Solving Gaussian Random Variable Expected Value: CDF & Expectation

[Jaggis]

Hi,

I have trouble with the following problem:

Gaussian random variable is defined as follows

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

[tex]E(exp(G^2\lambda/2)).[/tex]

Hint:

Because [itex]\phi[/itex] is a cumulative distribution function, [itex]\phi(+\infty) = 1[/itex].

My attempt at solution:

I start with:

[tex] 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[/tex]
[tex]=1/\sqrt{2\pi} \int^{\infty}_{-\infty}(\int^{\sqrt{2/\lambda*lnt})}_{\sqrt{2/\lambda*lnt})}e^{-x^2/2}dx)dt.[/tex]

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 [itex]\phi(+\infty) = 1[/itex].

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

Any help?

 

[mathman]

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

As you can see [itex]\lambda < 1[/itex] is necessary.

 

[zinq]

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