how to find the expected value of a joint density function?

Question

I have to find the covariance of this joint density function $$f(x,y) = \begin{cases} \ xe^{-x(y+1)}, & \text{if } x > 0,\ y > 0 \\[2ex] 0, & \text{otherwise} \end{cases}$$ for $E[XY]$ do I integrate $\int_0^\infty \int_0^\infty xy f(x,y) \, dy\,dx$? and for $E[X]$ do I find the marginal of $X$ and integrate or do I integrate $\int_0^\infty\int_0^\infty x f(x,y) \, dy \, dx$?

Answer 1

To find the expected value $\operatorname{E}(G(X,Y))$ where $G$ is any function of two variables you can find $$ \int_0^\infty \int_0^\infty G(x,y) f(x,y)\,dy\,dx $$ and that's it. If $G(x,y)=xy$ then you have just the integral you wrote and if $G(x,y)=x$ then you get this: $$ \int_0^\infty \int_0^\infty x f(x,y)\,dy\,dx. \tag 1 $$ Notice that $(1)$ is $$ \int_0^\infty \left(\int_0^\infty x f(x,y)\,dy\right)\,dx $$ and in the inside integral $$ \int_0^\infty x f(x,y)\,dy $$ the variable $x$ does not change as $y$ goes from $0$ to $\infty$, so the inside integral is the same as $$ x\int_0^\infty f(x,y)\,dy $$ and that is $x$ times the marginal density of $X$. So finding $(1)$ is the same as integrating $x$ times the marginal density of $X$ from $0$ to $\infty$.

There is also the question of whether it is more efficient to find $$ \int_0^\infty \int_0^\infty x f(x,y)\,dy\,dx \quad\text{or}\quad \int_0^\infty \int_0^\infty x f(x,y)\,dx\,dy. $$ In the second form, you can't pull $x$ out of the inside integral the way you can in the first form.