This solution X, can also be written as 
foe} 
KS SG 9 Cae Cie, pt Gasp a? 2 (5.71) 
j=0 
Equation 5.71 shows the AR(2) process is expressed by MA( © ), as was the case for 
AR(1). 
5.2.3.2 Inverse Function for AR(2). To say the inverse function for the autoregres- 
sive process has little meaning was pointed out for AR(1). Formally, however, for AR(2), 
from 
X,+ a)Xj-1 + 22X12 = Er, 
(1 +.a;B + apB)X, = CW, _; 
j=0 
= (—Ip—1;B —InB?- - - 1; B')X, . (j > &) 
Therefore 
Ip =-1 
rg =-@Q) 
In =-a2 
I; =0 for j 2 3. (5.72) 
5.2.3.3 Stationality of AR(2). Returning to the general solution of X;, we must con- 
sider the behavior of the complementary function or the solution of the homogeneous 
Eq. 5.60 that is in the shape of 
Ay Mj +A 13. 
This complementary function represents the free oscillation of a forced oscillation system 
which should die out over time t and become asymptotically stationary. For this condi- 
tion it is necessary that 
Wyl<l, ol <1. 
This means that the quadrant, as expressed by Eq. 5.65 equated to zero, 
flZ) =Z?+a;Z+a,=0 (5.73) 
should have roots less than 1 in their absolute values. Therefore, {Z) = 0 must have 
roots inside the unit circle. These roots of f(Z) are the reciprocals of the roots of 
Q(Z) =1+a,Z+aZ’. (5.74) 
Accordingly, @(Z) must have roots outside the unit circle in order for AR(2) to be asymp- 
totically stationary. 
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