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Suppose we have $X,Y$ independent normally distributed r.v. $X \sim \mathcal N(a,\sigma^2_1)$, $Y \sim \mathcal N(a,\sigma^2_2)$, and $Z=\rho X+\sqrt{1-\rho^2}Y$ where $\rho$ is some constant.

How can I calculate the $\mathbb{E}[\max(0,e^Z-e^Y)]$?

Thanks.

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Is there a mistake? If $X,Y$ are independent then their correlation $\rho$ is 0, so $Z=Y$ and this is trivial. –  Nate Eldredge Oct 24 '11 at 18:07
    
I mean $\rho$ to be not the correlation, but just some constant –  user18174 Oct 24 '11 at 20:41
    
If you say two random variables are normally distributed and specify their means and variances, then you've said what their separate distributions are, but if you add to that their correlation, then that falls short of specifying their joint distribution. But then if you also say they're jointly normally distributed, then that fully specifies their distribution. –  Michael Hardy Oct 24 '11 at 21:59
    
If $X \sim \mathcal N(a,\sigma^2_1)$, $Y \sim \mathcal N(a,\sigma^2_2)$, and $Z=\rho X+\sqrt{1-\rho^2}Y$ and $X$ and $Y$ are independent then $Z \sim \mathcal N\left(\rho a + \sqrt{1-\rho^2}\ a,\ \ \rho^2\sigma_1^2 + (1-\rho^2)\sigma_2^2\right)$. –  Michael Hardy Oct 24 '11 at 22:02
    
...and you need to think about the covariance, and hence the correlation, between $Y$ and $Z$. –  Michael Hardy Oct 24 '11 at 22:03

1 Answer 1

Hint: $\mathbb E[ \max(0, \mathrm e^Z - \mathrm e^Y) ] = \mathbb E \left[ \mathrm e^Z - \mathrm e^Y | Z>Y \right] \mathrm P(Z>Y) $

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