R(r) =a R(r-1) when r ~0 (5.40) 
R(O) = a*R(O) +02 when r = 0, (5.41) 
because 
ages 
E[X;- €;]= hii (5.42) 
Ele?]=o2 i=], 
Equation 5.40 is the same type of homogeneous difference equation as that of the original 
process {X,} shown in Eq. 5.14. 
R(r)—aR(r—1)=0. (5.43) 
Using the backward shift operator B gives 
(1-aB)R(r) = 0. (5.44) 
Thus R(r) = (1—-aB)'R(r) = 0 (5.45) 
and R(r) = Aw’. (5.46) 
Here x is the root of f(Z) = Z—a =0 when n=a 
and thus R(r) = Aa’. (5.47) 
In order to be asymptotically stationary, R(r) — 0 as(r) — © when lal < 1 and f(Z) =0 
must have its roots inside the unit circle. 
From Eq. 5.41 
2 ae 
RO)=0;=Az= >: (5.48) 
l-a 
Therefore, the general solution is, from Eq. 5.43, 
o2 
RN) => a" = R(0)a’ = 020". (5.49) 
—a 
This equation is the same as Eq. 5.36. 
522.6 s(w) of AR(1). Here we assume u, = 0, lal <1. 
Then the spectrum s(w) is obtained as the Fourier transform of R(r) 
sw) == SY ROea 
r=-—-@ 
u = R(0) +2 Y. Re) cosrw 
r=1 
110 
