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According to Theorem 8.8 in Time Series A.W. van der Vaart an ARMA process $$\phi (L)X_t=\theta(L)\epsilon_t$$ has a unique stationary solution $X_t=\psi(L)\epsilon_t$ with $\psi=\theta/\phi$ if $\phi$ has no roots on the complex unit circle. This would imply that the explosive process, with $\rho>1$, is a stationary process $$X_t=\rho X_{t-1}+\epsilon_t$$ with stationary solution $X_t=\sum_{i=1}^\infty \rho^{-i}\epsilon_{t+i}$.

Now indeed $\sum_{i=1}^{\infty} \rho^{-i} < \infty$ so that weak stationarity can be proved by using this representation.

However, here on stackexchange I see a lot of question/answers that suggest that the process above is not stationary (see for example Are explosive ARMA(1, 1) processes stationary?, Non-Stationary: Larger-than-unit root). In particular, the accepted answer of the latter question claims that the process is non-stationary by simulating a series and showing it displays explosive trending behaviour.

I think the only way to reconcile the theorem I mention above and the plots in the accepted answer of (Non-Stationary: Larger-than-unit root) is the following: the explosive process is indeed stationary but non-ergodic, that is, we cannot find the statistical properties of $X_t$ such as $\mathbb{E}(X_t)=\mu$ by observing a single infinitely long sample path of the explosive process, mathematically: $$\lim_{t \to \infty}\frac{1}{t}\sum_{t=1}X_t \neq\mathbb{E}X_t$$

Is this reading correct?

Richard Hardy
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Joogs
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  • ARMA is stationary when the _roots of the autoregression polynomial lie outside the unit circle_. – corey979 Apr 02 '19 at 15:29
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    @corey979 I am baffled by the notion that an explosive process could be considered stationary, and would undoubtedly experience wonder and delight if I were shown that it is so. That said: the variances of explosive processes are functions of time, and the means of explosive processes are functions of time, and perturbations to explosive processes give stronger effects as more time passes since they occurred, so I am not understanding how an explosive process could be stationary in any sense. – Alexis Apr 02 '19 at 15:47
  • Also: Welcome to CV, Joogs! – Alexis Apr 02 '19 at 15:48
  • Might this be a duplicate of [this](https://stats.stackexchange.com/questions/400296/are-explosive-arma1-1-processes-stationary)? Or the other way around? – Richard Hardy Apr 02 '19 at 16:13
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    @Alexis, "variances of explosive processes are functions of time" - sure about this? – Aksakal Apr 02 '19 at 17:21
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    Excellent question on a subtle topic! – Aksakal Apr 02 '19 at 17:22
  • @Aksakal Am learning here, so I am placing my money and being willing to be wrong. :) Let me say that I would be surprised to learn that the variance of an explosive process is independent of time *without* invoking slight of hand like "if you assume a separate linear model at for each length of the time series". So, sure? :) The variance of $y_{t} = \rho y_{t-1} + \varepsilon_{t}$ grows with $t$ for $|\rho| \ge 1$ – Alexis Apr 02 '19 at 17:32
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    @Alexis, you have to turn time back – Aksakal Apr 11 '19 at 19:23
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    @Aksakal I spent some time starring at the solution, and it went click: you not only need a time machine for *some* point in the future, but for *all* futures. So you are quite correct when you say "I won't like it!" :) – Alexis Apr 12 '19 at 16:32
  • It seems me that this story not hold. The AR(1) model is stationary if, using the notation above, $-11$ at all. Read (also) my answer on this this “mirror” discussion: https://stats.stackexchange.com/questions/494135/the-explosive-ar1-process-with-varphi1-where-was-this-first-represented-a/494187#494187 – markowitz Oct 29 '20 at 13:51
  • @ Joogs ; take care with parameters. You said that “… an ARMA process has a unique stationary solution if $\phi$ has no roots on the complex unit circle. This would imply that the explosive process … is stationary”. Contradictory conclusion. I do not have your book at hand but I know that some presentations and software make different inversion/manipulation on the characteristic polynomials. – markowitz Oct 29 '20 at 13:52
  • @markowitz, when $|\rho|<1$ the process is not only stationary but also *causal*, when $\rho>1$ it is not *causal* but is still stationary. – Aksakal Nov 11 '21 at 23:32
  • @Aksakal; in some books AR processes with $|\rho| \geq 1$ are considered as non stationary and/or not considered at all. Moreover in some books the concept of “causal process” you refers on are not mentioned at all. I'm dubious, give me more time for think better about all concept used here and in related post. However I expressed one point of view in this strongly related discussion. Take a look: https://stats.stackexchange.com/questions/494135/the-explosive-ar1-process-with-varphi1-where-was-this-first-represented-a/494187#494187 – markowitz Nov 12 '21 at 10:57

2 Answers2

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Yes, there is a stationary solution for $\rho>1$ in AR(1) process: $$X_t=\rho X_{t-1}+\varepsilon_t$$ I'm not sure you'll like it though: $$X_t=-\sum_{k=1}^\infty\frac 1 {\rho^k}\varepsilon_{t+k}$$ Notice the index: $t+k$, you'd need DeLorean to use this in practice.

When $\rho>1$ the process is not invertible.

Aksakal
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  • Argh! I am so busy right now. I wanna go play with this in simulation to appreciate why that is stationary, but that is going to have to wait until at least tonight. Thank you, as always, for pushing my understanding on time series. – Alexis Apr 02 '19 at 17:38
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    @Alexis, look at 4.5.3 example to simulate here http://www.maths.qmul.ac.uk/~bb/TimeSeries/TS_Chapter4_5.pdf basically, it's stationary but not causal, i.e. your today depends on tomorrow. – Aksakal Apr 02 '19 at 17:41
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    Stangely appropriate to what I am currently busy with: time-varying confounding and g-estimation of causal effects in Hernán & Robins. :) – Alexis Apr 02 '19 at 17:42
  • Am I right about the non-ergodicity of the process? – Joogs Apr 02 '19 at 17:57
  • @Joogs, I don't think so. – Aksakal Apr 02 '19 at 18:02
  • But from the stationary solution we obtain $\mathbb{E}(X_t)=0$, but all the simulated series display trending behaviour which basically becomes unrevertible for values of the process high enough, why would the time average converge to something finite (and in particular 0)? – Joogs Apr 02 '19 at 18:08
  • @Joogs, I need to think about ergodicity. My gut feeling is that if we drop causality, then how is $\rho>1$ process is different from $\rho<1$? One is the mirror of another on time axis, i.e. you turn time back and ergodicity seems to be the same for both. I'll think more tonight, and may change answer later. However, my main answer stands correct in either case: AR(1) is stationary for $\rho>1$ – Aksakal Apr 02 '19 at 18:15
  • @Joogs, I thought about ergodicity, and couldnt find argument why solution for $\rho$>1 wouldn't be considered ergodic. So, no, I disagree with your statement on ergodicty. Again, the only problem with that solution is that it's non causal, which makes it useles in most poractical situations, but it's a valid stationary solution nevertheless – Aksakal Apr 03 '19 at 13:46
  • 1/2 @Aksakal "I couldn't find argument why solution for $\rho>1$ wouldn't be considered ergodic": I do not have the mathmatical statistics to put this formally, but when you look at simulated processes (graphically) with $|\rho|>1$ they diverge, and pretty quickly. Like, pick any time series of length $T$ to simulate, then simulate, say, $2T$, and you are **very** likely to see the period from $t=1 \to T$ represented as a flattish line, with 'the action' happening from $T+1 \to 2T$. ('Quickly' unless $|\rho| \approx 1$ but still $>1$) – Alexis Nov 11 '21 at 23:17
  • 2/2 @Aksakal So: would a possible argument against ergodicity be that the rate of divergence away from $X_t=0$ is happening much faster than the rate of decrease in probability of return to values near equilibrium? Just musing. – Alexis Nov 11 '21 at 23:18
  • @Aksakal: In your last sentence of the answer, as far as I know, the word you're looking for is "causal" not "invertible". Invertible is reserved for whether MA or ARMA processes can be written in terms of $X$. You can also expand the answer to include $\rho – ColorStatistics Nov 12 '21 at 00:48
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First we can write the model in reverse AR(1) form as:

$$X_{t} = \frac{1}{\rho} X_{t+1} - \frac{\epsilon_{t+1}}{\rho}.$$

Suppose you now define the observable values using the filter:

$$X_t = - \sum_{k=1}^\infty \frac{\epsilon_{t+k}}{\rho^k}.$$

You can confirm by substitution that both the original AR(1) form and the reversed form hold in this case. As pointed out in an excellent answer to a related question by Michael, this means that the model is not identified unless we exclude this solution by definition.

Ben
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