CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-14

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Almost-surely finite stopping does not imply integrable stopping

Statement

Assume AC. The first time $τ$ that a simple symmetric random walk started at zero hits $1$ is almost surely finite but satisfies $E τ = \infty$ .

Facts & Assumptions

Given: The hypotheses, objects, and conventions in the Statement.

[F1]

Optional stopping fails for an unbounded simple-random-walk hitting time proves $τ < \infty$ almost surely and computes $S_{τ} = 1$ , $S_{0} = 0$ .

[F2]

Optional stopping with integrable time and bounded increments would apply if $τ$ were integrable.

[F3]

The Axiom of Choice is inherited from the martingale results.

Proof

1.1

F1 proves that $τ < \infty$ almost surely. Suppose for contradiction that $E τ < \infty$ .

F1assume-contra

2.1

The walk is a martingale and $∣ S_{n} - S_{n - 1} ∣ = 1$ , so F2 would imply $E S_{τ} = E S_{0} .$ But F1 computes the two sides as $1$ and $0$ . This contradiction proves $E τ = \infty$ . AC has exactly the inherited role in F3.

F1F2F3step 1.1discharge-contradiction∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

10 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Durrett, Probability: Theory and Examples, 5th ed., optional stopping counterexamples in §4.8 (standard reference, not scraped)