LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-12

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Moments give derivatives of the characteristic function

Statement

Let $k$ be a nonnegative integer, and suppose $E ∣ X ∣^{k} < \infty$ , with $∣ X ∣^{0} = 1$ . Then $φ_{X} \in C^{k} (R)$ and $φ_{X}^{(j)} (t) = E [(i X)^{j} e^{i t X}], 0 \leq j \leq k .$ No converse is asserted.

Facts & Assumptions

Given: The hypotheses and conventions in the statement.

[F1]

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

[F2]

A specified dominated sequence has convergent integrals. Dominated convergence.

[F3]

Frequency increments factor by exponential addition. $\exp (z + w) = \exp z \exp w$ , and the complex exponential extends the real exponential.

[F4]

Sine and cosine are differentiable with the usual derivatives. The derivatives of sine and cosine are cosine and minus sine.

[F5]

Real-parameter differentiation of sine and cosine obeys the chain rule. The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g (c)$ , then $f \circ g$ is differentiable at $c$ with $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ .

[F6]

MVT bounds increments using bounded real derivatives. The mean value theorem, as the case $g (x) = x$ of Cauchy's: for $f$ continuous on $[a, b]$ with $a < b$ and differentiable on $(a, b)$ there is $c \in (a, b)$ with $f (b) - f (a) = f^{'} (c) (b - a)$ .

[F7]

Difference quotients commute with integrable linear combinations. The Lebesgue integral is linear on $L^{1} (μ)$ .

[F8]

Euler form and unit modulus control every frequency. $\exp (x + i y) = e^{x} (\cos y + i \sin y)$ , $∣ \exp (x + i y) ∣ = e^{x}$ , and $e^{i π} + 1 = 0$ .

Proof

technique · direct

1.1F1F2F3F4F6F8

For $0 \leq j \leq k$ , $∣ x ∣^{j} \leq 1 + ∣ x ∣^{k}$ , so $G_{j} (t) = E [(i X)^{j} e^{i t X}]$ exists. Applying MVT to sine and cosine gives $∣ e^{i v} - e^{i u} ∣ \leq \min (2, 2 ∣ v - u ∣)$ . For $∣ h ∣ \leq 1 / n$ it follows that $∣ G_{j} (t + h) - G_{j} (t) ∣ \leq E [∣ X ∣^{j} \min (2, 2 ∣ X ∣ / n)] .$ This prescribed nonnegative sequence tends pointwise to zero and is dominated by $2 ∣ X ∣^{j}$ . DCT makes the bound tend to zero, proving continuity of every $G_{j}$ without selecting an arbitrary sequence of frequencies.

2.1F2F3F4F5F6F7F8step 1.1

For real x and nonzero h put $Q_{h} (x) = (e^{i h x} - 1) / h - i x$ . MVT applied to $\cos (h x) - 1$ and $\sin (h x)$ yields numbers between zero and hx with $∣ \frac{\cos (h x) - 1}{h} ∣ \leq ∣ x ∣ \min (1, ∣ h x ∣), ∣ \frac{\sin (h x)}{h} - x ∣ \leq ∣ x ∣ \min (2, ∣ h x ∣) .$ Here the first bound uses $∣ \sin v ∣ \leq \min (1, ∣ v ∣)$ and the second $∣ \cos v - 1 ∣ \leq \min (2, ∣ v ∣)$ , each obtained from the same derivative bounds. Therefore $∣ Q_{h} (x) ∣ \leq \min (4 ∣ x ∣, 2 ∣ h ∣ ∣ x ∣^{2})$ , including x=0. For $1 \leq j \leq k$ , linearity and exponential addition give $∣ \frac{G_{j - 1} (t + h) - G_{j - 1} (t)}{h} - G_{j} (t) ∣ \leq E [∣ X ∣^{j - 1} \min (4 ∣ X ∣, 2 ∣ X ∣^{2} / n)]$ whenever $0 < ∣ h ∣ \leq 1 / n$ , interpreting the integrand as zero at X=0. The right side tends to zero by DCT, dominated by $4 ∣ X ∣^{j}$ . Hence $G_{j - 1}' = G_{j}$ .

3.1F1step 1.1step 2.1∎

Starting from $G_{0} = φ_{X}$ , the derivative identities in step 2.1 and continuity in step 1.1 establish the assertion through order k. If k=0 only the continuity conclusion of step 1.1 is required. At t=0 the formula becomes $φ_{X}^{(j)} (0) = i^{j} E [X^{j}]$ . All limits used prescribed majorants indexed by positive integers; this proof introduces no selection axiom.

Moments give derivatives of the characteristic function

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources