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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?

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    $\begingroup$ Note that $$ \Pr(M\leq m)=\Pr(X\leq m,Y\leq m) $$ $\endgroup$ – Chamberlain Foncha May 1 '18 at 7:42
  • $\begingroup$ As $X-Y\sim\mathcal N(0,2)$, $|X-Y|$ has a half-normal density. It's mean and variance is easily found. $\endgroup$ – StubbornAtom May 1 '18 at 10:23
  • $\begingroup$ Since $M+L=X+Y$, the variance of $M+L$ is $2.$ By symmetry, $M$ and $L$ have the same variance (after all, the distribution one is the distribution of the negative of the other) and it is now easily found by solving the equation $\operatorname{Var}(M+L)=\operatorname{Var}(M)+\operatorname{Var}(L)+2\operatorname{Cov}(M,L).$ $\endgroup$ – whuber May 1 '18 at 17:44
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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.

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    $\begingroup$ I think the part of this proof that is missing is that $var(M) = var(L)$. You may want to elaborate that point. $\endgroup$ – Greenparker May 1 '18 at 8:57
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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 $\mathrm{Var}(L)$.

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