what is the distribution of $XY/Z$?

Question

If $X$, $Y$, and $Z$ are independent random variables, each uniformly distributed over (0, 1), what is the distribution of $XY/Z$?

I want to solve this exercises by the transformation method, The answer suggests me to use the following transformation:

$U=X$, $V=Y$, $W=XY/Z$

In the end I must come to $f(W)=\left(\frac{1}{4} - \frac{1}{2}\ln(w)\right) I_{(0,1)} \hspace{.1cm}(w)+ \frac{1}{4w^2} I_{[1,\infty)} \hspace{0.1cm} (w).$

But I don't understand why it divides the result like that. Why does $w$ go from $(0,1),$ and also from $(1, \infty$)?

Answer 1

You might have already worked out that $T:=XY$ has PDF $-\ln t$ on $[0,\,1]$. Once you know that, we just seek a ratio distribution. So$$f_W(w)=-\int_0^{\min\{1/w,\,1\}}z\ln(wz)dz.$$Now consider the cases $w\le1,\,w>1$ separately.

Answer 2

Consider the equation $xy = zw$. If $0 < w < 1$, $z$ can be anywhere in $(0,1)$, and and then $zw < x < 1$ with $0 < y = zw/x < 1$. On the other hand, if $w > 1$, we need $0 < z < 1/w$.