6
$\begingroup$

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.

$\endgroup$
5
$\begingroup$

$2Z=V+|U|$ 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 $$4E(Z^2)=E(V^2)+E(U^2)=E((X+Y)^2)+E((X-Y)^2) =2E(X^2)+2E(Y^2)=4.$$

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – Harry Feb 1 '18 at 5:37
  • 2
    $\begingroup$ @Harry That's right, it's essential here that uncorrelated normal rvs are independent. $\endgroup$ – Lord Shark the Unknown Feb 1 '18 at 5:39
  • $\begingroup$ Sweet, awesome. Thanks a lot. $\endgroup$ – Harry Feb 1 '18 at 5:39

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.