# A proof for the stationarity of an AR(2)

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?

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

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)

• this is a very detailed explanation. – Marco Jul 29 '15 at 10:47
• @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" – shani Aug 28 '18 at 4:08
• Thanks, quite right! I was referring to the sqaured modulus, see the edit. – Christoph Hanck Aug 28 '18 at 8:53
• @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? – Richard Hardy May 20 '19 at 11:15
• 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. – Christoph Hanck May 20 '19 at 12:44