How could I compute the first order autocorrelation of the process $x_t = \delta + \phi x_{t-1} + \eta_t$? Could anyone give me some pointers?

I tried this:

$E(\delta + \phi x_{t-1} + \eta_t - \frac{\delta}{1- \phi})(\delta + \phi x_{t-2} + \eta_{t-1} - \frac{\delta}{1- \phi})$. But how could I compute for instance $E(x_{t-1}x_{t-2})$ here? Thanks.

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    $\begingroup$ I think this answer will be helpful stats.stackexchange.com/questions/68243/… $\endgroup$ – Alecos Papadopoulos Jun 15 '14 at 13:47
  • $\begingroup$ @AlecosPapadopoulos Your answer is indeed similar, but only in that example it's the case that $E(y_t)=0$, making for the fact that you can recognize $E(y^2_t)$ as the variance in the expression you find for the covariance. Here I cannot seem to do that, because of the non-zero mean (so I don't know what $E(x_{t-1}x_{t-2})$ will be. Would you know how I could apply a similar procedure here? $\endgroup$ – user3482499 Jun 15 '14 at 14:00
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    $\begingroup$ What exactly troubles you in the expression $$E[x_{t-1}x_{t-2}] = E[(\delta + \phi x_{t-2} + \eta_{t-1})x_{t-2}]$$ $$=\delta E(x_{t-2})+\phi E[x_{t-2}^2]+E[\eta_{t-1}x_{t-2}]$$ ? The variance of a AR(1) with drift is widely known and available, so... $\endgroup$ – Alecos Papadopoulos Jun 15 '14 at 14:46
  • $\begingroup$ @AlecosPapadopoulos The term $E[x^2_{t-2}]$. I thought there should be a nice way to quickly derive that (without making use of any results). $\endgroup$ – user3482499 Jun 15 '14 at 14:55
  • $\begingroup$ But don't you see that you can calculate it by rearranging the defining expression for the variance? $\endgroup$ – Alecos Papadopoulos Jun 15 '14 at 14:56

This seems to work. Write the process as $x_t- \mu = \phi(x_{t-1} - \mu) + \eta_t$, where $\mu = \frac{\delta}{1-\phi}$. Then it is easy to see that $\gamma_1 = \phi \gamma_0$. Also, from the original equation we get that $\gamma_0 = \phi^2 \gamma_0 + \sigma^2$ so that $\gamma_0=\frac{\sigma^2}{1-\phi^2}$. The result then follows.

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