# Exponential of Gaussian Distribution

1. May 20, 2013

### SeriousNoob

I'm looking for the expected value of an exponential Gaussian

$Y=\text{exp}(jX) \text{ where } X\text{~}N(\mu,\sigma^2)$

From wolframalpha, http://www.wolframalpha.com/input/?i=expected+value+of+exp%28j*x%29+where+x+is+gaussian

$E[Y]=\text{exp}(j^2\sigma^2/2+j\mu)$

If I were to use the expected value definition:
$E[Y]=\int_{-\infty}^\infty uf_Y(u)du$
then I would have to figure out the pdf of Y.

I'm having trouble remembering how to get the pdf of Y, is there a more explicit way to derive the expected value?

2. May 20, 2013

### mathman

You can do it directly by evaluating ∫exp(ix)f(x)dx where f(x) is the normal density function. Also note that this is simply φ(1) where φ(t) is the characteristic function of this particular normal distribution.

3. May 24, 2013

### NegativeDept

What mathman said is right - I just wanted to add a couple of things:

1) The Law of the Unconscious Statistician is a snarky name for the usual way to find the EV of a function of a random variable.

$E[y(X)] = \int y(x) f_X(x) dx$
$E[e^{jX}] = \frac{1}{\sigma\sqrt{2\pi}} \int e^{jx} e^{-\frac{(x-\mu)^2}{2 \sigma^2}} dx$

2) Exponentials of Gaussians show up often enough to have their own name. If $Y = \exp(X)$, then $Y$ has a lognormal distribution. The name 'lognormal' is a reminder that $\log(Y)$ is normally distributed.