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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?

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$\newcommand{\E}{\mathrm{E}}$ $\newcommand{\Var}{\mathrm{Var}}$ $\newcommand{\cov}{\mathrm{Cov}}$ $\newcommand{\Expect}{{\rm I\kern-.3em E}}$

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: $\cov (X,Y)= \E(XY)-\E(X)\E(Y)= 0 + \E(Y)\E(Y) = \frac{1}{\sqrt{\pi}}\frac{1}{\sqrt{\pi}}=\frac{1}{\pi}$.

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  • $\begingroup$ Should have a negative sign for $cov (X,Y) $ - you've used $E(Y)^2$ instead of $E (X)E (Y) $ in fact 4. Good answer though. $\endgroup$ – probabilityislogic Jan 9 '17 at 0:51
  • $\begingroup$ E(X)E(Y) would be the same as E^2(Y). The problem is that the sign changed from - to +. It is a shame to give such a nice proof and make one careless mistake at the very end. I overlooked it at first glance. $\endgroup$ – Michael R. Chernick Jan 9 '17 at 1:16
  • $\begingroup$ The sign changed as $\E(X) = -\E(Y)$ (fact 2). I posted the answer as I thought readers from What is cov(X,Y), where X=min(U,V) and Y=max(U,V) for independent uniform(0,1) variables U and V? could be interested. $\endgroup$ – Franck Dernoncourt Jan 9 '17 at 1:21
  • $\begingroup$ Sorry Franck somehow I got confused. it looks like you had it right and we had it wrong. $\endgroup$ – Michael R. Chernick Jan 9 '17 at 2:16

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