Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedaudited 2026-09-13
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Second-order characteristic-function expansion

Statement

If EX=0 and EX2=σ2<∞, then φX(t)=1−σ2t2/2+o(t2) as t→0. No third moment or choice axiom is assumed. The same scalar estimates give 0≤1−cos⁡u≤u2/2 for every real u. More precisely, for r(u)=eiu−1−iu+u2/2, ∣r(u)∣≤min⁡(∣u∣3/3,4u2),∣eiu−1−iu∣≤u2.

Facts & Assumptions

[F1]

Real Taylor remainders are bounded by the uniform next derivative bound. A uniform derivative bound gives a uniform Taylor remainder bound.

[F2]

Sine and cosine have derivatives of all orders bounded by one. The derivatives of sine and cosine are cosine and minus sine.

[F4]

DCT applies to the prescribed nonnegative majorant sequence below. Dominated convergence.

[F5]

Integrable real and complex linear combinations commute with integration. The Lebesgue integral is linear on L1(μ).

[F6]

The characteristic function is the expectation of the unit exponential. Characteristic function of a real random variable.

Proof

Given: If EX=0 and EX2=σ2<∞, then φX(t)=1−σ2t2/2+o(t2) as t→0. No third moment or choice axiom is assumed. The same scalar estimates give 0≤1−cos⁡u≤u2/2 for every real u. More precisely, for r(u)=eiu−1−iu+u2/2, ∣r(u)∣≤min⁡(∣u∣3/3,4u2),∣eiu−1−iu∣≤u2.

1.1F1F2F3

Taylor at zero through degree two for cosine and sine, with third derivatives bounded by one, gives ∣cos⁡u−1+u2/2∣≤∣u∣3/6 and ∣sin⁡u−u∣≤∣u∣3/6. Euler form and the triangle inequality give ∣r(u)∣≤∣u∣3/3. Taylor through degree one gives ∣cos⁡u−1∣≤u2/2 and ∣sin⁡u−u∣≤u2/2, hence ∣eiu−1−iu∣≤u2. Adding u2/2 bounds ∣r(u)∣≤3u2/2≤4u2 for all real u. At u=0 all remainders vanish.

2.1step 1.1F4

For 0<∣t∣≤1/n, the first step gives ∣r(tX)∣/t2≤X2min⁡(∣X∣/(3n),4). This is a measurable nonnegative sequence tending pointwise to zero, bounded by the integrable 4X2. DCT therefore makes its expectations tend to zero. The bound is uniform over all such t, so Er(tX)=o(t2) as a genuine two-sided real limit, without choosing a sequence of frequencies or requiring E∣X∣3<∞. Also ∣X∣≤1+X2 gives integrability of the linear term.

3.1step 1.1step 2.1F5F6∎

By linearity, φX(t)=1+itEX−t2EX2/2+Er(tX). Insert the stated moments and the preceding remainder limit. If σ=0, the same majorants have zero integral, so the remainder vanishes and the formula still holds. At t=0 the defining expectation is one. Every limiting integrand was explicitly specified; no choice principle enters.

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Sources