I've found in a book the following property about gaussian processes (with zero mean) that is not proven and I am having difficulties with the proof.
Let \(\displaystyle z\) be a Gaussian process with zero mean. Then \(\displaystyle <e^z>=e^{<z^2>/2}\)
So far my approach has been
\(\displaystyle <e^z>=\int_{-\infty}^{\infty}e^zf_G(z)dz=\sum_{i=0}^{\infty} \frac{1}{i!}\int_{\infty}^{\infty}z^if_G(z)dz=\sum_{i=0}^{\infty}\frac{1}{i!}<z^i>\)
and since \(\displaystyle z\) is a gaussian process with 0 mean the odd moments vanish and we have
\(\displaystyle <e^z>=1+\frac{<z^2>}{2!}+\frac{<z^4>}{4!}+...=\sum_{n=0}^{\infty}\frac{<z^{2n}>}{(2n)!}\)
Any idea on how to continue from here?
Let \(\displaystyle z\) be a Gaussian process with zero mean. Then \(\displaystyle <e^z>=e^{<z^2>/2}\)
So far my approach has been
\(\displaystyle <e^z>=\int_{-\infty}^{\infty}e^zf_G(z)dz=\sum_{i=0}^{\infty} \frac{1}{i!}\int_{\infty}^{\infty}z^if_G(z)dz=\sum_{i=0}^{\infty}\frac{1}{i!}<z^i>\)
and since \(\displaystyle z\) is a gaussian process with 0 mean the odd moments vanish and we have
\(\displaystyle <e^z>=1+\frac{<z^2>}{2!}+\frac{<z^4>}{4!}+...=\sum_{n=0}^{\infty}\frac{<z^{2n}>}{(2n)!}\)
Any idea on how to continue from here?