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Revision as of 23:09, 9 March 2015
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) $ .
$ 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). $
