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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.

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    $\begingroup$ 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 or here. 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). $\endgroup$ – Matthew Gunn Mar 4 '17 at 14:24
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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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  • $\begingroup$ 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 $\phi_2 \in ]-2/3;1[$? $\endgroup$ – Niko24 Mar 4 '17 at 16:06
  • $\begingroup$ @Niko24: Where do you have the condition that $-1 \leq \phi_2\leq 1 - |-1/3|$ from? $\endgroup$ – Roland Mar 4 '17 at 18:08
  • $\begingroup$ 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 $\phi_2$. $\endgroup$ – Niko24 Mar 5 '17 at 12:01
  • $\begingroup$ @Niko24: I've added a sketch of an answer. There's still some work to be done for the case where $z_-$ is real. $\endgroup$ – Roland Mar 5 '17 at 14:51

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