Ytukyg

Consider the joint pdf of $(X, Y)$ given by

$$f(x, y) = begin{cases} y^{-1}text{exp}left(-yright) & 0 < x < y \[1em]
0, & text{ otherwise.} end{cases} $$

a) Compute $mathbb{E}[X^{3} | Y = y]$

b) Are $X$ and $Y$ independent?

So, my attempt for $(a)$ was to first compute $f_{x mid y}(x mid y)$ as follows:

$$f_{x|y}(x|y) = frac{f(x,y)}{int_{-infty}^{infty}f(x,y) mathop{dx}} = frac{1}{y}.$$

So,

$$mathbb{E}[X^{3} mid Y = y] = int_{0}^{y} x^{3}frac{1}{y} mathop{dx} = boxed{y^3/4}$$

For $(b)$, I try to decompose $f(x, y)$ into the product of two marginals. We have

$$f_{Y}(y) = int_{0}^{y} f(x, y) mathop{dx} = e^{-y}.$$

But, I can't compute

$$f_{X}(x) = int_{x}^{infty} f(x,y) mathop{dy}. $$

Wolfram Alpha can't compute it in a nice way either. Is my answer for $(a)$ correct? How can I do $(b)$?

(a) is quite correct.

For (b) you have that $f_{X,Y}(x,y)=y^{-1}e^{-y}mathbf 1_{0leq xleq y}$ and $f_Y(y)=e^{-y}mathbf 1_{0leq y}$ .   Okay.

Don't worry about having trouble finding the marginal density function for $X$.  
Just ask, can there be any mono-variate function wrt $x$, $g(x)$, such that $y^{-1}e^{-y}mathbf 1_{0leq xleq y}=g(x)~e^{-y}mathbf 1_{0leq y}$ ? Therefore, can $X,Y$ be independent?

Also notice $f_{Xmid Y=2}(3)=0$ while $f_{Xmid Y=4}(3)>0$.