Expectation of the product of elements of a Gaussian Vector

Question

Suppose that $X \in \mathbb{R}^2,$ and $(X_1, X_2) \sim \mathcal{N}(0, \Sigma),$ for some valid covariance matrix $\Sigma.$ I want to compute $$\mathbb{E} \left[ X_1^+ X_2^+ \right],$$ where $y^+ = \max(0, y)$ is the positive part function Is there a good way to compute this?

Answer 1

Let $\sigma_1^2=\text{Var}(X_1)$, $\sigma_2^2=\text{Var}(X_2)$, and $\text{Cov}(X_1,X_2)=\rho\sigma_1\sigma_2$.

We can assume WLOG that $\sigma_1=\sigma_2=1$, otherwise we can consider $(Y_1,Y_2) = (X_1/\sigma_1, X_2/\sigma_2)$, and then use $\mathbb{E} \left[ Y_1^+ Y_2^+ \right]=\sigma_1\sigma_2\mathbb{E} \left[ X_1^+ X_2^+ \right]$.

Now let $$(Z_1, Z_2) = \left(X_1, \frac{X_2-\rho X_1}{\sqrt{1-\rho^2}}\right).$$

$Z_1$ and $Z_2$ are independent standard normal, since they are jointly Gaussian and $\text{Var} Z_1=\text{Var} Z_2 = 1$, and $\text{Cov}(Z_1,Z_2)=0$.

Letting $a=-\frac{\rho}{\sqrt{1-\rho^2}}$, we can write the expectation as \begin{align*} \mathbb{E} \left[ X_1^+ X_2^+ \right]&= \mathbb{E} \left[ Z_1^+ \left(\sqrt{1-\rho^2}Z_2+\rho Z_1\right)^+ \right]&\\ &= \frac{1}{2\pi}\int_0^\infty\int_{cz_1}^\infty\left(\rho z_1^2+\sqrt{1-\rho^2}z_1z_2\right) e^{-\frac{1}{2}z_1^2-\frac{1}{2}z_2^2}dz_2dz_1\\ &=\frac{1}{2\pi}\int_{\arctan(a)}^{\frac{\pi}{2}} \int_0^\infty \left(\rho \cos^2\theta + \sqrt{1-\rho^2}\cos\theta\sin\theta\right)r^3e^{-\frac{1}{2}r^2}drd\theta\\ &=\frac{1}{\pi}\int_{\arctan(a)}^{\frac{\pi}{2}} \left(\rho \cos^2\theta +\sqrt{1-\rho^2}\cos\theta\sin\theta\right)d\theta\\ &=\frac{\rho}{\pi}\left(\frac{\theta +\sin\theta\cos\theta}{2}\right)\Big|_{\arctan(a)}^{\frac{\pi}{2}} +\frac{\sqrt{1-\rho^2}}{2\pi}\left(\sin^2\theta\right)|_{\arctan(a)}^{\frac{\pi}{2}}\\ &= \frac{\rho}{4}-\frac{\rho\tan^{-1}\left(-\frac{\rho}{\sqrt{1-\rho^2}}\right)}{2\pi}+\frac{\sqrt{1-\rho^2}}{2\pi}. \end{align*}

We conclude that $$\mathbb{E} \left[ X_1^+ X_2^+ \right]=\frac{\rho\sigma_1\sigma_2}{4}-\frac{\rho\sigma_1\sigma_2\tan^{-1}\left(-\frac{\rho}{\sqrt{1-\rho^2}}\right)}{2\pi}+\frac{\sigma_1\sigma_2\sqrt{1-\rho^2}}{2\pi}.$$