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.)

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    $\begingroup$ If you are aware of $z$ is Cauchy, can you determine the distribution of $\frac{1}{1 + z}$ by applying the transformation formula? $\endgroup$
    – Zhanxiong
    Jan 8, 2018 at 16:54
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    $\begingroup$ You could try a transformation of the joint of $X,Y$ (which you know, assuming they're independent) to something like $X/(X+Y)$ and $X+Y$. $\endgroup$
    – aleshing
    Jan 8, 2018 at 17:11
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    $\begingroup$ Zhanxiong, I like this direction, thanks. I can easily see how to get the distribution of $z' = 1 + z$ which is also Cauchy, but I don't see how to obtain the distribution of the inverse $1/z'$. Any hints? $\endgroup$
    – user118967
    Jan 8, 2018 at 17:27
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    $\begingroup$ The distribution of $X/Y$ is only Cauchy if $X$ and $Y$ are independent and have zero means (in addition to being Normal). You do not state this requirement in your question. $\endgroup$
    – wolfies
    Jan 8, 2018 at 17:32
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    $\begingroup$ Hint: you can express $X$ as a linear combination of $X+Y$ and $X+\alpha Y$ where $\alpha$ is chosen to make $X+Y$ uncorrelated with $X+\alpha Y$. That will reduce the fraction $X/(X+Y)$ to a constant plus a multiple of a ratio of independent, zero-mean Normals. $\endgroup$
    – whuber
    Jan 8, 2018 at 18:18

1 Answer 1



$$Z = \frac{X}{X+Y} \implies ZX + ZY = X \implies (1-Z)X = ΖY$$

$$ \implies \frac{Z}{1-Z} = \frac {X} {Y} \equiv C$$

where $C$ is a standard Cauchy r.v. Applying a change of variables

$$\frac {\partial C}{\partial Z} = \frac {1}{(1-Z)^2}$$


$$f_Z(z) = \left|\frac {1}{(1-z)^2}\right|f_C[(z/(1-z)] = \frac {1}{(1-z)^2}\frac 1 {\pi}\frac {1}{1+[z/(1-z)]^2}$$

$$f_Z(z) = \frac 1{\pi}\frac {1}{1-2z+2z^2}, \;\;\;z\in (-\infty, \infty)$$

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

enter image description here

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

enter image description here

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

enter image description here

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    $\begingroup$ (1) Did you notice that $f_Z$ is merely a shifted, rescaled Cauchy density? (The explanation is buried in a comment I wrote to the question.) (2) You have assumed $X$ and $Y$ have equal variances. A similar result holds when their variances differ: the result is still in the Cauchy family. $\endgroup$
    – whuber
    Jan 8, 2018 at 22:16
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    $\begingroup$ @whuber I admit I didn't go into that much depth. Let me work this to enhance the post. $\endgroup$ Jan 8, 2018 at 23:02
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    $\begingroup$ @whuber I have trouble matching the density I found to the location-scale version of the Cauchy density. I need to set shape $\gamma =2$ which gives me location $z_0 = 1/2$ as it should. But then the constant term in the denominator is $\gamma (1+z_0^2)$ and equals $5/2$, not $1$ as in the density I found. What am I doing wrong? $\endgroup$ Jan 8, 2018 at 23:11
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    $\begingroup$ I obtain $1-2z+2z^2 = \frac{1}{2}(1 + (2z-1)^2).$ $\endgroup$
    – whuber
    Jan 8, 2018 at 23:27
  • $\begingroup$ @whuber Thanks, silly mistake, I wrote $\gamma/\gamma^2 = \gamma$ instead of $1/\gamma$. $\endgroup$ Jan 9, 2018 at 0:38

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