Finding distribution and density function of $X^2/(X^2+Y^2)$ where $X,Y∼N(0,1)$

Question

I have two random independent standard normal variables $X,Y∼N(0,1)$.

How can I find the distribution of $\,\dfrac{X^2}{X^2+Y^2}\;?$

I know that if we talk about only $X^2$ then it will be a Chi-Square distribution.

Answer 1

\begin{align} \Pr\left(\frac{X^2}{X^2+Y^2}\leq a\right)&=\Pr\left(\frac{Y^2}{X^2}\geq \frac{1-a}{a}\right)\\ &=1-\Pr\left(\frac{Y^2}{X^2}\leq \frac{1-a}{a}\right)\\ &=1-\frac{\int_0^{1-a}t^{-1/2}(1-t)^{-1/2}dt}{B(1/2,1/2)}\\ &=1-\frac{1}{\pi}\int_0^{1-a}t^{-1/2}(1-t)^{-1/2}dt \end{align}

Note: $Y^2/X^2$ is Fisher with $(1,1)$ degrees of freedom and its CDF you can find here. Also from the comments you see that $0\leq a\leq1$.

Furher, the pdf turns out to be $$f(a)=\frac{1}{\pi}(1-a)^{-1/2}a^{-1/2},$$ so the quantity you are analyzing is a $\beta(1/2,1/2)$ variate.