Calculate the covariance between X and Y when these variables have joint distribution:
$$f(x, y)=3\, \min\{x,\, y\}\, \text{ if $0<x,\, y<1$}$$
and $0$ otherwise.
Attempt:
A first step is to calculate $E(XY)$, which is given by:
$$E(XY)=\int_{0}^{1}\int_{0}^{1}\, 3xy \min\{ x,\ y \} \, dx\, dy$$
I don't know how to solve it! Then, it occurred to me that I could express it in the following way:
$$E(XY)=\int_{0}^{1}\int_{0}^{1}\, 3xy \frac{x + y - \mid x - y\, \mid}{2} \, dx\, dy$$
But now, I don't know how to remove the absolute value. I think I would have two cases, when $x \geq y$ and $x < y$. Everything else I've tried doesn't work.
Yes. Do that.
Since $\min\{x,y\}=\begin{cases}x&:& x<y\\y&:& y\leq x\end{cases}$ , therefore:
$$\begin{align} \mathsf E(XY)&=\quad\int_0^1\int_0^1 3xy\min\{x,y\}\,\mathrm d x\,\mathrm d y\\[1ex]&=\int_0^1\left(\int_0^y3x^2y\,\mathrm d x+\int_y^13xy^2\,\mathrm d x\right)\,\mathrm d y\end{align}$$