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}=? $
