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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:

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

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    $\begingroup$ 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? $\endgroup$ – whuber Apr 7 '16 at 20:58
  • $\begingroup$ What i had in mind was the general problem.. $\endgroup$ – ambushed Apr 7 '16 at 21:05
  • $\begingroup$ It looks like this question has been asked before. Appologies. $\endgroup$ – ambushed Apr 7 '16 at 21:31
  • $\begingroup$ 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. $\endgroup$ – whuber Apr 7 '16 at 21:42
  • $\begingroup$ Indeed, you are right! $\endgroup$ – ambushed Apr 7 '16 at 21:50
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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).

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About the distribution |X-Y|, the Folded normal distribution may be helpful.

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  • $\begingroup$ This doesn't seem to answer the question $\endgroup$ – kjetil b halvorsen Nov 17 '18 at 6:37

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