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.
