Notation: $\dot{Z}_t = Z_t – E(Z_t)$, so that it is centered at 0. $a_t$ stands for the residual and we assume the $a_t$ are independent and normally distributed with mean 0 and constant standard deviation $\sigma_a^2$. And, when I say stationarity, I mean weak stationarity and not strict stationarity.
In my time series class, we have been given that an MA(q) model is of the form
$$\dot{Z}_t = \theta_q(B) a_t$$
where
$$\theta_q(B) = 1 – \theta_1 B – \theta_2 B^2 – \cdots – \theta_q B^q.$$
We were told MA(q) is always stationary and it is invertible when all of the roots of $\theta_q(B)$ are outside the unit circle.
For an AR(p) model, we were told it is of the form
$$\phi_p(B) \dot{Z}_t = a_t$$
where
$$\phi_p(B) = 1 – \phi_1 B – \phi_2 B^2 – \cdots – \phi_p B^p.$$
We were told this model is stationary when all of the roots of $\phi_p(B)$ are outside the unit circle, but I don't remember hearing anything about when these are invertible. If I use the parallels from the MA(q) model, I might say the AR(p) model is always invertible. Is that correct?
Thanks
Best Answer
The answer to your question can be summarized as follows: