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Let $ \{X_t\}$ ~ AR(1): $$ X_t=2.62-0.84X_{t-1}+\epsilon_t, \ \ \ \epsilon_t\sim WN(0,2.27)$$

Compute the variance of $$ \overline{X}= \frac{1}{3}\sum_{t=1}^{3} X_t $$

The solution is: Var($\overline X$)=0.9

How do I get that result?

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  • $\begingroup$ Why don't you write down the explicit expression of $\overline{X}$? $\endgroup$ – Richard Hardy Jun 7 at 11:51
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First, you'll get that $\operatorname{var}(X_t)=v\approx7.71$ (Hint: take the variance of both sides of the AR equation). And, use the fact that $$\operatorname{var}(\bar{X})=\frac{1}{9}\left(2(\operatorname{cov}(X_3,X_2)+\operatorname{cov}(X_3,X_1)+\operatorname{cov}(X_1,X_2))+\sum_{t=1}^3\operatorname{var}(X_t)\right)$$ Specific values of $t$ doesn't matter: $$\operatorname{cov}(X_t,X_{t-1})=\operatorname{cov}(-0.84X_{t-1}+\epsilon_t,X_{t-1})=-0.84v$$ Similarly $$\operatorname{cov}(X_t,X_{t-2})=-0.84\operatorname{cov}(X_{t-1},X_{t-2})=0.84^2v$$ Substituting all into the formula of $\operatorname{var}(\bar{X})$ yields the answer you need, i.e. $\operatorname{var}(\bar{X})\approx 0.9$:

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By computing $\overline{X}$, we obtain $\overline{X} = \tfrac{1}{3}\left(X_1+X_2+X_3\right) = \tfrac{1}{3}\left(9.708672+0.8656\epsilon_1+0.16\epsilon_2+\epsilon_3\right)$

Then, $\mathbb{V}ar\left[\overline{X}\right] = \left[\left(\tfrac{0.8656}{3}\right)^2+\left(\tfrac{0.16}{3}\right)^2+\left(\tfrac{1}{3}\right)^2\right](2.27) = 0.44766$

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