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

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.

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


• I really can't see how this will help me get to the desired answer @BruceET – Will Dec 10 '15 at 19:55
• Sorry, maybe someone will post a complete answer. There are people on this site that are really good at displaying such answers. My job is to help you get started. To that end, how about trying out the basic method on a less-'gross' problem. Maybe just finding the joint distribution of $V = X_1$ and $W = X_1 + X_2$, then integrating out $V$ to find the 'marginal' dist'n of $W$. (With $X_1$ and $X_2$ indep exponentials (same mean) as in your problem. Notice that $W$ has a gamma distribution. – BruceET Dec 10 '15 at 20:10