**Unformatted text preview: **1 TIME SERIES ANALYSIS Assignment 2 answer keys
1.4 (a, c)
(a) = ; independent of
, = 0
is stationary. = 0; independent of Therefore,
(c) , = ! 1.5 1+)
&' ℎ = ()
0
When ) = 0.8, 1.64
&' ℎ = (0.8
0
(b) When ) = 0.8,
4 + 1 "# $ 1
*ℎ=0
and
,
ℎ
=
(
* ℎ = ±2
)/ 1 + )
'
ℎ
0 (a) The ACVF and ACF of ! ; independent of is NOT stationary except in the trivial case when is an integer multiple of %. Therefore, 3 ℎ=0
ℎ = ±1
ℎ = ±2
ℎ +
0 + + (c) When ) = −0.8, *ℎ=0
1
* ℎ = ±2 and ,' ℎ = (0.49
ℎ
0
5 *ℎ=0
* ℎ = ±2 .
ℎ are, respectively + 6 6 6 1
77
/4" =
16
9:4 8:4 ! 8, *ℎ=0
* ℎ = ±2 .
ℎ 9" = 1
;4&' 0 + 4&' 2 < = 0.61.
16 1
;4&' 0 + 4&' 2 < = 0.21.
16
The negative lag-2 correlation in
means that positive deviations of from zero tend to
be followed two time units later by a compensating negative deviation, resulting in smaller
variability in the sample mean than in part (b).
3 For > = ∑[email protected]:B
1.10 @ @ ! 4 + + , we have
A ∇> = 7 @:B @ @ 5 + A −7 @:B 6 /4" = @ −1 @ =D A AE4 AE +7 @:B @ @ 2 TIME SERIES ANALYSIS
B , ⋯ , AE for some constants −1 , since That is, ∇> is a polynomial of degree D − 1. A = A −D AE4 + ⋯. By successive application of the difference operator ∇, we deduce that ∇A 4 > = 0. 1.18
2.3 (a) & 0 = 1 + 0.3 + −0.4 = 1.25 & 1 = 1 0.3 + 0.3 −0.4 = 0.18
& 2 = 1 −0.4 = −0.4 & ℎ = 0 for ℎ = 3, 4, ⋯ Moreover, & −ℎ = & ℎ for ℎ = 1, 2, ⋯ (b) & 0 = 0.25 ;1 + −1.2 + −1.6 < = 1.25 & 1 = 0.25 ;1 −1.2 + −1.2 −1.6 < = 0.18
& 2 = 0.25 1 −1.6 = −0.4 & ℎ = 0 for ℎ = 3, 4, ⋯ Moreover, & −ℎ = & ℎ for ℎ = 1, 2, ⋯ Same ACVF as in part (a)
2.5
Absolute convergence:
Note that, L I7J) 9
9:4 L KE9 JM = 7|)|9 J = ! KE9 " 9:4 KE9 J since |)| < 1; the first inequality follow from that
! J 9
As a result, ∑L
9:4J) 9
That is, ∑T
9:4 ) For > > W, KE9 J" KE9 KE9 J −3 !J ≤ 7|)|9 P&' 0 + Q' < ∞ KE9 J" < ∞ with probability 1. L 9:4 ≤ ! KE9 " = &' 0 + Q' . converges absolutely (with probability 1) as > → ∞. Mean square convergence: 3 TIME SERIES ANALYSIS |#T − #@ | = T X 7 ) Y Y:@ 4
T
T = 7 KEY Z 7 ) Y:@ 4 [:@ 4 Y [ T = 7 &' T T 7 )Y Y:@ 4 [:@ 4 − 9
Therefore, ∑T
9:4 ) KE9 T + Q' ≤ 7 = &' 0 + Q' X 7 |)|Y Z → 0 Y
as W, > → ∞ (since ∑L
Y:B|)| < ∞). [ KEY KE[
T 7 |)|Y Y:@ 4 [:@ 4 Y:@ 4 converges in mean square as > → ∞. [ &' 0 + Q' ...

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