Communication, Networking, Signal and Image Processing (CS)

Question 1: Probability and Random Processes

August 2006

1

Let $\mathbf{U}_{n}$ be a sequence of independent, identically distributed zero-mean, unit-variance Gaussian random variables. The sequence $\mathbf{X}_{n}$ , $n\geq1$ , is given by $\mathbf{X}_{n}=\frac{1}{2}\mathbf{U}_{n}+\left(\frac{1}{2}\right)^{2}\mathbf{U}_{n-1}+\cdots+\left(\frac{1}{2}\right)^{n}\mathbf{U}_{1}.$

(a) (15 points)

Find the mean and variance of $\mathbf{X}_{n}$ .

i) Find $E\left[\mathbf{X}_{n}\right]$

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

ii) Find $E\left[\mathbf{X}_{n}^{2}\right]$

$E\left[\mathbf{X}_{n}^{2}\right]=E\left[\left(\sum_{k=0}^{n-1}\left(\frac{1}{2}\right)^{k+1}\mathbf{U}_{n-k}\right)^{2}\right]=E\left[\sum_{k=0}^{n-1}\sum_{j=0}^{n-1}\left(\frac{1}{2}\right)^{k+1}\left(\frac{1}{2}\right)^{j+1}\mathbf{U}_{n-k}\mathbf{U}_{n-j}\right] $$=E\left[\sum_{k=0}^{n-1}\left(\frac{1}{2}\right)^{2k+2}\mathbf{U}_{n-k}^{2}+\underset{k\neq j}{\sum_{k=0}^{n-1}\sum_{j=0}^{n-1}}\left(\frac{1}{2}\right)^{k+1}\left(\frac{1}{2}\right)^{j+1}\mathbf{U}_{n-k}\mathbf{U}_{n-j}\right]$$ =\sum_{k=0}^{n-1}\left(\frac{1}{2}\right)^{2k+2}E\left[\mathbf{U}_{n-k}^{2}\right]+\underset{k\neq j}{\sum_{k=0}^{n-1}\sum_{j=0}^{n-1}}\left(\frac{1}{2}\right)^{k+1}\left(\frac{1}{2}\right)^{j+1}E\left[\mathbf{U}_{n-k}\right]E\left[\mathbf{U}_{n-j}\right]$$=\sum_{k=0}^{n-1}\left(\frac{1}{2}\right)^{2k+2}=\sum_{k=1}^{n}\left(\frac{1}{2}\right)^{2k}=\frac{\left(\frac{1}{2}\right)^{2}\left(1-\left(\frac{1}{2}\right)^{2n}\right)}{1-\left(\frac{1}{2}\right)^{2}}=\frac{1}{3}\left(1-\left(\frac{1}{2}\right)^{2n}\right).$

iii) Find $Var\left[\mathbf{X}_{n}\right]$

$Var\left[\mathbf{X}_{n}\right]=E\left[\mathbf{X}_{n}^{2}\right]-\left(E\left[\mathbf{X_{n}}\right]\right)^{2}=\frac{1}{3}\left(1-\left(\frac{1}{2}\right)^{2n}\right).$

(b) (15 points)

Find the characteristic function of $\mathbf{X}_{n}$ .

Since $\mathbf{U}_{n}$ is a sequence of i.i.d. Gaussian random variables, $\mathbf{X}_{n}$ is a sequence of Gaussian random variables with zero mean and variance $\sigma_{\mathbf{X}_{n}}^{2}=\frac{1}{3}\left(1-\left(\frac{1}{2}\right)^{2n}\right)$ . Hence the characteristic function of $\mathbf{X}_{n}$ is $\Phi_{\mathbf{X}_{n}}\left(\omega\right)=\exp\left(i\mu_{\mathbf{X}_{n}}\omega-\frac{1}{2}\sigma_{\mathbf{X}_{n}}^{2}\omega^{2}\right)=\exp\left(-\frac{\omega^{2}}{6}\left(1-\left(\frac{1}{2}\right)^{2n}\right)\right).$

(c) (10 points)

Does the sequence $\mathbf{X}_{n}$ converge in distribution? A simple yes or no answer is not sufficient. You must justify your answer.

$\Phi=F_{\mathbf{X}_{n}}\left(x\right)=\int_{-\infty}^{x}\frac{1}{\sqrt{2\pi}\sigma_{\mathbf{X}_{n}}}\exp\left(-\frac{x'^{2}}{2\sigma_{\mathbf{X}_{n}}^{2}}\right)dx'$ where $\sigma_{\mathbf{X}_{n}}^{2}=\frac{1}{3}\left(1-\left(\frac{1}{2}\right)^{2n}\right)$ .

Since $\lim_{n\rightarrow\infty}\sigma_{\mathbf{X}_{n}}^{2}=\frac{1}{3} , \lim_{n\rightarrow\infty}F_{\mathbf{X}_{n}}=\int_{-\infty}^{x}\frac{1}{\sqrt{\frac{2\pi}{3}}}\exp\left(-\frac{x'^{2}}{2\sigma_{\mathbf{X}_{n}}^{2}}\right)dx'=F_{\mathbf{X}}\left(x\right).$

$\therefore$ The squance $\mathbf{X}_{n}$ converges in distribution.