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

Question

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?

Answer 1

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.

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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.

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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.

Answer 2

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.