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Is it possible to describe AR(1) as stable in terms of a unit triangle as in the explanation below or not because a lack of a second degree characteristic polynomial?

Consider the equation $$ \lambda^2-\phi_1\lambda-\phi_2=0 $$

If $z$ is a root of the "standard" characteristic equation $1-\phi_1 > z-\phi_2 z^2=0$ and setting $z^{-1}=\lambda$, the display obtains from rewriting the standard one as follows: \begin{eqnarray*} 1-\phi_1 > z-\phi_2 z^2&=&0\\ \Rightarrow z^{-2}-\phi_1 z^{-1}-\phi_2 &=&0\\ > \Rightarrow \lambda^2-\phi_1\lambda -\phi_2 &=&0 \end{eqnarray*} Hence, an alternative condition for stability of an $AR(2)$ is that all roots of the first display are inside the unit circle, $|z|>1 > \Leftrightarrow |\lambda|=|z^{-1}|<1$.

We use this representation to derive the stationarity triangle of an $AR(2)$ process, that is that an $AR(2)$ is stable if the following three conditions are met:

  1. $\phi_2<1+\phi_1$
  2. $\phi_2<1-\phi_1$
  3. $\phi_2>-1$

Recall that you can write the roots of the first display (if real) as $$ > \lambda_{1,2}=\frac{\phi_1\pm\sqrt{\phi_1^2+4\phi_2}}{2} > $$ to find the first two conditions.

Then, the $AR(2)$ is stationary iff $|\lambda|<1$, hence (if the $\lambda_i$ are real): \begin{eqnarray*} > -1<\frac{\phi_1\pm\sqrt{\phi_1^2+4\phi_2}}{2}&<&1\\ \Rightarrow -2<\phi_1\pm\sqrt{\phi_1^2+4\phi_2}&<&2 > \end{eqnarray*} The larger of the two $\lambda_i$ is bounded by $\phi_1+\sqrt{\phi_1^2+4\phi_2}<2$, or: \begin{eqnarray*} > \phi_1+\sqrt{\phi_1^2+4\phi_2}&<&2\\ \Rightarrow > \sqrt{\phi_1^2+4\phi_2}&<&2 - \phi_1\\ \Rightarrow > \phi_1^2+4\phi_2&<&(2 - \phi_1)^2\\ \Rightarrow \phi_1^2+4\phi_2&<&4 - > 4\phi_1+\phi_1^2\\ \Rightarrow \phi_2&<&1 - \phi_1 \end{eqnarray*} Analogously, we find that $\phi_2<1 + \phi_1$.

If $\lambda_i$ is complex, then $\phi_1^2<-4\phi_2$ and so $$\lambda_{1,2} = \phi_1/2\pm i\sqrt{-(\phi_1^2+4\phi_2)}/2.$$ The squared modulus of a complex number is the square of the real plus the square of the imaginary part. Hence, $$ \lambda^2 = (\phi_1/2)^2 + > \left(\sqrt{-(\phi_1^2+4\phi_2)}/2\right)^2 = > \phi_1^2/4-(\phi_1^2+4\phi_2)/4 = -\phi_2. $$ This is stable if $|\lambda|<1$, hence if $-\phi_2<1$ or $\phi_2>-1$, as was to be shown. (The restriction $\phi_2<1$ resulting from $\phi_2^2<1$ is redundant in view of $\phi_2<1+\phi_1$ and $\phi_2<1-\phi_1$.)

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2 Answers 2

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The second condition fails.

If you have an non-stable AR1 process then $\phi_1 = 1$. You could interpret it, if you like, as a AR2 process with $\phi_2 = 0$. Then you have that the second condition fails

$$0=\phi_2<1-\phi_1 = 0$$

The condition $0<0$ is false.


The characteristic equation of higher order but with coefficients zero will have the same roots but with some additional roots equal to zero. $\lambda^2-\lambda \theta_1- \theta_2$ and $\lambda-\theta_1$ are just the same when $\theta_2=0$ (except for that additional root $\lambda = 0$).

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Indeed, since there is only one coefficient, there is no two-dimensional figure to plot (unlike the answer A proof for the stationarity of an AR(2) you cite, which discusses the AR(2) case). Stability means that all solutions to the characteristic polynomial are outside the unit circle.

In this case, the polynomial is just $1-\phi z=0$, such that the condition becomes $$ |z|>1/|\phi| $$ or $|\phi|<1$.

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