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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


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