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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, $|J|=V$.


$f_U(u)=\frac{1}{2\pi}\int_{-\infty}^{\infty}ve^{\frac{-v^2}{2}(2u^2-2u+1)}dv$. 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?

  • 7
    $\begingroup$ You can do it more simply: the ratio of two standard Normals is Cauchy $\endgroup$
    – wolfies
    Commented Nov 22, 2019 at 4:36
  • 3
    $\begingroup$ Ratio mentioned in Wikipedia. $\endgroup$
    – BruceET
    Commented Nov 22, 2019 at 7:09

2 Answers 2


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:

These are all definitions:

  • 4
    $\begingroup$ (+1) Very intuitive explanation for the 1/2 location and scale parameters! $\endgroup$
    – Xi'an
    Commented Nov 22, 2019 at 15:47
  • 2
    $\begingroup$ @Xi'an Thank you for that perceptive comment: the impetus for this post was your answer that clearly pointed out the values of those parameters, which led me to wonder why they had those values and whether the result could be made intuitively obvious. $\endgroup$
    – whuber
    Commented Nov 22, 2019 at 16:07
  • $\begingroup$ Thank you for providing so many additional materials to study. I appreciate your in-depth help. $\endgroup$
    – Ron Snow
    Commented Nov 23, 2019 at 19:41

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.

  • 1
    $\begingroup$ Thank you for correcting my answer so that I can see where I messed up. This greatly helps, thanks! $\endgroup$
    – Ron Snow
    Commented Nov 23, 2019 at 19:39
  • 1
    $\begingroup$ Tks for the Pillai and Meng reference, Xi'an! Really interesting. $\endgroup$
    – Zen
    Commented Dec 6, 2019 at 3:32

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