TITLE: X,Y are independent standard normal distributed then what is the distribution of $\frac{X}{X+Y}$
QUESTION [6 upvotes]: X, Y are independent standard normal random variables, what is the distribution of
$$ \frac{X}{X+Y} $$
Could anyone help me with this? Thanks.

I have worked the problem by multivariable transformation:
Let $$Z=\frac{X}{X+Y} , W=X$$
Consider transformation $$(X,Y)\longrightarrow(Z,W)$$
Then $$X(Z,W)=W , Y(Z,W)=\frac{W(1-Z)}{Z}$$  defines the inverse transformation.
The Jacobian is $$J(Z,W)=\frac{w}{z^{2}} $$
So $$f_{Z,W}(z,w)=f_{X,Y}(w,\frac{w(1-z)}{z})\cdot\mid\frac{w}{z^{2}}\mid$$
As X  and Y  are independent. Then the marginal pdf of Z  is $$f_{Z}(z)=\intop_{0}^{\infty}\frac{w}{z^{2}}\cdot f_{X}(w)\cdot f_{Y}(\frac{w(1-z)}{z})dw+\intop_{-\infty}^{0}-\frac{w}{z^{2}}\cdot f_{X}(w)\cdot f_{Y}(\frac{w(1-z)}{z})dw$$ 
After calculation we get $$f_{Z}(z)=\frac{1}{\pi\cdot\frac{1}{2}\cdot(1+(\frac{z-\frac{1}{2}}{\frac{1}{2}})^{2})}$$
Hence $$Z\sim \mathrm{Cauchy}(\frac{1}{2},\frac{1}{2}).$$

REPLY [5 votes]: Since $X$ and $Y$ are independent standard gaussian random variables, the distribution of $Z=\frac{X}{X+Y}$ has density $f_Z$, where for every $z$ in $\mathbb R$,
$$
\color{red}{f_Z(z)=\frac1\pi\,\frac1{z^2+(1-z)^2}}.
$$
The direct way to prove this (as is now done by the OP) is to rely on the change of variables method expanded here.
One can deduce from the expression of $f_Z$ that $Z=\frac12(1+T)$, where $T$ is standard Cauchy, that is, the distribution of $T$ has density $f_T$, where for every $t$ in $\mathbb R$,
$$
\color{purple}{f_T(t)=\frac1\pi\,\frac1{1+t^2}}.
$$
But the formulas for $f_Z$ and $f_T$ are also direct consequences of two facts: 

The ratio of two independent standard gaussian random variables is a standard Cauchy random variable.
If $X$ and $Y$ are independent standard gaussian random variables, then the random variables $\frac1{\sqrt2} (X+Y)$ and $\frac1{\sqrt2}(X-Y)$  are independent standard gaussian random variables as well.