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Let's take an AR(p) model $\phi(L)y_t=z_t$ where $\phi(L)=1-\phi_1-...-\phi_pL^p$ and L is the lag operator. I have just studied that if there are no roots of the polynomial on the unit circle,

$1/\phi(L)=\sum_{j=-\infty}^\infty\psi_jL^j$

and

$\sum_{j=-\infty}^\infty|\psi_j|<\infty$.

But it is also true that if the above condition holds, then the process $y_t=\sum_{j=-\infty}^\infty\psi_jz_{t-j}$ is stationary, provided that $z_t$ is stationary.

So my question is: is it correct to say that all AR process such that the polynomial $\phi$ has no roots on the unit circle are stationary? This would imply that also a process like $y_t=2y_{t-1}+z_t$ is stationary (although non-causal).

This does not fit with what I have been told before, i.e. that an AR process is stationary if and only if it has all roots oustide the unit circle. So what is the correct stationarity condition for AR processes?

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  • $\begingroup$ Relation with causality is easy: correlation does not imply causation. $\endgroup$ – Tim♦ May 24 '18 at 17:59
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    $\begingroup$ @Tim, I think causality in the context of AR models is a different thing than causality as in cause-effect relation, so I wonder in which way your comment could be relevant. $\endgroup$ – Richard Hardy May 24 '18 at 18:36
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It might not be relevant for @MrJames anymore but my answer might help others.

Have a look at "Introduction to Time Series and Forecasting" by Brockwell and Davis (third edition). Here they say that a stationary solution of $\Phi(L)y_t=z_t$ exists and is unique if and only if $\Phi$ has no roots on the unit circle. It is causual if all roots of $\Phi$ are outside the unit circle.

Hence, causality implies stationarity but not the other way around.

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  • $\begingroup$ I am not sure whether arbitrary roots inside the unit circle are compatible with stationarity, hence I think this definition of stationarity is questionable. $\endgroup$ – Richard Hardy Jan 10 '19 at 8:53
  • $\begingroup$ I didn't give a definition for stationarity. This tehorem just states when a stationary solution exists and is unique. You can find the proof in the book I mentioned. $\endgroup$ – sunny Jan 10 '19 at 13:05
  • $\begingroup$ Sorry about the imprecise wording. I will try better: I am not sure whether arbitrary roots inside the unit circle are compatible with stationarity. I checked the book and at least on the surface, it indeed looks like your quote is correct. I am left wondering what to do with explosive processes such as $x_t=5x_{t-1}+\varepsilon_t$ – are they stationary? $\endgroup$ – Richard Hardy Jan 10 '19 at 14:07
  • $\begingroup$ In the case of explosive roots, it is still possible to solve the difference equation forwards. Then you are writing the current value as an infinite sum of future values. It remains stationary in the sense intended (weak / covariance stationarity). I think this is why causality is an interesting property - it is stationarity combined with the restriction that the current value only depends on prior values. See e.g. Brockwell and Davis, Example 2.2.1. for their discussion of this issue (although I only have the 2nd edition, so maybe the number has changed in the 3rd). $\endgroup$ – cfulton Mar 29 '19 at 1:33
  • $\begingroup$ @RichardHardy, yes, explosive processes like yours are stationary but not invertible, my answer here stats.stackexchange.com/a/400800/36041 $\endgroup$ – Aksakal Jun 6 at 4:14

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