4
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.