# Probability Question - Joint density function

1. Mar 4, 2015

### cpatel23

Before I begin, here is the question:

If the PDF of two independent random variables X and Y are:
f(x) = exp(-x)u(x)
f(y) = exp(-y)u(y)
Determine the join probability density function (JPDF) of Z&W defined by:
Z = X+Y
W = X/(X+Y).

So, I know how to solve this except for one thing. How do I get the expression for Z and W.
For Z do I literally just add f(x) + f(y) meaning z = exp(-x) + exp(-y) for (x,y) >0? Same with W?
Once I get the expressions I just find fxy(x,y) and divide by the determinant of the Jacobian. The problem is that the Jacobian depends on the derivative of Z and W which I do not know how to get an expression for.

2. Mar 4, 2015

### mathman

The density function for Z is the convolution of the density functions for X and Y. For W it is much more complicated.

3. Mar 4, 2015

### Ray Vickson

The formula for the density of a sum $X+Y$ of two independent random variables with densities $f_X(x)$ and $f_Y(y)$ is found in every probability textbook, as well as on-line. If you don't know it you can work it out from first principles; one way is to get density $f_Z(z)$ from the fact that
$$f_Z(z) \, \Delta z \doteq P(z < X+Y < z + \Delta z) \;\text{as} \: \Delta z \to 0$$
and to integrate the joint density $f_X(x) f_Y(y)$ over the region $\{z < x+y < z + \Delta z \}$ in $(x,y)-$ space.

To get the joint density of $(Z,W)$, use the standard transformation formulas that involve Jacobians, etc. The joint density of $(X,Y)$ is $f_X(x) f_Y(y)$, and certainly does not involve a sum $f_X(x) + f_Y(y)$, or anything like it.