If $X_1$ and $X_2$ are iid random variables from the exponential distribution with parameter $\lambda$. I need to find the pdf of $X_1/(X_1 + X_2)$.
As of now I have used $X_1=\lambda e^{-\lambda x}$ and $X_2=\lambda e^{-\lambda x}$ or essentially the same thing. If we plug in for $X_1/(X_1+X_2)$ we get $\lambda e^{-\lambda x}/(\lambda e^{-\lambda x}+\lambda e^{-\lambda x})=\lambda e^{-\lambda x}/(2\lambda e^{-\lambda x})=\frac{1}{2}$ and then we integrate. So $P(0\leq x \leq \infty)=\int\frac{1}{2}dx$ from 0 to $\infty$ which seems to be indeterminant.
Or it could be $\Gamma(\lambda,1)/(\Gamma(\lambda,2)$ and then integrating? I was thinking this is incorrect because the operation is addition and not mulitplication.
Best Answer
I assume you mean without using the properties of Poisson processes or any other tools which would make the result immediate.
You want to compute the CDF and then take derivative for the pdf. First step is always writing the probability of interest and applying simple algebra. For any $0<x<1$:
$P(\frac{X_1}{X_1+X_2}\leq x)=...=P(X_2\geq \frac{1-x}{x}X_1)$.
When you have two random variables in the same probability the safest bet is conditioning:
$P(X_2>\frac{1-x}{x}X_1)=\int_0^\infty f_{X_1}(u)P(X_2>\frac{1-x}{x}X_1|X_1=u)du$.
From there you can use the independence assumption. After computing the integral you should get a uniform distribution.