Difference between revisions of "ECE PhD QE CNSIP 2008 Problem1.1" - Rhea

Latest revision as of 10:08, 10 March 2015

1

The Weak Law of Large Numbers states that if $ \mathbf{X}_{1},\mathbf{X}_{2},\mathbf{X}_{3},\cdots $ is a sequence of i.i.d. random variables with finite mean $ E\left[\mathbf{X}_{i}\right]=\mu $ for every $ i $ , then the sample mean $ \mathbf{Y}_{n}=\frac{1}{n}\sum_{i=1}^{n}\mathbf{X}_{i} $ converges to $ \mu $ in probability. Suppose that instead of being i.i.d , $ \mathbf{X}_{1},\mathbf{X}_{2},\mathbf{X}_{3},\cdots $ each have finite mean $ \mu $ , and the covariance function of the sequence $ \mathbf{X}_{n} $ is $ Cov\left(\mathbf{X}_{i},\mathbf{X}_{j}\right)=\sigma^{2}\rho^{\left|i-j\right|} $ , where $ \left|\rho\right|<1 $ and $ \sigma^{2}>0 $ .

a. (13 points)

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

$ E\left[\mathbf{Y}_{n}\right]=E\left[\frac{1}{n}\sum_{i=1}^{n}\mathbf{X}_{i}\right]=\frac{1}{n}\sum_{i=1}^{n}E\left[\mathbf{X}_{i}\right]=\frac{n\mu}{n}=\mu. $

$ Cov\left(\mathbf{X}_{i},\mathbf{X}_{j}\right)=E\left[\left(\mathbf{X}_{i}-\mu\right)\left(\mathbf{X}_{j}-\mu\right)\right]=E\left[\mathbf{X}_{i}\mathbf{X}_{j}\right]-\mu^{2}. $

$ E\left[\mathbf{X}_{i}\mathbf{X}_{j}\right]=\begin{cases} \begin{array}{lll} \sigma^{2}+\mu^{2} & & ,\; i=j\\ \sigma^{2}\rho^{\left|i-j\right|}+\mu^{2} & & ,\; i\neq j. \end{array}\end{cases} $

$ E\left[\mathbf{Y}_{n}^{2}\right]=E\left[\frac{1}{n^{2}}\sum_{i=1}^{n}\sum_{j=1}^{n}\mathbf{X}_{i}\mathbf{X}_{j}\right]=\frac{1}{n^{2}}\left[\sum_{i=1}^{n}E\left[\mathbf{X}_{i}^{2}\right]+\underset{_{i\neq j}}{\sum_{i=1}^{n}\sum_{j=1}^{n}}E\left[\mathbf{X}_{i}\mathbf{X}_{j}\right]\right] $$ =\frac{1}{n^{2}}\left[n\left(\sigma^{2}+\mu^{2}\right)+2\sum_{i=1}^{n-1}\left(n-i\right)\left(\sigma^{2}\rho^{i}+\mu^{2}\right)\right]=\frac{1}{n^{2}}\left[n\sigma^{2}+2\sigma^{2}\sum_{i=1}^{n-1}\left(n-i\right)\rho^{i}\right]+\mu^{2}. $

$ Var\left[\mathbf{Y}_{n}\right]=E\left[\mathbf{Y}_{n}^{2}\right]-E\left[\mathbf{Y}_{n}\right]^{2}=\frac{1}{n^{2}}\left[n\sigma^{2}+2\sigma^{2}\sum_{i=1}^{n-1}\left(n-i\right)\rho^{i}\right]+\mu^{2}-\mu^{2}=\frac{1}{n^{2}}\left[n\sigma^{2}+2\sigma^{2}\sum_{i=1}^{n-1}\left(n-i\right)\rho^{i}\right]. $

Note

If the element $ m_{ij} $ of $ n\times n $ matrix is defined as $ \left|i-j\right| $ , then

$ \left[\begin{array}{ccccc} 0 & 1 & 2 & \cdots & n-1\\ 1 & 0 & 1 & \ddots & \vdots\\ 2 & 1 & 0 & \ddots & 2\\ \vdots & \ddots & \ddots & \ddots & 1\\ n-1 & \cdots & 2 & 1 & 0 \end{array}\right]. $

b. (12 points)

Does the sample mean still converge to $ \mu $ in probability? You must justify your answer.

ref. You can see the definition of convergence in probability.

If $ \mathbf{Y}_{n}\rightarrow\left(p\right)\rightarrow0 $ , then for any $ \epsilon>0 $ , $ P\left(\left\{ \left|\mathbf{Y}_{n}-\mu\right|>\epsilon\right\} \right)\rightarrow0 $ as $ n\rightarrow\infty $ . According to the Chebyshev inequality, $ P\left(\left\{ \left|\mathbf{Y}_{n}-\mu\right|>\epsilon\right\} \right)\leq\frac{\sigma_{\mathbf{Y}}^{2}}{\epsilon^{2}}. $ $ \lim_{n\rightarrow\infty}\sigma_{\mathbf{Y}}^{2} $ Therefore, we know that $ \lim_{n\rightarrow\infty}P\left(\left\{ \left|\mathbf{Y}_{n}-\mu\right|>\epsilon\right\} \right)=0 $ . Thus $ \mathbf{Y}_{n}\rightarrow\left(p\right)\rightarrow0 $ .