In practice, how to evaluate whether a AR(P) process is stationary or not?

How to determine the order for the AR and MA model?

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    $\begingroup$ For an AR process to be stationary the roots of the AR polynomial must be outside the unit circle. Thus if the model is an AR(1) the coefficient must be absolutely less than 1.0 . All AR processes are not stationary. $\endgroup$ – IrishStat Dec 13 '11 at 21:54
  • $\begingroup$ @IrishStat - yes, you're correct. I wasn't thinking straight. Perhaps you can post that as an answer. $\endgroup$ – Macro Dec 13 '11 at 22:01
  • $\begingroup$ @IrishStat: I do not understand your comment, particularly the last sentence. Is there a typo there? $\endgroup$ – cardinal Dec 14 '11 at 1:09
  • $\begingroup$ Perhaps I should have said "AR processes are not necessarily stationary" $\endgroup$ – IrishStat Dec 14 '11 at 16:42
  • $\begingroup$ @IrishStat: Ah. That makes more sense. :) $\endgroup$ – cardinal Dec 14 '11 at 17:31

Extract the roots of the polynomial. If all the roots are outside the unit circle then the process is stationary. Model identification aids can be found on the web. Fundamentally the pattern of the ACF's and the pattern of the PACF's are used to identify which model might be a good starting model. If there are more significant ACF's than significant PACF's then an AR model is suggested as the ACF is dominant. if the converse is true where the PACF is dominant then an MA model might be appropriate. The order of the model is suggested by the number of significant values in the subordinate.

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    $\begingroup$ Actualy the roots should not be on the unit circle. If the roots are inside the unit circle, the solution is stationary, but not invertible. $\endgroup$ – mpiktas Dec 14 '11 at 7:09
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    $\begingroup$ Where can I find the proof of such theorem (or at least an schema of the proof ?) $\endgroup$ – Antoni Apr 25 '16 at 15:27
  • $\begingroup$ Where can I find more info on acf vs pacf for AR vs MA model? $\endgroup$ – mlstudent Feb 29 at 5:34
  • $\begingroup$ autobox.com/pdfs/ForecastingSeminar.pdf starting at slide 161 would be a good practical place to start.. $\endgroup$ – IrishStat Feb 29 at 9:06
  • $\begingroup$ people.duke.edu/~rnau/arimrule.htm might also be useful IF AND ONLY IF your data is free of pulses , level/step shifts, seasonal pulses and/or local time trends AND has parameters that don't change over time AND has an error variance that doesn't change over time. Simple model identification tools work when the data is simple and uncomplicated. Otherwise Not So Much. ! $\endgroup$ – IrishStat Feb 29 at 9:27

If you have an AR(p) process like this:

$$ y_t = c + \alpha_1 y_{t - 1} + \cdots + \alpha_p y_{t - p} $$

Then you can build an equation like this:

$$ z^p - \alpha_1 z^{p - 1} - \cdots - \alpha_{p - 1} z - \alpha_p = 0 $$

Find the roots of this equation, and if all of them are less than 1 in absolute value, then the process is stationary.

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  • $\begingroup$ It's nice to see you contributing answers. Thanks! $\endgroup$ – whuber Dec 14 '11 at 17:14
  • $\begingroup$ Note, that you wrote less while it has to be greater ("outside the unit circle"). $\endgroup$ – Dmitrij Celov Dec 15 '11 at 12:52
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    $\begingroup$ @DmitrijCelov: No, I don't think so. Look carefully. It appears robbrit has used the $z$-transform and then multiplied by an additional $z^p$ factor, which will not change the location of the roots, except for adding one (of multiplicity $p$) at zero. If you factor out a $z^p$ and then substitute $B = z^{-1}$, you'll arrive at something that may look more familiar to you. The roots of the polynomial in $B$ must lie outside the unit circle. But, there is a simple correspondence between the roots of the polynomial in $B$ and the associated one in $z$. Cheers. :) $\endgroup$ – cardinal Dec 15 '11 at 13:58
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    $\begingroup$ @cardinal, you are right. robbrit, is not mentioning the $z$ transform, though yes he has made the one. The most of statistical packages however will return the roots for $1- \alpha_1 z - \dots -\alpha_p z^p = 0$ not for this one, so it could be a missleading suggestion for not so careful users (like me :D) if the $B= z^{-1}$ is not stressed. Thanks for explanation :) $\endgroup$ – Dmitrij Celov Dec 15 '11 at 14:12
  • $\begingroup$ @DmitrijCelov: It gave me a moment's pause on first reading as well. When I said "look carefully", it was not intended in any way as an admonishment (though I can see how it could be read that way!), but rather only as a sign that there was something subtle to be aware of. Cheers. :) $\endgroup$ – cardinal Dec 15 '11 at 14:20

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