Communication, Networking, Signal and Image Processing (CS)

Question 1: Probability and Random Processes

August 2004

## Question

2. (20 pts.)

Let $\mathbf{X}$ and $\mathbf{Y}$ be two independent identically distributed exponential random variables having mean $\mu$ . Let $\mathbf{Z}=\mathbf{X}+\mathbf{Y}$ . Find $f_{\mathbf{X}}\left(x|\mathbf{Z}=z\right)$ , the conditional pdf of $\mathbf{X}$ given the event $\left\{ \mathbf{Z}=z\right\}$ .

Note

This problem is very simlar to the example except that it deals with the exponential random variable rather than the Poisson random variable.

Solution

By using Bayes' theorem,

$f_{\mathbf{X}}\left(x|\mathbf{Z}=z\right)=\frac{f_{\mathbf{XZ}}\left(x,z\right)}{f_{\mathbf{Z}}\left(z\right)}=\frac{f_{\mathbf{Z}}\left(z|\mathbf{X}=x\right)f_{\mathbf{X}}\left(x\right)}{f_{\mathbf{Z}}\left(z\right)}=\frac{f_{\mathbf{Y}}\left(z-x\right)f_{\mathbf{X}}\left(x\right)}{f_{\mathbf{Z}}\left(z\right)}=?$

Acording to the definition of the exponential distribution, $f_{\mathbf{X}}\left(x\right)=\frac{1}{\mu}e^{-\frac{x}{\mu}}\text{ and }f_{\mathbf{Y}}\left(y\right)=\frac{1}{\mu}e^{-\frac{y}{\mu}}.$

$\Phi_{\mathbf{X}}\left(\omega\right)=\Phi_{\mathbf{Y}}\left(\omega\right)=\frac{1}{1-i\mu\omega}.$

$\Phi_{\mathbf{Z}}\left(\omega\right)=E\left[e^{i\omega\mathbf{Z}}\right]=E\left[e^{i\omega\left(\mathbf{X}+\mathbf{Y}\right)}\right]=E\left[e^{i\omega\mathbf{X}}\right]E\left[e^{i\omega\mathbf{Y}}\right]=\Phi_{\mathbf{X}}\left(\omega\right)\Phi_{\mathbf{Y}}\left(\omega\right)=\frac{1}{1-i\mu\omega}\cdot\frac{1}{1-i\mu\omega}=?$

## Alumni Liaison

has a message for current ECE438 students.

Sean Hu, ECE PhD 2009