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:
$\mathbb{E}\left[\text{max}(X,Y)\right] = \mathbb{E}\left[\frac{X+Y}{2}+\frac{|X-Y|}{2}\right] = \ ...$
 A: The solution is contained in the paper https://www.gwern.net/docs/conscientiousness/2008-nadarajah.pdf   cited by @Lucas in his answer at Distribution of the maximum of two correlated normal variables
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).  
A: About the distribution |X-Y|, the Folded normal distribution may be helpful.
