Communication, Networking, Signal and Image Processing (CS)

Question 1: Probability and Random Processes

January 2002

3. (20 pts)

Let $\mathbf{X}_{t}$ be a zero mean continuous parameter random process. Let $g(t)$ and $w\left(t\right)$ be measurable functions defined on the real numbers. Further, let $w\left(t\right)$ be even. Let the autocorrelation function of $\mathbf{X}_{t}$ be $\frac{g\left(t_{1}\right)g\left(t_{2}\right)}{w\left(t_{1}-t_{2}\right)}$ . From the new random process $\mathbf{Y}_{i}=\frac{\mathbf{X}\left(t\right)}{g\left(t\right)}$ . Is $\mathbf{Y}_{t}$ w.s.s. ?

## Solution 1

$E\left[\mathbf{Y}\left(t\right)\right]=E\left[\frac{\mathbf{X}\left(t\right)}{g\left(t\right)}\right]=\frac{1}{g\left(x\right)}E\left[\mathbf{X}\left(t\right)\right]=0.$

$E\left[\mathbf{Y}\left(t_{1}\right)\mathbf{Y}\left(t_{2}\right)\right]=E\left[\frac{\mathbf{X}\left(t_{1}\right)\mathbf{X}^{\star}\left(t_{2}\right)}{g\left(t_{1}\right)g\left(t_{2}\right)}\right]=\frac{1}{g\left(t_{1}\right)g\left(t_{2}\right)}E\left[\mathbf{X}\left(t_{1}\right)\mathbf{X}^{\star}\left(t_{2}\right)\right]$$=\frac{1}{g\left(t_{1}\right)g\left(t_{2}\right)}\times\frac{g\left(t_{1}\right)g\left(t_{2}\right)}{w\left(t_{1}-t_{2}\right)}=\frac{1}{w\left(t_{1}-t_{2}\right)},$

which depends on $t_{1}-t_{2}$ .
$\therefore\;\mathbf{Y}_{t}\text{ is wide-sense stationary.}$

## Alumni Liaison

Prof. Math. Ohio State and Associate Dean
Outstanding Alumnus Purdue Math 2008

Jeff McNeal