Communication, Networking, Signal and Image Processing (CS)

Question 1: Probability and Random Processes

January 2002

2. (20 pts)

Let $\mathbf{X}_{t}$ and $\mathbf{Y}_{t}$ by jointly wide sense stationary continous parameter random processes with $E\left[\left|\mathbf{X}\left(0\right)-\mathbf{Y}\left(0\right)\right|^{2}\right]=0$ . Show that $R_{\mathbf{X}}\left(\tau\right)=R_{\mathbf{Y}}\left(\tau\right)=R_{\mathbf{XY}}\left(\tau\right)$ .

## Solution 1

$E\left[\mathbf{X}\left(t\right)\left(\mathbf{X}^{\star}\left(t+\tau\right)-\mathbf{Y}^{\star}\left(t+\tau\right)\right)\right]=E\left[\mathbf{X}\left(t\right)\mathbf{X}^{\star}\left(t+\tau\right)\right]-E\left[\mathbf{X}\left(t\right)\mathbf{Y}^{\star}\left(t+\tau\right)\right]=R_{\mathbf{X}}\left(\tau\right)-R_{\mathbf{XY}}\left(\tau\right).$

$E\left[\left|\mathbf{X}\left(t\right)\right|^{2}\right]=E\left[\mathbf{X}\left(t\right)\mathbf{X}^{\star}\left(t\right)\right]=R_{\mathbf{X}}\left(0\right).$

$E\left[\left|\mathbf{X}\left(t+\tau\right)-\mathbf{Y}\left(t+\tau\right)\right|^{2}\right]=E\left[\left(\mathbf{X}\left(t+\tau\right)-\mathbf{Y}\left(t+\tau\right)\right)\left(\mathbf{X}^{\star}\left(t+\tau\right)-\mathbf{Y}^{\star}\left(t+\tau\right)\right)\right] $$=R_{\mathbf{X}}\left(0\right)-R_{\mathbf{YX}}\left(0\right)-R_{\mathbf{XY}}\left(0\right)+R_{\mathbf{Y}}\left(0\right)$$ =E\left[\mathbf{X}\left(0\right)\mathbf{X}^{\star}\left(0\right)\right]-E\left[\mathbf{Y}\left(0\right)\mathbf{X}^{\star}\left(0\right)\right]-E\left[\mathbf{X}\left(0\right)\mathbf{Y}^{\star}\left(0\right)\right]+E\left[\mathbf{Y}\left(0\right)\mathbf{Y}^{\star}\left(0\right)\right] $$=E\left[\mathbf{X}\left(0\right)\mathbf{X}^{\star}\left(0\right)-\mathbf{Y}\left(0\right)\mathbf{X}^{\star}\left(0\right)-\mathbf{X}\left(0\right)\mathbf{Y}^{\star}\left(0\right)+\mathbf{Y}\left(0\right)\mathbf{Y}^{\star}\left(0\right)\right]$$ =E\left[\left(\mathbf{X}\left(0\right)-\mathbf{Y}\left(0\right)\right)\left(\mathbf{X}\left(0\right)-\mathbf{Y}\left(0\right)\right)^{\star}\right]=E\left[\left|\mathbf{X}\left(0\right)-\mathbf{Y}\left(0\right)\right|^{2}\right].$

By Cauchy-Schwarz inequality, $\left|R_{\mathbf{X}}\left(\tau\right)-R_{\mathbf{XY}}\left(\tau\right)\right|^{2}\leq R_{\mathbf{X}}\left(0\right)E\left[\left|\mathbf{X}\left(0\right)-\mathbf{Y}\left(0\right)\right|^{2}\right]=0$ .

$\therefore\; R_{\mathbf{X}}\left(\tau\right)=R_{\mathbf{XY}}\left(\tau\right).$ Similarly,

$E\left[\left(\mathbf{X}\left(t\right)-\mathbf{Y}\left(t\right)\right)\mathbf{Y}^{\star}\left(t+\tau\right)\right]^{2}\leq E\left[\left|\mathbf{X}\left(t\right)-\mathbf{Y}\left(t\right)\right|^{2}\right]E\left[\left|\mathbf{Y}\left(t+\tau\right)\right|^{2}\right]$$\left|R_{\mathbf{XY}}\left(\tau\right)-R_{\mathbf{Y}}\left(\tau\right)\right|^{2}\leq E\left[\left|\mathbf{X}\left(0\right)-\mathbf{Y}\left(0\right)\right|^{2}\right]R_{\mathbf{Y}}\left(0\right)=0.$

$\therefore\; R_{\mathbf{XY}}\left(\tau\right)=R_{\mathbf{Y}}\left(\tau\right).$

Thus, $R_{\mathbf{X}}\left(\tau\right)=R_{\mathbf{Y}}\left(\tau\right)=R_{\mathbf{XY}}\left(\tau\right).$

## Alumni Liaison

Basic linear algebra uncovers and clarifies very important geometry and algebra.

Dr. Paul Garrett