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General-state Markov kernels, invariance, and reversibility

1 · Prerequisites

2 · Summary

For a probability measure π and a pointwise probability kernel P on any measurable state space, the joint flow defined in General-state reversibility by symmetric flow measure is a probability measure on the product sigma-algebra. Reversibility means that this measure is symmetric under swapping its coordinates; it does not require a density or assume invariance in advance. Integrating its nonnegative indicator functions over the two coordinate marginals proves that reversible kernels preserve π (General-state reversibility implies invariance). The existing detailed-balance and convergence results on the stationary Markov-chain page remain countable-state statements; this page makes no general-state ergodicity or convergence claim.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

General-state reversibility by symmetric flow measure

Definition

Let (X,X) be a measurable space, let P be a probability kernel on it, meaning that P(x,⋅) is a probability measure for each x and x↦P(x,A) is measurable for each A∈X (Measure kernel and probability kernel), and let π be a probability measure, that is a countably additive measure of total mass one (Probability measures and probability spaces). The product sigma-algebra X⊗X is generated by measurable rectangles (The product sigma-algebra and its finite iterates). The joint flow measure of (π,P) is the probability measure Λπ,P on (X×X,X⊗X) given, for each C∈X⊗X, by

Λπ,P(C):=∫XP(x,Cx) π(dx),Cx:={y∈X:(x,y)∈C}.

Existence, countable additivity, and the nonnegative iterated-integral identity for this measure follow from A prior and model kernel define a joint law. Let τ(x,y):=(y,x) be the coordinate-swap map. It is measurable because the preimage of a measurable rectangle A×B is B×A and such rectangles generate the product sigma-algebra. Define the transpose measure by Λπ,P⊤(C):=Λπ,P(τ−1(C)).

The kernel P is reversible with respect to π when Λπ,P=Λπ,P⊤. This definition uses equality of measures and requires no density, standard Borel assumption, or prior assumption that π is invariant.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-luna)Open item page →

General-state reversibility implies invariance

Statement

Let (X,X) be a measurable space, P a probability kernel (Measure kernel and probability kernel) on it, and π a probability measure (Probability measures and probability spaces) on it. If P is reversible with respect to π in the sense of General-state reversibility by symmetric flow measure, then π is invariant for P: for every A∈X,

∫XP(x,A) π(dx)=π(A).

Facts & Assumptions

Given: A measurable space (X,X), a pointwise probability kernel P, and a probability measure π for which the joint flow measure Λπ,P is symmetric.

[F1]

The joint flow measure is the probability measure on the product sigma-algebra defined by Λπ,P(C)=∫P(x,Cx) π(dx), and reversibility means Λπ,P=Λπ,P⊤ under coordinate swap. (General-state reversibility by symmetric flow measure)

[F2]

The joint-law construction also gives, for every nonnegative product- measurable h, ∫h dΛπ,P=∫[∫h(x,y)P(x,dy)]π(dx); in particular its rectangle formula holds. (A prior and model kernel define a joint law)

[F3]

A probability π is invariant for P exactly when ∫P(x,A) π(dx)=π(A) for every measurable A. (Invariant and stationary distribution for a Markov kernel)

[F4]

A probability kernel has probability-measure rows and measurable event evaluations, and a probability measure has total mass one. (Measure kernel and probability kernel, Probability measures and probability spaces)

Proof

Given: The hypotheses in the statement.

Proof technique: Apply the nonnegative kernel-Tonelli identity to the two marginals of the symmetric joint flow measure.

1.1F2F4given

Fix A∈X. By [F2] on 1X×A, whose section at each x is A, Λπ,P(X×A)=∫XP(x,A) π(dx), a nonnegative integral bounded by one.

1.2F1given

Symmetry [F1] and τ−1(X×A)=A×X give Λπ,P(X×A)=Λπ,P⊤(X×A)=Λπ,P(A×X).

1.3F2F4given

The rectangle formula [F2] gives Λπ,P(A×X)=∫AP(x,X) π(dx)=∫A1 π(dx)=π(A), since each kernel row has mass one by [F4].

2.1F3step 1.1step 1.2step 1.3given∎

Steps 1.1, 1.2, and 1.3 give ∫XP(x,A) π(dx)=π(A) for arbitrary A, which is invariance by [F3]. The proof uses only nonnegative indicators, with no density, countability, or standard-Borel hypothesis.

5 · Examples, counterexamples and false statements

None yet.

Sources