Studying AR models, I found that there are two properties that these models can have stationarity and causality.

For what concerns stationarity, I have studied that this condition is satisfied if the equation $\phi(B) = 0$ has all roots outside the unit circle, i.e. they are in modulus greater than one.

Instead, for what concerns causality, I am having some troubles: I mean, the conditions for causality seem to me the same of stationarity (at least for what concerns simple $AR(1)$, $AR(2)$). Moreover, I am not sure of having understood what does causality actually mean.

Increasing my doubts, I have also found that some people talk about an invertibility condition for $AR$ models, but I have understood that it was only concerning $MA$ models and that it is a kind of counterpart for stationarity, given the fact that $MA$ are always stationary.

Could you help me making a bit of order in my mind?

  • $\begingroup$ For $AR(p)$ model, stationarity and causality are satisfied if all roots of $\phi(B)=0$ are outside the unit circle. For invertibility, all $AR(p)$ models are invertible. $\endgroup$ – Patrick Li Jun 5 '16 at 13:19

A linear process $X_t$ is defined to be causal if $X_t=\psi(B)w_t$ where $w_t$ are white noises and $\sum_{j=1}^{\infty}|\psi(j)|<\infty$.

$X_t$ is defined to be invertible if we can write $w_t=\pi(B) X_t$ where $\pi(B)=\pi_0 + \pi_1 B+\pi_2 B^2 + \cdots$ and $\sum_{j=0}^{\infty}|\pi(j)|<\infty$.

Apparently, an arbitrary $\text{AR}(p)$ model, $$X_t-\phi_1 X_{t-1}-\phi_2 X_{t-2}-\cdots - \phi_p X_{t-p}=w_t$$ automatically satisfies the requirement to be invertible since $\pi(B)=1-\phi_1 B - \cdots - \phi_p B^p$ and $\sum_{j=0}^{\infty}|\pi(j)|=1+\sum_{j=1}^{p}|\phi_j|<\infty$.

You can take a look at this note.

  • $\begingroup$ Thanks for your answer! So is it right to look at invertibility and stationarity as two counterparts? I mean stationarity is always satisfied for $MA(q)$ and it has to be checked in $AR(p)$ and the other way around? $\endgroup$ – PhDing Jun 5 '16 at 19:33
  • 1
    $\begingroup$ @Alessandro That is correct. $\endgroup$ – Patrick Li Jun 6 '16 at 2:03

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