Covariance of Autocorrelated Error Terms 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!
 A: 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 $\text{cov} \left( \varepsilon_t,
\varepsilon_{t - 2} \right) = \rho^2 \frac{\sigma^2}{1 - \rho^2}$.
