# Can the characteristic function of a multivariate normal distribution be extended from a neighborhood of the origin?

Let $x$ be a scalar random variable. There is a theorem that states that if $E[\exp(ixs)]= \exp\Big( i{s}\mu - \tfrac{1}{2} {\sigma^2s^2} \Big)$ for some neighborhood around the origin (i.e. $|s|<\delta$ for some $\delta>0$) then ${x}$ is normal (e.g. Lukacs, 1970, Chapter 7).

The question is if this holds for multivariate $\mathbf{x}$. Specifically, if $E[\exp(i\mathbf{s}'\mathbf{x})]= \exp\Big( i\mathbf{s}'\boldsymbol\mu - \tfrac{1}{2} \mathbf{s}'\boldsymbol\Sigma \mathbf{s} \Big)$ for some neighborhood around the origin then does $\mathbf{x}$ have a multivariate normal distribution?

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## 1 Answer

Sure, because being multivariate normal means exactly that every linear combination of the entries is normal and the one-dimensional result you recalled shows that every linear combination of the entries of $\mathbf x$ is normal.

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Thanks. That makes sense. The proof of the one-dimensional case uses analytic extension. Would the proof need to even be changed for the multivariate case? (I don't think so) – user103828 Oct 30 '13 at 8:03
"Would the proof need to even be changed for the multivariate case?" The exact content of my answer is that the multi-dim proof would use the one-dim case and nothing else, so, no, "the proof would not need to be changed". – Did Oct 30 '13 at 8:06