Distribution of $Y= \frac{X_1}{|X_2|}$? If $X_1$ and $X_2$ are independent and identically distributed Gaussian random variables with parameters $0$ and $\sigma^2$, how do I find the distribution of $Y= \frac{X_1}{|X_2|}$?
I'm not supposed to use the method using the Jacobian but I'm not sure what the "easier" way to go about this is.  

If $X_1$ and $X_2$ are independent and identically distributed Gaussian random variables then their common distribution is circularly symmetrical around the origin. Let this distribution be denoted by $N(u,v)$. Then the cdf of Y can be calculated as follows
$$F_Y(y)=P\left(\frac{X_1}{|X_2|}<y\right)=P(X_1<y\ |X_2|)=\iint_{\color{red}{A_y}}N(v,u)\ dv \ du.$$
Here the red region is 
$$\color{red}{A_y=\{(v,u): u<y|v| \}}$$
as shown in the following figure:

So, we have to integrate the circularly symmetric $N(v,u)$ over the red region. The result is 
$$F_Y(y)=P\left(\frac{X_1}{|X_2|}<y\right)=1-\frac1{2\pi}\color{red}{\alpha}=1-\frac1{\pi}\left(\frac{\pi}2-\arctan(y)\right).\tag1 $$
Here I used the following two facts facts: 


*

*the integral over the white area is proportional to the angle $\alpha$ because of the circular symmetry of $N$ around the origin.

*the integral of $N$ over the whole plane is $1$ because $N$ is a probability density function.


The density of $Y$ can be determined by differentiating $F_Y$:
$$f_Y(y)=\frac1{\pi}\frac1{1+y^2}.$$
This is the standard Cauchy distribution. There are three notes to be taken.


*

*The distribution of $Y$ does not depend on $\sigma$.

*The distribution of $Y=\frac{X_1}{|X_2|}$ is the same as that of $Y'=\frac{X_1}{X_2}$ which is also Cauchy.

*If $X_1$ and $X_2$ were not independent then the argumentation based on the circular symmetry of $N(v,u)$ would be false.

