Alphabeta Math
ExampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-14
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Stopping a likelihood-ratio martingale

Statement

Assume AC. Let NN0, let (Fn)0nN be a filtration on a probability space (Ω,FN,P), and let Q be a probability measure on FN with QP. Let Ln=EP[dQ/dPFn],0nN. For every stopping time τN, Lτ is nonnegative, Fτ-measurable, and EPLτ=1. Moreover Q(A)=EP[1ALτ](AFτ).

Facts & Assumptions

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

[F2]

Conditional expectation process is a martingale makes Ln a nonnegative martingale.

[F3]

Optional sampling for bounded stopping times gives conditional and unconditional identities at τ.

[F5]

The Axiom of Choice is used exactly for Radon–Nikodym and conditional-expectation existence.

Proof

1.1

F1 and conditional positivity make every Ln nonnegative; F2 makes the process a martingale. F4 makes Lτ Fτ-measurable. Applying F3 between τ and deterministic N gives Lτ=EP[LNFτ],EPLτ=EPLN=Q(Ω)=1.

F1F2F3F4
2.1

For AFτ, the defining conditional-expectation identity in step 1.1 gives EP[1ALτ]=EP[1ALN]=Q(A), where the final equality is the Radon–Nikodym identity and AFτFN. AC has precisely the role in F5.

F1F3F5step 1.1

Depends on

Used by

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Sources