Expectation of $\exp(-yx)$ when $x$ and $y$ are jointly normal distributed

Question

Is there an analytical solution for $\operatorname{E}[\exp(-xy)]$ when $(x,y)$ is jointly distributed $N(\mu,\Sigma)$? $\mu$ is the mean vector and $\Sigma$ is the variance-covariance matrix. I know that $\operatorname{E}[\exp(x)]=\exp(\operatorname{E}(x)+\operatorname{Var}(x)/2)$ but don't know how to deal with bivariate distribution.

Answer 1

We assume that $\Sigma = \left[\Sigma_{i,j}\right]_{i,j = 1,2}$ is a $2 \times 2$ matrix and $\mu = 0$ for simplicity. Let $Z, W$ be independent standard normal random variables. Then \begin{equation} (X,Y) \stackrel{\text{d}}{=} \left(\sqrt{\Sigma_{1,1}} Z, \frac{\Sigma_{1,2}}{\sqrt{\Sigma_{1,1}}}Z + \sqrt{\Sigma_{2,2} - \frac{\Sigma_{1,2}^2}{\Sigma_{1,1}}} W\right) \end{equation} which you can check by calculating the variances and covariances. Plugging this in we get \begin{align} \mathbb{E}\left[ e^{-XY} \right] &= \mathbb{E}\left[ e^{- \Sigma_{1,2}Z^2 - \sqrt{\Sigma_{1,1} \Sigma_{2,2} - \Sigma^2_{1,2}} ZW} \right] \\ &= \mathbb{E}\left[ \mathbb{E}\left[\left. e^{- \Sigma_{1,2}Z^2 - \sqrt{\Sigma_{1,1} \Sigma_{2,2} - \Sigma_{1,2}^2} ZW} \right| Z\right] \right]\\ &= \mathbb{E}\left[e^{-(\Sigma_{1,2} - \frac{1}{2}\left(\Sigma_{1,1}\Sigma_{2,2} - \Sigma_{1,2}^2\right))Z^2}\right] \\ &= \left(1 + 2 \Sigma_{1,2} - \Sigma_{1,1}\Sigma_{2,2} + \Sigma_{1,2}^2 \right)^{-\frac{1}{2}} \end{align} provided that the last expression is well-defined, and is equal to $\infty$ otherwise.

Answer 2

$Ee^{-yx}$ need not be finite. If $x$ has standard normal distribution and $y=-x$ then $(x,y)$ has a joint normal distribution but $Ee^{-yx}=Ee^{x^{2}}=\infty$.

When the expectation exists you can use the following procedure to reduce the computation to the case when the variables are independent: There exists $a$ such that $cov(y-ax,x)=0$. This makes $y-ax$ and $x$ independent since they have a joint normal distribution. Now write $Ee^{-yx}$ as $Ee^{-yx}$ as $Ee^{-((y-ax)+ax)x}$. Can you proceed?