7
$\begingroup$

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?

$\endgroup$

1 Answer 1

10
$\begingroup$

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

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

$\endgroup$
4
  • $\begingroup$ 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. $\endgroup$ Jan 9, 2017 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. $\endgroup$ Jan 9, 2017 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. $\endgroup$ Jan 9, 2017 at 1:21
  • $\begingroup$ Sorry Franck somehow I got confused. it looks like you had it right and we had it wrong. $\endgroup$ Jan 9, 2017 at 2:16

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.