What is the distribution of $x/(x + y)$ for $x$, $y$ independent and normally distributed with mean $0$ What is the distribution of $x/(x + y)$ for independent and normally distributed $x$, $y$ with mean $0$? I'm aware $x/y$ has a Cauchy distribution but I don't know if there is a way to make use of it.
(This may seem like a homework question, but it is not; it's a highly abstracted piece of what I am trying to solve in this other question.)
 A: Set 
$$Z = \frac{X}{X+Y} \implies ZX + ZY = X \implies (1-Z)X = ΖY$$
$$ \implies \frac{Z}{1-Z} = \frac {X} {Y} \equiv C$$
where $C$ is a standard Cauchy r.v. Applying a change of variables
$$\frac {\partial C}{\partial Z} = \frac {1}{(1-Z)^2}$$
So
$$f_Z(z) = \left|\frac {1}{(1-z)^2}\right|f_C[(z/(1-z)] = \frac {1}{(1-z)^2}\frac 1 {\pi}\frac {1}{1+[z/(1-z)]^2}$$
$$f_Z(z) = \frac 1{\pi}\frac {1}{1-2z+2z^2}, \;\;\;z\in (-\infty, \infty)$$
One can verify that this integrates to unity. The density has its mode and median at $m=1/2$.  But no moments here, heavy lies the Cauchy legacy. In fact, to exploit @whuber's contribution in the comments, this is also a Cauchy density, but with non-zero location parameter and non-unity scale parameter. It can be written
$$f_Z(z) = \frac 1{\pi} \frac {1}{(1/2)\left [1+ \frac{[z- (1/2)]^2}{(1/2)^2}\right]}$$
and so it is a Cauchy distribution with location parameter $z_0 = 1/2$ and scale parameter $\gamma = 1/2$ also.
The density is symmetric around its median, with $P(Z\leq 0) = 1/4$, $P(Z<1) = 3/4$, so half of the probability mass is in $[0,1]$.
A graph based on calculating the density is

I also simulated the distribution drawing $200.000$ data points. In all cases, some of the values produced where extremely large (not surprising given the slow decay). Shedding approx $12.000$ observations I got the estimated density plot

while the relative empirical frequency graph looked like (applying the Freedman-Diaconis rule for optimal bin length)

