What is the PDF of random variable Z=XY? Given two independent random variables X and Y, how can I find the PDF of random variable $Z=XY$? 
*If their joint distribution is required, assume that we also have it.
 A: If $X$ and $Y$ are two random variables, you may find the product distribution as follows:
$$
f_Z(z)=\int_{-\infty}^{\infty} \frac{1}{|t|} f_X\left(t\right)f_Y\left(\frac{z}{t}\right)dt
$$
To see this, suppose that the distribution of $X$ is continuous at 0:
$$
P(Z\leq z)=P(XY\leq z)=
P(Y\leq \frac{z}{X}\big|X>0)P(X>0)+P(Y\geq \frac{z}{X}\big|X<0)P(X<0)=
$$
$$
=\int_{0}^{\infty} P(Y\leq \frac{z}{t}\big)f_X\left(t\right)dt+
\int_{-\infty}^{0} P(Y\geq \frac{z}{t}\big) f_X\left(t\right)dt
$$
We can find the derivation of both sides w.r.t. $z$ and we get:
$$
f_Z(z)=\int_{0}^{\infty} \frac{1}{t} f_Y(\frac{z}{t}\big)f_X\left(t\right)dt+
\int_{-\infty}^{0} \frac{-1}{t} f_Y(\frac{z}{t}\big) f_X\left(t\right)dt
$$
$$
=\int_{-\infty}^{\infty} \frac{1}{|t|} f_X\left(t\right)f_Y\left(\frac{z}{t}\right)dt
$$
The moral: it is not always possible to use this formula.
A: \begin{align}
{\cal P}\left(Z\right)
&=
{{\rm d} \over {\rm d}Z}\int_{-\infty}^{Z}{\cal P}\left(Z'\right)\,{\rm d}Z'
=
{{\rm d} \over {\rm d}Z}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}
{\rm P_{X}}\left(X\right)\,{\rm P_{Y}}\left(Y\right)\,
\Theta\left(Z - XY\right)\,{\rm d}X\,{\rm d}Y\,
\\[3mm]&=
\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}
{\rm P_{X}}\left(X\right)\,{\rm P_{Y}}\left(Y\right)\,
\delta\left(Z - XY\right)\,{\rm d}X\,{\rm d}Y\,
=
\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}
{\rm P_{X}}\left(X\right)\,{\rm P_{Y}}\left(Y\right)\,
{\delta\left(Y - Z/X\right) \over \left\vert X\right\vert}\,{\rm d}X\,{\rm d}Y\,
\end{align}
$$
\begin{array}{|c|}\hline\\
\color{#ff0000}{\large\quad%
{\cal P}\left(Z\right)
=
\int_{-\infty}^{\infty}
{\rm P_{X}}\left(X\right)\,{\rm P_{Y}}\left(Z \over X\right)\,
\,{{\rm d}X \over \left\vert X\right\vert}
\quad}
\\ \\ \hline
\end{array}
$$
