Covariance of two uniform order statistics ($\mathrm{Cov}(U,V)=1/36$)

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 $\mathbb{E}U=1/3$. Since $U+V=X+Y$ and $UV=XY$, $\mathbb{E}V=2/3$ 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 $(b-a{)}^2$.