# Determine if AR(p) model is causal stationary or invertible

I was going through these problems and think I figured out most of them both, but am having some troubles at one of the last steps.

The question is for each of the following models:

Express them using the backshift operator, state the autoregressive or moving average polynomial and use it to determine whether they are causal stationary or invertible. $$(1) X_t = 0.7X_{t-1} + \epsilon_t$$ $$(2) X_t = 1.8X_{t-1}-0.81X_{t-2}+\epsilon_t$$

For Question (1), I expressed it using the backshift operator $$(1-0.7B)X-t=\epsilon_t$$ And because it is an AR(1) model, it is causal stationary when $$|\phi_1|<1$$. In this case, $$|0.7| < 1$$ Meaning this model is causal stationary. Is that correct?

For (2) expressed using backshift is $$(1-1.8B+0.81B^2)X_t = \epsilon_t$$ The root of this polynomial is $$z_{1,2}=\frac{1.8±\sqrt{0}}{1.62}=1.111$$

Except as this is not an AR(1) model like before I can't just check if $$\phi_1 < 1$$, so where do I go from here to determine causal stationarity/invertibility?

For causality, you check if the roots are outside of the unit circle. Take your root $$z^*$$, take it's modulus $$\bar{z}^*$$, then check to make sure it's outside of the unit circle. These all are, so for all of these models, you have causality.
These are roots of a complex polynomial. You are not solving for the parameters, you are solving for the $$z$$. When you looked at an AR(1) and found conditions for the parameter $$\phi$$: that was the old way. This is the new way. But note that in that special case that these two conditions are the same because if the root of $$(1 - \phi z)$$ is $$z^* = 1/\phi$$, then $$|\phi| < 1 \text{ if and only if }\bar{z}^* > 1.$$