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?

  • $\begingroup$ Just by curiosity, this come of a book? $\endgroup$
    – Masacroso
    Jan 26, 2016 at 7:27
  • 1
    $\begingroup$ @Masacroso Yes, it is stated as an elementary fact in: amazon.com/Processes-Stochastic-Calculus-Cambridge-Mathematics/…, but unfortunately I don't fint it elementary. $\endgroup$
    – user119615
    Jan 26, 2016 at 7:29
  • $\begingroup$ You know what the derivative of your function is, so you can use Lagrange, and all you need to show is that the derivative is dominated uniformly in u by an integrable function. $\endgroup$
    – Kore-N
    Jan 26, 2016 at 7:46
  • $\begingroup$ @Cornelis Thank you, but which theorem are you referring to? Is it the mean-value theorem?, i have seen something like that before used in a situation like this, but the function is complex now, do we still have the result then?, for complex functions? $\endgroup$
    – user119615
    Jan 26, 2016 at 10:33
  • $\begingroup$ The complex numbers should not bother you. You should think after it as a function from R to R^2. In fact Lagrange only holds in Form of inequality (you do it in each entrance of the vector). I will write the explicit calculations as an answer. $\endgroup$
    – Kore-N
    Jan 26, 2016 at 11:00

1 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).


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