[Math] Characteristic function of a random vector

Question

We consider the random vector $ X\colon \Omega \to \mathbb {R}^n$ defined on the probability space $(\Omega, \mathfrak F, P)$. Let denote by $\Phi_{X}(x) = \mathbb E(e^{i\left<x, X\right>})$ its characteristic function.

I would like to show the following equivalence: $X$ is a Gaussian vector if and only if $\Phi_{X}(x)$ is given by $$\Phi_{X}(x)= e^{i\left<m, x\right> -\frac{1}{2}\left<A x, x\right>} \qquad (*),$$
where $m=(\mathbb E(X_1), \dots, \mathbb E(X_n))$ and $A=Cov(X)$.

I showed the direct sense (i.e. if $X$ is a Gaussian vector then $\Phi_{X}(x)$ is given as $(*)$). In fact, I have used the following $\Phi_{X}(x)=\Phi_{Z_x}(1) = \exp\{im_{x} – \frac{1}{2}\sigma_{x}^2\} = …$, where $Z_x=\sum_{j=1}^{n}x_j X_j$ is a random variable in $N(m, \sigma^2)$ ….

Now, I need help for the opposite direction.

Thank you in advance

Answer 1

Hint: fix $a := (a_1,\ldots,a_n)$ and find the characteristic function of $\langle a, X\rangle.$ Hover below for a full answer.

Then the characteristic function of $a_1X_1 + \cdots + a_n X_n$ is \begin{align*}\Phi_{\langle a, X \rangle}(t) &= \mathbb{E}(e^{it\langle a, X \rangle}) \\&= \Phi_X(ta) \\&= \exp\left(i \langle m, ta\rangle - \frac{1}{2}\langle A(ta),ta\rangle \right) \\&= \exp\left(i \langle m,a\rangle t - \frac{1}{2}\langle Aa,a\rangle t^2\right).\end{align*} This last equation is the characteristic function of $N(\langle m,a\rangle, \langle Aa,a\rangle).$