What is the distribution of $\ln(\frac{X_1+X_2}{X_1})$ when $X_1, X_2 \sim \text{IID Exp}(1)$? I know it is $\text{Exp}(1)$ but I cannot prove this. (EDIT: I figured out an argument, I will draft a proof and share it here.) 
 A: Taking $X_1,X_2 \sim \text{IID Exp}(1)$ we have: 
$$R \equiv \ln \bigg( \frac{X_1+X_2}{X_1} \bigg) \sim  \text{Exp}(1).$$
There are various ways to demonstrate this.  In cases of difficulty, the simplest way is to derive the CDF.  For all $r \geqslant 0$ we have:
$$\begin{equation} \begin{aligned}
F_R(r) = \mathbb{P}(R \leqslant r)
&= 1- \mathbb{P}( R > r ) \\[6pt]
&= 1-\mathbb{P}( e^R > e^r ) \\[6pt]
&= 1-\mathbb{P}( X_1+X_2 > X_1 \cdot e^r ) \\[6pt]
&= 1-\mathbb{P}( X_2 > X_1 \cdot (e^r-1) ) \\[6pt]
&= 1-\int \limits_0^\infty (1-F_{X_2}( x \cdot (e^r-1))) f_{X_1}(x) \ dx \\[6pt]
&= 1-\int \limits_0^\infty \exp(-x \cdot (e^r-1)) \exp(-x) \ dx \\[6pt]
&= 1-\int \limits_0^\infty \exp(-x \cdot e^r) \ dx \\[6pt]
&= 1-\Big[ -e^{-r} \exp(-x \cdot e^r) \Big]_{x=0}^{x \rightarrow \infty} \\[6pt]
&= 1- e^{-r}. \\[6pt]
\end{aligned} \end{equation}$$
This confirms that $R \sim \text{Exp}(1)$.
A: Observe that $ln(\frac{X_1+X_2}{X_1})$ = $-ln(\frac{X_1}{X_1+X_2})$. Since $-ln({U})$ is exp(1) when U ~ Uniform(0,1). If we can prove $\frac{X_1}{X_1+X_2}$ is standard uniform then we will have completed the proof. 
Observe that
$P(\frac{X_1}{X_1+X_2}\leq t) = P(\frac{X_1+X_2}{X_1}\leq \frac{1}{t}) = P\{X_2\leq X_1(\frac{1}{t}-1)\}= \int_0^\infty f_{X_1}(x_1)P(X_2 \leq x_1(\frac{1}{t}-1))dx = \int_0^{\infty}e^{-x_1}e^{-x_1(\frac{1}{t}-1)}dx_1 = \int_0^{\infty}e^\frac{x_1}{t}dx_1=t$
This implies $(\frac{X_1+X_2}{X_1})$ is uniform, completing the proof.
