Density function of summed variables Let $X$ and $Y$ be two independent and exponentially distributed random variables with parameter $\lambda$. 
Let $U := \frac{X}{X + Y}$.
Find the density function $f_U$ of $U$.
So we basically know: 
$$f_X(x) = \lambda e^{-\lambda x}\\
f_Y(y) = \lambda e^{-\lambda y}
$$
But I really have no clue how to find $f_{\frac{X}{X+Y}}(u)$.
Does anyone have some ideas and could help me?
 A: Use the CDF method. The support of $U$ is in $(0, 1)$ since you're dividing a positive random variable $X$ by a larger positive random variable $X+Y$, so let $u \in (0, 1)$. Then
$$\mathbb{P}(U \leq u) = \mathbb{P}\left(\dfrac{X}{X+Y} \leq u\right) = \mathbb{P}\left(\dfrac{1-u}{u}\cdot X \leq Y\right)\text{.}$$
Since $u \in (0, 1)$, then $\dfrac{1-u}{u}\cdot X$ will be a positively-sloped line in the $X$-$Y$ plane:

and 
$$\mathbb{P}\left(\dfrac{1-u}{u}\cdot X \leq Y\right) = \int_{0}^{\infty}\int_{(1-u)x/u}^{\infty}f_{X, Y}(x, y)\text{ d}y\text{ d}x\text{.}$$
By independence, $$f_{X, Y}(x, y) = \lambda^2e^{-\lambda x}e^{-\lambda y}$$
for $x, y > 0$. Hence,
$$\begin{align}
\int_{0}^{\infty}\int_{(1-u)x/u}^{\infty}f_{X, Y}(x, y)\text{ d}y\text{ d}x &= \lambda\int_{0}^{\infty}e^{-\lambda x}e^{-\lambda (1-u)x/u}\text{ d}x \\
&= \lambda\int_{0}^{\infty}e^{-\lambda x/u}\text{ d}x \\
&= \lambda \cdot \dfrac{u}{\lambda} \\
&= u\text{.}
\end{align}$$
There are several shortcuts that I used above that you should be able to identify. If you don't understand how I got from one step to the next, please let me know. Take the derivative of this with respect to $u$ to get $$f_{U}(u) = \begin{cases}
1, & u \in (0, 1) \\
0, & \text{otherwise.}
\end{cases}$$
Hence, $U$ is uniform in $(0, 1)$.
