I have this process: $X_t-0.5X_{t-1}=\epsilon_t-1.3\epsilon_{t-1}+0.4\epsilon_{t-2 }$

I'm wondering if this ARMA(1,2) is stationary and/or invertible. I know we can rewrite the process as follows: $$(1-0.5L)X_t=(1-1.3L+0.4L^2)\epsilon_t$$where $L$ is the backward operator.

  • It's stationary cause the MA part is by definition, and AR part has $-0.5$ that's $|-0.5|<1$
  • But I have some problem understanding the invertibility part. By definition the AR part is invertible, so we need to check the MA part, so we pick the polynomial: $(1-1.3L+0.4L^2)$ and we find the roots, if they are $|L_{(1,2)}| > 1$ my process is invertible. The roots are:$1.25$ and $2$
    So I would say that the process is invertible.

But now come my question: I know that in a process (AR(2) or MA(2)) we can easily prove stationairy or invertibilty looking at the coefficients we these restrictions : \begin{cases} |\theta_2|<1\\ \theta_2+\theta_1<1\\ \theta_2-\theta_1<1 \end{cases} If if put inside my coefficients, I get:\begin{cases} |0.4|<1 \quad True\\ 0.4+1.3<1 \quad False\\ 0.4-1.3 <1\quad True \end{cases} So if I apply these rule my process is not invertible. Why do I get this contradiction?


2 Answers 2


The confusion comes from the fact that these conditions (that you state under the label "can be easily proven") pertain to the $Y_t=\varepsilon_t-\theta_1\varepsilon_{t-1}-\theta_2\varepsilon_{t-2}$ formulation. In your case this means $\theta_1=1.3$, but $\theta_2=-0.4$ (not $0.4$!). Substituting these to the conditions, you'll see that all of them is fulfilled.

  • $\begingroup$ Yes, I shouldn't write "easily proven", I don't know the proof so I can't state this. Maybe it's very difficult, anyway now I see the problem! So glad! $\endgroup$ Oct 30, 2017 at 19:41
  • 2
    $\begingroup$ @Mario For a proof see stats.stackexchange.com/a/145652/77222 $\endgroup$ Oct 30, 2017 at 19:48
  • $\begingroup$ @MarioMigliaccio : Don't get me wrong, I didn't intend to blame your for writing "can be easily proven", I just used it to clearly identify which conditions I refer to at that part of my answer :) Nevertheless, thanks Jarle Tufto for linking the proof! $\endgroup$ Oct 30, 2017 at 20:18
  • $\begingroup$ ^ looks like I was right, then. $\endgroup$
    – Taylor
    Oct 30, 2017 at 20:44

It's causal and stationary because the AR roots are outside the unit circle.

polyroot(c(1,-.5))  # 2+0i

It's invertible because the MA roots are outside the unit circle.

> polyroot(c(1,-1.3,.4)) #1.25-0i 2.00+0i

You got these already, so nice.

The other restrictions you're writing down must be equivalent to the above. I haven't checked, but my guess is that because quadratic polynomials have explicit formulas for the roots (the quadratic formula), so you can just set those roots to be greater than $1$, and then voila.

However, in this case, your AR polynomial is linear. So there's no reason why that should apply. And for the MA part, well, it's probably the above thing I mentioned. Although, you should check, because I haven't.


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