4
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
0
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.