Let X and Y be independent standard normal random variables.
What is the distribution of $\large \frac{X}{|Y|}$?
Attempt:
Let $\large U = \frac{X}{|Y|}$ and $ V = |Y|$.
This transformation is not one-to-one. We can make it one-to-one by restricting consideration to either positive or negative values of y. Let
$A_1 = $ {$(x,y): y > 0$} $, A_2 = ${$(x,y): y < 0$}, $A_0 = $ {$(x,y): y = 0$}.
$B = ${$ (u,v): v > 0$} is the image of both $A_1$ and $A_2$. The inverse transformations are given by:
$y = v, x = uv$ for B to $A_1$, and $ y = -v, x = -uv$ for B to $A_2$.
Both Jacobians $J_1$ and $J_2$ are equal to v.
Then the joint distribution of U,V is given by:
$f_{U,V}(u,v) = \Large \frac{1}{2\pi}e^{-uv^2/2}e^{-v^2/2}*v+\frac{1}{2\pi}e^{-uv^2/2}e^{-(-v^2/2)}*v $
$= \Large \frac{v}{\pi}e^{-(u^2+1)v^2/2}$, $-\infty < u < \infty, 0 < v < \infty$.
To find the marginal pdf of U, integrate out V:
$ f_{U}(U) = \Large \int_0^\infty\frac{v}{\pi}e^{-(u^2+1)v^2/2}dv$
$ = \Large \frac{1}{2\pi}\int_0^\infty e^{-(u^2+1)z/2}dz$, where $z = v^2$
$ = \Large \frac{1}{2\pi} \frac{2}{(u^2+1)}$ (integrand is kernel of exponential ($\beta = 2/(u^2+1)))$
$= \Large \frac{1}{\pi (u^2+1)},$ $-\infty < u < \infty$, which is a Cauchy random variable.
Notice, $|Y|=\sqrt{\frac{Y^2}{1}}$. So you have $\frac{Z_1}{\sqrt{\frac{{Z_2}^2}{1}}}$. So it is $t_1$, which is a Cauchy distribution.