Difference between revisions of "ECE PhD QE CNSIP 2004 Problem1.2" - Rhea

Latest revision as of 00:36, 10 March 2015

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}=? $