proposal $q = δ_{0}$ . The posterior assigns positive mass to state $1$ , but the proposal assigns it zero mass.

1.1F1F2given

Since $q ({1}) = 0$ while $π ({1}) = 1 / 2$ , $π ≪̸ q$ . Therefore the Radon–Nikodym weight required by [F2] does not exist on the full target support, so the stated conditional particle filter hypotheses fail.

2.1F1F2step 1.1given∎

Starting from reference state $0$ , every free particle is also $0$ because it is drawn from $q = δ_{0}$ . The particle filter therefore cannot represent the positive posterior mass at state $1$ by a weighted proposal path. The exact conditional particle filter in [F1] is not defined with the required target-to-proposal weight on the full posterior support. Thus the support-free exactness claim is false as stated.

Particle MCMC: PMMH and particle Gibbs — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

Pseudo-marginal rejection retains the current likelihood estimate

Example

Facts & Assumptions

Verification

Refreshing the current estimate can destroy exactness

Statement refuted

Facts & Assumptions

Counterexample

An invariant particle Gibbs chain can be sticky

Example

Facts & Assumptions

Verification

A particle Gibbs proposal that misses posterior path support

Statement refuted

Facts & Assumptions

Counterexample

Sources