Lets say there exists a random vector $(X,Y)$ evenly distributed in the unit square $[0,1]^2$. Now lets introduce the following two new random variables $U=min(X,Y)$ and $V=max(X,Y)$, what is the value of $Cov(U,V)$.
I have an idea of what I should, use directly the definition of the covariance,
$Cov(U,V)=EUV-EUEV = EXY-EUEV=EXEY-EUEV=\frac{1}{4}-EUEV$
But the real problem is I have no idea how to get the density functions of $X$ and $Y$. If anybody has a hint or an idea it would be great.
By definition, $$ U = \min(X,Y) = \begin{cases} X & ,~ \text{in the upper triangle of the unit square} \\ Y & ,~ \text{in lower triangle} \end{cases} $$ With the upper triangle region being the first integral (and the lower region being the 2nd integral), \begin{align} E[U] &= \int_{y = 0}^1 \left[ \int_{x = 0}^y 1\cdot x\, \mathrm{d}x \right]\, \mathrm{d}y + \int_{x = 0}^1 \left[ \int_{y = 0}^x 1\cdot y\, \mathrm{d}y \right]\, \mathrm{d}x \\ &= \int_{y = 0}^1 \frac{y^2}2\, \mathrm{d}y + \int_{x = 0}^1 \frac{x^2}2\, \mathrm{d}x \\ & = \frac13 \end{align} Similarly, for $V = \max(X,Y)$, we have \begin{align} E[V] &= \int_{y = 0}^1 \left[ \int_{x = 0}^y 1\cdot \color{magenta}{y}\, \mathrm{d}x \right]\, \mathrm{d}y + \int_{x = 0}^1 \left[ \int_{y = 0}^x 1\cdot \color{magenta}{x}\, \mathrm{d}y \right]\, \mathrm{d}x \\ &= \int_{y = 0}^1 y^2\, \mathrm{d}y + \int_{x = 0}^1 x^2\, \mathrm{d}x \\ & = \frac23 \end{align} Let me know if you need anything clarified.