Let (X,Y) be a continuous random vector with probability density p where
$p(x,y)= \frac{3}{2} x^{2} y^{-2}$ when $0
I know that X has the density $p_{1}(x) = \frac{3}{8} x^{2}$ when $0
and Y has the density $p_{2}(y) = 4y^{-2}$ when $2
I now define V=XY and my question is how can I show that V has variance?
I tried to find the probability density function for V and ended up with:
$q(z) = 0$ for $z\leq 0$
$q(z) = \frac{3}{2} z^{2}\cdot$log(z) for $0
$q(z) = \frac{3}{2} z^{2}\cdot$log(2) for 2$\leq$z
and then i want to calculate $\int_\infty^\infty z^{2}q(z) dz$
in order to show that $\int_\infty^\infty z^{2}q(z) dz$
But I end up with $\infty$ when I try to calculate $\int_\infty^\infty z^{2}q(z) dz$
What is wrong?