Difference between revisions of "ECE PhD QE CNSIP 2006 Problem1.1" - Rhea

Latest revision as of 09:36, 10 March 2015

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.