The partial derivative of a characteristic function (exercise).

Question

Assume that you have a probability space $(\Omega, \mathcal{F},P)$ and a random varaible $X: (\Omega, \mathcal{F})\rightarrow(\mathbb{R}^d,\mathcal{B}(\mathbb{R}^d))$. Define the characteristic function $\phi_X: \mathbb{R}^d \rightarrow \mathbb{C}$, by $\phi_X(u)=\int_\Omega e^{i<u,X(\omega)>}dP$.

I want to prove this theorem:

If $X=(X_1,\ldots,X_d)$ and $\mathbb{E}(|X_j^n|)<\infty$ for some $1\le j\le d$ and $n \in \mathbb{N}$ then:

$\mathbb{E}(X_j^n)=i^{-n}\frac{\partial^n}{\partial u_j^n}\phi_X(u) |_{u=0}$

I tried two things, but I get problems with both approaches:

approach 1:

It is not hard to show this "naively", if you just differentiate like you learned in calculus, and go under the integral sign. then you will get the disered result. But there are two problems with this approach:

1. Why is it partial differentiable at 0, and so why can we differantiate at all?

2. Why can we differentiate under the integral sign?

approach 2(dominated convergence theorem):

Here I will try to solve both problems directly, but I still get problems: I start with n=1 for simplicity:

Let $h_n$ be any sequence of real numbers converging to 0, but which is never zero, let $u_n=(0,0,0,h_n,0,0)$, where $h_n$ is in position j. Then if the partial derivative exits at 0, we must have that

$\frac{\int_\Omega e^{i<u_n,X>}dP-\int_\Omega e^{i<0,X>}dP}{h_n}=\int_\Omega\frac{e^{i<u_n,X>}-1}{h_n}dP$ converges.

Now the problem is reduced to taking a limit outside the integral, to inside the integral, if we can move the limit inside the integral, we see that we then get our derivative inside the integral. but why can we move the limit inside the integral?

If we denote: $g_n=\frac{e^{i<u_n,X>}-1}{h_n}$, the problem will be solved if all $|g_n|$ will be bounded by an integrable function?(the dominated convergence theorem). But can you see a dominating function here? Do you guys have any tips or help?

Answer 1

I will write the answer for $X$ one dimensional.

Your approach of bringing the derivative inside the integral is right.

You need to estimate: $ \frac {e^{iux}-1}{u}.$ Writing down this function you find: $$\frac {\cos (ux) -1}{u} + i \frac {\sin (ux)}{u} = x \sin (\xi_1 x) - ix \cos (\xi_2 x)$$ by the mean value theorem.

So in norm this can be estimated by $2x$, and you find the uniform bound you needed, to apply dominated convergence. Note that the estimate is global (w.r.t. u). In general it is sufficient to have a local estimate (e.g. for u small enough).