Variance of random variables involving two independent standard Normals Let $X$ and $Y$ be two independent standard Normal variables. Let $M := \max(X, Y)$ and $L := \min(X, Y)$. It is given that the covariance between $M$ and $L$ is given by $\text{Cov}(M, L) = 1 / \pi$ How do I find the variance of $M$ and $L$?
We first note that $M - L = |X - Y|$ and $M + L = X + Y$. Since $X$ and $Y$ are independent we have $X + Y \sim \mathcal{N}(0, 2)$. Therefore $\text{Var}(X + Y) = 2 = \text{Var}(M) + \text{Var}(L) + 2 \text{Cov}(M, L)$. Hence $\text{Var}(M) + \text{Var}(L) + \frac{2}{\pi} = 2$.
I could not find another equation involving $\text{Var}(M)$ and $\text{Var}(L)$ using $M - L = |X - Y|$, as I could not find the variance of $|X - Y|$. How do I continue from this point? Also,are there any other approaches which use the symmetry property of the standard normals?
 A: As edited from the earlier version of the question, we have that
$$\mathbb{E}[M]=\underbrace{\mathbb{E}[X]}_{=0}+\underbrace{\mathbb{E}[(Y-X)\mathbb{I}_{Y-X\ge 0}]}_{Y-X\sim{\cal N}(0,2)}=\frac{1}{\sqrt\pi}$$
implies
$$\text{cov}(M,L)=\mathbb{E}[ML]-\mathbb{E}[M]\mathbb{E}[L]=\underbrace{\mathbb{E}[XY]}_{=0}+\underbrace{\mathbb{E}[M]^2}_{L\sim-M}=\frac{1}{\pi}$$
where $L\sim -M$ means that $L$ and $-M$ have the same (marginal) distribution since
$$M\sim\max(X,Y)\sim\max(-X,-Y)=-\min(X,Y)\sim -L$$
Now, the derivation proposed in the second paragraph is that
\begin{align}
\text{var}(M+L)&=\text{var}(X+Y)\\&=\underbrace{\text{var}(M)+\text{var}(L)}_{\text{identical: }L\sim -M}+2\text{cov}(M,L)\\&=2\left\{\text{var}(M)+\text{cov}(M,L)\right\}\\
&=2\left\{\text{var}(M)+\frac{1}{\pi}\right\}=2\\
\end{align}
Therefore
$$\text{var}(M)=\dfrac{\pi-1}{\pi}$$
which somewhat surprisingly implies$$\mathbb{E}[M^2]=1=\mathbb{E}[X^2]$$
Except that$$2\mathbb{E}[M^2]=\mathbb{E}[M^2+L^2]=\mathbb{E}[X^2+Y^2]=2\mathbb{E}[X^2]=2$$explains the identity.
A: I don't think one needs to know what $\mathrm{Cov}(M,L)$ is to find the variances.
As $X$ and $Y$ are independent, $X-Y\sim\mathcal{N}(0,2)$, so that $E|X-Y|=\sqrt\frac{2}{\pi}\cdot \sqrt{2}=\frac{2}{\sqrt{\pi}}$.
So we have $E(M)=E(\frac{1}{2}(X+Y+|X-Y|))=\frac{1}{2}E(|X-Y|)=\frac{1}{\sqrt{\pi}}$.
Again, $2M=X+Y+|X-Y|=U+|V|$, say, where $U=X+Y,V=X-Y$
Then $E(UV)=E(X^2-Y^2)=1-1=0$, which implies $\mathrm{Cov}(U,V)=0$.
This in turn means that $U$ and $V$ are independent as $(U,V)$ is jointly normal. 
So, $4E(M^2)=E(U+|V|)^2=E(U^2)+E(V^2)+2E(U)E(|V|)$
$\qquad\qquad\quad=E(X+Y)^2+E(X-Y)^2=4$
Finally we have $E(M^2)=1$ which means $\mathrm{Var}(M)=1-\frac{1}{\pi}$.
Try doing something similar to find $\mathrm{Var}(L)$.
