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.} $
