Let's take an AR(p) model $\phi(L)y_t=z_t$ where $\phi(L)=1-\phi_1-…-\phi_pL^p$ and L is the lag operator. I have just studied that if there are no roots of the polynomial on the unit circle,
$1/\phi(L)=\sum_{j=-\infty}^\infty\psi_jL^j$
and
$\sum_{j=-\infty}^\infty|\psi_j|<\infty$.
But it is also true that if the above condition holds, then the process $y_t=\sum_{j=-\infty}^\infty\psi_jz_{t-j}$ is stationary, provided that $z_t$ is stationary.
So my question is: is it correct to say that all AR process such that the polynomial $\phi$ has no roots on the unit circle are stationary? This would imply that also a process like $y_t=2y_{t-1}+z_t$ is stationary (although non-causal).
This does not fit with what I have been told before, i.e. that an AR process is stationary if and only if it has all roots oustide the unit circle. So what is the correct stationarity condition for AR processes?
Best Answer
It might not be relevant for @MrJames anymore but my answer might help others.
Have a look at "Introduction to Time Series and Forecasting" by Brockwell and Davis (third edition). Here they say that a stationary solution of $\Phi(L)y_t=z_t$ exists and is unique if and only if $\Phi$ has no roots on the unit circle. It is causual if all roots of $\Phi$ are outside the unit circle.
Hence, causality implies stationarity but not the other way around.