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Suppose I have the bivariate normal variable $[X \ \ Y]'$ which has mean $[\mu_X \ \ \mu_Y]'$ and covariance matrix $ \left[ \begin{array}{cc} \sigma_X ^2 & \sigma_Y \ \sigma_X \ \\ \sigma_Y \ \sigma_X \ & \sigma_Y^2 \end{array} \right]$.

I am trying to figure out what is $\text{Cov}(e^X, e^Y)$.

I'm aware of Stein's lemma which says $\text{Cov}(g(X), Y) = \text{Cov}(X,Y)E[g'(X)]$, but I don't know how it's derived, or how to extend it to functions on both $X$ and $Y$. Presumably this case should be simpler since it's the exponential function.

Otherwise I'm unsure what other steps to take. I don't need to prove this, but the answer is necessary for a calculation I am doing.

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    – Clement C.
    Commented Mar 5, 2021 at 3:23
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    $\begingroup$ Thanks, I appreciate the tips! I've updated the question with more information, I hope this improves it. $\endgroup$
    – linkspan
    Commented Mar 5, 2021 at 3:30

1 Answer 1

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$$\operatorname{Cov}[e^X, e^Y] = \operatorname{E}[e^{X}e^{Y}] - \operatorname{E}[e^X]\operatorname{E}[e^Y] = \operatorname{E}[e^{X+Y}] - M_X(1) M_Y(1),$$ where $M_X, M_Y$ are the moment-generating functions of the marginal distributions of $X$ and $Y$. The handling of $\operatorname{E}[e^{X+Y}]$ requires an additional step, which is to compute the mean and variance of $X+Y$, since their sum is normal whenever $(X,Y)$ are bivariate normal. Specifically, we obviously have $$\mu = \operatorname{E}[X+Y] = \mu_X + \mu_Y$$ and $$\sigma^2 = \operatorname{Var}[X+Y] = \sigma_X^2 + 2\sigma_{XY} + \sigma_Y^2.$$ This results in a third MGF to compute. These are all straightforward calculations if one is familiar with bivariate normal distributions.

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