# CDF and expectation

1. Oct 31, 2015

### Jaggis

Hi,

I have trouble with the following problem:

Gaussian random variable is defined as follows

$$\phi(t) = P(G \leq t)= 1/\sqrt{2\pi} \int^{t}_{-\infty} exp(-x^2/2)dx.$$
Calculate the expected value

$$E(exp(G^2\lambda/2)).$$

Hint:

Because $\phi$ is a cumulative distribution function, $\phi(+\infty) = 1$.

My attempt at solution:

$$E(exp(G^2\lambda/2)) = \int^{\infty}_{-\infty}P(exp(G^2\lambda/2) \geq t)dt = \int^{\infty}_{-\infty}P(-\sqrt{2/\lambda*lnt}) \geq G \geq \sqrt{2/\lambda*lnt})dt$$
$$=1/\sqrt{2\pi} \int^{\infty}_{-\infty}(\int^{\sqrt{2/\lambda*lnt})}_{\sqrt{2/\lambda*lnt})}e^{-x^2/2}dx)dt.$$

Then my instinct would be to use Fubini theorem because I'd like to get rid of the integral of exp(-x^2/2) by $\phi(+\infty) = 1$.

However, because both bounds are functions of t, it wouldn't work.

Any help?

2. Oct 31, 2015

### mathman

General formula: Let h(G) be a function of G. Let f(x) be the probability density function for G. Then:
$E(h(G))=\int_{-\infty}^{\infty}h(x)f(x)dx$.
In your case $h(G)=exp(G^2 \lambda /2)$ and $f(x)=\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}$.

As you can see $\lambda < 1$ is necessary.

3. Nov 2, 2015

### zinq

Jaggis, I suggest you double-check your first equals sign.