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Notation: $\dot{Z}_t = Z_t - E(Z_t)$, so that it is centered at 0. $a_t$ stands for the residual and we assume the $a_t$ are independent and normally distributed with mean 0 and constant standard deviation $\sigma_a^2$. And, when I say stationarity, I mean weak stationarity and not strict stationarity.

In my time series class, we have been given that an MA(q) model is of the form $$\dot{Z}_t = \theta_q(B) a_t$$ where $$\theta_q(B) = 1 - \theta_1 B - \theta_2 B^2 - \cdots - \theta_q B^q.$$ We were told MA(q) is always stationary and it is invertible when all of the roots of $\theta_q(B)$ are outside the unit circle.

For an AR(p) model, we were told it is of the form $$\phi_p(B) \dot{Z}_t = a_t$$ where $$\phi_p(B) = 1 - \phi_1 B - \phi_2 B^2 - \cdots - \phi_p B^p.$$ We were told this model is stationary when all of the roots of $\phi_p(B)$ are outside the unit circle, but I don't remember hearing anything about when these are invertible. If I use the parallels from the MA(q) model, I might say the AR(p) model is always invertible. Is that correct?

Thanks

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    $\begingroup$ It would help to define your notation a little more, even though many readers will be able to make educated guesses. What is $a_t$? What is the significance of the dot over $\dot Z_t$, etc? Are you considering a Gaussian model so that there is some chance of strict stationarity or a more general model wherein only second-order stationarity can be hoped for? Also, does the definition of invertibility you are using implicitly assume causal invertibility or does it admit noncausal inversion? $\endgroup$ – cardinal Feb 28 '12 at 20:57
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    $\begingroup$ Yes, good points. I believe I have taken care of most of what you asked about. I don't know what you mean by a Gaussian model but it may not matter since I now have made it clear I mean weak stationarity. For invertibility, I have no idea what you're asking. For the MA(q) model, the point of invertibility is you can write it in the form of an infinite AR, and the coefficients decrease fast enough that the infinite AR converges. $\endgroup$ – GeoffDS Feb 28 '12 at 21:03
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    $\begingroup$ The point regarding invertibility is: Are you considering only solutions that involve an infinite series of previous terms or also potentially of future terms. The former is often said to have a causal structure and the latter is noncausal, for obvious reasons. :) $\endgroup$ – cardinal Feb 28 '12 at 21:14
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    $\begingroup$ @cardinal only previous terms $\endgroup$ – GeoffDS Feb 29 '12 at 18:18
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The answer to your question can be summarized as follows:

  • Pure MA models are always stationary (since they contain no AR terms).
  • Pure MA models may or may not be invertible.
  • Pure AR models are always invertible (since they contain no MA terms).
  • Pure AR models may or may not be stationary.
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  • $\begingroup$ When you say 'invertible' does that mean 'invertible to an autoregressive model'? Or does the third point actually contain the much stronger statement that pure autoregressive models are always invertible to a moving average process? $\endgroup$ – Omar Haque May 22 '17 at 13:12
  • $\begingroup$ Walsh, is the above answer valid for p=infinity and q=infinity for AR(p) and MA(q)? Why or why not? $\endgroup$ – Erdogan CEVHER Aug 22 '19 at 13:22
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AR processes are always invertible and stationary iff the root of the characteristic polynomial are outside the unit circle.

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    $\begingroup$ Do you have a reference for this statement that the OP can use? This statement as an answer is a bit short to make something out of it. At least you could add a reason as to why your statement is true. $\endgroup$ – Andy Aug 22 '15 at 16:46
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    $\begingroup$ This seems to be a restatement of (part of) the information given in the question. $\endgroup$ – Silverfish Aug 22 '15 at 21:39
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    $\begingroup$ This statement is incorrect -- the question correctly states what you need roots outside the unit circle for and correct information relating to invertibility is already in Graeme Walsh's answer.... If you disagree, please provide some kind of argument for or reference for your statement. (also NB @Silverfish, I suspect you simply misread the above) $\endgroup$ – Glen_b Aug 23 '15 at 1:36
  • $\begingroup$ @Glen_b you're quite right - I misread it as a regurgitation of the AR stationary condition given in the question, when actually it is attempting to answer the question albeit incorrectly. $\endgroup$ – Silverfish Aug 23 '15 at 7:46

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