Covariance of two gamma distributions.

Question

I'm struggling to find $E[XY]$ as part of calculating the covariance of two gamma RV's. The joint distribution is $f(x,y) = 8x^2 e^{-2y}, 0<x<y$, which gives marginal distributions of $f(x) \sim Ga(3,2) $ and $f(y) \sim Ga(4,2)$.

I understand that $$ E(XY) = \int \int xy f(x,y) \text{ } dx dy$$but when I work this out with the limits I used for the first part, I get divergent integrals. The limits used in the first part were $x \in (0,y), y \in (x,\infty) $.

Answer 1

If you integrate over $x$ the limits for x are $x\in (0,y)$. It looks like

What are then limits for $y$ ?

$y$ has to be just greater than $0$. Thus the term for the expection of the product is

$$ \mathbb E(X\cdot Y)=\int_0^{\infty} \int_0^y x\cdot y\cdot 8x^2\cdot e^{-2y} \, dx \, dy$$

Answer 2

You want to write your expectation as an iterated or nested integral. On the inside, integrate with respect to $x$ over the range $[0,y]$. This results in a function of $y$, namely $\frac 8 3 y^4 \exp(-2y)$. Now integrate that (this is the outside integral) with respect to $y$ over the range $[0,\infty)$, to get your answer.