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Consider a mean-centred AR(2) process $$X_t=\phi_1X_{t-1}+\phi_2X_{t-2}+\epsilon_t$$ where $\epsilon_t$ is the standard white noise process. Just for sake of simplicity let me call $\phi_1=b$ and $\phi_{2}=a$. Focusing on the roots of the characteristics equation I got $$z_{1,2}=\frac{-b\pm\sqrt{b^2+4a}}{2a}$$ The classical conditions in the textbooks are the following: $$\begin{cases}|a|<1 \\ a\pm b<1 \end{cases}$$ I tried to solve manually (with the help of Mathematica) the inequalities on the roots, i.e the system $$\begin{cases}|\frac{-b-\sqrt{b^2+4a}}{2a}|>1 \\ |\frac{-b+\sqrt{b^2+4a}}{2a}|>1\end{cases}$$ obtaining just $$a \pm b<1$$ Can the third condition ($|a|<1$) be recover adding the previous two solutions to each other getting $a+b+a-b<2 \Rightarrow a<1$ that through some sign considerations becomes $|a|<1$? Or am I missing a solution?

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My guess is that the characteristic equation you are departing from is different from mine. Let me proceed in a couple of steps to see whether we agree.

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 $\phi_2<1-\phi_1$.)

Plotting the stationarity triangle, also indicating the line that separates complex from real roots, we get

enter image description here

Produced in R using

phi1 <- seq(from = -2.5, to = 2.5, length = 51) 
plot(phi1,1+phi1,lty="dashed",type="l",xlab="",ylab="",cex.axis=.8,ylim=c(-1.5,1.5))
abline(a = -1, b = 0, lty="dashed")
abline(a = 1, b = -1, lty="dashed")
title(ylab=expression(phi[2]),xlab=expression(phi[1]),cex.lab=.8)
polygon(x = phi1[6:46], y = 1-abs(phi1[6:46]), col="gray")
lines(phi1,-phi1^2/4)
text(0,-.5,expression(phi[2]<phi[1]^2/4),cex=.7)
text(1.2,.5,expression(phi[2]>1-phi[1]),cex=.7)
text(-1.75,.5,expression(phi[2]>1+phi[1]),cex=.7)
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  • $\begingroup$ this is a very detailed explanation. $\endgroup$ – Marco Jul 29 '15 at 10:47
  • $\begingroup$ @Christoph: Is there a typo in the answer? Look at equation for $\lambda^2$. Also, what do you mean by square of a complex number? If $z = a + bi$ then $z^2 = a^2 - b^2 + 2iab$. How do you say square of a complex number is "square of the real plus the square of the imaginary part" $\endgroup$ – shani Aug 28 '18 at 4:08
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    $\begingroup$ Thanks, quite right! I was referring to the sqaured modulus, see the edit. $\endgroup$ – Christoph Hanck Aug 28 '18 at 8:53
  • $\begingroup$ @ChristophHanck, what is your take on Aksakal's answers in these two threads: 1 and 2? Are they in conflict with your answer, and if so, what is the correct answer? $\endgroup$ – Richard Hardy May 20 at 11:15
  • $\begingroup$ I think he is quite right when defining weak stationarity as constancy of the first two moments. Often, and also in the present thread, "stationarity" and "existence of a causal representation", i.e., a summable $MA(\infty)$ representation without dependence on the future, are conflated. What my answer therefore more precisely shows is conditions for the existence of the latter. $\endgroup$ – Christoph Hanck May 20 at 12:44

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