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Conditional sub-Gaussian increments generate exponential test supermartingales

Statement

Assume AC and let (Ω,F,(Ft)t∈N0,P), (Xt)t≥1, μ∈R, and σ2>0 satisfy Conditionally sub-Gaussian increments. For each fixed λ∈R, set S0(μ):=0,St(μ):=∑i=1t(Xi−μ),Ltλ(μ):=exp⁡ ⁣(λSt(μ)−λ2tσ22). Then (Ltλ(μ))t≥0 is a nonnegative test supermartingale for the singleton null family {P}, with L0λ(μ)=1.

Facts & Assumptions

Given: Assume AC, the conditional sub-Gaussian model, and a fixed real λ.

[F1]

Conditionally sub-Gaussian increments: for each t≥1 and fixed λ, eλ(Xt−μ) is integrable and its conditional expectation given Ft−1 is at most eλ2σ2/2 almost surely.

[F2]

Taking out what is known: a bounded measurable factor can be taken out of a conditional expectation; the identity also holds for finite measurable factors when both products are integrable.

[F3]

Monotone convergence for the integral: an increasing sequence of nonnegative measurable functions has integrals increasing to the integral of its pointwise limit.

[F4]

Martingale submartingale and supermartingale: an adapted integrable process is a supermartingale when E[Lt∣Ft−1]≤Lt−1 almost surely at every adjacent pair.

[F5]

Nonnegative test supermartingales: under AC, an adapted, integrable, almost surely nonnegative supermartingale with initial expectation at most one is a nonnegative test supermartingale.

[A1]

The Axiom of Choice: AC supplies the conditional-expectation convention used in the supermartingale and test-supermartingale definitions.

Proof

1.1givenF1F2F3construct

Fix t≥1, put Yt:=exp⁡ ⁣(λ(Xt−μ)−λ2σ2/2) and Qt:=E[Yt∣Ft−1]. By [F1], Yt is integrable and 0≤Qt≤1. Starting from L0λ(μ)=1, suppose Lt−1λ(μ) is integrable and let Zn:=Lt−1λ(μ)∧n. The bounded-factor clause of [F2] gives E[ZnYt]=E[ZnQt]≤EZn≤ELt−1λ(μ). Since ZnYt↑Ltλ(μ), [F3] shows Ltλ(μ) is integrable and ELtλ(μ)≤ELt−1λ(μ). Thus induction establishes integrability at every time.

2.1F2F4F5A1step 1.1∎

The process is adapted and nonnegative by its definition. Since Lt−1λ(μ)Yt=Ltλ(μ) and both products are integrable by step 1.1, the unbounded-factor clause of [F2] gives E[Ltλ(μ)∣Ft−1]=Lt−1λ(μ)Qt≤Lt−1λ(μ). Also EL0λ(μ)=1. Therefore [F4] and [F5] give the asserted test-supermartingale property.

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