Transformation of Exponential Random Variables I came across this example, and I was completly thrown off! Hopefully someone will be able to help me start it or even explain it all to me! Thanks for any help!
X and Y are two independent random variables with exponential distribution and parameter $\lambda = 1$
Suppose that $Z=\frac{X}{X+Y}$ . 
a)Explain why ImZ = (0,1)
b) By computing Fz(z) = P(Z ≤ z), show that the random variable Z has a uniform distribution on the interval (0, 1)
I think maybe I have to use something like the Convolution Formula for b) but not sure how to

For (a), it suffices to prove that for $x,y > 0$, you have $0 < \frac{x}{x+y} < 1$. This should be straightforward.
For (b): after (a), you can assume $z\in(0,1)$. Rewrite
$$\begin{align}
\mathbb{P}\{Z \leq z\} &= \mathbb{P}\left\{X \leq \frac{z}{1-z} Y\right\} = \int_{0}^\infty\int_{0}^\infty \mathbb{1}_{\{x \leq \frac{z}{1-z}y\}} dydx e^{-y}e^{-x} \\
&= \int_{0}^\infty dy e^{-y} \int_{0}^{\frac{z}{1-z}y}dx e^{-x}
\end{align}$$
and you can compute this last quantity by direct integration: also, to conclude recall that for a uniform distribution, $F(z) = z$ for $z\in(0,1)$.
