# Relation between AR(p) stationarity and causality

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?

• Relation with causality is easy: correlation does not imply causation. – Tim May 24 '18 at 17:59
• @Tim, I think causality in the context of AR models is a different thing than causality as in cause-effect relation, so I wonder in which way your comment could be relevant. – Richard Hardy May 24 '18 at 18:36

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.
• Sorry about the imprecise wording. I will try better: I am not sure whether arbitrary roots inside the unit circle are compatible with stationarity. I checked the book and at least on the surface, it indeed looks like your quote is correct. I am left wondering what to do with explosive processes such as $x_t=5x_{t-1}+\varepsilon_t$ – are they stationary? – Richard Hardy Jan 10 '19 at 14:07