- #1
Jaggis
- 36
- 0
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?
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?