# Somewhere else, part 111

Un article intéressant en français, ces derniers jours (venant de l’autre côté de la montagne)

et un peu de lecture en danois, pour les plus courageux (via Philippe Réka, a.k.a. @visionscarto),

« Prix moyen de l’électricité pour les ménages (à gauche) et pour l’industrie (à droite) »

et « Prix moyen du gas pour les ménages (à gauche) et pour l’industrie (à droite) »

Did I miss something?

# Triangle for Parameters of AR(2) Stationary Processes

We’ve seen yesterday conditions on $(\phi_1,\phi_2)$ so that the canonical $AR(2)$ process, $(X_t)$, satisfying

$X_t=\phi_1 X_{t-1}+\phi_2 X_{t-2}+\varepsilon_t$

The condition is rather simple, since $(\phi_1,\phi_2)$ should be a triangular region. But the proof is a bit more tricky…

Recall that we want to parametrize the region

$\{(\phi_1 ,\phi_2)\in\mahtbb{R}^2: 1-\phi_1z-\phi_1z^2\neq 0,\forall z\in\mathbb{C},\vert\vert z\vert\vert \leq 1\}$

Since we have a true $AR(2)$ process, then $\phi_2\neq 0$. Our polynomial is here

$\Phi(z)=1-\phi_1z-\phi_1z^2=\left(1-\frac{z}{\lambda_1}\right)\left(1-\frac{z}{\lambda_2}\right)$

where $\lambda_i$‘s are the roots – in $\mathbb{C}$ – of $\Phi(\cdot)$. Consider now some kind of dual version of that polynomial,

$\tilde\Phi(z)=\left(1-{z}{\lambda_1}\right)\left(1-{z}{\lambda_2}\right)=1+\frac{\phi_1}{\phi_2}z+\frac{1}{\phi_2}z^2$

Having the roots of $\Phi(\cdot)$ outside the unit circle is the same as having the roots of $\tilde\Phi(\cdot)$ inside the unit circle. Obserse that we can write

$\tilde\Phi(z)=\frac{1}{\phi_2}(\underbrace{z^2-\phi_1 z-\phi_2}_{\bar{\Phi}(z)})$

Roots of $\bar{\Phi}(\cdot)}$ are then

$\xi = \frac{1}{2}\left(\phi_1\pm\sqrt{\phi_1^2+4\phi_2}\right)$

From this point, we should discuss a little bit, depending on the value of $\Delta=\phi_1^2+4\phi_2$.

• if $\Delta=\phi_1^2+4\phi_2=0$

Then there is one root, and only one. So we need to have $\vert\phi_1\vert <2$ or equivalently $\phi_2>-1$.

• if $\Delta=\phi_1^2+4\phi_2>0$

Then we got roots in $\mathbb{R}$, and

$-1< \frac{1}{2}\left(\phi_1\pm\sqrt{\phi_1^2+4\phi_2}\right)< 1$

means, equivalently, that

$\phi_2>-1 \ ; \ \phi_2-\phi_1<1 \ ; \ \phi_2+\phi_1<1$

• if $\Delta=\phi_1^2+4\phi_2<0$

Then we have two (conjugate) roots in $\mathbb{C}$, and the square of norm of those roots is $\vert\vert \xi\vert\vert^2=-\phi_2$. Thus, $\phi_2>-1$.

We get what was mention in the course: the canonical $AR(2)$ has a stationary solution if, and only if

$\left\{\begin{array}{l} \phi_2-\phi_1<1 \\\phi_2+\phi_1<1\\ \vert\phi_2\vert<1\end{array}\right.$

which is a triangular region, see