Triangle for Parameters of AR(2) Stationary Processes

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

http://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 http://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

http://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 http://latex.codecogs.com/gif.latex?AR(2) process, then http://latex.codecogs.com/gif.latex?\phi_2\neq%200. Our polynomial is here

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

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

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

http://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 http://latex.codecogs.com/gif.latex?\Delta=\phi_1^2+4\phi_2.

  • if http://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 http://latex.codecogs.com/gif.latex?\vert\phi_1\vert%20%3C2 or equivalently http://latex.codecogs.com/gif.latex?\phi_2%3E-1.

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

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

http://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

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

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

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

http://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


One thought 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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