Probability Distributions – Distribution of the Ratio of Two i.i.d Standard Normals

Question

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?

Answer 1

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)