# Expectation of the maximum of two correlated normal variables

I am curious what the derivation for the expectation of the maximum of two jointly normal random variables $X$ and $Y$ with correlation coefficient $\rho$.

I could start with the following but the absolute value sign under expectation doesn't look like a walk in the park:

$\mathbb{E}\left[\text{max}(X,Y)\right] = \mathbb{E}\left[\frac{X+Y}{2}+\frac{|X-Y|}{2}\right] = \ ...$

• The distribution of $\max(X,Y)$ (for $X$ and $Y$ with equal variances and equal means) is given at stats.stackexchange.com/questions/139072. From that you can compute the expectation (it looks like numerical methods have to be used). The general problem (for arbitrary variances and means) looks difficult: do you need a solution in that case?
– whuber
Commented Apr 7, 2016 at 20:58
• What i had in mind was the general problem.. Commented Apr 7, 2016 at 21:05
• It looks like this question has been asked before. Appologies. Commented Apr 7, 2016 at 21:31
• I'm not sure of that: I could not find an exact duplicate. The link in my comment was found after conducting three or four keyword searches of this site and inspecting several likely threads; it's the closest I could come. The link you found concerns the maximum of independent normal variables and its answer relies fundamentally on that assumption.
– whuber
Commented Apr 7, 2016 at 21:42
• About the distribution |X-Y|, the Folded normal distribution may be helpful.
– MyNT
Commented Nov 17, 2018 at 3:50

I will give the answer here, maybe I come back to add a proof ... Let $$(X_1,X_2)$$ be a bivariate random vector with a binormal distribution, with means $$\mu_1, \mu_2$$, standard deviations $$\sigma_1, \sigma_2$$ and correlation coefficient $$\rho$$. Then $$X=\max(X_1, X_2)$$ has probability density function $$f(x) = f_1(-x)+f_2(-x)$$ where $$f_1(x)= \frac1{\sigma_1}\phi(\frac{x+\mu_1}{\sigma_1})\cdot \Phi\left( \frac{\rho(x+\mu_1)}{\sigma_1\sqrt{1-\rho^2}}-\frac{x+\mu_2}{\sigma_2\sqrt{1-\rho^2}} \right) \\ f_2(x)= \frac1{\sigma_2}\phi(\frac{x+\mu_2}{\sigma_2})\cdot \Phi\left( \frac{\rho(x+\mu_2)}{\sigma_2\sqrt{1-\rho^2}}-\frac{x+\mu_1}{\sigma_1\sqrt{1-\rho^2}} \right)$$ where $$\phi, \Phi$$ are the density and cumulative distribution function of the standard normal.
This paper also gives an exact expression for the expectation: $$\DeclareMathOperator{\E}{\mathbb{E}} \E X = \mu_1 \Phi\left( \frac{\mu_1-\mu_2}{\theta} \right) + \mu_2 \Phi\left( \frac{\mu_2-\mu_1}{\theta} \right) + \theta \phi\left( \frac{\mu_1-\mu_2}{\theta} \right)$$ where $$\theta = \sqrt{\sigma_1^2 +\sigma_2^2 - 2\rho\sigma_1\sigma_2}$$. (the paper contains more, like the variance and moment generating functions).