Tag Archives: roots

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 https://latex.codecogs.com/gif.latex?a, https://latex.codecogs.com/gif.latex?b and https://latex.codecogs.com/gif.latex?c denote three real-valued coefficients, under which assumption on those three coefficients does https://latex.codecogs.com/gif.latex?ax^2+bx+c has a real-valued root ?

Everyone aswered https://latex.codecogs.com/gif.latex?b^2-4ac\geq%200, but no one mentioned that it is necessary to have a proper quadratic equation, first. For instance, if both https://latex.codecogs.com/gif.latex?a and https://latex.codecogs.com/gif.latex?b are null, there are no roots.

Continue reading What happens if we forget a trivial assumption ?

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