Given the joint density function of $X$ and $Y$, find the probability density function of $Z = XY$

The joint density function of $$X$$ and $$Y$$ is given by $$f(x,y) = xe^{-x(y+1)}$$, $$x > 0, y > 0$$.

(a) Find the conditional density of $$X$$, given $$Y = y$$, and that of $$Y$$, given $$X = x$$.

(b) Find the density function of $$Z = XY$$.

MY SOLUTION

In the first place, we should determine the marginal distributions: \begin{align*} f_{X}(x) = \int_{0}^{\infty}f_{X,Y}(x,y)\mathrm{d}y = \int_{0}^{\infty}xe^{-x(y+1)}\mathrm{d}y = xe^{-x} \end{align*}

where $$x > 0$$. Analogously, we have \begin{align*} f_{Y}(y) = \int_{0}^{\infty}f_{X,Y}(x,y)\mathrm{d}x = \int_{0}^{\infty}xe^{-x(y+1)}\mathrm{d}x = \frac{1}{(y+1)^{2}} \end{align*}

where $$y > 0$$.

(a) Based on the previous results, it comes

$$\begin{cases} f_{X|Y}(x|y) = \displaystyle\frac{f_{X,Y}(x,y)}{f_{Y}(y)} = x(y+1)^{2}e^{-x(y+1)}\\\\ f_{Y|X}(y|x) = \displaystyle\frac{f_{X,Y}(x,y)}{f_{X}(x)} = e^{-xy} \end{cases}$$

(b) Finally, we have \begin{align*} F_{Z}(z) & = \textbf{P}(XY \leq z) = \int_{0}^{\infty}\int_{0}^{z/x}f_{X,Y}(x,y)\mathrm{d}y\mathrm{d}x = \int_{0}^{\infty}\int_{0}^{z/x}xe^{-x(y+1)}\mathrm{d}y\mathrm{d}x\\\\ & = \int_{0}^{\infty}(e^{-x} - e^{-x-z})\mathrm{d}x = 1 - e^{-z} \Rightarrow f_{Z}(z) = \frac{\mathrm{d}}{\mathrm{d}z}(1-e^{-z}) = e^{-z} \end{align*}

where $$z > 0$$.

The joint density $$f(x,y)$$ factors in the following ways:

$$f(x,y)=xe^{-xy}\mathbf1_{y>0}\,e^{-x}\mathbf1_{x>0}\tag{1}$$

$$f(x,y)=(1+y)^2xe^{-(1+y)x}\mathbf1_{x>0}\,\frac{1}{(1+y)^2}\mathbf1_{y>0}\tag{2}$$

From $$(1)$$ it follows that the conditional density of $$Y\mid X$$ is

$$f_{Y\mid X=x}(y)=xe^{-xy}\mathbf1_{y>0}$$

, and the marginal density of $$X$$ is $$f_X(x)=e^{-x}\mathbf1_{x>0}$$

From $$(2)$$, your answer for the conditional and marginal densities is correct.

So your answer for $$f_X(x)$$, though seemingly okay as it integrates to unity, is not correct. As a result, $$f_{Y\mid X=x}(y)$$ (the variable is $$y$$ here because it is a function of $$y$$ for a given $$x$$, not $$(x,y)$$ as you have written) is also out of line. The rest looks okay.

• In the first place, thank you very much for the contribution. As to the notation of conditional probability, I think the most common way to denote it is $f_{Y|X}(y|x)$. Feb 3, 2019 at 16:34
• That is the notation used in books, to emphasize the "given $x$" part. Feb 3, 2019 at 16:36
• (+1) A minor quibble: the notation $f_{Y\mid X}(y\mid x)$ is common, one could even argue it is preferable to $f_{Y\mid X=x}(y)$.
– Did
Feb 3, 2019 at 16:36