Relation of distribution of $\frac{X}{X+Y}$ where $X,Y$ are iid. rv s.t exponential distribution and parameter $\alpha>0$ and $\beta'(\alpha,\alpha)$.

Question

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?

Answer 1

Hint:

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

and Beta prime