Unformatted text preview: 1 TIME SERIES ANALYSIS Assignment 5 answer keys
3.2 (for Problem 3.1(b, d), as 3.1(b) is not causal)
Lag h
ℎ
ℎ 1
-0.994
-0.994 2
0.980
-0.810 3
-0.959
0 4
0.932
0 5
-0.900
0 6
0.866
0 7
-0.830
0 8
0.792
0 9
-0.753
0 10
0.715
0 3.3 (for Problem 3.1(b, d), as 3.1(b) is not causal)
=1 = −1.8 = −1.8
= −1.8 = −1.8
3.4 = −1.8 = −1.8 + −0.81 + −0.81 + −0.81
+ −0.81 ACF: Multiply both sides of = 2.43 = −2.916 = 3.2805 = −3.54294
− 0.8 = by , ℎ ≥ 0, and take expectations. We have 0 − 0.8 2 = ! 1 − 0.8 1 = 0 2 − 0.8 0 = 0
⋮ ℎ − 0.8 ℎ − 2 = 0 for ℎ ≥ 3 2 TIME SERIES ANALYSIS
0 = Solve the first 3 equations, we obtain
Moreover, with ) 1 = 0, and , ℎ = 0.8 ℎ − 2 for ℎ ≥ 3, we obtain
0.8 ℎ =* It follows that ACF . ℎ =/ PACF: By definition,
component of
/ that is, '( 0 = 1, . 0
. 1 . 1
2
. 0 2 = 0.8 = 6 . 3.8 9
0 0.8
0 25!
/ 1 =0 0 1
/ ℎ = 0 for ℎ > 2. On the other hand, / 2 = '( ) . |ℎ| = 0,2, ⋯ |ℎ| = 1,3, ⋯ |ℎ| = 0,2, ⋯
|ℎ| = 1,3, ⋯ 1 = . 1 = 0, and . 1
1 0
2=3
4
. 2
0 1 3 0
0
4 = 3 4;
0.8
0.8 2 is the last By equation (2.2.11) (textbook, page 54), the stationary solution of the difference equations,
=6
+ with |6| > 1 & 9 :~<= 0, ! , is given by
A It follows that C Now, < = −6 Then, C < = C = 0 and
. −6 FGH < @ , < = FGH C = 1+6 Thus, 9< :~<= 0, E( .
'( = −>6 D @ ?B ℎ = E( 6 '( = 0 and −6 D @ ℎ −6 | | , D ? @? . . −6 ℎ−1 −6 D !
ℎ + 1 = I6
0 ℎ = 0.
ℎ≠0 Note that, the causal representation has a smaller white noise variance than the noncausal
representation.
3.9
(a) The ACVF is given by 3 TIME SERIES ANALYSIS K ℎ = FGH +L
@ +L
@
= ! 1 + L + L 19
+ ! L 19 B± : B , +L
+L
+
!
L
1
+
! L L 19
:
9 B± : @ or ! 1+L +L
Q
! L
O
ℎ
=
K
! LL
P
! L
O
N
0
(b) From R, the sample mean is 28.831 and
Lag h
X ℎ
Lag h
X ℎ
Lag h
X ℎ 0
152670
7
-31165
14
17758 So, LZ = )[ (c) Matching 1 , K [\] K 1
-54327
8
-1088
15
-6200 11 , K = −0.586, LZ 2
-15072
9
15277
16
-9656 2.16 ℎ=0
ℎ = ±1
ℎ = ±11
ℎ = ±12
GRℎSTUVWS 3
14585
10
-12435
17
27981 4
-17178
11
29802
18
-29456 5
6340
12
-50867
19
3693 6
17421
13
13768
20
7569 12 with XK 1 , XK 11 , XK 12 , respectively, we have
= That is, the model for 9^ : = ∇∇
^ = 28.831 + : B± ! L = −54327
I! L L = 29802.
! L = −50867 )[ : is − 0.586 = −0.549, !X =
− 0.549 . [\ = 92740. , 9 :~<= 0, 92740 . ...
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