4
$\begingroup$

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.)

$\endgroup$
7
  • 3
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 3
    $\begingroup$ 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
  • 2
    $\begingroup$ 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

10
$\begingroup$

Set

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

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

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

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

So

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

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

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

$\endgroup$
5
  • 3
    $\begingroup$ (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
  • 1
    $\begingroup$ @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
  • 1
    $\begingroup$ @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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.