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Suppose you have a process $x_{t} = \mu +w_{t} -0.8*w_{t-1}$ where $w_{t}$ ~ $wn(0,\sigma_{w}^2)$. How do I calculate the standard error of $var(\bar{x})$ for estimating the mean. I know: $var(\bar{x}) = \frac{1}{n}\sum_{j=-n}^n (1-\frac{|h|}{n})\gamma_{x}(h)$. I am stuck on what to do next. I know that $var(x_{t}) = \sigma_{w}^2 + 0.64*\sigma_{w}^2$. I am not sure how to use $|h|$ in the above expression. I know that $\gamma(0) = 1.64\sigma_{w}^2 , \gamma(1) = -0.8\sigma_{w}^2$ and $\gamma(h>=2) = 0$. So when I evaluate this sum I get $\frac{1}{n}\sum_{j=-1}^1(1-\frac{|h|}{n})\gamma_{x}(h) = \frac{1}{n}[(1-\frac{1}{n})(-0.8\sigma_{w}^2) + 1.64\sigma_{w}^2 + (1-\frac{1}{n})(-0.8\sigma_{w}^2) $ . Is this correct?

EDIT: The solutions manual states that the answer is $var(\bar{x}) = \frac{\sigma_{w}^2}{n}(1-2\frac{n-1}{n}0.8)$ but I am not sure how this was arrived at. I get $\frac{\sigma_{w}^2}{n}(1.68-2\frac{n-1}{n}0.8)$ using the approach above.

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I updated my answer. – lord12 Jan 17 '13 at 5:52
I am still not sure if my answer is correct. – lord12 Jan 19 '13 at 21:20
What is $\mu$ ? – Ewan Delanoy Jan 22 '13 at 14:07
It the overall population mean of the process. It is a constant – lord12 Jan 22 '13 at 18:18
Or is the solution manual wrong. Because I can't see any other way. – lord12 Jan 22 '13 at 18:31
up vote 2 down vote accepted

Yes, this is correct.

I assume that $$ \bar x = \frac{1}{n} (x_{t+n-1} + x_{t+n-2} + \dots + x_t), $$ for some $t>0$. In this case \begin{eqnarray*} {\rm Var\ } \bar x&=&\frac{1}{n^2} {\rm Var \ } (x_{t+n-1} + x_{t+n-2} + \dots + x_t)\\ &=& \frac{1}{n^2} {\rm Var \ }\left( (w_{t+n-1}-0.8 w_{t+n-2})+\dots+(w_t-0.8w_{t-1})\right), \end{eqnarray*} since the variance does not change upon adding a constant \begin{eqnarray*} &=&\frac{1}{n^2} {\rm Var \ }( w_{t+n-1}+0.2 w_{t+n-2}+\dots+0.2 w_t-0.8 w_{t-1})\\ &=& \frac{\sigma_{w}^2}{n^2} (1 + 0.2^2 + \dots + 0.2^2 + 0.8^2)\\ &=& \frac{\sigma_w^2}{n^2} (1 + (n-1) 0.04 + 0.64) \\ &=& \frac{\sigma_w^2}{n} (0.04 + \frac{1.6}{n}). \end{eqnarray*} This can also be obtained by your formula above: $$ \gamma_x(0)=(1+0.8^2) \sigma_w^2, \ \ \gamma_x(\pm 1)=-0.8 \sigma_w^2, \ \ \gamma_x(h)=0 {\rm \ \ otherwise}, $$ so \begin{eqnarray*} \frac{1}{n}\sum_h (1-\frac{|h|}{n})\gamma_{x}(h) &=&\frac{\sigma_w^2}{n} ((1+0.8^2)+2(1-\frac{1}{n})(-0.8))\\ &=&\frac{\sigma_w^2}{n} (0.04+\frac{1.6}{n}). \end{eqnarray*} If the solutions manual says that $$ {\rm Var\ }{\bar x}=\frac{\sigma_{w}^2}{n}(1-2\frac{n-1}{n}0.8),\qquad (*) $$ then this is obviously wrong as the right-hand side of (*) is negative for all $n\ge 3$.

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