ExampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-14

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Simple random-walk transition kernel

Statement

Assume Choice. Let $X_{0} = x_{0} \in Z$ and $X_{n} = x_{0} + \sum_{k = 1}^{n} ξ_{k},$ where the $ξ_{k}$ are IID with $P (ξ_{k} = 1) = P (ξ_{k} = - 1) = 1 / 2$ . Then $X$ is a Markov chain on $Z$ with $p (x, x + 1) = p (x, x - 1) = \frac{1}{2},$ all other entries zero, and generator $L f (x) = \frac{f (x + 1) - 2 f (x) + f (x - 1)}{2} .$

Facts & Assumptions

Given: Choice and the displayed IID signs.

[F1]

Disjoint blocks of an independent family generate independent sigma-algebras. (Disjoint groups of an independent sigma-algebra family remain independent)

[F2]

The Markov condition is the conditional transition identity. (Time-homogeneous Markov chain with transition kernel)

[F3]

The discrete generator is $L f (x) = \sum_{y} p (x, y) (f (y) - f (x))$ for bounded $f$ . (Discrete generator of a countable-state transition matrix)

Verification

1.1

The natural past $σ (X_{0}, \dots, X_{n})$ equals [F1, F2] $σ (ξ_{1}, \dots, ξ_{n})$ because $x_{0}$ is fixed and $ξ_{k} = X_{k} - X_{k - 1}$ . By [F1], $ξ_{n + 1}$ is independent of this past. Thus for every $A \subseteq Z$ , $P (X_{n + 1} \in A ∣ F_{n}) = \frac{1}{2} 1_{A} (X_{n} + 1) + \frac{1}{2} 1_{A} (X_{n} - 1) = p (X_{n}, A) .$ This includes $A = \emptyset$ , $A = Z$ , and $n = 0$ , and verifies [F2]. Choice is used only for the displayed conditional expectation.

F1F2

2.1

Only $y = x + 1$ and $y = x - 1$ contribute to [F3], so [F3, step 1.1] $\begin{aligned} L f (x) & = \frac{1}{2} (f (x + 1) - f (x)) + \frac{1}{2} (f (x - 1) - f (x)) \\ = \frac{1}{2} {f (x + 1) - 2 f (x) + f (x - 1)} . \end{aligned}$ The sum is finite, and constants give $L f = 0$ .

F3step 1.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources

Roch, Markov Chains: Martingale Methods, Exercise 24.1 (standard reference, not scraped)