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

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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Symmetric bounded-increment Azuma bound

Statement

Assume AC. If $(M_{k}, F_{k})_{k = 0}^{n}$ is a martingale and $∣ M_{k} - M_{k - 1} ∣ \leq c_{k}$ almost surely for deterministic $c_{k} \geq 0$ , then for every $t > 0$ , $P (∣ M_{n} - M_{0} ∣ \geq t) \leq 2 \exp ⁣ (- \frac{t^{2}}{2 \sum_{k = 1}^{n} c_{k}^{2}}),$ again interpreting the right exponential as $0$ when every $c_{k} = 0$ .

Facts & Assumptions

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

[F1]

Azuma-Hoeffding inequality supplies both one-sided bounds with predictable endpoints.

[F2]

The finite union bound combines the upper and lower deviations.

[F3]

The Axiom of Choice is the exact inherited conditional-expectation dependence from F1.

Proof

1.1

Apply F1 with $A_{k} = - c_{k}$ and $B_{k} = c_{k}$ . Their width is $2 c_{k}$ , so each one-sided probability is at most $\exp ⁣ (- \frac{2 t^{2}}{\sum_{k} (2 c_{k})^{2}}) = \exp ⁣ (- \frac{t^{2}}{2 \sum_{k} c_{k}^{2}}) .$

2.1

The event ${∣ M_{n} - M_{0} ∣ \geq t}$ is the union of the upper and lower tail events. F2 gives twice the bound in step 1.1. If all $c_{k} = 0$ , all increments vanish almost surely and the event is empty. Zero-width individual terms otherwise simply contribute zero to the sum. AC is used exactly as stated in F3.

F2F3step 1.1∎

Depends on

Used by

Azuma bound for simple random walk Example

Dependency tree · two levels

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Sources

Roch, Notes 20: Azuma's Inequality, Theorem 20.8, pp. 3–4 (standard reference, not scraped)