$X,Y$ are independent exponentially distributed: What is the distribution of $X/(X+Y)$ I'm a business major, so really distribution theory is not my strong suit. I know how to get the distribution of a ratio of exponential variables and of the sum of them, but i can't piece everything together.
The exercise goes as this:
If $X,Y$ are independent exponentially distributed with $\beta=1$ (parameter of the exponential distribution $= 1$) then what is the distribution of $X/(X+Y)$
Any ideas?
Thanks a lot.
 A: Obviously $X/(X+Y)$ is between $0$ and $1$.  Let (lower-case) $x$ be a number between $0$ and $1$.
$$
\frac{X}{X+Y} \le x \text{ if and only if } (1-x)X \le xY.
$$
We will find this probability.
The conditional probability that $Y\ge\dfrac{(1-x)}{x}X$, given the value of $X$, is
$$
e^{-(1-x)X/x}.
$$
The probability we seek is then the expected value of that:
\begin{align}
\mathbb E e^{-(1-x)X/x} & = \int_0^\infty e^{-(1-x)w/x} \Big(e^{-w}\,dw\Big) \\[10pt]
& = \int_0^\infty e^{-w/x} \, dw \\[10pt]
& = x.
\end{align}
In other words, this random variable is uniformly distributed between $0$ and $1$.
Second method:  The equality $Y=\frac{1-w}{w}X$ is a straight line through the $(X,Y)$-plane, passing through $(0,0)$ and having positive slope.
We can let $y$ go from $0$ to $\infty$ and then for any fixed value of $y$, let $x$ go from $0$ to $\frac{w}{1-w}y$
\begin{align}
& {}\qquad\int_0^\infty \left(\int_0^{wy/(1-w)} e^{-y} e^{-x} \,dx \right) \, dy \\[10pt]
& = \int_0^\infty e^{-y}\left(1-e^{-wy/(1-w)}\right) \, dy \\[10pt]
& = \int_0^\infty e^{-y} - e^{-y/(1-w)} \, dy  \\[10pt]
& = 1 - (1-w) \\[10pt]
& = w.
\end{align}
