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.
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 $(-1,-1/36]$.
