An easy way to find the PDF of $XY/Z$ Suppose that we know the joint distribution $f_{X,Y,Z}$. I want to find the PDF of $W = XY/Z$ in a way that does not involve too much calculation.
I have thought about first finding the Marginalized PDF $f_{X,Y}$, since we have $XY$ in the numerator. Then I have heard that there was a way to, for example, substitute a random variable $M$, so that we can get to the Joint PDF of $f_{XY, M}$, and after that find $f_{XY, Z}$. Then we introduce another random variable $N$, find  $f_{XY/Z, N}$ and finally marginalize out to get our desired distribution $f_{XY/Z}$.
I have all these hints, but I still have no idea how to perform this. In particular, the introduction of random variables $M$ and $N$ confuse me. Any help would be appreciated. Thank you!
 A: Let $f(x,y,z)$ denote the density of $(X,Y,Z)$. Consider the change of variables from $(x,y,z)$ to $(x,\overbrace{xy}^w,z)$, with Jacobian
$$J_1=\frac{\text{d}(x,y,z)}{\text{d}(z,w,z)}=\frac{\text{d}y}{\text{d}w}=\frac{1}{x}$$
Then $(XY,Z)=(W,Z)$ has density
$$\tilde f(w,z)=\int_{\mathfrak X} f(x,w/x,z)\,\frac{\text{d}x}{x}$$
Now, consider the change of variables from $(w,z)$ to $(\overbrace{w/z}^s,z)$ with Jacobian
$$J_2=\frac{\text{d}(w,z)}{\text{d}(s,z)}=\frac{\text{d}w}{\text{d}s}=z$$
Then $S=W/Z=XY/Z$ has density
$$\hat f(s)=\int_{\mathfrak Z} \tilde f(sz,z) z\,\text{d}z
=\int_{\mathfrak Z}\int_{\mathfrak X} f(x,sz/x,z)\,\frac{\text{d}x}{x} z\,\text{d}z$$
For instance, if $X$ is an exponential variate, $Y$ and $Z$ standard normal variates, all independent, then $R=Y/Z$ is a standard Cauchy variate and the product $S=XR$ has density
$$g(s)=\int_0^\infty e^{-x}\frac{1}{\pi}\frac{1}{1+s^2/x^2}\frac{\text{d}x}{x}$$
which is well-defined when $s\ne 0$ even though $f(0,0,0)>0$. One can define $g(0)$ arbitrarily.
