Get (5.128) 
Go=1 
Goa, + G = by ae SEG (5.129) 
Goa2 + a1G; + G2 = 0 G2 = —a,(bj—a)— a2 (5.130) 
Gj =—a1G;1-a2Gji2, j = 2. (5.131) 
Egs. 5.128 — 5.130 are the initial conditions. 
The solution of the homogeneous Eq. 5.126 is the solution of an nth order differen- 
tial equation, that is, a linear combination of exponential functions, u/ w being the 
characteristic roots, i.e., the roots of the characteristic equation f(Z). Equation 5.73 in this 
case is of second order. 
Therefore for this case, 
Gj = 81 Mi +e wh. (5.132) 
Using the initial conditions given by Eqs. 5.128 and 5.129, 
8it+g2=1 (5.133) 
Sill + 822 = Mi +2 + dr, (5.134) 
where 
(41 + M2 = —a)}). 
Solving gives 
_Mitdi 
Hi-#2 (5.135) 
_ Hatbi (5.136) 
(Lira 
2-1 
Inserting these values into Eq. 5.132, gives the same expression of Green’s function Eq. 
5.125 for ARMA(2.1). As will be discussed in some detail in the next section, in order 
that the original process X;, which can be considered as an output of the input €, , be sta- 
tionary, 4; 2 should be less than 1 in absolute value, | “4; | <1, |“21< 1, wherew 1,2 
are the roots of the characteristic Eq. 5.73. 
fiZ) =Z* +a\Z+ ay =0 
-a\ Ee ¥a?—4a, 
41, Mee aa 2 
for “1 +M2 =—a, and yz = a2. When a?—4a, <0, then 4, “, are complex and con- 
jugate to each other as 
SSG Vv 4a2—a? 
Ee ie ee 
(5.137) 
(5.137) 
siz 
