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Sub-Gaussian mixture confidence sequences — Examples

1 · Prerequisites

2 · Summary

Examples

Independent Gaussian increments satisfy the conditional mgf bound, including when the supplied proxy is larger than their variance. By contrast, repeating one Rademacher draw at every time leaves each marginal centered and sub-Gaussian but makes the running mean constant. The displayed interval radius tends to zero, so the resulting intervals eventually exclude the true mean on every outcome.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

Marginal sub-Gaussianity does not justify an anytime confidence sequence

Statement refuted

For every α∈(0,1) and ρ>0, marginal centered sub-Gaussianity with proxy 1 alone makes the intervals in Gaussian-mixture confidence-sequence boundary a (1−α) confidence sequence for the common mean, regardless of dependence.

Facts & Assumptions

Given: Fix α∈(0,1) and ρ>0. Let Ω={−1,+1} with equal probabilities, let Z(ω)=ω, let F0={∅,Ω} and Ft=σ(Z) for every t≥1, and set Xt=Z, μ=0, and σ2=1.

[F1]

Conditional Hoeffding bound for bounded martingale differences: if D is conditionally centered and lies between measurable endpoints A≤D≤B of deterministic width at most c, then E[eλD∣G]≤eλ2c2/8 for every real λ.

[F2]

Sub-Gaussian variables and vectors, and sub-exponential variables: a centered real random variable is sub-Gaussian with proxy 1 when EeλX≤eλ2/2 for every λ∈R.

[F3]

Conditioning a known variable and an independent variable: an integrable variable independent of a sigma-algebra has conditional expectation equal almost surely to its ordinary expectation.

[F4]

Conditional expectation given a sigma algebra: a conditional-expectation version is a measurable integrable function with the same integral as the input on every event in the conditioning sigma-algebra.

[F5]

Conditional expectation is unique almost surely: any two versions of the same conditional expectation agree almost surely.

[F6]

Conditionally sub-Gaussian increments: the required sequential premise bounds the conditional mgf of each increment given Ft−1 for every fixed real λ.

[F7]

Gaussian-mixture confidence-sequence boundary: for σ2=1, the radius at time t≥1 is t−1(t+ρ)log⁡((t+ρ)/(α2ρ)).

[A1]

The Axiom of Choice: AC is assumed by the conditional-expectation convention in [F1].

Counterexample

1.1givenF1F2F3A1

We have EZ=0 and −1≤Z≤1. Since F0 is trivial, [F3] gives E[Z∣F0]=0 and, for every λ, E[eλZ∣F0]=EeλZ; the latter variable is integrable because Z has finite support. Apply [F1] with D=Z, A=−1, B=1, and c=2 to get E[eλZ∣F0]≤eλ2/2, hence EeλZ≤eλ2/2 for every λ∈R. Thus every marginal Xt=Z is centered and sub-Gaussian with proxy 1 by [F2].

1.2givenF4F5F6algebra

At time t=2, X2=Z is F1-measurable. Its own version satisfies [F4], so [F5] gives E[eX2∣F1]=eZ almost surely. On the event {Z=1}, which has probability 1/2, this equals e>e1/2. Hence the conditional sub-Gaussian premise [F6] with proxy 1 fails at t=2 and λ=1.

1.3givenF7F8algebra

Every sample mean is Xˉt=Z. For all sufficiently large t, (t+ρ)/(α2ρ)≥1; using [F8] and u−1≤u−3/4 for u≥1 gives (uα(t)/t)2≤4(t+ρ)5/4/((α2ρ)1/4t2)=4t−3/4(1+ρ/t)5/4/(α2ρ)1/4→0. Thus the radius is eventually less than 1.

2.1givenstep 1.3F7algebra∎

Since ∣Xˉt∣=∣Z∣=1, step 1.3 implies that 0 is outside every boundary interval for all sufficiently large t, on both outcomes. Therefore P(0∈Ct for all t≥1)=0<1−α, refuting the statement.

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

Independent Gaussian increments satisfy the conditional proxy

Example

Let Θ⊆R, let (Ω,F,(Ft)t∈N0) be a common discrete filtered measurable space, let (Pθ)θ∈Θ be probability laws on it, and let (Xt)t≥1 be a common real-valued process adapted under every Pθ. Under each Pθ, assume the residuals Yt(θ):=Xt−θ have law N(0,τ2) for a known τ2>0 and, for every t≥1, every Borel B⊆R, and every A∈Ft−1, Pθ({Yt(θ)∈B}∩A)=Pθ(Yt(θ)∈B)Pθ(A). Let the reported proxy be any known σ2≥τ2, and fix ρ>0 and α∈(0,1). The Gaussian-mixture intervals with proxy σ2 are a (1−α) confidence sequence for θ.

Facts & Assumptions

Given: Assume AC, τ2>0, σ2≥τ2, ρ>0, α∈(0,1), a common discrete filtered measurable space with laws (Pθ)θ∈Θ, a common real-valued process (Xt) adapted under every Pθ, and, under each Pθ, residuals Yt(θ)=Xt−θ∼N(0,τ2) satisfying the displayed event-factorization identity for every t≥1, Borel B, and A∈Ft−1.

[F1]

Standard normal and normal laws: N(0,τ2) is the law of an affine scaling of the standard normal law.

[F2]

Closed form for a Gaussian exponential mixture: for ρ′>0, s∈R, and v≥0, the Gaussian mixture integral equals ρ′/(v+ρ′)es2/[2(v+ρ′)].

[F3]

Conditioning a known variable and an independent variable: if Borel events of an integrable random variable factor with every event in G, its conditional expectation given G is its unconditional expectation almost surely.

[F4]

Conditionally sub-Gaussian increments: conditional sub-Gaussianity requires the conditional mgf bound for every fixed real λ at every time.

[F5]

A Gaussian mixture confidence sequence for conditional sub-Gaussian means: under that conditional proxy at every time and for every λ, the stated closed intervals cover the true parameter simultaneously with probability at least 1−α.

[A1]

The Axiom of Choice: AC is used by the normal-law, conditional-expectation, and confidence-sequence suppliers.

Verification

1.1givenF1F2algebra

If Θ=∅ the conclusion is vacuous. Otherwise fix θ∈Θ, t≥1, and λ∈R, and write Yt=Yt(θ). Apply [F2] with ρ′=1/τ2, v=0, and s=λ. Since N(0,τ2)=N(0,1/ρ′) by [F1], this gives EeλYt=eλ2τ2/2 under Pθ, finite for every real λ.

2.1givenF3F4step 1.1

For the fixed θ,t,λ, the map y↦eλy is Borel, so for each Borel D⊆R its preimage is Borel; the stated factorization for Yt(θ) therefore gives the same factorization for eλYt. This variable is integrable by step 1.1, so [F3] yields E[eλYt∣Ft−1]=E[eλYt] under Pθ. Since Xt−θ=Yt, step 1.1 now gives EPθ[eλ(Xt−θ)∣Ft−1]=eλ2τ2/2≤eλ2σ2/2 almost surely. As t and λ were arbitrary, [F4] applies for each θ, time, and fixed tuning value.

3.1F5A1step 2.1∎

For each θ, the common filtered model and adapted process together with the conditional proxy verified in step 2.1 and [A1] meet the hypotheses of [F5]. Therefore the intervals have simultaneous coverage at least 1−α under each Pθ.