(i), ˆ

STAT 3830: Time Series and Forecasting
Reference answer for Assignment 2
April 27, 2007
1,
Using Mathematical induction:
(i),Ẑn (1) = α[Zn + (1 − α)Zn−1 + . . .]
Ẑn (2) = α[Ẑn (1) + (1 − α)Zn + (1 − α)2 Zn−1 + . . .]
= αẐn (1) + (1 − α)[α(Zn + (1 − α)Zn−1 + . . .)]
= αẐn (1) + (1 − α)Ẑn (1)
= Ẑn (1)
(ii), letẐn (i) = Ẑn (1)f ori ≤ k ≤ l
we have Ẑn (k + 1) = α[Ẑn (k) + (1 − α)Ẑn (k − 1) + (1 − α)2 Ẑn (k − 2)
+ . . . + (1 − α)k−1 Ẑn (1) + (1 − α)k Zn + . . .]
1 − (1 − α)k
] + (1 − α)k [αZn + (1 − α)Zn−1 + . . .]
= α[Ẑn (1)
1 − (1 − α)
= Ẑn (1) − (1 − α)k Ẑn (1) + (1 − α)k Ẑn (1)
= Ẑn (1)
Consider (i) and (ii), we can conclude from MI that Ẑn (l) = Ẑn (1), for l > 0
2,
Ẑ30 (2) = Ẑ30 (1) = 102.5
Ẑ31 (1) = Ẑ30 (1) + α[Z31 − Ẑ30 (1)]
= 102.5 + 0.1(105 − 102.5)
= 102.75
3, β̂0,24 = 30, β̂1,24 = 2, α = 0.1, w = 0.9
(
[1]
[2]
[1]
2S24 − S24 = 30
S24 = 12
⇒
[1]
[2]
[2]
0.1
[S − S24 ] = 2
S24 ] = −6
0.9 24
1



[1]
[2]
β̂0,25 = 2S25 − S25 = 31.24
[1]
[1]
S25 = 0.1Z25 + 0.9S24 = 2.8 + 0.9 × 12 = 13.6
[1]
[2]
⇒ β̂1,25 = wa (S25
− S25 ) = 1.96
[2]

S
=
0.1
×
13.6
+
0.9
×
(−6)
=
−4.04
 25
Ẑ25 (1) = 31.24 + 1.96 = 33.2



[1]
[2]
β̂0,26 = 2S26 − S26 = 32.782
[1]
[1]
S26 = 0.1Z26 + 0.9S25 = 3.1 + 0.9 × 13.6 = 15.34
[1]
[2]
⇒ β̂1,26 = wa (S26
− S26 ) = 1.938
[2]

 S26 = 0.1 × 15.34 + 0.9 × (−4.04) = −2.102
Ẑ26 (1) = 32.782 + 1.938 = 34.72



[1]
[2]
β̂0,27 = 2S27 − S27 = 34.9632
[1]
[1]
S27 = 0.1Z27 + 0.9S26 = 3.6 + 0.9 × 15.34 = 17.406
[2]
[1]
⇒ β̂1,27 = wa (S27
− S27 ) = 1.9508
[2]

 S27 = 0.1 × 17.406 + 0.9 × (−2.102) = −0.1512
Ẑ27 (1) = 34.9632 + 1.9508 = 36.887
4,








1
1
1 0 0
1 0 0
a, f (j + 1) =  sin π2 (j + 1)  =  cos π2 j  = f (j)  0 0 1  ⇒ L =  0 0 1 
cos π2 (j + 1)
− sin π2j
0 −1 0
0 −1 0


1


j+1


b, f (j + 1) = 
=
π
 sin 2 (j + 1) 
cos π2 (j + 1)

1

1
= f (j) 
0
0
0 0
1 0
0 0
0 −1

0

0
⇒L=
1
0
2

1

1

0
0


1


 j+1 


 cos π2 j 
− sin π2j
0 0
1 0
0 0
0 −1

0

0

1
0

1


j+1




π
(j
+
1)
sin


2
c, f (j + 1) = 
=


cos π2 (j + 1)


π
 (j + 1)sin 2 (j + 1) 
(j + 1)cos π2 (j + 1)

1


j+1




π
j
cos


2


 −sinπ2j 


 (j + 1)cos π2 j 
−(j + 1)sin π2 j


1
1


0
= f (j) 
0

0
0
0 0 0 0
1 0 0 0
0 0 1 0
0 −1 0 0
0 0 1 0
0 −1 0 −1

0
0


0
⇒L=
0

1
0

1
1


0

0

0
0



1


j+1




(j
+
1)
 sin 2π

2
c, f (j + 1) = 
=
2π
 cos 2 (j + 1) 


 sin 4π
(j + 1) 
2
cos 4π
(j + 1)
2

1

1

0
= f (j) 
0


0
0
0 0
1 √0
0 23
0 − 21
0 0
0 0
0
0
1
√2
3
2
0
0
0
0
0
0
1
2√
−
3
2
0
0
0
0
√


1



1



0
⇒L= 

0




3
0
2 
1
0
2
3

0
0


0

0

1
0
0 0 0 0
1 0 0 0
0 0 1 0
0 −1 0 0
0 0 1 0
0 −1 0 −1

1

j+1
√
 3
 2 sin 2πj
+ 12 cos 2πj
12
12
 √3
 cos 2πj − 1 sin 2πj
 2
12 √ 2
12
1
4πj
4πj
3
sin
+
cos
2
12
2
12
√
4πj
4πj
3
1
cos
−
sin
2
12
2
12

0 0
1 √0
0 23
0 − 12
0 0
0 0

0
0
1
√2
3
2
0
0
0
0
0
0
1
2√
−
3
2
0
0
0
0
√
















3 
2 
1
2