Triangle for Parameters of AR(2) Stationary Processes

We’ve seen yesterday conditions on https://latex.codecogs.com/gif.latex?(\phi_1,\phi_2) so that the canonical https://latex.codecogs.com/gif.latex?AR(2) process, https://latex.codecogs.com/gif.latex?(X_t), satisfying

https://latex.codecogs.com/gif.latex?X_t=\phi_1%20X_{t-1}+\phi_2%20X_{t-2}+\varepsilon_t

The condition is rather simple, since https://latex.codecogs.com/gif.latex?(\phi_1,\phi_2) should be a triangular region. But the proof is a bit more tricky…

Recall that we want to parametrize the region

https://latex.codecogs.com/gif.latex?\{(\phi_1%20,\phi_2)\in\mahtbb{R}^2:%201-\phi_1z-\phi_1z^2\neq%200,\forall%20z\in\mathbb{C},\vert\vert%20z\vert\vert%20\leq%201\}

Since we have a true https://latex.codecogs.com/gif.latex?AR(2) process, then https://latex.codecogs.com/gif.latex?\phi_2\neq%200. Our polynomial is here

https://latex.codecogs.com/gif.latex?\Phi(z)=1-\phi_1z-\phi_1z^2=\left(1-\frac{z}{\lambda_1}\right)\left(1-\frac{z}{\lambda_2}\right)

where https://latex.codecogs.com/gif.latex?\lambda_i‘s are the roots – in https://latex.codecogs.com/gif.latex?\mathbb{C} – of https://latex.codecogs.com/gif.latex?\Phi(\cdot). Consider now some kind of dual version of that polynomial,

https://latex.codecogs.com/gif.latex?\tilde\Phi(z)=\left(1-{z}{\lambda_1}\right)\left(1-{z}{\lambda_2}\right)=1+\frac{\phi_1}{\phi_2}z+\frac{1}{\phi_2}z^2

Having the roots of https://latex.codecogs.com/gif.latex?\Phi(\cdot) outside the unit circle is the same as having the roots of https://latex.codecogs.com/gif.latex?\tilde\Phi(\cdot) inside the unit circle. Obserse that we can write

https://latex.codecogs.com/gif.latex?\tilde\Phi(z)=\frac{1}{\phi_2}(\underbrace{z^2-\phi_1%20z-\phi_2}_{\bar{\Phi}(z)})

Roots of https://latex.codecogs.com/gif.latex?\bar{\Phi}(\cdot)} are then

https://latex.codecogs.com/gif.latex?\xi%20=%20\frac{1}{2}\left(\phi_1\pm\sqrt{\phi_1^2+4\phi_2}\right)

From this point, we should discuss a little bit, depending on the value of https://latex.codecogs.com/gif.latex?\Delta=\phi_1^2+4\phi_2.

  • if https://latex.codecogs.com/gif.latex?\Delta=\phi_1^2+4\phi_2=0

Then there is one root, and only one. So we need to have https://latex.codecogs.com/gif.latex?\vert\phi_1\vert%20%3C2 or equivalently https://latex.codecogs.com/gif.latex?\phi_2%3E-1.

  • if https://latex.codecogs.com/gif.latex?\Delta=\phi_1^2+4\phi_2%3E0

Then we got roots in https://latex.codecogs.com/gif.latex?\mathbb{R}, and

https://latex.codecogs.com/gif.latex?-1%3C%20\frac{1}{2}\left(\phi_1\pm\sqrt{\phi_1^2+4\phi_2}\right)%3C%201

means, equivalently, that

https://latex.codecogs.com/gif.latex?\phi_2%3E-1%20\%20;%20\%20\phi_2-\phi_1%3C1%20\%20;%20\%20\phi_2+\phi_1%3C1

  • if https://latex.codecogs.com/gif.latex?\Delta=\phi_1^2+4\phi_2%3C0

Then we have two (conjugate) roots in https://latex.codecogs.com/gif.latex?\mathbb{C}, and the square of norm of those roots is https://latex.codecogs.com/gif.latex?\vert\vert%C2%A0\xi\vert\vert^2=-\phi_2. Thus, https://latex.codecogs.com/gif.latex?\phi_2%3E-1.

We get what was mention in the course: the canonical https://latex.codecogs.com/gif.latex?AR(2) has a stationary solution if, and only if

https://latex.codecogs.com/gif.latex?\left\{\begin{array}{l}%20\phi_2-\phi_1%3C1%20\\\phi_2+\phi_1%3C1\\%20\vert\phi_2\vert%3C1\end{array}\right.

which is a triangular region, see


2 thoughts on “Triangle for Parameters of AR(2) Stationary Processes”

  1. 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.

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