# Determine the distribution of $Y$ and the conditional distribution of $X$ given that $Y=y$.

The density function of the two-dimensional random variable $(X, Y)$ is $$f_{X,Y}(x,y)=\begin{cases} \frac{x^2}{2y^3} \cdot e^{-\frac{x}{y}} & \text{for } 0<x<\infty,\; 0<y<1, \\[8pt] 0, & \text{otherwise}.\end{cases}$$

(a) Determine the distribution of $Y$.

(b) Find the conditional distribution of $X$ given that $Y=y$.

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This is phrased in the language appropriate for use by an instructor assigning something to students ((a) Find this. (b) Find that.) Instead of passing on to us a question that someone else wrote, can you ask your own questions about it. In particular, what have you done so far and where did you get stuck? –  Michael Hardy Feb 19 '13 at 20:31
In regards to part (a), I think I need to integrate the joint density function with respect to x. For part (b), I'm assuming I need to compute $f_{Y|X=x}=\frac{f_{X,Y}(x,y)}{f_{X,Y}(x,z)}$. But with both, I struggle with knowing what values to integrate on. –  user4 Feb 19 '13 at 20:43
When integrating with respect to $x$, $y$ is a constant, and thus you can find $\int ax^2e^{-bx}\,dx$ via integration by parts or you can use a substitution to relate the value to $\Gamma(3)$. –  Dilip Sarwate Feb 19 '13 at 21:01
Thank you, I didn't realize it was a $\Gamma(3)$ function. I was able to find that Y has a $U(0,1)$ distribution. –  user4 Feb 19 '13 at 21:27