Find the CDF of $XY$ given CDF's of noncontinuous $X, Y$. For $X, Y$ independent random variables with CDF's $F_X, F_Y$ how would one go about finding the CDF of $XY$? I ask because the PDF's aren't necessarily defined (wikipedia says $X$ needs to be continuous), and the every source I find online for finding $F_{XY}$ goes through the PDF's.
 A: In theory, if $X$ and $Y$ are independent and singular continuous random variables, then I would try to work it out with the characteristic functions $\phi_X$ and $\phi_Y$, which always exist.
Then by conditioning like here:
$$\phi_{XY}(t)=E[e^{itXY}]=E[E[e^{itXY}|Y]]=E[\phi_{X}(tY)]$$
And if I got that far, I could try to calculate $F_{XY}$ by using the inversion formulae.
But if $F_X$ and $F_Y$ are just mixed continuous and discrete random variables (no singular continuous component), you can decompose $X$ and $Y$ into their continuous and discrete parts like here, and work out $XY$ on components.
For example if:
$$F_X(t)=F_Y(t)=
\begin{cases}
0 &\text{ if } t<0\\
\frac{t}{2}+\frac{1}{2} &\text{ if } 0\le t<1\\
1 &\text{ if } t\ge1
\end{cases}$$
Then you could write $X$ as:
$$X=
\begin{cases}
0 &\text{ with probability } \frac{1}{2}\\
U_X &\text{ with probability } \frac{1}{2}\\
\end{cases}
$$
And the same for $Y$:
$$Y=
\begin{cases}
0 &\text{ with probability } \frac{1}{2}\\
U_Y &\text{ with probability } \frac{1}{2}\\
\end{cases}
$$
With $U_X$ and $U_Y$ i.i.d. uniformly distributed on $[0,1]$
Then:
$$XY=
\begin{cases}
0 &\text{ with probability } \frac{3}{4}\\
U_XU_Y &\text{ with probability } \frac{1}{4}\\
\end{cases}
$$
And $F_{XY}$ follows easily.
