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Confidence sequences by test-supermartingale inversion

Statement

Assume AC and fix α∈(0,1). Let Θ be a parameter set, (Pθ)θ∈Θ a family of laws on a common discrete filtered measurable space, and, for each θ∈Θ, let Eθ=(Etθ)t≥0 be a nonnegative test supermartingale for the singleton null family {Pθ}. Define Ct(ω)={θ∈Θ:Etθ(ω)<1/α}. Then (Ct)t≥0 is a (1−α) confidence sequence Confidence sequences.

No measurable structure on Θ or joint measurability across θ is needed for this pointwise membership and coverage statement.

Facts & Assumptions

Given: Assume AC, α∈(0,1), the parameter family, and the candidate test-supermartingale Eθ for each θ. If Θ is empty the conclusion is vacuous; otherwise fix an arbitrary θ when proving its coverage bound.

[F1]

Nonnegative test supermartingales: for each fixed θ, Eθ is adapted and nonnegative under Pθ.

[F2]

A test supermartingale gives anytime-valid p-values: transforming a test supermartingale by pt=1 when Et≤0 and pt=min⁡(1,1/Et) otherwise gives an anytime-valid p-value process under its null law.

[F3]

Confidence sequences: each fixed-parameter membership event must be adapted and the true parameter must be covered simultaneously with probability at least 1−α.

[A1]

The Axiom of Choice: AC is required by the conditional-expectation convention in the test-supermartingale definition and its validity theorem.

Proof

1.1F1F2A1given

Fix θ∈Θ. Under Pθ, define ptθ=1 when Etθ≤0 and ptθ=min⁡(1,1/Etθ) otherwise. By adaptedness in F1 and the Borel transformation used in F2, each ptθ is Ft-measurable. Hence {θ∈Ct}={Etθ<1/α}={ptθ>α} is Ft-measurable.

2.1F2A1step 1.1

For this fixed θ, F2 gives Pθ(∃t≥0:ptθ≤α)≤α. The complement is exactly {θ∈Ct for every t≥0}, whose probability is therefore at least 1−α.

3.1F3step 1.1step 2.1∎

Steps 1.1 and 2.1 hold for every θ. They give the membership measurability and simultaneous-coverage clauses in F3, so (Ct) is a confidence sequence.

Depends on

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