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?
Thanks in advance!!
Best Answer
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 $$ |\phi| < 1 \text{ if and only if }\bar{z}^* > 1. $$