From the joint pdf below, I need to find $E(X)$.
$$f(x,y) = \begin{cases} 8x^2e^{-2y}, & 0<x<y\\\ 0, &\text{otherwise} \end{cases}$$
I know that $$E(X)=\int xf(x)dx$$ so I calculate $f(x)$ first. Since $0<x<y,\,$ I set the boundary $y>0$ or $0<y<\infty$. But I'm not sure I got it right. $$f(x)=\int_0^\infty 8x^2e^{-2y} \,dy= 4x^2$$
I subtitute this result to $E(X)$ by setting the boundary from $0$ to $\infty$. But I got diverge result. $$E(X)=\int_0^\infty x4x^2\,dx=\int_0^\infty 4x^3\,dx=\text{diverge}$$ I guess I set the boundary wrong. But I'm not be able to figure it out.
How am I supposed to set the right boundary? Did I get the step correctly?
The boundary should be $x<y<\infty$
Therefore:
$f(x)= \int^{\infty}_{x}{8x^2e^{-2y}}dy=8x^2 \int_{x}^{\infty}e^{-2y}dy=8x^2[-\frac{1}2e^{-2y}|^\infty_x]=4x^2e^{-2x}$
Now you should be able to find $E(X)$ (Which should be $\frac{3}2$)