Communication, Networking, Signal and Image Processing (CS)

Question 1: Probability and Random Processes

August 2003

4. (25% of Total)

Let $\mathbf{X}_{n},\; n=1,2,\cdots$ , be a zero mean, discrete-time, white noise process with $E\left(\mathbf{X}_{n}^{2}\right)=1$ for all $n$ . Let $\mathbf{Y}_{0}$ be a random variable that is independent of the sequence $\left\{ \mathbf{X}_{n}\right\}$ , has mean $0$ , and has variance $\sigma^{2}$ . Define $\mathbf{Y}_{n},\; n=1,2,\cdots$ , to be an autoregressive process as follows: $\mathbf{Y}_{n}=\frac{1}{3}\mathbf{Y}_{n-1}+\mathbf{X}_{n}.$

a. (20 %)

Show that $\mathbf{Y}_{n}$ is asymptotically wide sense stationary and find its steady state mean and autocorrelation function.

$\mathbf{Y}_{n}=\frac{1}{3}\mathbf{Y}_{n-1}+\mathbf{X}_{n}=\frac{1}{3}\left(\frac{1}{3}\mathbf{Y}_{n-2}+\mathbf{X}_{n-1}\right)+\mathbf{X}_{n}=\left(\frac{1}{3}\right)^{2}\mathbf{Y}_{n-2}+\mathbf{X}_{n}+\frac{1}{3}\mathbf{X}_{n-1} $$=\cdots=\left(\frac{1}{3}\right)^{n}\mathbf{Y}_{0}+\sum_{k=0}^{n-1}\left(\frac{1}{3}\right)^{k}\mathbf{X}_{n-k}. E\left[\mathbf{Y}_{n}\right]=E\left[\left(\frac{1}{3}\right)^{n}\mathbf{Y}_{0}+\sum_{k=0}^{n-1}\left(\frac{1}{3}\right)^{k}\mathbf{X}_{n-k}\right]=\left(\frac{1}{3}\right)^{n}E\left[\mathbf{Y}_{0}\right]+\sum_{k=0}^{n-1}\left(\frac{1}{3}\right)^{k}E\left[\mathbf{X}_{n-k}\right]$$ =\left(\frac{1}{3}\right)^{n}\cdot0+\sum_{k=0}^{n-1}\left(\frac{1}{3}\right)^{k}\cdot0=0.$

$E\left[\mathbf{Y}_{m}\mathbf{Y}_{n}\right]=\left(\frac{1}{3}\right)^{m+n}E\left[\mathbf{Y}_{0}^{2}\right]+\left(\frac{1}{3}\right)^{m}E\left[\mathbf{Y}_{0}\sum_{k=0}^{n-1}\left(\frac{1}{3}\right)^{k}\mathbf{X}_{n-k}\right] $$\qquad+\left(\frac{1}{3}\right)^{n}E\left[\mathbf{Y}_{0}\sum_{k=0}^{m-1}\left(\frac{1}{3}\right)^{k}\mathbf{X}_{n-k}\right]+\sum_{i=0}^{m-1}\sum_{j=0}^{n-1}\left(\frac{1}{3}\right)^{i+j}E\left[\mathbf{X}_{m-i}\cdot\mathbf{X}_{n-j}\right]$$ =\left(\frac{1}{3}\right)^{m+n}\cdot\left(\sigma^{2}+0^{2}\right)+\sum_{k=1}^{\min\left(m,n\right)}\left(\frac{1}{3}\right)^{m+n-2k}$$=\left(\frac{1}{3}\right)^{m+n}\cdot\sigma^{2}+\sum_{k=1}^{\min\left(m,n\right)}\left(\frac{1}{3}\right)^{m+n-2k}.$

$\because\;\sum_{i=0}^{m-1}\sum_{j=0}^{n-1}\left(\frac{1}{3}\right)^{i+j}E\left[\mathbf{X}_{m-i}\cdot\mathbf{X}_{n-j}\right]=\sum_{i=1}^{m}\sum_{j=1}^{n}\left(\frac{1}{3}\right)^{m-i}\left(\frac{1}{3}\right)^{n-j}E\left[\mathbf{X}_{i}\cdot\mathbf{X}_{j}\right]=\sum_{k=1}^{\min\left(m,n\right)}\left(\frac{1}{3}\right)^{m+n-2k}.$

b. (5%)

For what choice of $\sigma^{2}$ is the process wide sense stationary; i.e., not just asymptotically wide sense stationary?

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