What is $Z:=XY$ for random vector $(X,Y)$?

Question

The stochastic vector $(X,Y)$ has a continuous distribution with pdf: $$f(x,y) = \begin{cases} xe^{-x(y+1)} & \text{if $x,y>0$} \\[2ex] 0 & \text{otherwise} \end{cases}$$
Define $Z:=XY$.
I would like to know what exactly $XY$ is. It seems to me that the function $f(x,y)$ has but one output, so what does $XY$ mean here? I feel like I'm missing something obvious. Thanks in advance.

Answer 1

Hint

$$\mathbb P\{Z\leq t\}=\iint_{\{(x,y)\mid xy\leq t\}}f_{(X,Y)}(u,v)dudv.$$

Answer 2

Possibly clarifying some confusion. You are not telling us what is $X$ and what is $Y$. You are telling us their joint density $$\mathbb P[(X,Y)\in (dx,dy)]:=f_{(X,Y)}:=f: \mathbb R\times\mathbb R\to[0,\infty).$$ Quite fundamentally, a density has to have only one output. Knowing $Z:=XY$ means knowing its law, which is a consequence of knowing its density $f_{Z}$ (when it exists). As $X$ and $Y$ are real valued, $Z$ is real valued, so you should expect that $f_Z:\mathbb R\to [0,\infty)$. You can compute $F_Z$ via the hint of Surb above (extra hint, you know what is $\mathbb E [\phi(X,Y)]$, then pick $\phi(x,y)=\mathbf 1_{xy\le t}$).