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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

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    $\begingroup$ What definition of Gassian Vector are you using? There are many different (equivalent) definitions; in fact, this can be used as a definition of a Gaussian vector. From some definitions, this statement will be easier to get than from others. $\endgroup$ – Marcus M Mar 28 '18 at 16:11
  • $\begingroup$ $X$ is Gaussian if any linear combination $a_1 X_1 +\dots + a_n X_n$ follows a Gaussian law, where $a_j\in \mathbb R$ and $n\in \mathbb N$. $\endgroup$ – Z. Alfata Mar 28 '18 at 16:16
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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 $N(\langle m,a\rangle, \langle Aa,a\rangle).$

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  • $\begingroup$ yes it is ca what I did to show the direct sense, but I need the other direction (reverse). $\endgroup$ – Z. Alfata Mar 28 '18 at 16:30
  • $\begingroup$ I'm not sure you're catching what I'm saying: to show that $X$ is Gaussian, we need to show $\langle a, X \rangle$ is univariate Gaussian for each $a$, so we calculate the characteristic function of $\langle a, X \rangle$ for each $a$. $\endgroup$ – Marcus M Mar 29 '18 at 1:32
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The converse follows immediately from Levy's Continuity Theorem, which in particular implies that any two random variables with the same characteristic function are equal in distribution.

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