Alphabeta Math
ExampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)
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Characteristic function of a gaussian law

Example

Assume AC. For mR and σ0, the law N(m,σ2) has characteristic function φ(t)=exp(imtσ2t2/2),tR, including σ=0, when the law is δm.

Facts & Assumptions

Given: The hypotheses and conventions in the example.

[F1]

The general normal law is the affine pushforward of the standard law. Standard normal and normal laws.

[F2]

Under AC the standard Gaussian density is a probability density. The standard normal density has total mass one.

[F3]

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

[F4]

Affine maps change the transform by scaling frequency and multiplying by a phase. Characteristic functions under affine maps and independent sums.

[F5]

Dominated limits pass through integrals. Dominated convergence.

[F7]

Compact Riemann integrals agree with Lebesgue integrals under countable choice. A bounded Riemann integrable function on a closed bounded interval is Lebesgue measurable and has the same integral.

[F8]

AC covers Gaussian normalization and the compact integration bridge. The Axiom of Choice.

[F9]

The real exponential differentiates to itself. The exponential function is smooth and (exp)=exp.

[F10]

Sine and cosine derivatives give the derivative of exp(itx) componentwise. The derivatives of sine and cosine are cosine and minus sine.

[F15]

Nonnegative density integrals are integrals of products. Integrating against a density agrees with integrating the product.

[F16]

Nonnegative truncations increasing to a function recover its integral. Monotone convergence for the integral.

[F17]

A finite first absolute moment justifies differentiating the transform. Moments give derivatives of the characteristic function.

Verification

technique · direct
1.1

Let g(x)=ex2/2/2π and let Z have its law on the canonical real probability space. Normalization is supplied by the Gaussian-density lemma. Exponential differentiation and the chain rule give g(x)=xg(x). FTC, reflection substitution and the bridge yield RRxg(x)dx=22π(1eR2/2)(R>0). Monotone convergence over positive integer R gives EZ=2/2π< by density integration. Apply the moments lemma at order one: φ(t)=ixeitxg(x)dx, converting real positive/negative parts of the density integral separately.

F1F2F3F7F9F11F12F14F15F16F17
2.1

On [R,R] both g and the real and imaginary parts of eitx are continuously differentiable. Integration by parts, applied componentwise, gives RRixeitxg(x)dx=i[eitxg(x)]RRtRReitxg(x)dx. The boundary term has modulus at most 2g(R)0. The left integrand is dominated by the integrable xg(x), and the last integral by g(x); DCT along integer R therefore gives φ(t)=tφ(t) for every real t. No imaginary displacement of an integration contour is involved.

step 1.1F5F6F7F10F11
3.1

The real and imaginary components of H(t)=et2/2φ(t) are differentiable. The product and chain rules and step 2.1 give H(t)=et2/2(tφ(t)+φ(t))=0. The zero-derivative theorem applied to each component on R makes H constant. Its value at zero is φ(0)=g=1, so φ(t)=et2/2. Finally affine scaling gives φm+σZ(t)=eimtφZ(σt)=eimtσ2t2/2. If σ=0, the random variable is constantly m and its transform is directly eimt, agreeing with the formula. The stated AC is inherited from normalization and the compact integration bridge (including their countable-choice prerequisites).

step 1.1step 2.1F1F3F4F8F9F11F13

Depends on

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