Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

1 Answer 1

up vote 1 down vote accepted

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.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.