STK320 / WST321 Time Series Analysis exercise 1 memo

1


EXERCISE 1 SUGGESTED SOLUTION
1.(a)
E (12 Z1 + 13 Z 2 ) = 12 E (Z1 ) + 13 E (Z 2 )
= 12 µ1 + 13 µ 2
= 12 × 4 + 13 × 3
=3

1.(b)
Cov(Z1 , Z 2 ) = γ 1, 2
γ 1, 2
= ρ1, 2 γ 1, 1γ 2, 2 since ρ1, 2 =
γ 1, 1γ 2, 2
1
= 2
16 × 9
=6

1.(c)
Var (Z1 + Z 2 ) = Var (Z1 ) + Var (Z 2 ) + 2Cov(Z1 , Z 2 )
= γ 1, 1 + γ 2, 2 + 2γ 1, 2
= 9 + 16 + 2 × 6
= 37

1.(d)
Var (7 + 5Z1 − 2 Z 2 ) = 52Var (Z1 ) + (−2) 2Var (Z 2 )
+ 2 × 5 × (−2)Cov(Z1 , Z 2 )
= 25 × 16 + 4 × 9 − 20 × 6
= 316

1.(e)
Cov(Z1 − Z 2 , Z1 + Z 2 ) = Var (Z1 ) + Cov(Z1 , Z 2 )
− Cov(Z1 , Z 2 ) − Var (Z 2 )
= Var (Z1 ) − Var (Z 2 )
= 16 − 9
=7

1.(f)
Cov(1 + Z 2 − 3Z1 , Z1 + 2 Z 2 ) = Cov(Z1 , Z 2 ) + 2Var (Z 2 )
− 3Var (Z1 ) − 3 × 2Cov(Z1 , Z 2 )
= 6 + 2 × 9 − 3 × 16 − 6 × 6
= −60




WST321

, 2


2.(a) (i)
Var (Yt ) = Var (at + 13 at −1 )

= Var (at ) + (13 ) Var (at −1 )
2


= σ a2 + 19 σ a2
= 109 σ a2

Cov(Yt , Yt +1 ) = Cov (at + 13 at −1 , at +1 + 13 at )
= 13 Var (at )
= 13 σ a2

Cov(Yt , Yt + k ) = Cov(at + 13 at −1 , at + k + 13 at + k −1 )
=0 for k = 2, 3, ...

Cov(Yt , Yt +1 )
Corr (Yt , Yt +1 ) =
Var (Yt )Var (Yt +1 )
1 2
σ
3 a
=
( )(
10 2 10 2
σ σ
9 a 9 a
)
= 0 .3

Cov(Yt , Yt + k )
Corr (Yt , Yt + k ) =
Var (Yt )Var (Yt + k )
=0 for k = 2, 3, ...

2.(a) (ii) Var ( X t ) = 10σ a2

Cov ( X t , X t +1 ) = 3σ a2

Cov ( X t , X t + k ) = 0 for k = 2, 3, ...

Corr ( X t , X t +1 ) = 0.3

Corr ( X t , X t + k ) = 0 for k = 2, 3, ...

2.(b) Yes, the variance of {Yt } is much less than that of {X t } .




WST321