# 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.)

• it would be the same as demonstrating that $X_1/(X_1+X_2)$ is standard uniform. – Glen_b -Reinstate Monica Oct 31 '19 at 6:07

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{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}

This confirms that $$R \sim \text{Exp}(1)$$.

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.