~ 2u sin (vAt) cosh (uAt)—2v sinh (uAt) cos (vAt) 
Then, 
u [2P sin (vAt) cosh (uAt)+ sin (2vAr)] (6.93) 
v= ooo. 5 
2P sinh (uAr) cos (vAt)+ sinh (2uAr) 
From Egs. 6.90 and 6.81 
we eA — [ay (6.94) 
Therefore, from Eqs. 6.91 and 6.94 
2 Vaz 
\— (ai — 4a) 
sin (vAr) = Ales 
2 
cos (vAf) = 
(6.95) 
1+ 
cosh (uAr) = 22 
1-ap 
2Vay 
sinh (uAr) = 
With Eqs. 6.95 and 6.85, Eq. 6.93 can be transformed into an equation with a), a2, and 5), 
— ik —2a,;+(1+a)(b) ++ 
we Sa) V— (a3 — 4a) 22+ Ctan(ritd) (6.96) 
2Ar 2(1 — a3) —ay(1 —a)(b, +2) 
If we look at the first equation of Eq. 6.95, v is derived as 
14 [ee (6.97) 
1 
vV=— cos” : 
At 2 Va, 
That value is not unique, since it does not include the MA parameter b,. However, 
Eq. 6.96 shows that v is uniquely determined from aj, a2, and by, since it includes the 
MA parameter b, as well as a; and a2. This is another reason why generally we prefer 
ARMA(2.1) over AR(2) as the discrete process corresponding to A(2) and u is expressed 
by Eq. 6.91. 
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