distributionsexpected valueextreme valuejoint distributionnormal distribution
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:
$\mathbb{E}\left[\text{max}(X,Y)\right] = \mathbb{E}\left[\frac{X+Y}{2}+\frac{|X-Y|}{2}\right] = \ …$
According to Nadarajah and Kotz, 2008, Exact Distribution of the Max/Min of Two Gaussian Random Variables, the PDF of $X = \max(X_1, X_2)$ appears to be
$$f(x) = 2 \cdot \phi(x) \cdot \Phi\left( \frac{1 - r}{\sqrt{1 - r^2}} x\right),$$
where $\phi$ is the PDF and $\Phi$ is the CDF of the standard normal distribution.
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I don't think the formula given in the question can be correct in all cases, it is developed using joint normality. Without joint normality we can use copulas. For $X,Y$ random variables with joint distribution with cumulative distribution function $F(x,y)$ and joint density $f(x,y)$ define the transformed random variables (rv) $U=F_X(X), V=F_Y(Y)$ where $F_X, F_Y$ denotes the marginal cumulative distribution functions (cdf). Then the joint distribution of $U;V$ $$ \DeclareMathOperator{\P}{\mathbb{P}} \P(U \le u, V \le v)=C(u,v) $$ is the copula of $X$ and $Y$, with copula density $c(u,v)$ (when it exists). So, in your setup let us assume that the copula density exists, and for simplicity I will take both $X$ and $Y$ as standard normals. So what possibility exists for the conditional expectation of $X$ given $Y=y$ ? Using Sklar's theorem we can write the joint density as $$ f(x,y) = c(\Phi(x),\Phi(y)) \phi(x) \phi(y) $$ where $\Phi, \phi$ are the standard normal cdf, pdf, respectively. Then the conditional density is given by $$ f(x \mid y) = \frac{f(x,y)}{f(y)}=\frac{c(\Phi(x),\Phi(y))\phi(x)\phi(y)}{\phi(y)}= c(\Phi(x),\Phi(y))\phi(x) $$ Then we can look at this with various copula functions, see https://en.wikipedia.org/wiki/Copula_(probability_theory) .
There is a general inequality for copulas $$ W(u,v)=\max(u+v-1,0) \le C(u,v) \le M(u,w)=\min(u,v) $$ where both upper and lower limits are copulas (This is the Frechet-Hoeffding bounds). The upper limit isn't very interesting, since it gives $\P(U=V)=1$ so gives correlation equal 1. The lower limit similarly corresponds to $U=1-V$ with probability one, so correlation is -1. But for these two extremal copula the conditional expectation function certainly is linear!
Lets look at some intermediate cases. I will use the R package copula and some numerical integration to find the conditional expectation function, for the case of the gumbel copula. The code can be simply adapted for other copulas.
library(copula)
C <- gumbelCopula(2)
make_cond <- Vectorize(function(y) {
function(x) dCopula( cbind(pnorm(x), pnorm(y)), C)*dnorm(x)
} )
the last command makes a function representing the conditional density given $Y=y$. Let us look at how this looks like for three different values of $y$:
cond_dens <- make_cond(c(-1, 0, 1))
plot(cond_dens[[1]], from=-3, to=3, col="blue", ylim=c(0, 0.7))
plot(cond_dens[[2]], from=-3, to=3, col="orange", add=TRUE)
plot(cond_dens[[3]], from=-3, to=3, col="red", add=TRUE)
title("conditional densities for y=-1, 0, 1")
showing clearly that the conditional distributions now are non-normal. We can also see clearly that the conditional variance is non-constant.
For more examples using the copula package see https://www.r-bloggers.com/modelling-dependence-with-copulas-in-r/ and Generating values from copula using copula package in R . Then we can find the conditional expectation function using numerical integration:
plot(function(y) sapply(make_cond(y), FUN=function(fun)
integrate(function(x) x*fun(x) ,
lower=-Inf, upper=Inf)$value),
from=-3, to=3,
ylab="conditional expectation given y", xlab="y")
title("conditional expectation of X given Y=y")
and it is quite clear that the conditional expectation function is not linear!
Best Answer
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).