Distribution of the ratio of two i.i.d standard normals I am reading an article in which the author states that given two i.i.d random variables $X,Y\sim\mathcal{N}(0,1)$, we have $\mathbb{P}(\frac{X}{Y}\le t)=\mathbb{P}(\frac{X}{|Y|}\le t)$ since the random varaibles $\frac{X}{Y}$ and $\frac{X}{|Y|}$ are identically distributed by the symmetry of the standard normal distribution.
But I don't really understand why the random variables $\frac{X}{Y}$ and $\frac{X}{|Y|}$ are identically distributed. Can you provide a reason for why these two random variables are identically distributed?
 A: If you look at a Cartesian plane, then given any t, the points where $\frac{x}{y}$ = t will be a line with slope $\frac{1}{t}$. That is, the equation of the line will be x=ty. The points where $\frac{x}{|y|}$ = t will be that line with the bottom half reflected horizontally.  That is, for negative y, the equation will become x= -ty. So by including that absolute value, we are, for negative y values, going from looking for P(x=2y0, y=y0) to looking for P(x= -2y0, y=y0). But because the normal distribution is symmetric, the probabilities are the same.
A: Setting $B=\frac{X}{|X|}$ and $C=\frac{Y}{|Y|}$ we have $B,C,|X|,|Y|$ independent.
Observe that $\frac{X}{Y}=\frac{B}{C}\frac{|X|}{|Y|}$ and $\frac{X}{|Y|}=B\frac{|X|}{|Y|}$.
It is not difficult to prove that  $\frac{B}{C}$ and $B$ have the same distribution.
A: There are different way to see this. 
1
The top side (y positive) is a (point symmetric) mirror image of the bottom side (y negative). So you can express $P(\frac{x}{y}|y<0)$ by $P(\frac{x}{y}|y>0)$ and similarly $P(\frac{x}{|y|}|y<0)$ by $P(\frac{x}{|y|}|y>0)$ 
2


*

*$\frac{x}{y}=b$ occurs when $\frac{x}{y}=b$ for $y>0$ and $\frac{x}{y}=b$ for $y<0$

*$\frac{x}{|y|}=b$ occurs when $\frac{x}{y}=b$ for $y>0$ and $\frac{x}{y}=-b$ for $y<0$


thus effectively you flip the bottom half from $\frac{x}{y}=b$ to $\frac{x}{y}=-b$ but this half is symmetric in $P(\frac{x}{y}|y<0) = P(-\frac{x}{y}|y<0)$
3
a conversion to cyclic coordinates may also work. $P(\frac{x}{y}=a) \propto   2 P(\theta=tan^{-1}(a))$ and also $P(\frac{x}{|y|}=a) \propto 2 P(\theta=tan^{-1}(a))$ 

image: 1000 points x,y distributed i.i.d. N(0,1)


