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Let $X,Y,Z$ be i.i.d. $U(0,1)$ distributed. How can I determine the distribution of

$$ \frac{X}{X+Y+Z}?$$

I have no idea how to go about this problem. Obviously this expression also has values between $0$ and $1$ but is there a way of finding the exact distribution?

flawr
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1 Answers1

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Obviously $W=\frac{X}{X+Y+Z}$ belongs to $[0,1]$. Moreover, for any $\alpha\in[0,1]$ we have: $$ \mathbb{P}[W\leq\alpha] = \mathbb{P}[X\leq \alpha(X+Y+Z)]=\mathbb{P}[X\leq\frac{\alpha}{1-\alpha}(Y+Z)]. $$ Can you compute the last probability?

It is useful to recall that the ratio distribution between two uniformly distributed random variables over $[0,1]$ has pdf: $$ f_R(r)=\left\{\begin{array}{rcl}0&\text{if}&r\leq 0,\\\frac{1}{2}&\text{if}&0< r\leq 1,\\\frac{1}{2r^2}&\text{if}&r\geq 1.\end{array}\right.$$

The pdf of $W$ has quite a strange shape:

$\hspace2in$enter image description here

Such a function is $\frac{1}{(1-w)^2}$ for $w\in[0,\frac{1}{3}]$, $\frac{1-w}{3w^3}$ for $w\in[\frac{1}{2},1]$ and $\frac{-1 + 6 w - 9 w^2 + 3 w^3}{3w^3(1-w)^2}$ for $w\in[\frac{1}{3},\frac{1}{2}]$, zero otherwise.

Jack D'Aurizio
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  • Thank you very much for your answer! I understand what you did so far but I do not know how to proceed (I know next to nothing about probability theory.) Intuitively $Y+Z$ must have a 'triangle' distribution with support $(0,2)$ but I do not know how to justify this or how to use this. Could you give me an additional hint on how to go on from there? – flawr May 26 '15 at 21:44
  • @flawr: Hint: you have to compute the probability that $\frac{Y+Z}{X}$ exceeds $\frac{1-\alpha}{\alpha}$, and you have a general result about the ratio distribution. – Jack D'Aurizio May 26 '15 at 21:50
  • So I think I can use the formula on your wikipedia link, for determining this, thanks! As a followup question: Is there a general receipe for calculating the pdf when you 'combine' multiple distributions like in my example? – flawr May 26 '15 at 22:02
  • @flawr: no general recipe, just the fact that the pdf is the derivative of the cdf, and the hope that the cdf is not too difficult to compute (this problem is border-line). – Jack D'Aurizio May 26 '15 at 22:06