# What happens if we forget a trivial assumption ?

Last week, @dmonniaux published an interesting post entitled l’erreur n’a rien d’original  on  his blog. He was asking the following question : let $a$, $b$ and $c$ denote three real-valued coefficients, under which assumption on those three coefficients does $ax^2+bx+c$ has a real-valued root ?

Everyone aswered $b^2-4ac\geq 0$, but no one mentioned that it is necessary to have a proper quadratic equation, first. For instance, if both $a$ and $b$ are null, there are no roots.

# 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