Communication, Networking, Signal and Image Processing (CS)

Question 1: Probability and Random Processes

August 2008

2

Let $\mathbf{X}_{1},\mathbf{X}_{2},\mathbf{X}_{3},\cdots$ be a sequence of i.i.d Bernoulli random variables with $p=1/2$ , and let $\mathbf{Y}_{n}=2^{n}\mathbf{X}_{1}\mathbf{X}_{2}\cdots\mathbf{X}_{n}$ .

a. (15 points)

Does the sequence $\mathbf{Y}_{n}$ converge to $0$ almost everywhere? • You can see the definition of convergence almost everywhere.

• range of $\mathbf{X}_{i}=\left\{ 0,1\right\}$ .

• range of $\mathbf{Y}_{n}=\left\{ 0,2^{n}\right\}$ .

• the probability of $\mathbf{Y}_{n}=2^{n}\Longrightarrow P\left(\mathbf{Y}_{n}=2^{n}\right)=P\left(\left\{ \mathbf{X}_{1}=1\right\} \cap\cdots\cap\left\{ \mathbf{X}_{n}=1\right\} \right)=\left(\frac{1}{2}\right)^{n}.$

$\lim_{n\rightarrow\infty}P\left(\mathbf{Y}_{n}=2^{n}\right)=0$ . $\lim_{n\rightarrow\infty}P\left(\mathbf{Y}_{n}=0\right)=1$ .

• Thus, $\left\{ \mathbf{Y}_{n}\right\} \rightarrow0$ with probability 1. This is incorrect!

b. (15 points)

Does the sequence $\mathbf{Y}_{n}$ converge to 0 in the mean square sense?

Note

You can see the definition of convergence in mean-square.

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

$E\left[\left|\mathbf{Y}_{n}-0\right|^{2}\right]=E\left[\mathbf{Y}_{n}^{2}\right]=E\left[2^{2n}\mathbf{X}_{1}^{2}\mathbf{X}_{2}^{2}\cdots\mathbf{X}_{n}^{2}\right]=2^{2n}E\left[\mathbf{X}_{1}^{2}\mathbf{X}_{2}^{2}\cdots\mathbf{X}_{n}^{2}\right]=\left(4E\left[\mathbf{X}^{2}\right]\right)^{n}=2^{n}.$

$\lim_{n\rightarrow\infty}E\left[\mathbf{Y}_{n}^{2}\right]=\lim_{n\rightarrow\infty}2^{n}=\infty.$

Thus, $\mathbf{Y}_{n}$ does not converge to $0$ in the mean square sense.

## Alumni Liaison

Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.

Buyue Zhang