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I have the following multiple regression model with autocorrelated error terms:

$Y_t=\beta_0+\beta_1X_{t1}+\beta_2X_{t2}+...+\beta_{p-1}X_{t,p-1}+\epsilon_t$

$\varepsilon_t=\rho\varepsilon_{t-1}+u_t$

where $u_t$'s are independent $N(0,\sigma^2)$ disturbance terms and $|\rho|<1$. Also, although the $\varepsilon_t$'s are correlated over time, they still have mean 0 and constant variance: $E(\varepsilon_t)=0, \sigma^2(\varepsilon_t)=\frac{\sigma^2}{1-\rho^2}$ with the $\sigma^2$ in the variance, the variance of the $u_t$ disturbance terms.

I need to derive and simplify the term for $Cov(\varepsilon_t,\varepsilon_{t-2})$

This is what I have so far:

$Cov(\varepsilon_t,\varepsilon_{t-2})=E[(\varepsilon_t-\mu_t)(\varepsilon_{t+2}-\mu_{t+2})]=E(\varepsilon_t\varepsilon_{t-2})=E[(\rho\varepsilon_{t-1}+u_t)(\rho\varepsilon_{t-3}+u_{t-2})]=E(\rho^2\varepsilon_{t-1}\varepsilon_{t-3})+E(\rho\varepsilon_{t-1}u_t)+E(\rho\varepsilon_{t-3}u_{t-2})+E(u_tu_{t-2})=\rho^2E(\varepsilon_{t-1}\varepsilon_{t-3})+\rho E(\varepsilon_{t-1}u_t)+\rho E(\varepsilon_{t-3}u_{t-2})+0$

I'm not sure where to go from here. With the $E(\varepsilon_{t-1}\varepsilon_{t-3})$ term, is this not a similar situation as in the beginning, with $E(\varepsilon_t\varepsilon_{t-2})$?

I believe that I should end up with $Cov(\varepsilon_t,\varepsilon_{t-2})=\rho^2(\frac{\sigma^2}{1-\rho^2})$. Am I taking the right approach? Any guidance would be greatly appreciated!

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1 Answer 1

up vote 2 down vote accepted

Applying the formula $\text{cov} \left( \varepsilon_t, \varepsilon_{t - k} \right) = E \left[ \varepsilon_t \varepsilon_{t - k} \right] - E \left[ \varepsilon_t \right] E \left[ \varepsilon_{t - k} \right]$ for any $k > 0$ and using the fact that $E \left[ \varepsilon_t \right] = E \left[ \varepsilon_{t - k} \right] = 0$, we deduce \begin{eqnarray*} \text{cov} \left( \varepsilon_t, \varepsilon_{t - 2} \right) & = & E \left[ \left( \rho \varepsilon_{t - 1} + u_t \right) \varepsilon_{t - 2} \right]\\ & = & \rho E \left[ \varepsilon_{t - 1} \varepsilon_{t - 2} \right] + E \left[ u_t \varepsilon_{t - 2} \right]\\ & = & \rho E \left[ \varepsilon_{t - 1} \varepsilon_{t - 2} \right] \end{eqnarray*} where we used the fact that $u_t$ is independent of $\varepsilon_t$ (innovation process). We need $E \left[ \varepsilon_{t - 1} \varepsilon_{t - 2} \right]$. We know by stationarity that it is equal to $E \left[ \varepsilon_t \varepsilon_{t - 1} \right] = \rho E \left[ \varepsilon_{t - 1}^2 \right] + E \left[ u_t \varepsilon_{t - 1} \right] = \rho \frac{\sigma^2}{1 - \rho^2}$.

Thus the answer is $\text{cov} \left( \varepsilon_t, \varepsilon_{t - 2} \right) = \rho^2 \frac{\sigma^2}{1 - \rho^2}$.

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1  
@Jess. Do you understand (or know how to prove) why $u_t$ is independent of $\varepsilon_t$? –  Learner Mar 13 '13 at 11:54
    
Yes. I don't know why I didn't see that initially. Thank you so much for the help! It makes everything a lot clearer. –  Jess Mar 14 '13 at 20:10

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