Covariance of two uniform order statistics
= Covariance of two uniform order statistics
{title2=$\operatorname{Cov}(U,V)=1/36$}
For two <independent random variables> with the <uniform distribution> on $[0,1]$, put $U=\min(X,Y)$ and $V=\max(X,Y)$. Then $P(U>u)=(1-u)^2$, so $\mathbb EU=1/3$. Since $U+V=X+Y$ and $UV=XY$, $\mathbb EV=2/3$ and $\mathbb E(UV)=1/4$. Thus their <covariance> is $1/36$, although the original variables are independent. Rescaling the interval to $[a,b]$ multiplies the covariance by $(b-a)^2$.