# Variance of Z = max(X,Y) where X Y are jointly bivariate normal

I have a question about the bivariate normal r.v.'s

Given $X, Y \sim \operatorname{Normal}(0,1)$ with correlation coefficient $\rho$. Let $Z=\max(X,Y)$. Show that $\operatorname E Z^2=1$.

My attempt:

I found that $\operatorname E Z=\frac{1}{2}[\operatorname E(X+Y+|X-Y|)]=\frac{1}{2} \operatorname E(|X-Y|)=\sqrt{\frac{1-\rho}{\pi}}$. I found this by obtaining the distribution of $U=X-Y$, which is Normal as well, $U\sim \operatorname{Normal}(0,2(1-\rho))$.

Now, $\operatorname E Z^2=\operatorname{Var} Z+(\operatorname E Z)^2 = \frac{1}{4} \operatorname{Var}(X+Y+|U|) + \frac{1-\rho}{\pi}$.

Then, $\operatorname{Var}(X+Y+|U|)=\operatorname{Var} X+\operatorname{Var} Y + \operatorname{Var}|U| + 2\operatorname{Cov}(X,Y)+2\operatorname{Cov}(X,|U|) + 2\operatorname{Cov}(Y,|U|)$.

$$\operatorname{Var}(X+Y+|U|) = 1 + 1 + \operatorname{Var}|U| + 2\rho + 2\operatorname{Cov}(X,|U|) + 2\operatorname{Cov}(Y,|U|).$$

My question is, is there any way to calculate these three covariances because I couldn't find a way to tackle them without doing messy integration? Any other approaches are highly appreciated.

## 1 Answer

$2Z=V+|U|$ where $V=X+Y$ and $U=X-Y$. Then $(U,V)$ form a bivariate normal distribution with means zero and whose variances you can calculate. But $E(UV)=E(X^2-Y^2)=0$ since $X$ and $Y$ are $N(0,1)$. Therefore $U$ and $V$ are independent normal variables. $$4E(Z^2)=E(V^2)+E(U^2)+2E(V|U|).$$ But $E(V|U|)=E(U)E(|V|)=0$, since $U$ and $|V|$ are independent. Therefore $$4E(Z^2)=E(V^2)+E(U^2)=E((X+Y)^2)+E((X-Y)^2) =2E(X^2)+2E(Y^2)=4.$$

• Ah, I see. One question, you get independence only because they are jointly normal r.v., right? In general, this is not true even when the covariance is 0. – Harry Feb 1 '18 at 5:37
• @Harry That's right, it's essential here that uncorrelated normal rvs are independent. – Lord Shark the Unknown Feb 1 '18 at 5:39
• Sweet, awesome. Thanks a lot. – Harry Feb 1 '18 at 5:39