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Show that for a Gaussian process $z$ with zero mean we have $<e^z>=e^{<z^2>/2}$.

If we denote by $f_G(z)$ the Gaussian process PDF

$<e^z>=\int_{-\infty}^{\infty}e^zf_G(z)dz=\sum_{i=0}^{\infty}\frac{1}{i!}\int_{-\infty}^{\infty}z^if_G(z)dz=\sum_{i=0}^{\infty}\frac{1}{i!}<z^i>$

Since our Gaussian process has 0 mean, all odd moments vanish and we have

$<e^z>=1+\frac{<z^2>}{2!}+\frac{<z^4>}{4!}+...= \sum_{n=0}^{\infty}\frac{<z^{2n}>}{(2n)!}$

but I can't see how to continue from here to obtain the exponential.

Jesús Ros
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1 Answers1

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It helps to know that $\langle z^{k}\rangle =\sigma^{k}(k-1)!!$ for even $k$, where $\sigma^2=\langle z^2\rangle$. This is easily proven by integration by parts and induction. Also, the double factorial for even numbers $p$ is $(p-1)!!=\frac{p!}{2^n(p/2)!}$

Alex R.
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  • Ok. I got to the result using your suggestion. Now I just have to prove it^^ – Jesús Ros Dec 21 '13 at 19:10
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    @gunbl4d3: You will need to relate the $k-2$ moment to the $k$ moment when $k$ is even. You'll need to perform integration by parts on $z^{k-2}e^{-z^2/2\sigma}$. As a hint, take $dv=z^{k-2}$, so $u=...$. – Alex R. Dec 21 '13 at 19:13
  • In your previous comment, there's a missing $\sigma$ in the Normal distribution. I copied yours for the proof and I did not know why I was missing one ;P. Apart from that, I suppose that the same proof applies to the multivariant normal distribution, right? – Jesús Ros Dec 21 '13 at 19:29
  • Whoops, dropped a $\sigma^2$, yep. In the multivariate case, the moments factor when everything is independent and has 0 covariance. Things become rather tricky if you have nonzero covariances. You'll need something akin to Isserlis's theorem: http://en.wikipedia.org/wiki/Isserlis%E2%80%99_theorem

    and even then you'll definitely have many cross-terms to deal with and I doubt something as elegant as the above would hold.

     – Alex R. Dec 21 '13 at 19:53
  • So the result (not this specific proof) $<e^z>=e^{<z^2>/2}$ only holds in the univariant case, and the independent and 0 covariance multivariant case? – Jesús Ros Dec 21 '13 at 20:19