Distribution of X/(XY-WZ) when all RVs are distributed (iid) U(0,1) Assume that we have four random variates $W, X, Y, Z$ such that $W, X, Y, Z\overset{iid}{\sim} U(0,1)$. I wish to determine the distribution of the following:
\begin{equation}
 Q = \frac{X}{XY-WZ}
\end{equation}
From prior questions, we can determine the distribution of $XY$ and $WZ$, however, I am not quite sure how to handle the rest.
If anyone could assist, I would be very thankful.
BTW - This is not for a course. I am just rusty at my mathematical statistics.
 A: Let $g(y) = x/(x y - z w), \, y_0 = z w/x + 1/q$, then $g(y_0) = q$. We have
$$f_Q(q) = \int_{[0, 1]^4} \delta(g(y) - q) \, dx dy dz dw =
\int_{[0, 1]^3} \frac {[0 < y_0 < 1]} {|g'(y_0)|} \, dx dz dw = \\
\frac 1 {q^2} \int_{[0, 1]^3} [-x/q < z w < x (1 - 1/q)] \, dx dz dw = \\
\frac 1 {q^2} \int_0^1 (F_{W Z}(x(1 - 1/q)) - F_{WZ}(-x/q)) \, dx.$$
$F_{WZ}$ is known:
$$F_{WZ}(r) = r (1 - \ln r) \, [0 < r < 1] + [1 \leq r].$$
For $0 < q < 1$, we get $f_Q(q) = 0$ because both terms in the integrand are zero. For $1 < q$, $F_{WZ}(-x/q) = 0$ and the argument of $F_{WZ}(x(1 - 1/q))$ is inside the interval $[0, 1]$. For $q < 0$, we split the integration range into intervals where the arguments of $F_{WZ}$ are less/greater than $1$. After some calculations,
$$f_Q(q) = \cases {
\frac 1 {4 q^3} \left( \frac {3 q^2 - q - 3} {q - 1} + \ln q^2 \right) &
 $q < - 1$ \\
\frac 1 {4 (1 - q)} & $-1 < q < 0$ \\
0 & $0 < q < 1$ \\
\frac {q - 1} {4 q^3} \left( 3 - 2 \ln \frac {q - 1} q \right) & $1 < q$}.$$
A: Hint. One approach that may simplify matters is to let $R\equiv1/Q$ so $R=Y-WZ/X$.
Then,
$$
F_{Q}(q)=\mathbb{P}(Q\leq q)=\mathbb{P}(R\geq1/q)=1-\mathbb{P}(R<1/q)=1-\mathbb{P}(R\leq1/q)=1-F_{R}(1/q)
$$
and hence $f_{Q}(q)=f_{R}(1/q)q^{-2}$.
