Assume a $K$-dimensional VAR($p$) process given by $$y_t=\nu+A_1y_{t-1}+\ldots+A_py_{t-p}+u_t$$ This process is called stable if the roots of the reverse characteristic polynomial are bigger than 1 in terms of the Euclidean norm. Therefore one has to check whether: $$\det(I_k-A_1z-\ldots-A_pz^p)\neq0 \text{ for } |z|\leq1.$$

Is there a way to find the values of $z$ without computing the polynomial $\det(I_k-A_1z-\ldots-A_pz^p)$ by hand? I see that there exist some implementations for the standard statistical programs, but I would like to understand how such a procedure works.

  • $\begingroup$ it needs to be $|z|\leq 1$ to rule out unit roots! $\endgroup$ – Christoph Hanck Oct 16 '15 at 12:58
  • $\begingroup$ I am not sure I understand your question - you do not want to do it by hand, but also not rely on statistical programs? $\endgroup$ – Christoph Hanck Oct 16 '15 at 12:59
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    $\begingroup$ Thank you for your commentd Christoph Hanck. As long as I did not include a typo in my question $|z|>1$ should be correct (note that $z$ is the inverse of the eigenvalues computed by solving for the roots of $\det (zI-A-\ldots)$.) $\endgroup$ – muffin1974 Oct 17 '15 at 17:25
  • $\begingroup$ Your second comment summarizes my question correct, I just want to know wether there is any more general formula out there that makes computation by hand easier. $\endgroup$ – muffin1974 Oct 17 '15 at 17:26
  • $\begingroup$ I still think my version is correct: boil it down to an AR(1) - it then reads that $1-\phi z=0$ must not have solutions on or inside the unit circle (equivalently, must have all its solutions outside the unit circle. Algebraically, $1-\phi z\neq0$ for all $|z|\leq1$. $\endgroup$ – Christoph Hanck Oct 17 '15 at 17:49

The eigenvalues of $$ F = \left[\begin{array}{ccccc} A_1 & A_2 & \cdots & A_{p-1} & A_p\\ I_k & O & \cdots & O & O\\ \vdots & \vdots & \ddots & \vdots &\vdots\\ O & O & \cdots & I_k & O \end{array}\right] $$ have to lie inside the unit circle, i.e. $$ \mathrm{det}(I_k\lambda^p - A_1\lambda^{p-1} - \cdots - A_p) = 0\qquad(\star) $$ for a stable VAR(p) process. Or equivalently: All $z\in\mathbb{R}^k$ that satisfy $$ \mathrm{det}(I_k - A_1z - \cdots - A_pz^p) = 0\qquad\qquad(\star\star) $$ lie outside the unit circle.

So, from my (basic) knowledge of Numerics, there are two possible ways of checking the stable condition: You either calculate the eigenvalues of $F$ or you solve the non-linear equation $(\star)$. I guess that most statistical software relies on the latter attempt since there are many algorithms that focus on solving non-linear equations as @Carlos Dutra already pointed out.

But to answer your question: In either way, there is no "simple" method to check the stable condition.

  • $\begingroup$ Nice. This question about difference between "inside" and "outside" the unit circle always creates confusion. And the methods are not really "ease" or "simple". $\endgroup$ – Carlos Dutra Oct 24 '16 at 21:18
  • $\begingroup$ I actually believe this is the initial reason for confusion. ($\star\star$) is basically OPs statement. However, it is bad way of formulating the problem since how shall a program find "all $z\in\mathbb{R}^k$"?! There are infinitely many values to consider. A better way (and that is the reason why I rewrote the inside/outside story) is to start with an equality like in ($\star$). $\endgroup$ – Syd Amerikaner Oct 24 '16 at 22:39

This is a root finding problem, and is independent of the AR study.

In practice, you will not calculate on hand the analytical solution, which is very complicated to higher orders. What algorithms do is search for a numerical approximation of the solution. To do this there are several methods, you can see the explanation of some of the most common, from the simplest to the most complex here and here. The Newton Method is a nice one to study.

  • $\begingroup$ That does not really help since the OP asked for a way of getting a better understanding rather than just using a computer solution. $\endgroup$ – mdewey Oct 24 '16 at 13:02
  • $\begingroup$ A numerical solution is not necessarily a computer solution. $\endgroup$ – Carlos Dutra Oct 24 '16 at 13:12

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