Joint Density of two Cauchy Distribution $\mathbf {Problem:}$ If $X, Y, Z$ are i.i.d. from standard normal, find the joint density of
$W=X/Y$ and $V=Y /Z$?
$\mathbf {Attempt:}$ All I know is $W=X/Y$ and $V=Y /Z$ both follows Cauchy Distribution. But how do I find out the joint density of $W$ and $V$? I think $W$ and $Z$ are not independent, so can't use $$ f_{W,V}(w,v) = f_{W}(w) \cdot f_V(v)$$ Where $f$ denote density.
Any help would be much appreciated.
 A: There could be an easier way but you could try applying the usual change of variables technique.
Joint density of $(X,Y,Z)$ is $$f_{X,Y,Z}(x,y,z)=\frac{1}{(\sqrt{2\pi})^3}e^{-\frac{1}{2}(x^2+y^2+z^2)}\qquad,\,(x,y,z)\in\mathbb R^3$$
Transform $(X,Y,Z)\to(U,V,W)$ such that $$U=\frac{X}{Y},\quad V=\frac{Y}{Z},\quad W=Z$$
We have the inverse solutions $$x=uvw,\quad y=vw,\quad z=w$$
Then $(x,y,z)\in\mathbb R^3\implies (u,v,w)\in\mathbb R^3$.
A quick calculation gives the absolute value of Jacobian as $|J|=|v|w^2$.
So the joint density of $(U,V,W)$ is $$f_{U,V,W}(u,v,w)=\frac{|v|w^2}{(\sqrt{2\pi})^3}e^{-\frac{w^2}{2}(u^2v^2+v^2+1)}\qquad,\,(u,v,w)\in\mathbb R^3$$
Integrating over $w$ you get the joint pdf of $(U,V)$ as
\begin{align}
f_{U,V}(u,v)&=\frac{|v|}{(\sqrt{2\pi})^3}\int_{\mathbb R}w^2e^{-\frac{w^2}{2}(u^2v^2+v^2+1)}\,dw
\\&=\frac{2|v|}{(\sqrt{2\pi})^3}\int_0^\infty w^2e^{-\frac{w^2}{2}(u^2v^2+v^2+1)}\,dw
\\&=\frac{|v|}{(\sqrt{2\pi})^3}\int_0^\infty\sqrt{t}e^{-\alpha t}\,dt\qquad\left[w^2=t;\quad (u^2v^2+v^2+1)/2=\alpha\right]
\\&=\frac{|v|}{(\sqrt{2\pi})^3}\frac{\Gamma(3/2)}{\alpha^{3/2}}
\end{align}
So finally, 
$$f_{U,V}(u,v)=\frac{|v|}{2\pi(u^2v^2+v^2+1)^{3/2}}\qquad,\,(u,v)\in\mathbb R^2$$
Integrating this over the support yields $1$, so looks like this is the correct density.
