# Function of a random gaussian variable

1. Apr 25, 2004

### Jply

I'm having trouble showing the following relation:

E(exp(z)) = exp(E(z^2)/2)

where z is a zero-mean gaussian variable and E() is the avg

anyone can help?

2. Apr 26, 2004

### HallsofIvy

Staff Emeritus
Wasn't this posted just recently?

For the standard Gaussian Normal distribution,
$$E(f(z))= \int_{-\inf}^{\inf}{f(z)e^{-\frac{z^2}{2}}}dz$$

In this case,
$$f(x)= e^{\frac{z^2}{2}}$$
so the integral becomes
$$E(e^{\frac{z^2}{2}} )= \int_{-\inf}^{\inf}{e^{\frac{z^2}{2}} e^{-\frac{z^2}{2}}}dz$$

can you do that?