Computing product distribution

Question

If $X$ and $Y$ have the following joint density function $$f(x,y)=xe^{-x(y+1)}$$ For $0<x,0<y$, compute the density function of $Z=XY$.
I am having some problems with this question because $f_X=e^{-x}$ and $f_Y=\frac{1}{(y+1)^2}$, implying that $X$ and $Y$ are not independent.
Trying to integrate $$\iint_{xy\le a} xe^{-x(y+1)} \mathrm{d}(x,y)$$ I ran into some difficult integrands that made me doubt whether this was the right approach.
I thought also about finding the distribution of the sum of $\ln X$ and $\ln Y$ but this also gave some difficult integrands.
I am unsure as to how to proceed with this question. What approach could I take for this problem?

Answer 1

While your answer is good, note that it is more straightforward to switch the order of integration. For $a>0$ we have $$P(XY \leq a)= \int_A x e^{-x(y+1)}dydx$$ where $A=\{(x,y):x,y>0,xy \leq a\}$. Writing this as an iterated integral: $$\begin{align}P(XY \leq a) &=\int_0^{\infty}\int_0^{a/x}xe^{-x(y+1)}\mathrm{d}y\mathrm{d}x \\ & =\int_0^{\infty} xe^{-x} \int_0^{a/x} e^{-xy} \mathrm{d}y \mathrm{d}x \\ &=\int_0^\infty xe^{-x} \big[\frac{-1}{x}e^{-yx}\big]_{y=0}^{a/x} \mathrm{d}x \\ & =\int_0^\infty e^{-x}(1-e^{-a}) \mathrm{d}x \\ & =(1-e^{-a}) \big[-e^{-x} \big]_{x=0}^\infty \\ & =1-e^{-a} \end{align}$$ Since if $a \leq 0$ we have $P(XY \leq a)=0$, you may recognize this as the CDF of the exponential distribution with parameter 1. If not, simply differentiate it to get the density of XY: $$f_{XY}(a)=\frac{d}{da} P(XY \leq a)=\frac{d}{da} \big[ 1-e^{-a}\big] =e^{-a}$$ which holds for $a>0$. For $a \leq 0$, $f_{XY}(a)=0$.

Answer 2

It seems I made a mistake in my calculation of the integral. I am posting this answer so that if anyone else runs into this question in Sheldon Ross' Book, First Course in Probability, they will not have the same problems: simply calculate $$\begin{align}\iint_{xy \le a}xe^{-x(y+1)}\mathrm{d}x\mathrm{d}y &=\int_0^{\infty}\int_0^{a/y}xe^{-x(y+1)}\mathrm{d}x\mathrm{d}y \\ I & =\int_0^{\infty} \frac{1-\exp\left[-\frac{a}{y}(y+1)\right]\left(\frac ay(y+1)+1\right)}{(y+1)^2} \mathrm{d}y\end{align}$$ We use a modified version of integration by parts, namely: $$\int \frac{u\mathrm{d}v}{v^2}=\int \frac{\mathrm{d}u}{v}-\frac uv $$ And set $v=y+1$ and $u=1-\exp\left[-\frac{a}{y}(y+1)\right]\left(\frac ay(y+1)+1\right)$ which implies that $\mathrm{d}v=\mathrm{d}y$ and $\mathrm{d}u =-e^{-a}e^{-\frac ay} \frac {a^2}{y^3} (y+1) \mathrm{d}y $ Substituting into the modified integration by parts we get $$I=-e^{-a}\int e^{-\frac ay} \frac {a^2}{y^3} \mathrm{d}y-\frac {1-\exp\left[-\frac{a}{y}(y+1)\right]\left(\frac ay(y+1)+1\right)}{y+1}$$ After setting $w= \frac ay$, the first integral in this new expression becomes $$e^{-a} \int w e^{-w} \mathrm{d}w$$ Integrating by parts and substituting into $I$ we obtain $$I=-\exp \left[-\frac ay (y+1) \right]\left( \frac ay +1 \right)-\frac {1-\exp\left[-\frac{a}{y}(y+1)\right]\left(\frac ay(y+1)+1\right)}{y+1}$$ Simplifying yields: $$I=-\frac{1+y\exp\left[-\frac{a}{y}(y+1)\right]}{y+1}$$ Taking limits we get that $$\lim_{y \to 0^+} I = -1$$ $$\lim_{y \to \infty} I = -e^{-a}$$ Therefore, we obtain that $\mathbb{P}(XY \le a)=1-e^{-a}=F_Z(a)$ and after taking derivatives, we get that $f_Z(a)=e^{-a}$.