Transformation of probability density function Having two independent random variables $X$ with pdf $f_X$ and $Y$ with pdf $f_Y$ what is the correct way to derive the formula for $f_Z$ where $Z = X/Y$ and $f_W$ where $W = XY$ ?
I know that the classical convolution formula for (V = X+Y) is derived in the following way:
$$F_V(x) = P(V \lt x) = P(X+Y \lt x) = \iint_{X+Y \lt x} df_x df_y = \int_{-\infty}^{\infty}df_x\int_{-\infty}^{z-x}df_y = \int_{-\infty}^{\infty}f_x(z)f_y(z-x) dz$$
pdf is then given by $f_V = \frac{dF_V}{dx}$
If I go through the same process with $Z = X/Y$ and $W = XY$ what is the correct domain of integration in the last step? 
What would be the approach using characteristic functions?
 A: Try this systematic approach which avoids the unnecessary complications which arise when one uses the détour through CDFs you seem to be mentioning.
For example, to compute the density $f_Z$ of $Z$, note that, for every bounded measurable function $u$,
$$
\mathbb E(u(Z))=\iint u(x/y)f_X(x)f_Y(y)\mathrm dx\mathrm dy.
$$
The change of variables $x=zt$, $y=t$, yields $\mathrm dx\mathrm dy=t\mathrm dz\mathrm dt$ hence
$$
\mathbb E(u(Z))=\iint u(z)f_X(zt)f_Y(t)t\mathrm dz\mathrm dt.
$$
This identity holds for every test function $u$ hence
$$
f_Z(z)=\int f_X(zt)f_Y(t)t\mathrm dt.
$$
If one writes correctly the density functions $f_X$ and $f_Y$, there is no problem of domain of integration. For example, if $X$ and $Y$ are i.i.d. uniform on $(0,1)$, then $f_X=f_Y=\mathbf 1_{(0,1)}$ hence, for every $z\gt0$,
$$
f_Z(z)=\int \mathbf 1_{0\leqslant zt\leqslant1}\mathbf 1_{0\leqslant t\leqslant1}t\mathrm dt=\int_0^{\min(1,1/z)}t\mathrm dt=\frac{\min(1,1/z)^2}2,
$$
that is,
$$
f_Z(z)=\begin{cases}\frac12&\text{if}\ z\leqslant1,\\ \frac1{2z^2}&\text{if}\ z\geqslant1.\end{cases}
$$
