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.
Best Answer
$$\operatorname{Cov}[e^X, e^Y] = \operatorname{E}[e^{X}e^{Y}] - \operatorname{E}[e^X]\operatorname{E}[e^Y] = \operatorname{E}[e^{X+Y}] - M_X(1) M_Y(1),$$ 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.