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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.2011-10-24
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    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.2011-10-24
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    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 $Z \sim \mathcal N\left(\rho a + \sqrt{1-\rho^2}\ a,\ \ \rho^2\sigma_1^2 + (1-\rho^2)\sigma_2^2\right)$.2011-10-24
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    ...and you need to think about the covariance, and hence the correlation, between $Y$ and $Z$.2011-10-24
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    I mean $\rho$ to be not the correlation, but just some constant2011-10-24

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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) $