- #1

dionysian

- 53

- 1

## Homework Statement

Let X and Y be two independent random variables each exponentially distributed with parameter 1. Define a new random variable:

[tex]z = \frac{x}{{x + y}}[/tex]

Find the PDF of Z

## Homework Equations

## The Attempt at a Solution

[tex]\begin{array}{l}

{F_Z}(z) = P(Z < z) = P\left( {\frac{x}{{x + y}} < z} \right) = P\left( {x \le \frac{{zy}}{{1 - z}}} \right) \\

{F_Z}(z) = \int\limits_0^\infty {\int_0^{\frac{{zy}}{{1 - z}}} {{f_{xy}}(x,y)dxdy} } \\

{f_{xy}}(x,y) = {f_x}(x){f_y}(y) \\

{F_Z}(z) = \int\limits_0^\infty {\int_0^{\frac{{zy}}{{1 - z}}} {{f_x}(x){f_y}(y)dxdy = } } \int\limits_0^\infty {\int_0^{\frac{{zy}}{{1 - z}}} {{e^{ - x}}{e^{ - y}}dxdy} } = \int\limits_0^\infty {{e^{ - y}}\left[ {\int_0^{\frac{{zy}}{{1 - z}}} {{e^{ - x}}dx} } \right]} dy \\

{F_Z}(z) = \int\limits_0^\infty {{e^{ - y}}\left[ { - {e^{ - \frac{{zy}}{{1 - z}}}} + 1} \right]} dy = \int\limits_0^\infty { - {e^{ - y}}{e^{ - \frac{{zy}}{{1 - z}}}} + {e^{ - y}}} dy = \int\limits_0^\infty { - {e^{ - \frac{{y(1 - z) - zy}}{{1 - z}}}} + {e^{ - y}}} dy \\

{F_Z}(z) = \int\limits_0^\infty { - {e^{ - \frac{y}{{1 - z}}}} + {e^{ - y}}} dy = (1 - z){e^{ - \frac{y}{{1 - z}}}}|_0^\infty - {e^{ - y}}|_0^\infty = z \\

\end{array}[/tex]

Now i know that if i take the derivative of this i will get the "pdf" but its obviously wrong. Any thoughts?