# Whether a AR(P) process is stationary or not?

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?

• 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. Dec 13, 2011 at 21:54
• @IrishStat - yes, you're correct. I wasn't thinking straight. Perhaps you can post that as an answer. Dec 13, 2011 at 22:01
• @IrishStat: I do not understand your comment, particularly the last sentence. Is there a typo there? Dec 14, 2011 at 1:09
• Perhaps I should have said "AR processes are not necessarily stationary" Dec 14, 2011 at 16:42
• @IrishStat: Ah. That makes more sense. :) Dec 14, 2011 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.

• 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. Dec 14, 2011 at 7:09
• Where can I find the proof of such theorem (or at least an schema of the proof ?) Apr 25, 2016 at 15:27
• Where can I find more info on acf vs pacf for AR vs MA model? Feb 29, 2020 at 5:34
• autobox.com/pdfs/ForecastingSeminar.pdf starting at slide 161 would be a good practical place to start.. Feb 29, 2020 at 9:06
• 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. ! Feb 29, 2020 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.

• It's nice to see you contributing answers. Thanks!
– whuber
Dec 14, 2011 at 17:14
• Note, that you wrote less while it has to be greater ("outside the unit circle"). Dec 15, 2011 at 12:52
• @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. :) Dec 15, 2011 at 13:58
• @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 :) Dec 15, 2011 at 14:12
• @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. :) Dec 15, 2011 at 14:20