How can I obtain a Cauchy distribution from two standard normal distributions? I am interested in 

Let $X\sim N(0,1), Y \sim N(0,1)$ independently. Show $\frac{X}{X+Y}$
  is a Cauchy random variable.

My work:
$f_{X,Y}(x,y)=\frac{1}{2\pi} e^{\frac{-1}{2}(x^2+y^2)}, -\infty<x,y<\infty$ by independence
Let $U=\frac{X}{X+Y},V=X+Y$. (Is there a better $V$ to choose for this bivariate transformation?)
Then, $X=UV, Y=V - UV$. So, $|J|=V$.
$f_{U,V}(u,v)=f_{X,Y}(uv,v-uv)|J|=\frac{v}{2\pi}e^{\frac{-v^2}{2}(2u^2-2u+1)},-\infty<u<\infty$
$f_U(u)=\frac{1}{2\pi}\int_{-\infty}^{\infty}ve^{\frac{-v^2}{2}(2u^2-2u+1)}dv$. Let $y=\frac{v^2}{2}(2u^2-2u+1)$, so $dy=v(1+2u^2-2u)dv$. Then,
$f_U(u)=\frac{1}{2\pi(2u^2-2u+1)}\int_0^{\infty}e^{-y}dy=(\pi[(\frac{u-1/2}{2})^2+1])^{-1}, -\infty<u<\infty$, 
which is not exactly a Cauchy distribution. Where did I mess up? More importantly, how would you proceed in solving this problem?
 A: This can be done with a minimum of computation, relying only on (a) simple algebra and (b) basic knowledge of distributions associated with statistical tests.  As such, the demonstration may have substantial pedagogical value--which is a fancy way of saying it's worth studying.

Let $Z=X/(X+Y),$ so that
$$Z - \frac{1}{2} = \frac{X}{X+Y} - \frac{X/2+Y/2}{X+Y} = \frac{1}{2}\frac{X-Y}{X+Y} = \frac{1}{2}\frac{(X-Y)/\sqrt{2}}{(X+Y)/\sqrt{2}} = \frac{1}{2}\frac{U}{V}$$
where $$(U,V) = \left(\frac{X-Y}{\sqrt{2}}, \frac{X+Y}{\sqrt{2}}\right).$$ Because $(U,V)$ is a linear transformation of the bivariate Normal variable $(X,Y),$ it too is bivariate Normal, and an easy calculation (ultimately requiring, apart from arithmetical definitions, only the fact that $1+1=2$) shows the variances of $U$ and $V$ are unity and $U$ and $V$ are uncorrelated: that is, $(U,V)$ also has a standard Normal distribution.
In particular, $U$ and $V$ are both symmetrically distributed (about $0$), implying $U/V$ has the same distribution as $U/|V|.$  But $|V| = \sqrt{V^2}$ has, by definition, a $\chi(1)$ distribution.  Since $U$ and $V$ are independent, so are $U$ and $|V|,$ whence (also by definition) $U/|V| = U/\sqrt{V^2/1}$ has a Student t distribution with one degree of freedom.
The conclusion, after no integration and only the simplest of algebraic calculations, is

$W = 2Z-1 = U/V$ has a Student t distribution with one degree of freedom.

That's just another name for the (standard) Cauchy distribution.  Since $Z = W/2 + 1/2$
is just a rescaled and shifted version of $W,$, $Z$ has a Cauchy distribution (once again by definition), QED.

Summary of facts used
Every one of the facts used in the foregoing analysis is of interest and well worth knowing.
These are basic theorems:

*

*Linear transformations of bivariate Normal variables are bivariate Normal.  (This could also be considered a definition.)


*Uncorrelated bivariate Normal variables are independent.


*The covariance is a quadratic form.  (This, too, can be part of the definition of covariance, but that would be a little unusual.)


*When two variables are independent, functions of each of them (separately) are also independent.
These are all definitions:

*

*A sum of $n$ independent standard Normal variables has a $\chi^2(n)$ distribution.


*The ratio of a standard Normal variable to the square root of $1/n$ times a $\chi^2(n)$ independent variable has a Student t distribution with $n$ degrees of freedom.  See also A normal divided by the $\sqrt{\chi^2(s)/s}$ gives you a t-distribution -- proof.


*A Cauchy distribution is a scaled, translated version of the Student t distribution with 1 degree of freedom.
A: Correction: the Jacobian of the transform is $|V|$, not $V$, which implies that
$$f_{U,V}(u,v)=f_{X,Y}(uv,v-uv)|J|=\frac{|v|}{2\pi}\exp\left\{\frac{-v^2}{2}(2u^2-2u+1)\right\}$$
Hence
\begin{align}f_U(u)&=\frac{1}{2\pi}\int_{-\infty}^{\infty}|v|e^{\frac{-v^2}{2}(2u^2-2u+1)}\text{d}v\\
&=\frac{2}{2}\frac{1}{\pi}\int_{0}^{\infty}ve^{-\overbrace{\frac{v^2}{2}(2u^2-2u+1)}^y}\text{d}v\\
&=\frac{1}{\pi(2u^2-2u+1)}\int_0^{\infty}e^{-y}\text{d}y\\
&=\frac{1}{\pi}\frac{1}{2u^2-2u+1}\\
&=\frac{1}{\pi}\frac{1}{2(u-½)^2+½}\\
&=\frac{1}{½\pi}\frac{1}{4(u-½)^2+1}\\
&=\frac{1}{½\pi}\frac{1}{(2[u-½])^2+1}\\
&=\frac{1}{½\pi}\left(\left[\frac{u-½}{½}\right]^2+1\right)^{-1}\end{align}
which is the density of a Cauchy distribution with location ½ (which is also the median) and scale ½ (which is also the MAD). (The last equality in the question is erroneously using 2 instead of ½ as scale and missing the ½ in the first fraction denominator.)

Check Pillai and Meng (2016) for further surprising properties of
  the Cauchy distribution.

