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

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A stopped martingale is a martingale

Statement

Assume AC. If $M$ is a martingale and $τ$ is a stopping time, then $M_{n}^{τ} = M_{n \land τ}$ is a martingale with respect to the original filtration.

Facts & Assumptions

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

[F1]

Stopped random variable and stopped process gives a finite formula for every $M_{n \land τ}$ .

[F2]

Equivalent event tests for a discrete stopping time gives ${τ \geq n} \in F_{n - 1}$ .

[F3]

Discrete martingale transform interprets the stopped increments as a bounded predictable transform.

[F4]

Martingale submartingale and supermartingale supplies the conditional mean-zero increments.

[F5]

The Axiom of Choice states AC, assumed here because the martingale and conditional-expectation interfaces require it.

[F6]

Taking out what is known permits the bounded $F_{n - 1}$ -measurable indicator in step 1.2 to be taken outside conditional expectation.

Proof

1.1

The finite formula in F1 makes $M_{n \land τ}$ $F_{n}$ -measurable and integrable: it is a finite sum of $M_{k} 1_{{τ = k}}$ for $k < n$ and $M_{n} 1_{{τ \geq n}}$ .

F1F2

1.2

Pathwise, $M_{n \land τ} - M_{(n - 1) \land τ} = 1_{{τ \geq n}} (M_{n} - M_{n - 1}) .$ The indicator is bounded and $F_{n - 1}$ -measurable by F2, so this is the $n$ th increment of the predictable transform in F3.

F2F3

2.1

Taking the conditional expectation of step 1.2 and pulling out the indicator by F6 gives zero by F4. Together with adaptedness and integrability from step 1.1, this is exactly the martingale condition. The stopped process never evaluates a value at infinity; AC has only the role in F5.

F4F5F6step 1.1step 1.2∎

Depends on

Used by

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Sources

van der Vaart, Martingales, Diffusions and Financial Mathematics, §§2.3 and 2.8 (standard reference, not scraped)