Are explosive AR(MA) processes stationary?

Question

According to Theorem 8.8 in Time Series by 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 Stack Exchange 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? and 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?

Answer 1

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.

Answer 2

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.