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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?

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    $\begingroup$ @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. $\endgroup$
    – Alexis
    Apr 2, 2019 at 15:47
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    $\begingroup$ @Alexis, "variances of explosive processes are functions of time" - sure about this? $\endgroup$
    – Aksakal
    Apr 2, 2019 at 17:21
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    $\begingroup$ Excellent question on a subtle topic! $\endgroup$
    – Aksakal
    Apr 2, 2019 at 17:22
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    $\begingroup$ @Alexis, you have to turn time back $\endgroup$
    – Aksakal
    Apr 11, 2019 at 19:23
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    $\begingroup$ @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!" :) $\endgroup$
    – Alexis
    Apr 12, 2019 at 16:32

2 Answers 2

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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.

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  • $\begingroup$ 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. $\endgroup$
    – Alexis
    Apr 2, 2019 at 17:38
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    $\begingroup$ @Alexis, look at 4.5.3 example to simulate here maths.qmul.ac.uk/~bb/TimeSeries/TS_Chapter4_5.pdf basically, it's stationary but not causal, i.e. your today depends on tomorrow. $\endgroup$
    – Aksakal
    Apr 2, 2019 at 17:41
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    $\begingroup$ Stangely appropriate to what I am currently busy with: time-varying confounding and g-estimation of causal effects in Hernán & Robins. :) $\endgroup$
    – Alexis
    Apr 2, 2019 at 17:42
  • $\begingroup$ Am I right about the non-ergodicity of the process? $\endgroup$
    – Joogs
    Apr 2, 2019 at 17:57
  • $\begingroup$ @Joogs, I don't think so. $\endgroup$
    – Aksakal
    Apr 2, 2019 at 18:02
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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.

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