3
$\begingroup$

Let $X=\min(U,V)$ and $Y=\max(U,V)$ for independent uniform(0,1) variables $U$ and $V$. What's the covariance of $X$ and $Y$? Could you develop some calculations, especially regarding the computation of $\mathbb{E}XY$?

$\endgroup$
10
$\begingroup$

Obviously $XY=UV$ therefore the calculation of $\mathbb{E} XY$ is really easy !

$\endgroup$
  • 2
    $\begingroup$ +1. $\mathbb{E} X$ and $\mathbb{E} Y$ are harder $\endgroup$ – Henry Apr 6 '12 at 9:56
  • 2
    $\begingroup$ +1. @Henry But since $E[U+V]=E[X+Y]=1$ so that $E[U]=1-E[V]$, only one "hard" calculation is necessary. $\endgroup$ – Dilip Sarwate Apr 6 '12 at 12:50
3
$\begingroup$

I see that we still need to find E[X] and E[Y].

From Franck's Question and Answer, we see that

E[X] + E[Y] = E[U] + E[V] = 1

and

E[Y - X] = E[Y] - E[X] = E[ |U - V| ]

So, we calculate E[ | U - V | ]

$$\int_{0}^{1} \int_{0}^{1} |u - v| dvdu = 2\int_{0}^{1} \int_{0}^{u} (u - v) dvdu = \int_{0}^{1} u^2 du = \frac{1}{3} $$

So, E[Y] - E[X] = 1/3 and E[Y] + E[X] = 1. There are two equations and two unknowns so we solve and get

$$E[Y] = \frac{2}{3} \quad and \quad E[X] = \frac{1}{3}$$

Finally,

$$cov[X, Y] = E[XY] - E[X]E[Y] = \frac{1}{4} - \frac{1}{3}*\frac{2}{3} = \frac{1}{36}$$

$\endgroup$

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.