Compute the density of $XY$ 
Let be $X,Y$ two independent real valued random variables, with density functions $f_X$ and $f_Y$. Compute the density function of the product $XY$, where $X\geq 0$.


My approach:
First, I think that the professor forgot the assumption that $f_Y$ must be continuous in order to be able to take the derivative later. We assume $X>0$ and keep in mind that both random variables are independent:
\begin{align*}
&P(XY\leq z)=P\left(Y\leq \frac{z}{X},X>0\right)\underset{\text{independence of } XY}{=}P\left(Y\leq \frac{z}{X}\right)P\left(X>0\right)\\
&=P\left(Y\leq \frac{z}{X}\right)P(-X<0)\\
&=\int\limits_{-\infty}^{\frac{z}{x}}f_Y(y)dy\cdot\int\limits_{-\infty}^{0}f_X(-x)dx\\
&=\int\limits_{-\infty}^{z}\frac{1}{x}f_Y\left(\frac{y}{x}\right)dy\cdot\int\limits_0^{\infty}f_X(x)dx
\end{align*}
If we take the derivative with respect to $z$ by applying the fundamental theorem of calculus (differentiability follows from continuity of $f_Y$) then we see that
\begin{align*}
& \frac{1}{x}f_Y\left(\frac{z}{x}\right)\int\limits_0^{\infty}f_X(x)dx.
\end{align*}
is a density function of $XY$.

The sample solution says $\int\limits_0^{\infty}f_X(x)\frac{1}{x}f_Y\left(\frac{z}{x}\right)dx$. Where is my mistake?
 A: Note that, since $X$ is absolutely continuous, $P(X=0)=0$. And, since $X \ge 0$ is given, we have that $X>0$ a.s.
Therefore,
$$P(XY \le z)=P\left(Y\le \frac{z}{X}\right)=\int_0^{\infty} \int_{-\infty}^{z/x}f_X(x)f_Y(y)dydx.$$
Can you take from here?
There are two fundamental mistakes in your approach.
First, it is not true em general that the events $\left(Y \le \frac{z}{X}\right)$ and $(X>0)$ will be independent. Although $X$ and $Y$ are independent, the event $\left(Y \le \frac{z}{X}\right)$ will depend of $X$ and will not be independent of $(X>0)$ in general. This will be only true in this case because $(X>0)$ has probability $1$.
Second, your computation of P$\left(Y \le \frac{z}{X}\right)$ is wrong. You seem to imply that
$$P\left(Y \le \frac{z}{X}\right)=\int_{-\infty}^{z/x}f_Y(y)dy,$$
and this is not true. Note that the LHS is a number depending only of $z$, and the RHS depends on $z$ and $x$. You forgot to take account of the possible values that $X$ can assume, and this is done by integrating over $x$.
