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✓ 5 results · all verified · 4 also independently AI-judged
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Sub-Gaussian mixture confidence sequences

1 · Prerequisites

2 · Summary

Objective

This page derives a time-uniform confidence sequence for a scalar mean from a discrete-time conditional moment-generating-function bound. At every step and for every real tuning value, the increment around the fixed mean must satisfy the same known sub-Gaussian proxy bound given the past. Independence and identical distribution are not required.

The proof builds the fixed-parameter exponential supermartingales, mixes them over a centered normal law with precision ρ>0, evaluates the Gaussian integral, and converts the crossing boundary into closed intervals. The simultaneous coverage guarantee follows from Ville's inequality. The proxy is an upper mgf bound, not a claim that it equals the variance.

Marginal sub-Gaussianity alone does not supply the conditional premise, as the repeated-Rademacher example shows; finite variance alone need not even give a marginal sub-Gaussian bound (Finite variance does not imply a sub-Gaussian tail). This is a discrete-time statement; no continuous-time extension or conditional-on-covariates guarantee is asserted.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

Conditionally sub-Gaussian increments

Definition

Assume the Axiom of Choice (AC) The Axiom of Choice for the conditional-expectation existence convention. Let (Ω,F,(Ft)t∈N0,P) be a discrete filtered probability space Filtration and filtered probability space, and let (Xt)t≥1 be a real-valued adapted process Adapted and integrable stochastic process. Fix μ∈R and a proxy σ2>0. The increments are conditionally sub-Gaussian around μ with proxy σ2 when, for every integer t≥1 and every fixed λ∈R, eλ(Xt−μ) is integrable and EP[eλ(Xt−μ)∣Ft−1]≤eλ2σ2/2P-almost surely, where conditional expectation means the class of integrable versions of Conditional expectation as an ae class represented as in Conditional expectation given a sigma algebra. The exceptional null set may depend on t and λ; no common null set over the uncountable set R is assumed.

The integrability of Xt follows from the two cases λ=1 and λ=−1: conditional expectation preserves the integral, so both eXt−μ and e−(Xt−μ) have finite expectation, and ∣Xt−μ∣≤eXt−μ+e−(Xt−μ). The proxy is an upper bound in the conditional moment-generating-function inequality; it is not asserted to be a variance. Independence and identical distribution of the increments are not part of this definition.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

Gaussian-mixture confidence-sequence boundary

Definition

Let σ2>0 be a supplied scale, let ρ>0 be a tuning parameter, and let α∈(0,1). For v≥0 define uα(v):=(v+ρ)log⁡ ⁣(v+ρα2ρ). The expression is real because (v+ρ)/(α2ρ)>1. For a real data sequence x1,x2,…, set xˉt=t−1∑i=1txi and define C0:=R,Ct:=[xˉt−uα(tσ2)t, xˉt+uα(tσ2)t](t≥1). The endpoints are included. This definition specifies an interval rule only; it makes no coverage or time-uniform validity assertion.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passOpen item page →

Closed form for a Gaussian exponential mixture

Statement

Assume AC. For ρ>0, v≥0, and s∈R, let πρ=N(0,1/ρ) under the normal-law convention of Standard normal and normal laws. Then ∫Rexp⁡ ⁣(λs−vλ22) πρ(dλ)=ρv+ρexp⁡ ⁣(s22(v+ρ)).

Facts & Assumptions

Given: Assume AC, ρ>0, v≥0, and s∈R.

[F1]

Standard normal and normal laws: N(m,σ2) is the law of the affine image z↦m+σz of the standard normal law, whose density is e−z2/2/2π.

[F2]

Change of variables for expectation: Integrating a nonnegative measurable function against a random element's law equals its expectation on the underlying space.

[F3]

The standard normal density has total mass one: The standard normal density e−z2/2/2π integrates to one.

[F4]

A C^1 diffeomorphism satisfies the change-of-variables formula for nonnegative Lebesgue measurable functions: If T:U→V is a C1 diffeomorphism between open subsets of Rn and f:V→[0,∞] is Lebesgue measurable, then, assuming countable choice, ∫Vf(y) dλn(y)=∫Uf(T(x))∣det⁡DT(x)∣ dλn(x).

[F5]

Continuous functions on Euclidean spaces are Borel measurable and Assuming countable choice, every Borel subset of Rn is Lebesgue measurable: Under countable choice, every continuous function on R is Borel measurable, and every Borel set is Lebesgue measurable.

[A1]

The Axiom of Choice: AC asserts that every family of nonempty sets has a choice function; it is assumed here.

[A2]

The Axiom of Countable Choice (ACω): AC implies countable choice, so the countable-choice premise of [F4] follows from the assumed AC.

Proof

1.1givenF1F2F5F6A1A2

Let Z have the standard normal law. By [F1] and [F6], Z/ρ has law πρ; [F2] and the density in [F1] give I=∫Rexp⁡ ⁣(sz/ρ−vz2/(2ρ))e−z2/22π dz, with a continuous nonnegative integrand, which [F5] makes Lebesgue measurable.

2.1F3F4F5F6A2step 1.1algebra∎

Set a=(v+ρ)/ρ>0 and c=sρ/(v+ρ). Completing the square in [step 1.1] gives I=es2/[2(v+ρ)]∫Re−a(z−c)2/22π dz. The map T(y)=c+y/a has the C1 inverse T−1(z)=a(z−c), so it is a diffeomorphism from R to R, and ∣T′(y)∣=1/a. The function g(z)=e−a(z−c)2/2/2π is continuous and nonnegative, hence Lebesgue measurable by [F5]. By [A2], AC supplies the countable-choice hypothesis of [F4]; applying [F4] to g and T gives ∫Rg(z) dz=1a∫Re−y2/22π dy=1a by [F3]. Therefore I=ρ/(v+ρ)es2/[2(v+ρ)].

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-luna)Open item page →

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.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-luna)Open item page →

The Gaussian-mixture threshold is the stated square-root boundary

Statement

Let α∈(0,1), ρ>0, v≥0, and s∈R. Define M(v,s):=ρv+ρexp⁡ ⁣(s22(v+ρ)). For uα from Gaussian-mixture confidence-sequence boundary, M(v,s)≥1α⟺∣s∣≥uα(v).

Facts & Assumptions

Given: α∈(0,1), ρ>0, v≥0, and s∈R.

[F1]

Gaussian-mixture confidence-sequence boundary: uα(v)=(v+ρ)log⁡((v+ρ)/(α2ρ)).

[F2]

Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm: the natural logarithm is strictly increasing on (0,∞) and satisfies log⁡(xy)=log⁡x+log⁡y for positive x,y.

Proof

1.1givenF2algebra

Since M(v,s)>0, applying the strictly increasing logarithm [F2] shows M(v,s)≥1/α iff s2/[2(v+ρ)]−12log⁡((v+ρ)/ρ)≥log⁡(1/α). Multiplying by 2(v+ρ)>0 and combining logarithms gives s2≥(v+ρ)log⁡((v+ρ)/(α2ρ)).

2.1F1F3step 1.1∎

The ratio (v+ρ)/(α2ρ) exceeds 1, so its logarithm is positive and [F3] supplies the nonnegative root in [F1]. Thus the last inequality in step 1.1 is exactly ∣s∣≥uα(v).

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-luna)Open item page →

The Gaussian mixture is a sub-Gaussian test supermartingale

Statement

Assume AC. Let (Ω,F,(Ft)t∈N0,P) and (Xt)t≥1 satisfy Conditionally sub-Gaussian increments for μ∈R with known proxy σ2>0. Fix ρ>0. Define S0(μ)=0, St(μ)=∑i=1t(Xi−μ), Vt=tσ2, and Mt(μ):=ρVt+ρexp⁡ ⁣(St(μ)22(Vt+ρ))(t∈N0). Then M(μ) is a nonnegative test supermartingale for {P}, with M0(μ)=1.

Facts & Assumptions

Given: Assume AC and the conditional sub-Gaussian model for fixed μ, σ2>0, and ρ>0.

[F1]

Conditional sub-Gaussian increments generate exponential test supermartingales: for each fixed λ∈R, Ltλ(μ)=exp⁡(λSt(μ)−λ2tσ2/2) is a nonnegative test supermartingale with initial value one.

[F2]

Mixtures of null test supermartingales: if each component is a nonnegative test supermartingale under every null law and (λ,ω)↦Ltλ(ω) is product-measurable, its probability mixture is a test supermartingale; the theorem uses a real-valued modification only where the integral is infinite.

[F3]

Tonelli's theorem for nonnegative measurable functions on a sigma-finite product: for sigma-finite measure spaces and nonnegative product-measurable functions, section integrals are measurable and equal the corresponding iterated integrals.

[F4]

Closed form for a Gaussian exponential mixture: for v≥0, s∈R, and ρ>0, the mixture of exp⁡(λs−vλ2/2) under N(0,1/ρ) equals ρ/(v+ρ)exp⁡(s2/[2(v+ρ)]).

[F5]

Standard normal and normal laws: N(0,1/ρ) is a probability law under the affine-image normal-law convention.

[A1]

The Axiom of Choice: AC is required by the conditional-expectation and test-supermartingale conventions in [F1] and [F2].

Proof

1.1givenF1construct

For each fixed λ, [F1] supplies the component process Ltλ(μ). At fixed t, St(μ) is Ft-measurable, so (λ,ω)↦exp⁡(λSt(μ)(ω)−λ2tσ2/2) is measurable on B(R)⊗Ft: it is the composition of the measurable coordinate maps with continuous arithmetic and exponential operations. Each component is nonnegative.

2.1F2F3F4F5A1step 1.1∎

Mix these components using the probability law N(0,1/ρ) from [F5]. Both this prior and P are probability measures, hence sigma-finite. The product-measurability in step 1.1 and the component conditions in [F1] meet [F2]; [F3] gives the needed measurable nonnegative section integrals. Therefore the mixture process is a nonnegative test supermartingale. By [F4] its value is the displayed finite formula for every finite St(μ), so the mixture theorem's infinite-integral modification is never applied. At t=0 the formula gives M0(μ)=1.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-luna)Open item page →

A Gaussian mixture confidence sequence for conditional sub-Gaussian means

Statement

Assume AC. Let Θ⊆R, let (Ω,F,(Ft)t∈N0) be a common discrete filtered measurable space, and let (Pθ)θ∈Θ be a family of probability laws on it. Let (Xt)t≥1 be a common real-valued process that is adapted under each Pθ, and suppose that for every θ∈Θ its increments are conditionally sub-Gaussian around θ with the same known proxy σ2>0 under Pθ in the sense of Conditionally sub-Gaussian increments. Fix ρ>0 and α∈(0,1). For t≥1, set Xˉt:=1t∑i=1tXi,Ct:=Θ∩[Xˉt−uα(tσ2)t, Xˉt+uα(tσ2)t], where uα is from Gaussian-mixture confidence-sequence boundary, and set C0:=Θ. Then (Ct)t≥0 is a (1−α) confidence sequence for θ: Pθ(θ∈Ct for every t≥0)≥1−α(θ∈Θ). The claim is pointwise in θ and requires no measurable structure on Θ, independence, or identical distribution of the observations.

Facts & Assumptions

Given: Assume AC and all data, law, filtration, proxy, and tuning hypotheses in the Statement.

[F1]

Conditionally sub-Gaussian increments: for each fixed θ, time t≥1, and λ∈R, the conditional mgf of Xt−θ given Ft−1 is at most eλ2σ2/2 almost surely.

[F2]

The Gaussian mixture is a sub-Gaussian test supermartingale: under each Pθ, Mt(θ)=ρ/(tσ2+ρ)exp⁡(St(θ)2/[2(tσ2+ρ)]) is a nonnegative test supermartingale with M0(θ)=1.

[F3]

The Gaussian-mixture threshold is the stated square-root boundary: Mt(θ)≥1/α exactly when ∣St(θ)∣≥uα(tσ2).

[F4]

Discrete Ville inequality and anytime threshold validity: under AC, for every null law and each α∈(0,1), a nonnegative test supermartingale has P(sup⁡t≥0Mt≥1/α)≤α.

[F5]

Confidence sequences: a (1−α) confidence sequence has Ft-measurable membership events and covers each true parameter at all integer times with probability at least 1−α.

[A1]

The Axiom of Choice: AC supplies the conditional-expectation and test-supermartingale conventions used in [F2] and [F4].

Proof

1.1givenF1F5

Fix θ∈Θ. The process is adapted under Pθ, so each Xi for i≤t and hence Xˉt is Ft-measurable. Therefore {θ∈Ct} is Ft-measurable for every t≥1, and it is the whole space at t=0.

2.1givenF3step 1.1

For t≥1, if θ∉Ct, the interval in the Statement has closed endpoints, so ∣St(θ)∣>uα(tσ2). By [F3], this implies Mt(θ)≥1/α. Consequently the event that θ is ever excluded from Ct is contained in {sup⁡t≥0Mt(θ)≥1/α}.

3.1F2F4F5A1step 2.1∎

By [F2] and Ville's inequality [F4], the latter event has Pθ-probability at most α. Its complement therefore has probability at least 1−α and is contained in {θ∈Ct for all t≥0}. Since [F5] requires exactly these membership and coverage clauses, (Ct) is a confidence sequence.

5 · Examples, counterexamples and false statements

None yet.

Sources