What is cov(X,Y), where X=min(U,V) and Y=max(U,V) for independent Normal(0,1) variables U and V? Let:


*

*$U, V \overset{i.i.d.}{\sim} \mathcal{N}(0,1)$, i.e. independent standard Normal random variables.

*$X=\min(U,V)$

*$Y=\max(U,V)$


What is the covariance of $X$ and $Y$? 

Related: What is cov(X,Y), where X=min(U,V) and Y=max(U,V) for independent uniform(0,1) variables U and V?
 A: $\newcommand{\E}{\mathrm{E}}$ $\newcommand{\Var}{\mathrm{Var}}$ $\newcommand{\cov}{\mathrm{Cov}}$ $\newcommand{\Expect}{{\rm I\kern-.3em E}}$
As a direct consequence of the definition of covariance, $\cov (X,Y)= \E(XY)-\E(X)\E(Y)$.
Fact 1:
$U, V \overset{i.i.d.}{\sim} \mathcal{N}(0,1)$
$\Rightarrow  U - V \sim \mathcal{N}(0,2)$  (sum of normally distributed random variables)
$ \Rightarrow |U - V|$ is a half-normal random variable with parameter $\sigma = \sqrt2$
$ \Rightarrow \E (|U - V|) = \frac{\sigma\sqrt{2}}{\sqrt{\pi}} = \frac{\sqrt{2}\sqrt{2}}{\sqrt{\pi}} = \frac{2}{\sqrt{\pi}}$
Fact 2:
$\E(X)+\E(Y) = \E(X+Y)$ (linearity of the expectation).
We have $\E(X+Y) = \E (\min(U,V)+\max(U,V))= \E(U+V) = \E(U)+\E (V) = 0 + 0 = 0$. As a result,  $\E(Y) = -\E(X)$.
Fact 3:
Since $Y-X = |U - V|$:
$2\E(Y) = \E(Y)-\E(X) = \E(Y-X) = \E (|U - V|)= \frac{2}{\sqrt{\pi}}$, hence $\E(Y)= \frac{2}{2\sqrt{\pi}}= \frac{1}{\sqrt{\pi}}$
Fact 4:
Since $XY=UV$, we have $\E(XY)=\E(UV)=\E (U)\E (V)=0$
Using these facts: $\cov (X,Y)= \E(XY)-\E(X)\E(Y)= 0 + \E(Y)\E(Y) = \frac{1}{\sqrt{\pi}}\frac{1}{\sqrt{\pi}}=\frac{1}{\pi}$.
