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:

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

  • 4
    $\begingroup$ 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

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).


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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.