We’ve seen yesterday conditions on so that the canonical
process,
, satisfying
The condition is rather simple, since should be a triangular region. But the proof is a bit more tricky…
Recall that we want to parametrize the region
Since we have a true process, then
. Our polynomial is here
where ‘s are the roots – in
– of
. Consider now some kind of dual version of that polynomial,
Having the roots of outside the unit circle is the same as having the roots of
inside the unit circle. Obserse that we can write
Roots of are then
From this point, we should discuss a little bit, depending on the value of .
- if
Then there is one root, and only one. So we need to have or equivalently
.
- if
Then we got roots in , and
means, equivalently, that
- if
Then we have two (conjugate) roots in , and the square of norm of those roots is
. Thus,
.
We get what was mention in the course: the canonical has a stationary solution if, and only if
which is a triangular region, see
OpenEdition suggests that you cite this post as follows:
Arthur Charpentier (January 24, 2014). Triangle for Parameters of AR(2) Stationary Processes. Freakonometrics. Retrieved February 9, 2025 from https://doi.org/10.58079/outs
there is a little typo of $\phi_2$ on the parameter set
Can you explain more of how you arrived at “some kind of dual version of that polynomial”?
Also, should the signs be – instead of + in that dual version? If not, I’m not sure how factoring out \phi_{2} flips the signs here.