How to find a closed-form expression for $E[\max(X,Y)]$?

Question

Let $X$ and $Y$ be two random variables, with

$X\sim N(a,b)$ and $Y\sim N(c,d)$. Furthermore, $X$ and $Y$ are correlated with a correlation coefficient equal to $p$.

How do I find a closed-form expression for $E[\max(X,Y)]$?

Whichever way I look at it, I am not able to get rid of the randomness in the solution.

My solution.

$E[\max(X,Y)] = XP(X>Y) + YP(Y>X)$

But if I do this I will get an answer in terms of $X$ and $Y$ which is still a random quantity.

Answer 1

Take the joint density $$\exp\left(-z/2(1-p^2)/(2\pi bd \sqrt{(1-p^2})\right)$$ where

$z=(x-a)^2/b^2 -2p(x-a)(y-b)/(bd) + (y-c)^2/d^2$.

Multiply it by $x$ over the region where $x>y$ and by $y$ over the region where $y>x$ and compute the double integral over those regions. This will not have a closed form but can be integrated numerically.

Answer 2

First, technically speaking, given only $X \sim N(\mu_1, \sigma_1^2), Y \sim N(\mu_2, \sigma_2^2)$ and $\operatorname{Corr}(X, Y) = \rho$ does not imply the joint distribution of $(X, Y)$ is bivariate normal $N_2(\mu_1, \mu_2, \sigma_1^2, \sigma_2^2, \rho)$. Check this answer for many counterexamples. Therefore, the accepted answer actually imposed additional conditions that were not originally posted in the question.

Secondly, if we do have $(X, Y) \sim N_2(\mu_1, \mu_2, \sigma_1^2, \sigma_2^2, \rho)$, then $E[\max(X, Y)]$ indeed admits a closed-form expression, which is deduced below.

Write $E[\max(X, Y)] = \frac{1}{2}E[(X + Y) + |X - Y|]$. Under the joint normality condition, we have \begin{align} & X + Y \sim N(\mu_1 + \mu_2, \sigma_1^2 + \sigma_2^2 + 2\rho\sigma_1\sigma_2), \\ & X - Y \sim N(\mu_1 - \mu_2, \sigma_1^2 + \sigma_2^2 - 2\rho\sigma_1\sigma_2). \end{align}
It thus suffices to compute $E[Z]$ and $E[|Z|]$ for a univariate r.v. $Z \sim N(\mu, \sigma^2)$. Clearly, $E[Z] = \mu$, while $E[|Z|] = \sigma E[|Z'|]$ with $Z' \sim N(\mu\sigma^{-1}, 1)$, and \begin{align} & E[|Z'|] = \int_{-\infty}^\infty |x|\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}(x - \mu\sigma^{-1})^2}dx \\ =& \frac{1}{\sqrt{2\pi}}\int_0^\infty xe^{-\frac{1}{2}(x - \mu\sigma^{-1})^2}dx - \frac{1}{\sqrt{2\pi}}\int_{-\infty}^0 xe^{-\frac{1}{2}(x - \mu\sigma^{-1})^2}dx \\ =& \left[\frac{1}{\sqrt{2\pi}}e^{-\mu^2\sigma^{-2}/2} + \mu\sigma^{-1}\Phi(\mu\sigma^{-1})\right] - \left[-\frac{1}{\sqrt{2\pi}}e^{-\mu^2\sigma^{-2}/2} + \mu\sigma^{-1}\Phi(-\mu\sigma^{-1})\right] \\ =& \frac{2}{\sqrt{2\pi}}e^{-\mu^2\sigma^{-2}/2} + \mu\sigma^{-1}(2\Phi(\mu\sigma^{-1}) - 1). \end{align} Denote $\sqrt{\sigma_1^2 + \sigma_2^2 - 2\rho\sigma_1\sigma_2}$ by $\sigma_0$, then \begin{align*} E[|X - Y|] = \sigma_0 \left(\frac{2}{\sqrt{2\pi}}e^{-(\mu_1 - \mu_2)^2\sigma_0^{-2}/2} + (\mu_1 - \mu_2)\sigma_0^{-1}(2\Phi((\mu_1 - \mu_2)\sigma_0^{-1}) - 1)\right) \end{align*} and \begin{align} & E[\max(X, Y)] \\ =& \frac{1}{2}E[X + Y] + \frac{1}{2}E[|X - Y|] \\ =& \frac{\mu_1 + \mu_2}{2} + \frac{\sigma_0}{\sqrt{2\pi}}e^{-(\mu_1 - \mu_2)^2\sigma_0^{-2}/2} + \frac{1}{2}(\mu_1 - \mu_2)(2\Phi((\mu_1 - \mu_2)\sigma_0^{-1}) - 1) \\ =& \mu_2 + \frac{\sigma_0}{\sqrt{2\pi}}e^{-\frac{(\mu_1 - \mu_2)^2}{2\sigma_0^2}} + (\mu_1 - \mu_2)\Phi\left(\frac{\mu_1 - \mu_2}{\sigma_0}\right). \end{align}