Determine the pdf of $U = \dfrac{X}{(X+Y)}$

Question

Let $X$ and $Y$ be two independent random variables and $X,Y \sim Exp(1)$.

Determine the pdf of $U = \dfrac{X}{(X+Y)}$

I've only just started doing transforms and I'm not sure how to do $U = \dfrac{X}{(X+Y)}$.

Answer 1

At the first step, we are going to compute the CDF. Next, we compute the PDF from the CDF.

\begin{align} F_{U}(u) &= P(U\leq u) \\&= P(\frac{X}{X+Y}\leq u) \\&= P\big(X(1-u)\leq uY\big) \\&= P(X\leq \frac{u}{1-u}Y) \\&= \int_{0}^{\infty}\int_{0}^{\frac{u}{1-u}y}f_{X,Y}(x,y)dxdy \quad (1) \\&= \int_{0}^{\infty}\int_{0}^{\frac{u}{1-u}y}f_{X}(x)f_{Y}(y)dxdy \quad (2) \end{align}

We can reach from Relation $(1)$ to Relation $(2)$ because $X$ and $Y$ are two independent random variables.

You must put the PDF of $X$ and $Y$ into Relation $(2)$ and after integration, you have the CDF of $U$. Finally, PDF = $\frac{d}{du}CDF$.

Answer 2

Note that $$\mathbb{P}(U\leq u) = \mathbb{P}\left(X\leq \frac{u}{1-u}Y\right) = \mathbb{P}(X\leq Z)$$ if $Z\sim\text{Exp}((1-u)/u)$ independent to $X$. Moreover: $$\mathbb{P}(X_1\leq Z_1) = \frac{\mu}{\lambda+\mu}$$ for $X_1\sim\text{Exp}(\lambda)$ and $Z_1\sim\text{Exp}(\mu)$ independent. See here why. Note that the notation of $U$ is quite suggestive.