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=Example. Mean of i.i.d. random variables= | =Example. Mean of i.i.d. random variables= | ||
| − | Let <math>\mathbf{X}_{1},\mathbf{X}_{2},\cdots,\mathbf{X}_{n}</math> be <math>M</math> jointly distributed i.i.d. random variables with mean <math>\mu</math> and variance <math>\sigma^{2}</math> . Let <math>\mathbf{Y}_{M}=\frac{1}{M}\sum_{n=0}^{M}\mathbf{X}_{n}</math> . | + | Let <math class="inline">\mathbf{X}_{1},\mathbf{X}_{2},\cdots,\mathbf{X}_{n}</math> be <math class="inline">M</math> jointly distributed i.i.d. random variables with mean <math class="inline">\mu</math> and variance <math class="inline">\sigma^{2}</math> . Let <math class="inline">\mathbf{Y}_{M}=\frac{1}{M}\sum_{n=0}^{M}\mathbf{X}_{n}</math> . |
| − | (a) Find the variance of <math>\mathbf{Y}_{M}</math> . | + | (a) Find the variance of <math class="inline">\mathbf{Y}_{M}</math> . |
| − | <math>Var\left[\mathbf{Y}_{M}\right]=E\left[\mathbf{Y}_{M}^{2}\right]-\left(E\left[\mathbf{Y}_{M}\right]\right)^{2}. </math> | + | <math class="inline">Var\left[\mathbf{Y}_{M}\right]=E\left[\mathbf{Y}_{M}^{2}\right]-\left(E\left[\mathbf{Y}_{M}\right]\right)^{2}. </math> |
| − | <math>E\left[\mathbf{Y}_{M}\right]=E\left[\frac{1}{M}\sum_{n=0}^{M}\mathbf{X}_{n}\right]=\frac{1}{M}\sum_{n=0}^{M}E\left[\mathbf{X}_{n}\right]=\frac{1}{M}\cdot M\cdot\mu=\mu.</math> | + | <math class="inline">E\left[\mathbf{Y}_{M}\right]=E\left[\frac{1}{M}\sum_{n=0}^{M}\mathbf{X}_{n}\right]=\frac{1}{M}\sum_{n=0}^{M}E\left[\mathbf{X}_{n}\right]=\frac{1}{M}\cdot M\cdot\mu=\mu.</math> |
| − | <math>E\left[\mathbf{Y}_{M}^{2}\right]=E\left[\frac{1}{M^{2}}\sum_{m=1}^{M}\sum_{n=1}^{M}\mathbf{X}_{m}\mathbf{X}_{n}\right]=\frac{1}{M^{2}}\sum_{m=1}^{M}\sum_{n=1}^{M}E\left[\mathbf{X}_{m}\mathbf{X}_{n}\right].</math> | + | <math class="inline">E\left[\mathbf{Y}_{M}^{2}\right]=E\left[\frac{1}{M^{2}}\sum_{m=1}^{M}\sum_{n=1}^{M}\mathbf{X}_{m}\mathbf{X}_{n}\right]=\frac{1}{M^{2}}\sum_{m=1}^{M}\sum_{n=1}^{M}E\left[\mathbf{X}_{m}\mathbf{X}_{n}\right].</math> |
| − | Now <math>E\left[\mathbf{X}_{m}\mathbf{X}_{n}\right]=\begin{cases} | + | Now <math class="inline">E\left[\mathbf{X}_{m}\mathbf{X}_{n}\right]=\begin{cases} |
\begin{array}{ll} | \begin{array}{ll} | ||
E\left[\mathbf{X}_{m}^{2}\right] ,m=n\\ | E\left[\mathbf{X}_{m}^{2}\right] ,m=n\\ | ||
E\left[\mathbf{X}_{m}\right]E\left[\mathbf{X}_{n}\right] ,m\neq n | E\left[\mathbf{X}_{m}\right]E\left[\mathbf{X}_{n}\right] ,m\neq n | ||
| − | \end{array}\end{cases}</math> because when <math>m\neq n</math> , <math>\mathbf{X}_{m}</math> and <math>\mathbf{X}_{n}</math> are independent <math>\Rightarrow \mathbf{X}_{m}</math> and <math>\mathbf{X}_{n}</math> are uncorrelated. | + | \end{array}\end{cases}</math> because when <math class="inline">m\neq n</math> , <math class="inline">\mathbf{X}_{m}</math> and <math class="inline">\mathbf{X}_{n}</math> are independent <math class="inline">\Rightarrow \mathbf{X}_{m}</math> and <math class="inline">\mathbf{X}_{n}</math> are uncorrelated. |
| − | <math>E\left[\mathbf{Y}_{M}^{2}\right]=\frac{1}{M^{2}}\left[M\left(\mu^{2}+\sigma^{2}\right)+M\left(M-1\right)\mu^{2}\right]=\frac{\left(\mu^{2}+\sigma^{2}\right)+\left(M-1\right)\mu^{2}}{M}=\frac{M\mu^{2}+\sigma^{2}}{M}.</math> | + | <math class="inline">E\left[\mathbf{Y}_{M}^{2}\right]=\frac{1}{M^{2}}\left[M\left(\mu^{2}+\sigma^{2}\right)+M\left(M-1\right)\mu^{2}\right]=\frac{\left(\mu^{2}+\sigma^{2}\right)+\left(M-1\right)\mu^{2}}{M}=\frac{M\mu^{2}+\sigma^{2}}{M}.</math> |
| − | <math>Var\left[\mathbf{Y}_{M}\right]=\frac{M\mu^{2}+\sigma^{2}-M\mu^{2}}{M}=\frac{\sigma^{2}}{M}.</math> | + | <math class="inline">Var\left[\mathbf{Y}_{M}\right]=\frac{M\mu^{2}+\sigma^{2}-M\mu^{2}}{M}=\frac{\sigma^{2}}{M}.</math> |
| − | (b) Now assume that the <math>\mathbf{X}_{1},\mathbf{X}_{2},\cdots,\mathbf{X}_{n}</math> are identically distributed with with mean <math>\mu</math> and variance <math>\sigma^{2}</math> , but they are only correlated rather than independent. Find the variance of <math>\mathbf{Y}_{M}</math> . | + | (b) Now assume that the <math class="inline">\mathbf{X}_{1},\mathbf{X}_{2},\cdots,\mathbf{X}_{n}</math> are identically distributed with with mean <math class="inline">\mu</math> and variance <math class="inline">\sigma^{2}</math> , but they are only correlated rather than independent. Find the variance of <math class="inline">\mathbf{Y}_{M}</math> . |
| − | Again, <math>Var\left[\mathbf{Y}_{M}\right]=\frac{\sigma^{2}}{M}</math> , because only uncorrelatedness was used in part (a). | + | Again, <math class="inline">Var\left[\mathbf{Y}_{M}\right]=\frac{\sigma^{2}}{M}</math> , because only uncorrelatedness was used in part (a). |
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Revision as of 07:14, 1 December 2010
Example. Mean of i.i.d. random variables
Let $ \mathbf{X}_{1},\mathbf{X}_{2},\cdots,\mathbf{X}_{n} $ be $ M $ jointly distributed i.i.d. random variables with mean $ \mu $ and variance $ \sigma^{2} $ . Let $ \mathbf{Y}_{M}=\frac{1}{M}\sum_{n=0}^{M}\mathbf{X}_{n} $ .
(a) Find the variance of $ \mathbf{Y}_{M} $ .
$ Var\left[\mathbf{Y}_{M}\right]=E\left[\mathbf{Y}_{M}^{2}\right]-\left(E\left[\mathbf{Y}_{M}\right]\right)^{2}. $
$ E\left[\mathbf{Y}_{M}\right]=E\left[\frac{1}{M}\sum_{n=0}^{M}\mathbf{X}_{n}\right]=\frac{1}{M}\sum_{n=0}^{M}E\left[\mathbf{X}_{n}\right]=\frac{1}{M}\cdot M\cdot\mu=\mu. $
$ E\left[\mathbf{Y}_{M}^{2}\right]=E\left[\frac{1}{M^{2}}\sum_{m=1}^{M}\sum_{n=1}^{M}\mathbf{X}_{m}\mathbf{X}_{n}\right]=\frac{1}{M^{2}}\sum_{m=1}^{M}\sum_{n=1}^{M}E\left[\mathbf{X}_{m}\mathbf{X}_{n}\right]. $
Now $ E\left[\mathbf{X}_{m}\mathbf{X}_{n}\right]=\begin{cases} \begin{array}{ll} E\left[\mathbf{X}_{m}^{2}\right] ,m=n\\ E\left[\mathbf{X}_{m}\right]E\left[\mathbf{X}_{n}\right] ,m\neq n \end{array}\end{cases} $ because when $ m\neq n $ , $ \mathbf{X}_{m} $ and $ \mathbf{X}_{n} $ are independent $ \Rightarrow \mathbf{X}_{m} $ and $ \mathbf{X}_{n} $ are uncorrelated.
$ E\left[\mathbf{Y}_{M}^{2}\right]=\frac{1}{M^{2}}\left[M\left(\mu^{2}+\sigma^{2}\right)+M\left(M-1\right)\mu^{2}\right]=\frac{\left(\mu^{2}+\sigma^{2}\right)+\left(M-1\right)\mu^{2}}{M}=\frac{M\mu^{2}+\sigma^{2}}{M}. $
$ Var\left[\mathbf{Y}_{M}\right]=\frac{M\mu^{2}+\sigma^{2}-M\mu^{2}}{M}=\frac{\sigma^{2}}{M}. $
(b) Now assume that the $ \mathbf{X}_{1},\mathbf{X}_{2},\cdots,\mathbf{X}_{n} $ are identically distributed with with mean $ \mu $ and variance $ \sigma^{2} $ , but they are only correlated rather than independent. Find the variance of $ \mathbf{Y}_{M} $ .
Again, $ Var\left[\mathbf{Y}_{M}\right]=\frac{\sigma^{2}}{M} $ , because only uncorrelatedness was used in part (a).
