Exchangeability of differentiation and integration to show a characteristic function

Question

I am having a hard time to understand how the exchangeability of integration and differentiation is justified under $E[|X|^{k}]<\infty$ for a characteristic function in the context of showing $E[X^{k}]=i^{-k}\phi^{(k)}(0)$, where $\phi^{\left(k\right)}(0)$ is the k-th differentiation of $\phi$ with respect to $t = 0$.

I must use the following exchangeability to show this.

$$\frac{d^{k}}{dt^{k}}\int e^{itx}d\mu(x)=\int \frac{d^{k}}{dt^{k}}e^{itx}d\mu(x)$$

However, I am not sure how $E[|X|^{k}]<\infty$ is used to show the exchangeability.

I already saw characteristic function differentiation

but it does not help me to understand the importance of a finite moment.

I would appreciate your help!

Answer 1

The condition $\mathbb{E}[|X|^k]<\infty$ is necessary to guarantee the existence and finiteness of $\mathbb{E}[X^k]$. Also, it is used to justify that interchanging integration and differentiation is valid. (In general, you can't interchange the order of two limiting operators without extra assumptions.)

In doing so, the relevant theorem is as follows:

Theorem. Let $\nu$ be a signed measure such that $\int_{\mathbb{R}} |\nu(\mathrm{d}x)| < \infty$ and $\int_{\mathbb{R}} |x| |\nu(\mathrm{d}x)| < \infty$. Then

$$ \frac{\mathrm{d}}{\mathrm{d}t} \int_{\mathbb{R}} e^{itx} \, \nu(\mathrm{d}x) = \int_{\mathbb{R}} \frac{\mathrm{d}}{\mathrm{d}t} e^{itx} \, \nu(\mathrm{d}x). $$

Once this has been proved, we can inductively check that $\mathbb{E}_{\mu}[|X|^k] < \infty$ implies

\begin{align*} \frac{\mathrm{d}^k}{\mathrm{d}t^k} \int_{\mathbb{R}} e^{itx} \, \mu(\mathrm{d}x) &= \frac{\mathrm{d}^{k-1}}{\mathrm{d}t^{k-1}} \int_{\mathbb{R}} \frac{\mathrm{d}}{\mathrm{d}t}e^{itx} \, \mu(\mathrm{d}x) \tag{Thm, $\nu = \mu$} \\ &= \frac{\mathrm{d}^{k-2}}{\mathrm{d}t^{k-2}} \int_{\mathbb{R}} \frac{\mathrm{d}^2}{\mathrm{d}t^2} e^{itx} \, \mu(\mathrm{d}x) \tag{Thm, $\nu=x\mu$} \\ &= \frac{\mathrm{d}^{k-3}}{\mathrm{d}t^{k-3}} \int_{\mathbb{R}} \frac{\mathrm{d}^3}{\mathrm{d}t^3} e^{itx} \, \mu(\mathrm{d}x) \tag{Thm, $\nu=x^2\mu$} \\ &\qquad \vdots \\ &= \int_{\mathbb{R}} \frac{\mathrm{d}^k}{\mathrm{d}t^k} e^{itx} \, \mu(\mathrm{d}x) \tag{Thm, $\nu=x^{k-1}\mu$} \end{align*}

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It remains to prove the theorem.

Proof. Write $\phi_{\mu}(t) = \int_{\mathbb{R}} e^{itx} \, \mu(\mathrm{d}x)$ and $\psi_{\mu}(t) = \int_{\mathbb{R}} ix e^{itx} \, \mu(\mathrm{d}x)$. Then for $h \neq 0$,

\begin{align*} \left| \frac{\phi_{\mu}(t+h) - \phi_{\mu}(t)}{h} - \psi_{\mu}(t) \right| &= \left| \int_{\mathbb{R}} \left( \int_{0}^{1} ix (e^{ihxs} - 1) \, \mathrm{d}s \right) e^{itx} \, \mu(\mathrm{d}x) \right| \\ &\leq \int_{\mathbb{R}} \left( \int_{0}^{1} \left| e^{ihxs} - 1\right| \, \mathrm{d}s \right) |x| \, |\mu(\mathrm{d}x)| \end{align*}

Since $x \mapsto \int_{0}^{1} \left| e^{ihxs} - 1\right| \, \mathrm{d}s$ is uniformly bounded by $2$ and tends to $0$ as $h \to 0$ pointwise, the above bound converges to $0$ as $h \to 0$ by the dominated convergence theorem. This proves that $\phi_{\mu}(t)$ is differentiable with $\phi_{\mu}'(t) = \psi_{\mu}(t)$ as desired.