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

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.