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Finite-horizon crossing bound for a nonnegative test supermartingale

Statement

Assume AC. Let P0 be a family of probability laws and let M=(Mt)t∈N0 be a nonnegative test supermartingale for P0. For every P∈P0, every a>0, and every N∈N0, P ⁣(max⁡0≤t≤NMt≥a)≤EPM0a≤1a.

Facts & Assumptions

Given: Assume AC. Let P0 be a null family and let M=(Mt)t≥0 be a nonnegative test supermartingale for that family. Fix P∈P0, a>0, and a finite horizon N∈N0.

[F1]

Nonnegative test supermartingales: under every null law, M is adapted and integrable, is nonnegative almost surely at every time, is a supermartingale, and satisfies EPM0≤1.

[F2]

First hitting time of an adapted process is a stopping time: for an adapted process and a Borel set B, its first time in B is a stopping time.

[F3]

Discrete stopping time: a discrete stopping time is a map into N0∪{∞} whose events {τ≤n} belong to Fn; it is bounded when it is bounded almost surely by a deterministic integer.

[F4]

Optional sampling for bounded stopping times: for bounded stopping times σ≤τ, optional sampling for a supermartingale gives EPMτ≤EPMσ.

[F5]
[A1]

AC is assumed by the test-supermartingale definition and the bounded optional-sampling theorem (The Axiom of Choice); it is used only through those suppliers.

Proof

1.1givenF1F2F3A1choose

Define τa=inf⁡{t≥0:Mt≥a}, with inf⁡∅=∞, and AN={τa≤N}={max⁡0≤t≤NMt≥a}. By [F1], M is adapted; applying [F2] with the Borel set [a,∞) makes τa a stopping time. Thus τa∧N and the constant time 0 are bounded stopping times, with 0≤τa∧N.

2.1F1F4A1step 1.1

Apply [F4] to 0≤τa∧N to get EPMτa∧N≤EPM0≤1.

2.2F1step 1.1

On AN, τa∧N=τa and Mτa≥a; on ANc, τa∧N=N and MN≥0 almost surely by [F1]. Hence Mτa∧N≥a1AN almost surely.

3.1F5step 2.1step 2.2algebra∎

By [F5] and step 2.1, aP(AN)=EP[a1AN]≤EPMτa∧N≤EPM0≤1. Dividing by a>0 proves both bounds.

Remarks

The finite-horizon event is exactly the first-crossing event by time N. No assumption that an infinite-horizon supremum is attained is used here.

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