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A systematic Gibbs sweep preserves the target

Statement

Let $m \geq 1$ and let $P_{1}, \dots, P_{m}$ be exact block-update kernels for the same target probability $π$ . Under the chronological composition convention of Composition of probability kernels, the systematic full-sweep kernel $P_{1} \dots P_{m}$ leaves $π$ invariant. It need not be reversible.

Facts & Assumptions

Given: The target probability and finite list of exact block updates in the Statement.

[F1]

Each exact block update leaves $π$ invariant. Each exact Gibbs block update is reversible and invariant.

[F2]

The systematic sweep is the chronological composition $P_{1} \dots P_{m}$ . Random and systematic Gibbs scan schemes.

[F3]

Kernel composition is defined by $(K L) (x, A) = \int L (y, A) K (x, d y)$ and is associative. Composition of probability kernels, Kernel composition is well defined and associative.

[F4]

Invariance means equality of the measure-kernel product with $π$ . Invariant and stationary distribution for a Markov kernel.

[F5]

A probability measure and probability kernel have the iterated joint-law integral identity. A prior and model kernel define a joint law.

[F6]

A supplied probability full-conditional kernel satisfies the block conditional event identity; the exact update refreshes that block while retaining the other coordinates. Proper full conditional kernel, Exact Gibbs block update.

[F7]

Reversibility requires equal two-time flow on every pair of measurable sets. Reversibility for a probability kernel.

Proof

Given: The hypotheses in the Statement and facts [F1]–[F5].

1.1F3F5

For two probability kernels $P, Q$ , the iterated-kernel identity [F5] gives $\int (P Q) (x, A) π (d x) = \int Q (y, A) (π P) (d y)$ . This follows by expanding $(P Q) (x, A) = \int Q (y, A) P (x, d y)$ and applying the joint-law iterated-integral identity to the bounded nonnegative function $Q (y, A)$ .

1.2F6constructalgebra

To prove that reversibility can fail, give $S = {0, 1}^{2}$ its full sigma-algebra and target masses $π (0, 0) = π (0, 1) = π (1, 0) = 1 / 5$ and $π (1, 1) = 2 / 5$ . Supply the $X$ -conditional kernel $K_{X} (0, {0}) = K_{X} (0, {1}) = 1 / 2$ , $K_{X} (1, {0}) = 1 / 3$ , $K_{X} (1, {1}) = 2 / 3$ , and the $Y$ -conditional kernel with these same respective probabilities when conditioning on $X = 0$ or $X = 1$ . These are probability kernels because their nonnegative rows sum to one and every map between the finite spaces is measurable. For every $x, y$ , $π (x, y) = π_{Y} (y) K_{X} (y, {x}) = π_{X} (x) K_{Y} (x, {y})$ by the displayed masses. Summing these identities over arbitrary subsets verifies every conditional event identity in [F6]. They are therefore exact full-conditionals without invoking the downstream Gaussian example.

2.1F1F2F3F4step 1.1

Define $μ_{0} = π$ and $μ_{k} = μ_{k - 1} P_{k}$ for $1 \leq k \leq m$ . If $μ_{k - 1} = π$ , then [F1] gives $μ_{k} = π P_{k} = π$ . Induction from $μ_{0} = π$ yields $μ_{m} = π$ ; associativity in [F3] identifies $μ_{m} = π (P_{1} \dots P_{m})$ , and [F4] calls this invariance.

3.1F3F6F7step 2.1step 1.2algebra∎

Let $P_{X}$ refresh $X$ and $P_{Y}$ refresh $Y$ , using the kernels in step 1.2, and set $P = P_{X} P_{Y}$ in chronological order. By [F3] and [F6], $P ((0, 0), {(0, 1)}) = K_{X} (0, {0}) K_{Y} (0, {1}) = 1 / 4$ , whereas $P ((0, 1), {(0, 0)}) = K_{X} (1, {0}) K_{Y} (0, {0}) = 1 / 6$ . Thus the forward stationary flow between these singleton states is $1 / 20$ and its reverse is $1 / 30$ . These unequal values violate [F7]. Step 2.1 nevertheless proves invariance for this sweep, since its two updates are exact by step 1.2. This independently supplies the nonreversibility witness and completes both claims.

Depends on

Used by

A systematic Gaussian Gibbs sweep need not be reversible Example

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Sources

Art B. Owen, Monte Carlo Theory, Methods and Examples, Chapter 12 (standard reference, not scraped)
Gareth O. Roberts and Jeffrey S. Rosenthal, General State Space Markov Chains and MCMC Algorithms (standard reference, not scraped)