AR(2)모델의 condition 유도하기

출처 : https://stats.stackexchange.com/questions/118019/a-proof-for-the-stationarity-of-an-ar2

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)

Consider the equation

λ2−ϕ1λ−ϕ2=0$${\lambda }^2-{\varphi }_1\lambda -{\varphi }_2=0$$

If z$z$ is a root of the "standard" characteristic equation 1−ϕ1z−ϕ2z2=0$1-{\varphi }_{1}z-{\varphi }_2{z}^2=0$ and setting z−1=λ$z^{-1}=\lambda$, the display obtains from rewriting the standard one as follows:

1−ϕ1z−ϕ2z2⇒z−2−ϕ1z−1−ϕ2⇒λ2−ϕ1λ−ϕ2===000$$\begin{array}{rcl}1-{\varphi }_{1}z-{\varphi }_2{z}^2& =& 0\\ \Rightarrow z^{-2}-{\varphi }_1{z}^{-1}-{\varphi }_2& =& 0\\ \Rightarrow {\lambda }^2-{\varphi }_1\lambda -{\varphi }_2& =& 0\end{array}$$

Hence, an alternative condition for stability of an AR(2)$AR(2)$ is that all roots of the first display are insidethe unit circle, |z|>1⇔|λ|=|z−1|<1$|z|>1\iff |\lambda |=|z^{-1}|<1$.

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

1. ϕ2<1+ϕ1${\varphi }_2<1+{\varphi }_1$
2. ϕ2<1−ϕ1${\varphi }_2<1-{\varphi }_1$
3. ϕ2>−1${\varphi }_2>-1$

Recall that you can write the roots of the first display (if real) as

λ1,2=ϕ1±ϕ21+4ϕ2−−−−−−−√2$${\lambda }_{1,2}=\frac{{\varphi }_1\pm \sqrt{{\varphi }_1^2+4{\varphi }_2}}{2}$$

to find the first two conditions.Consider the equation

λ2−ϕ1λ−ϕ2=0$${\lambda }^2-{\varphi }_1\lambda -{\varphi }_2=0$$

If z$z$ is a root of the "standard" characteristic equation 1−ϕ1z−ϕ2z2=0$1-{\varphi }_{1}z-{\varphi }_2{z}^2=0$ and setting z−1=λ$z^{-1}=\lambda$, the display obtains from rewriting the standard one as follows:

1−ϕ1z−ϕ2z2⇒z−2−ϕ1z−1−ϕ2⇒λ2−ϕ1λ−ϕ2===000$$\begin{array}{rcl}1-{\varphi }_{1}z-{\varphi }_2{z}^2& =& 0\\ \Rightarrow z^{-2}-{\varphi }_1{z}^{-1}-{\varphi }_2& =& 0\\ \Rightarrow {\lambda }^2-{\varphi }_1\lambda -{\varphi }_2& =& 0\end{array}$$

Hence, an alternative condition for stability of an AR(2)$AR(2)$ is that all roots of the first display are insidethe unit circle, |z|>1⇔|λ|=|z−1|<1$|z|>1\iff |\lambda |=|z^{-1}|<1$.

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

1. ϕ2<1+ϕ1${\varphi }_2<1+{\varphi }_1$
2. ϕ2<1−ϕ1${\varphi }_2<1-{\varphi }_1$
3. ϕ2>−1${\varphi }_2>-1$

Recall that you can write the roots of the first display (if real) as

λ1,2=ϕ1±ϕ21+4ϕ2−−−−−−−√2$${\lambda }_{1,2}=\frac{{\varphi }_1\pm \sqrt{{\varphi }_1^2+4{\varphi }_2}}{2}$$

to find the first two conditions.

Then, the AR(2)$AR(2)$ is stationary iff |λ|<1$|\lambda |<1$, hence (if the λi${\lambda }_i$ are real):

−1<ϕ1±ϕ21+4ϕ2−−−−−−−√2⇒−2<ϕ1±ϕ21+4ϕ2−−−−−−−√<<12$$\begin{array}{rcl}-1<\frac{{\varphi }_1\pm \sqrt{{\varphi }_1^2+4{\varphi }_2}}{2}& <& 1\\ \Rightarrow -2<{\varphi }_1\pm \sqrt{{\varphi }_1^2+4{\varphi }_2}& <& 2\end{array}$$

The larger of the two λi${\lambda }_i$ is bounded by ϕ1+ϕ21+4ϕ2−−−−−−−√<2${\varphi }_1+\sqrt{{\varphi }_1^2+4{\varphi }_2}<2$, or:

ϕ1+ϕ21+4ϕ2−−−−−−−√⇒ϕ21+4ϕ2−−−−−−−√⇒ϕ21+4ϕ2⇒ϕ21+4ϕ2⇒ϕ2<<<<<22−ϕ1(2−ϕ1)24−4ϕ1+ϕ211−ϕ1$$\begin{array}{rcl}{\varphi }_1+\sqrt{{\varphi }_1^2+4{\varphi }_2}& <& 2\\ \Rightarrow \sqrt{{\varphi }_1^2+4{\varphi }_2}& <& 2-{\varphi }_1\\ \Rightarrow {\varphi }_1^2+4{\varphi }_2& <& (2-{\varphi }_1{)}^2\\ \Rightarrow {\varphi }_1^2+4{\varphi }_2& <& 4-4{\varphi }_1+{\varphi }_1^2\\ \Rightarrow {\varphi }_2& <& 1-{\varphi }_1\end{array}$$

Analogously, we find that ϕ2<1+ϕ1${\varphi }_2<1+{\varphi }_1$.

If λi${\lambda }_i$ is complex, then ϕ21<−4ϕ2${\varphi }_1^2<-4{\varphi }_2$ and so

λ1,2=ϕ1/2±iϕ21+4ϕ2−−−−−−−√/2.$${\lambda }_{1,2}={\varphi }_1/2\pm i\sqrt{{\varphi }_1^2+4{\varphi }_2}/2.$$

The squared modulus of a complex number is the square of the real plus the square of the imaginary part. Hence,

λ2=(ϕ1/2)2−(ϕ21+4ϕ2−−−−−−−√/2)2=ϕ21/4−(ϕ21+4ϕ2)/4=−ϕ2.$${\lambda }^2=({\varphi }_1/2{)}^2-{\left(\sqrt{{\varphi }_1^2+4{\varphi }_2}/2\right)}^2={\varphi }_1^2/4-({\varphi }_1^2+4{\varphi }_2)/4=-{\varphi }_2.$$

This is stable if |λ|<1$|\lambda |<1$, hence if −ϕ2<1$-{\varphi }_2<1$ or ϕ2>−1${\varphi }_2>-1$, as was to be shown. (The restriction ϕ2<1${\varphi }_2<1$ resulting from ϕ22<1${\varphi }_2^2<1$ is redundant in view of ϕ2<1+ϕ1${\varphi }_2<1+{\varphi }_1$ and ϕ2<1−ϕ1${\varphi }_2<1-{\varphi }_1$.)

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