STK320/ WST 321 Exercise 7 memo

1


EXERCISE 7 SUGGESTED SOLUTION
1.(a) (i)
Zˆt (1) = E (Zt +1 | Zt )
= E (θ 0 + φ1Z t + φ2 Z t −1 + at +1 | Zt )
= θ 0 + φ1Z t + φ2 Zt −1

Zˆ t (2) = E (Z t + 2 | Z t )
= E (θ 0 + φ1Z t +1 + φ2 Z t + at + 2 | Z t )
= θ + φ Zˆ (1) + φ Z
0 1 t 2 t


The general linear process representation of Z t + h is:

Z t + h = µ + ω0at + h + ω1at + h −1 + ω2 at + h − 2 + ...

θ0
with µ = , ω0 = 1 , ω1 = φ1 and ωk = φ1ωk −1 + φ2ωk − 2 for k = 2, 3, ... .
1 − φ1 − φ2

Zˆ t (h) = E (Z t + h | Zt )
= E (µ + ω0 at + h + ω1at + h −1 + ω2 at + h − 2 + ... + ωh −1at +1
+ ωh at + ωh +1at −1 + ωh + 2 at − 2 + ... | Zt )
= µ + ωh at + ωh +1at −1 + ωh + 2 at − 2 + ...
∞
= µ + ∑ ωk at + h − k
k =h


1.(a) (ii)
et (1) = Z t +1 − Zˆt (1)
= [θ 0 + φ1Z t + φ2 Z t −1 + at +1 ] − [θ 0 + φ1Z t + φ2 Z t −1 ]
= at +1

Var[et (1)] = Var(at +1 )
= σ a2

et (2) = Z t + 2 − Zˆ t (2)
[
= [θ 0 + φ1Z t +1 + φ2 Z t + at + 2 ] − θ 0 + φ1Zˆt (1) + φ2 Z t ]
= a + φ Z − Zˆ (1)
t +2 1 [ t +1 t ]
= at + 2 + φ1at +1

Var[et (2)] = Var (at + 2 + φ1at +1 )
(
= σ a2 1 + φ12 )
WST321

, 2


et (h) = Z t + h − Zˆt (h)
= [µ + ω0 at + h + ω1at + h −1 + ω2 at + h − 2 + ... + ωh −1at +1
+ ωh at + ωh +1at −1 + ωh + 2 at − 2 + ...]
− [µ + ωh at + ωh +1at −1 + ωh + 2 at − 2 + ...]
= ω0 at + h + ω1at + h −1 + ω2 at + h − 2 + ... + ωh −1at +1
h −1
= ∑ ωk at + h − k
k =0


 h −1 
Var[et (h)] = Var ∑ ωk at + h − k 
 k =0 
h −1
= σ a2 ∑ ωk2
k =0


1.(a) (iii) 95% prediction interval for Z t +1 :

(Zˆ (1) − 1.96
t Var[et (1)] , Zˆ t (1) + 1.96 Var[et (1)] )
= (θ 0 + φ1Z t + φ2 Z t −1 − 1.96σ a , θ 0 + φ1Z t + φ2 Z t −1 + 1.96σ a )

95% prediction interval for Z t + 2 :

(Zˆ (2) − 1.96
t Var[et (2)] , Zˆt (2) + 1.96 Var[et (2)] )
(
= θ 0 + φ1Zˆ t (1) + φ2 Zt − 1.96σ a 1 + φ12 , θ 0 + φ1Zˆ t (1) + φ2 Zt + 1.96σ a 1 + φ12 )
95% prediction interval for Zt + h for h = 3, 4, K :

(Zˆ (h) − 1.96
t Var[et (h)] , Zˆt (h) + 1.96 Var[et (h)] )
 ∞ h −1 ∞ h −1 

= µ + ∑ ωk at + h − k − 1.96σ a
 ∑ ωk2 , µ + ∑ ωk at + h − k + 1.96σ a ∑ ωk2 
 k =h k =0 k =h k =0 

1.(b)
Zˆ t +1 (1) = E (Z t + 2 | Zt +1 )
= E (θ 0 + φ1Z t +1 + φ2 Z t + at + 2 | Zt +1 )
= θ 0 + φ1Z t +1 + φ2 Z t
[
= θ 0 + φ1Z t +1 + Zˆt (2) − θ 0 − φ1Zˆt (1) ]
t 1 [
= Zˆ (2) + φ Z − Zˆ (1) t +1 t ]



WST321