**Unformatted text preview: **1 Assignment 2
Textbook problems: 1.4, 1.5, 1.7, 1.10. 2 Answer keys:
= ; independent of 1.4 (30 points; 5 point per part)
(a) , (c) (d) (e) 0
is stationary. , Therefore, = , Therefore, = , Therefore, = , = , =* ℎ=0
ℎ = ±1
ℎ = ±2
ℎ is stationary.
! + 1 "# $ is NOT stationary except in the trivial case when is an integer multiple of %. = ; independent of ; independent of = 0; independent of is stationary.
&' ( ! 2 +ℎ "+ ℎ ) is NOT stationary except in the trivial case when is an integer multiple of %. = 0; independent of Therefore, ; independent of ℎ ; independent of = 0; independent of Therefore,
(f) = = 0; independent of Therefore,
(b) +
0 0 + ℎ=0
; independent of
|ℎ| > 0 is stationary. Moreover, 1.5 (20 points; 10+5+5)
1+4
12 ℎ = 34
0
When 4 = 0.8, (a) The ACVF and ACF of 1.64
12 ℎ = 30.8
0 ~/0 0, + . 1
5 ℎ = 0
5 ℎ = ±2 and 72 ℎ = 34/ 1 + 4
ℎ
0
are, respectively 5 ℎ = 0
1
and
7
ℎ
=
3
5 ℎ = ±2
0.49
2
ℎ
0 5 ℎ = 0
5 ℎ = ±2 .
ℎ 5 ℎ = 0
5 ℎ = ±2 .
ℎ 3 (b) When 4 = 0.8,
> ! ? + + (c) When 4 = −0.8, @ + + + + 1
/4" =
AA
16 ! B, CD? BD? C" = 1
E412 0 + 412 2 F = 0.61.
16 1
E412 0 + 412 2 F = 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).
> ! ? + + @ + /4" = + + H = I2 + IJ , which is independent of . 1.7 (5 points) Since
K , H = 0 for all
independent of . +H and , , Therefore,
+ H is weakly stationary by definition, and 12
shown above.
For M = ∑PODQ
1.10 (5 points) O O for some constants , we have
P ∇M = A ODQ Q , ⋯ , PT O O P −A , since ODQ O −1 That is, ∇M is a polynomial of degree S − 1. P −1 = P O =S −S P PT? + H = 12 ℎ + 1J ℎ , which is
J PT? ℎ = 12 ℎ + 1J ℎ , ∀ ℎ, as is PT +A + ⋯. ODQ O O By successive application of the difference operator ∇, we deduce that ∇P ? M = 0. ...

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