CorollaryStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-14

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Bounded harmonic functions yield Markov-chain martingales

Statement

Assume Choice. If $X$ is a countable-state $p$ -chain and bounded $f : S \to R$ is harmonic, meaning $P f = f$ , then $(f (X_{n}))_{n \geq 0}$ is a bounded martingale.

Facts & Assumptions

Given: Choice, a $p$ -chain $X$ , and bounded $f$ with $P f = f$ .

[F1]

The generator is $L = P - I$ . (Discrete generator of a countable-state transition matrix)

[F2]

For a $p$ -chain and bounded $f$ , $f (X_{n}) - \sum_{m < n} L f (X_{m})$ is a martingale. (Countable-state martingale-problem characterization)

Proof

1.1

Harmonicity and [F1] give $L f = P f - f = 0$ pointwise.

2.1

Hence the compensator in [F2] is the zero sum at every $n$ , and [F2] says that $f (X_{n})$ is a martingale. [F2, step 1.1] Moreover $∣ f (X_{n}) ∣ \leq ∥ f ∥_{\infty}$ , so it is bounded and integrable. The cases $f = 0$ , $f = 1$ , $n = 0$ , and a one-point chain are included. Choice is used only through [F2]'s conditional expectations; there is no converse claim.

F2step 1.1∎

Depends on

Used by

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Dependency tree · two levels

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Sources

Roch, Markov Chains: Martingale Methods, Note 24, Section 1 (standard reference, not scraped)