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I've become a bit confused by texts and answers that seem to contradict each other but I think I am just not understanding it quite right.

For an ARMA(p,q) process $$ X_{t} = Z_{t} + \phi(1)X_{t-1} + \dots \phi(p) X_{t-p} + \theta(1)Z_{t-1} + \dots \theta(q)Z_{t-q} $$

I have seen the condition for stationarity being that the roots of $\phi(x)$ are not on the unit circle.

And for $AR(p)$ processes $$ X_{t} = Z_{t} + \phi(1)X_{t-1} + \dots \phi(p) X_{t-p} $$ that they are outside of the unit circle.

A comment in this accepted answer mentions that its stationary but not invertible and it has some upvotes. Yet in my homework problems we have a few that are inside the unit circle and they are not stationary. I also checked them visually and they dont' seem to be.

Yet, an AR(p) process is just an ARMA(p, 0) process, so I don't understand how the conditions could differ?

Any insights are appreciated, I've been reading a lot of different sources trying to understand this and I'm still confused after a few weeks.

User1865345
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Oliver
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1 Answers1

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This is the basic fact that stationarity of an $\sf ARMA$ process depends solely on the autoregressive parameters.

Notice for $\sf ARMA(p, q), $

$$\left(1-\phi_1L-\phi_2L^2-\cdots-\phi_pL^p\right)X_t=\left(1+\theta_1L+\theta_2L^2+\cdots+\theta_qL^q\right)Z_t,$$ the roots of $1-\phi_1 x-\phi_2x^2-\cdots-\phi_px^p=0\tag 1\label 1$ must lie outside the unit circle for both sides of $\eqref 1$ to be divided by the compound lag operator $\phi(L)$ to reach $$ X_t =\psi(L)Z_t$$ where $\sum_{i=0}^\infty |\psi_i|<\infty.$

Reference:

$[\rm I] $ Time Series Analysis, James Douglas Hamilton, Princeton University Press, $1994, $ sec. $3.5, $ pp. $59-60.$

User1865345
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  • Thanks a lot, so that top comment in the other answer is wrong right? I've also seen this in endless lecture texts (upenn and UCLA), and I was utterly confused. Brockwell Davis p. 49 also notes that it is stationary if not on the unit circle. I would say I have found a 60/40 split in references. Though interestingly the references that provide AR stationary conditions often don't provide it for ARMA, which led me to this question. – Oliver Jan 23 '23 at 17:30
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    Technically, BD is talking about causality. So, yes, if the roots of $\phi(L) $ don't lie on the unit circle, they are stationary but only when they lie outside the unit circle, they are stationary and causal. But to answer your title question, it still depends on the $\sf AR$ part. – User1865345 Jan 23 '23 at 17:59
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    That's what I happened to come across while googling casually - check this note. – User1865345 Jan 23 '23 at 18:14
  • thank you for your patience. I am still confused. This is precisely my question. If not being on the unit circle is enough, then being inside it is ok, and from what I am reading, the comment, BD and the note you linked as well as your comment above say that if the $\phi(L)$ are not on the unit circle an $ARMA(p,q)$ process is stationary, this only relies don't he $AR$ part, but to me it contradicts that for $AR$ the $\phi(L)$ have to lie outside the unit circle. This is my whole confusion, they seem to be different conditions when they shouldn't be. – Oliver Jan 24 '23 at 08:19
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    See, Oliver you are near and yet far. Yes, if you ask what should be the condition for stationarity? The answer is straight forward: not on the unit circle. But then if you want real practical stationarity ie. along with causality, then the roots must lie outside the unit circle. Hamilton and others didn't mention explicitly causality but for all practical purpose, we would entertain the causal stationarity. – User1865345 Jan 24 '23 at 09:11
  • Thanks again for being so patient with me. I will have to ponder more on what you mean. Though for many simpler examples, where the roots are inside the unit circle we get explosive behaviour with trend. And this clearly violates the constant mean condition of stationarity, I am probably missing something here. – Oliver Jan 24 '23 at 09:37
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    (+1), especially for the nice reference, I love that book! @Oliver, it seems to me that this answers your query, and if so, please consider upvoting & accepting it. – utobi Jan 27 '23 at 18:13
  • @utobi I upvoted it (and all the comments, as I am grateful for the attempt) but it didn't answer my query. From the discussion, the answerer provides the same contradiction that led me to this question. Seemingly different conditions for $AR$ and $ARMA$ processes, but also the same. Above the answerer says not on the unit circle => stationary. Just not practical. But I have countless examples that aren't stationary and are inside the unit circle. – Oliver Jan 28 '23 at 09:53
  • @utobi if you or anyone else agrees that I am just not getting it and it answers it, I guess then I'll accept, to not keep this questions lingering around. Also mentioning that the reference given, is one of the texts I mention in my question that contradicts other texts leading to my confusion – Oliver Jan 28 '23 at 09:55