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Let $(\Omega, \mathcal{F}, P)$ be a probability space. A random variable (r.v) $X: \Omega \to \mathbb{R}$ is called a Gaussian r.v. with mean $\mu \in \mathbb{R}$ and variance $\sigma^{2}$ if its associate distribution $\mu$ has density $$\frac{1}{\sqrt{2\pi \sigma}}e^{-\frac{(x-\mu)^{2}}{2\sigma^{2}}}$$ In other words, $X$ is Gaussian if its distribution is a Gaussian measure in $\mathbb{R}^{n}$.

Now, I have the following definition for Gaussian random vectors: A random vector $X: \Omega \to \mathbb{R}^{n}$, given by $X(\omega) = (X_{1}(\omega),...,X_{n}(\omega))$ is called Gaussian if, for every $a_{1},..,a_{n}\in \mathbb{R}$, the random variable $\tilde{X}:= a_{1}X_{1}+\cdots+a_{n}X_{n}$ is Gaussian. Thus, we define Gaussian random vectors in terms of Gaussian random variables.

My question is: how can I prove, using the above definition, that a random vector is Gaussian if, and only if its characterist function has the form: $$e^{i\langle \theta, \mu\rangle -\frac{1}{2}\langle \theta, K\theta\rangle}$$ for some nonnegative $n \times n$ matrix $K$ and $\mu \in \mathbb{R}^{n}$? I've tried to compute the characteristic function for $\tilde{X}$ but I didn't get anywhere because I don't know, in principle, how each $X_{j}$ is distributed.

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Your definition of Gaussian RV at the beginning is incomplete because you miss the (degenerate) case where $\sigma=0$ and the measure has no density. Also the variance is $\sigma^2$ not $\sigma$. A better definition is a RV whose characteristic function is of the form $$ \mathbb{E}\ e^{itX}=e^{i\mu t-\frac{\sigma^2 t^2}{2}} $$ for some $\mu\in\mathbb{R}$ and $\sigma\ge 0$. Now for any $t\in\mathbb{R}$, $$ \mathbb{E}\ e^{it\bar{X}}=\mathbb{E}\ e^{it(a_1X_1+\cdots+a_nX_n)}=\mathbb{E}\ e^{i\langle \theta,X\rangle}= e^{i\langle \theta,\mu\rangle-\frac{1}{2}\langle \theta,K\theta\rangle} $$ where $\theta=t(a_1,\ldots,a_n)$. So it is of the wanted form. The main step you seem to have trouble with is the first equality which is an application of the abstract change of variable formula for push-forward measures.

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  • $\begingroup$ This is what I intended to do, too. My problem is: how can I conclude the last equality from my assumbtions? I mean, I know $\tilde{X}$ is Gaussian, but last equality seems to consider $X_{j}$ being Gaussian. Is that so? $\endgroup$ Commented Jan 17, 2020 at 20:28
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    $\begingroup$ You are moving the goalposts. I thought your question was about computing the characteristic function of $\bar{X}$ given that of $X$. I guess you are now asking about the proof of equivalence. The above shows that $X$ Gaussian implies that $\forall a$, $\bar{X}$ is Gaussian. The converse needs a little bit more work. $\endgroup$ Commented Jan 17, 2020 at 20:37

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