Vtyjkyuk

$mathbf Problem:$ 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.

There could be an easier way but you could try applying the usual change of variables technique.

Joint density of $(X,Y,Z)$ is $$f_X,Y,Z(x,y,z)=frac1(sqrt2pi)^3e^-frac12(x^2+y^2+z^2)qquad,,(x,y,z)inmathbb R^3$$

Transform $(X,Y,Z)to(U,V,W)$ such that $$U=fracXY,quad V=fracYZ,quad W=Z$$

We have the inverse solutions $$x=uvw,quad y=vw,quad z=w$$

Then $(x,y,z)inmathbb R^3implies (u,v,w)inmathbb R^3$.

A quick calculation gives the absolute value of Jacobian as $|J|=|v|w^2$.

So the joint density of $(U,V,W)$ is $$f_U,V,W(u,v,w)=frac(sqrt2pi)^3e^-fracw^22(u^2v^2+v^2+1)qquad,,(u,v,w)inmathbb R^3$$

Integrating over $w$ you get the joint pdf of $(U,V)$ as

beginalign
f_U,V(u,v)&=fracv(sqrt2pi)^3int_mathbb Rw^2e^-fracw^22(u^2v^2+v^2+1),dw
\&=fracv(sqrt2pi)^3int_0^infty w^2e^-fracw^22(u^2v^2+v^2+1),dw
\&=fracv(sqrt2pi)^3int_0^inftysqrtte^-alpha t,dtqquadleft[w^2=t;quad (u^2v^2+v^2+1)/2=alpharight]
\&=fracv(sqrt2pi)^3fracGamma(3/2)alpha^3/2
endalign

So finally,

$$f_U,V(u,v)=fracv2pi(u^2v^2+v^2+1)^3/2qquad,,(u,v)inmathbb R^2$$

Integrating this over the support yields $1$, so looks like this is the correct density.