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I have been given the following AR(2) process:

$X_t + \phi_1 X_{t-1} + \phi_2 X_{t-2} = \epsilon_t$

and I need to figure out for which values of $\phi_2$ the process is stationary, when I have been given that $\phi_1=-1/3$. I hope that someone can maybe help me with this, since I have no idea about where to start.

Matthew Gunn
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Niko24
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    Using the lag operator, your process is $(1 + \phi_1L + \phi_2L^2)X_t = \epsilon_t)$ where $(1 + \phi_1L + \phi_2L^2)$ is your lag polynomial. Compute the (possibly complex) roots of your lag polynomial, and check if the roots lie outside the unit circle. Some discussion is [here](https://kochanova.files.wordpress.com/2016/01/lecture2.pdf) or [here](http://socserv.mcmaster.ca/magee/761_762/other%20material/unit%20and%20char%20roots.pdf). There are equivalent formulations (eg. Roland's below) where you check if roots lie inside unit circle (it's basically a $\frac{1}{z} = w$ kinda thing). – Matthew Gunn Mar 04 '17 at 14:24

1 Answers1

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As described in this answer, determining if your AR(2) process is stationary breaks down to the question if all (complex) roots of the polynomial $$p(z) = z^2 - \left(-\frac{1}{3}\right)z - \phi_2 $$

lie inside the unit disk, i.e. have an absolute smaller than $1$. Can you take it from here?

It's worth noting that you get the polynomial by transformation from the characteristic polynomial of the process, but I stuck to the formulation of the linked answer for consistency's sake.

Here's what I'd do (but not a complete solution): The (possibly complex) solutions of $p(z) = 0$ are $$z_{\pm}=-\frac{1}{6}\pm \sqrt{\frac{1}{36}+\phi_2}.$$

If $\phi_2 > -\frac{1}{36}$, then both solutions are real. Then $|z_+|<1$ iff $-1<z_+$ and $z_+<1$. For $z_+ < 1$ consider

$$1 > -\frac{1}{6} + \sqrt{\frac{1}{36}+\phi_2} \Leftrightarrow \frac{7}{6}> \sqrt{\frac{1}{36}+\phi_2} \Leftrightarrow \frac{4}{3} >\phi_2.$$

Furthermore, if real, $z_+$ is always positive. Hence, $z_+$ lies inside the disc if $\phi_2 > \frac{4}{3}$. Similarly, $z_-$ needs to be considered. And we arrive at an admissable set for $\phi_2$.

If $\phi_2 < -\frac{1}{36}$, then the solitions are complex and the way I wrote them is not precise, but rather

$$z_{\pm}=-\frac{1}{6}\pm i \sqrt{-\frac{1}{36}-\phi_2}.$$

Here, $z_+$ and $z_-$ are complex conjugate and we have $|z_+|^2=|z_-|^2=z_+z_-$. So they lie in the unit disc iff

$$1> |z_+|^2 = \frac{1}{36} +(-\frac{1}{36} -\phi_2)=-\phi_2,$$

i.e. if $\phi_2 > -1$. We get another admissable set for $\phi_2$: the open interval $(-1,-1/36)$.

Considering that for $\phi_2 = -1/36$ we have that $z_+=z_-=-1/6$ which is inside the unit disc, we can amend that set to $(-1,-1/36]$.

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Roland
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  • So is it completely wrong to say that the stationarity condition requires that $-1 \leq \phi_2 \leq 1-|-1/3|$, and then solving for $\phi_2$ and getting something like $\phi_2 \in ]-2/3;1[$? – Niko24 Mar 04 '17 at 16:06
  • @Niko24: Where do you have the condition that $-1 \leq \phi_2\leq 1 - |-1/3|$ from? – Roland Mar 04 '17 at 18:08
  • I found it in some slides online, but now I actually cannot find it again. But I am still not quite sure how to do this - if I knew the value of $\phi_2$ I get how you solve it with the second order polynomial, but I don't know how to approach it when I don't know the value of $\phi_2$. – Niko24 Mar 05 '17 at 12:01
  • @Niko24: I've added a sketch of an answer. There's still some work to be done for the case where $z_-$ is real. – Roland Mar 05 '17 at 14:51