Find absolate value and division of normal distribution

Question

I am looking for at hint solve this:

$X_1, X_2 \sim \mathcal{N}(0,1)$

Find the distribution of $\frac{X_1+X_2}{|X_1-X_2|}$

I have gotten so far that $X_1+X_2 \sim \mathcal{N}(0,2)$ and $X_1-X_2 \sim \mathcal{N}(0,2)$

But I have no idea how to handle the absolute and division

Answer 1

Hint: represent graphically $\{(x,y) \in \mathbb{R}^2: \frac{x+y}{|x-y|}\}$ and study how to calculate the integral in the area: $\{(x,y) \in \mathbb{R}^2: \frac{x+y}{|x-y|} \leq t\}=F_Z(z)$. Then you can derivate $F_Z(z)$ (where it is possible) to get what you want (i.e. the distribution of $Z=\frac{X_1+X_2}{|X_1-X_2|}$).

It can be usefull to compute also the joint distribution.

Answer 2

By #Correlations_and_independence, $X_1+X_2$ and $X_1-X_2$ are independent. As @StubbornAtom mentioned,By distribution of $\frac{X}{|Y|}$

$$ \frac{\frac{X_1+X_2}{\sqrt{2}}}{\frac{\mid X_1-X_2 \mid}{\sqrt{2}}}\sim Cauchy(0,1)$$