At first I calculate the distribution of $X/(X+Y)$. Please correct me if I am wrong. My idea is as follows:

I construct the function $f: (\mathbb{R}^2,\mathcal B^2)\to (\mathbb{R},\mathcal B),\ (x,y)\mapsto \frac{x}{x+y}$. As $X$ and $Y$ are independent, their distribution on the product space is given by the pdf $\rho(x):=\alpha^2 e^{-\alpha x}e^{-\alpha y} 1_{[0,\infty)^2}(x,y)$ (this is just the product of the pdfs of two independent exponentially-distributed r.v.). The distribution of $\frac{X}{X+Y}$ is thus the distribution of $P\circ f^{-1}(-\infty,c]$ where $P$ is the measure of the product space $\mathbb{R}^2$ whose pdf is $\rho(x)$.

Now we need to figure out what $f^{-1}(-\infty,c]$ means: \begin{equation*} \frac{x}{x+y}\leq c\iff y\geq x\cdot\frac{1-c}{c} \end{equation*} Thus we get $f^{-1}(-\infty,c]=\{(x,y)\in \mathbb{R}^2:y\geq \frac{x(1-c)}{c}\}$ and for our final distribution we would get: \begin{equation*} P\circ f^{-1}(-\infty,c]=\int_0^\infty\int_{x(1-c)/c}^\infty \alpha e^{-\alpha y}dy\ \alpha e^{-\alpha x} dx=\int _0^\infty e^{-\alpha x(1-c)/c}\cdot \alpha e^{-\alpha x}\ dx=\alpha \int_0^\infty e^{-\alpha x/c}dx=c \end{equation*} This however, would imply that its distribution would be the uniform distribution.

I would like to ask the following: According to Wikipedia the desired distribution should be a Beta Prime distribution with PDF $x^{\alpha-1}(1-x)^{\alpha-1} B(\alpha,\alpha)^{-1}$. For the love of God, I could not verify that the resulting CDF would be the previously calculated CDF. Is it actually the same as my calculated distribution function and can we interpret the Beta Prime distribution as some sort of generalized uniform distribution? Does that yield anything interesting?

  • $\begingroup$ Signs of $x+y$ and $c$ seem to be important. $\endgroup$
    – Bob Dobbs
    Commented Oct 15, 2022 at 15:20
  • 1
    $\begingroup$ As $X,Y$ are exponentially distributed, it is implied that $X,Y,c> 0$ (due to the pdf) having the factor $1_{(0,\infty)}(x)$ at the end (though it could be with exception of some null sets that $c=0$ or something, however, I just disregarded that. Was I wrong in doing so? $\endgroup$ Commented Oct 15, 2022 at 15:26
  • $\begingroup$ $X$ and $Y$ are non-negative, thus $x/(x+y) \le c \le 1$ $\endgroup$
    – G Cab
    Commented Oct 15, 2022 at 16:58

1 Answer 1



Be careful: the numerator ($X$)and denominator ($X+Y$) are dependent

Computing first $Y/X$ and then $\frac{1}{1+Y/X}$

$$ \eqalign{ & X,Y \sim \;Exp(\lambda ) \sim Gamma\left( {1,\lambda } \right)\quad \Rightarrow \quad Y/X \sim Beta'(1,1)\quad \Rightarrow \quad 1/\left( {1 + Y/X} \right) \sim Beta(1,1) \cr & Beta(1,1) = U(0,1) \cr} $$ from Wikipedia article on Beta distribution beta_wiki

and Beta prime beta_wiki_3

  • $\begingroup$ @Snoop: right, amended $\endgroup$
    – G Cab
    Commented Oct 15, 2022 at 16:53
  • $\begingroup$ This answer looks a little terse to me, but should be understandable by the OP. $\endgroup$ Commented Oct 15, 2022 at 17:30
  • $\begingroup$ Thank you for answering my question (I don't know how I have overseen the dependence part), though I believe there is a tiny error and please correct me if I am wrong: The conversion is $Z\sim \beta(a,b)\implies \frac{1}{1-Z}\sim\beta'(a,b)$. You used the other direction and we would obtain that $1/(1-Y/X)\sim\beta(1,1)$. Right? But due to your first remark I was wrong to assume the beta prime distribution in the first place or am I missing something? $\endgroup$ Commented Oct 20, 2022 at 12:15
  • $\begingroup$ @EdwardTeach: what do you indicate with Z? $\endgroup$
    – G Cab
    Commented Oct 20, 2022 at 21:56
  • $\begingroup$ In this case $Z=Y/X$. You used $Y/X\sim\beta'(1,1)\implies 1/(1+Y/X)\sim\beta(1,1)$. I am not sure how you do this as (as far as I know) we can use the implication $Z\sim\beta(a,b)\implies 1/(1-Z)\sim\beta'(a,b)$ and not the other way around. Specifically there is also the leftover minus to address (in the case $Z=Y/X$). Right? $\endgroup$ Commented Oct 21, 2022 at 16:42

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