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

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Expected duration of symmetric gambler's ruin

Statement

Assume AC. In the symmetric gambler's ruin setting with $S_{0} = i$ and absorbing levels $0, N$ , the exit time satisfies $E τ = i (N - i) .$

Facts & Assumptions

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

[F1]

Gambler's ruin hitting probability from optional stopping gives almost-sure finiteness and $P (S_{τ} = N) = i / N$ .

[F2]

Square minus predictable quadratic variation is a martingale makes $S_{n}^{2} - n$ a martingale for unit-variance increments.

[F3]

Optional sampling for bounded stopping times applies at $τ \land n$ .

[F4]

Dominated convergence and Monotone convergence for the integral pass the two stopped terms separately.

[F5]

The Axiom of Choice is inherited from the martingale and optional-sampling interfaces.

Proof

1.1

The predictable quadratic variation of the symmetric walk is $n$ , since every increment has conditional square one. Thus F2 and bounded optional sampling give $E [S_{τ \land n}^{2}] - E (τ \land n) = i^{2} .$

F2F3

1.2

The stopped position lies in $[0, N]$ and converges almost surely to $S_{τ}$ by F1. DCT gives $E [S_{τ \land n}^{2}] \to E S_{τ}^{2} = N^{2} P (S_{τ} = N) = N i .$ Meanwhile $τ \land n ↑ τ$ , so MCT gives $E (τ \land n) ↑ E τ$ , initially allowing infinity.

F1F4

2.1

Taking limits in step 1.1 forces the latter value to be finite and gives $N i - E τ = i^{2}, E τ = i (N - i) .$ The argument does not assume integrability of $τ$ before proving it. AC has exactly the role in F5.

F5step 1.1step 1.2∎

Depends on

Used by

Expected duration of simple gambler's ruin Example

Dependency tree · two levels

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Sources

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