ECE Ph.D. Qualifying Exam in "Communication, Networks, Signal, and Image Processing" (CS)
Question 1, August 2011, Part 2
- Part 1,2]
$ \color{blue} \text{Show that if a continuous-time Gaussian random process } \mathbf{X}(t) \text{ is wide-sense stationary, it is also strict-sense stationary.} $
$ \color{blue}\text{Solution 1:} $
$ \mathbf{X}(t) \text{ is SSS if } F_{(t_1+\tau)...(t_n+\tau)}(x_1,...,x_n) \text{ does not depend on } \tau. \text{ To show that, we can show that } \Phi_{(t_1+\tau)...(t_n+\tau)}(\omega_1,...,\omega_n) \text{ does not depend on } \tau: $
$ \Phi_{(t_1+\tau)...(t_n+\tau)}(\omega_1,...,\omega_n) = E \left [e^{i\sum_{j=1}^{n}{\omega_jX(t_j+\tau)}} \right ] $
$ \text{Define } Y(t_j+\tau) = \sum_{j=1}^{n}{\omega_jX(t_j+\tau)} \text{, so} $
$ \Phi_{(t_1+\tau)...(t_n+\tau)}(\omega_1,...,\omega_n) = E \left [e^{Y(t_j+\tau)} \right ] = \Phi_{(t_1+\tau)...(t_n+\tau)}(1) $
$ \text{Since } Y(t) \text{ is Gaussian, it is characterized just by its mean and variance. So, we just need to show that mean and variance of } Y(t) \text{ do not depend on . Since } Y(t) \text{ is WSS, its mean is constant and does not depend on . For variance} $
$ \color{blue}\text{Solution 2:} $
here put sol.2
"Communication, Networks, Signal, and Image Processing" (CS)- Question 1, August 2011
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- Part 1: solutions and discussions
- Part 2: solutions and discussions
