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?
Best Answer
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.