Probability Theory - Transformation (of two variables) of continuous random variables

Question

Let $X_1$ and $X_2$ be independent and identically distributed continuous random variables, with probability density function

$$p(x)=\begin{cases} \exp(-x), & \text{if}\ x>0 \\ 0, & \text{otherwise}. \end{cases}$$

Let

$$Z=\frac{X_1}{X_1+X_2}+2X_2$$

Derive the probably density function of $Z$. It is sufficient to give the required pdf in the form of an integral of a joint pdf.

This is pretty gross and I cannot see how to start this question.

Answer 1

Comments:

To answer your second question, I'm guessing you're expected to 'integrate out' the unwanted variable(s) to give the pdf.

The standard method is to transform to variables $Y$ and $Z$ (where $Y$ is chosen for convenience), and then integrate $Y$ out of the joint density. As you work, be sure to keep in mind the support of each random variable involved.

Below is a brief simulation in R that gives 100,000 realizations of $Z$, and some clues about the answer. 'ECDF' is the empirical CDF of the realizations.

 m = 10^5;  x1 = rexp(m);  x2 = rexp(m)
 z = x1/(x1 + x2) + 2*x2
 mean(z);  sd(z)
 ## 2.499098    # Does your density give E(Z) = 2.5?
 ## 1.849711