Calculation of ACVF for ARMA Process

54 Pages·2013·1.38 MB·English

Calculation of ACVF for ARMA Process: I
consider causal ARMA(p, q) defined by
•
2
�(B)X = ✓(B)Z , Z WN(0, � )
t t t
{ } ⇠
want to determine ACVF �(h) for this process, which can
• { }
be done using four complementary methods
first method is based on MA( ) representation
• 1
1
X = Z = (B)Z ,
t j t j t
�
j=0
X
where (B) = ✓(B)/�(B)
have noted (overhead VII–5) that ACVF can be expressed as
•
1
2
�(h) = �
j j+ h
| |
j=0
X
BD–88, CC–56, SS–27 IX–1
Calculation of ACVF for ARMA Process: II
can use recursive scheme to compute ’s (overhead VIII-12):
j
•
p
= � + ✓ , j = 0, 1, 2, . . . ,
j k j k j
�
k=1
X
but in general need ’s for infinite number of integers j
j
since 0 as j , could compute ’s out to, say,
j j
• ! ! 1
j = J + h and use
| |
J
1
2 2
�(h) = � � ,
j j+ h j j+ h
⇡
| | | |
j=0 j=0
X X
with approximation getting better with increasing J
if we have a manageable expression for ’s (true for some
j
•
processes), can get analytic expression for �(h)
BD–88, CC–79, SS–101 IX–2
Example – ARMA(1,1) Process: I
for an ARMA(1,1) process, homework exercise says that
•
1 1
j 1
X = Z + (� + ✓) � Z Z ,
t t � t j j t j
⌘
� �
j=1 j=0
X X
j 1
so = 1 and = (� + ✓)� for j 1
0 j �
�
j 1
armed with x = (valid for x < 1), away we go:
1
j=0 1 x
• | |
�
�(0) P
1 1
2 2 2j 2
= = 1 + (� + ✓) �
j �
2
�
j=0 j=1
X X
2
(� + ✓)
1
2 2j
= 1 + (� + ✓) � = 1 +
2
1 �
j=0 �
X
BD–89, CC–78, SS–104 IX–3
Example – ARMA(1,1) Process: II
j 1
for h > 0, with = (� + ✓)� for j 1, have
j �
• �
�(h)
1 1
= = +
j j+h h j j+h
2
�
j=0 j=1
X X
1
h 1 2 2j+h 2
= (� + ✓)� + (� + ✓) �
� �
j=1
X
h 2
� (� + ✓)
h 1
= (� + ✓)� +
�
2
1 �
�
2
�(� + ✓)
h 1
= � � + ✓ +
�
2
1 �
!
�
note: �(h) = ��(h 1) for h 2
• � �
BD–89, CC–78, SS–104 IX–4
Example – ARMA(1,1) Process: III
overheads that follow show ACVFs (circles) for ARMA(1,1)
•
processes X with
t
{ }
2
� = 1
�
AR parameter � = 0.9
�
MA parameter ✓ ranging from 0.99 down to 0.99
� �
have � (h) = �� (h 1) for h 2, but not for h = 1
X X
• � �
h
casual AR(1) process Y has ACVF � (h) = � � (0)
t Y | | Y
• { }
have � (h) = �� (h 1) for h 1
Y Y
• � �
overheads also show ACVFs (asterisks) for Y with � = 0.9
t
• { }
and with � (0) set such that � (1) = � (1)
Y Y X
note: for some ✓, not possible to do!
�
IX–5
ACVF for ARMA(1,1) Process, � = 0.9, ✓ = 0.99
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0 5 10 15 20
h (lag)
IX–6
ACVF for ARMA(1,1) Process, � = 0.9, ✓ = 0.5
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IX–7
ACVF for ARMA(1,1) Process, � = 0.9, ✓ = 0
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IX–8
ACVF for ARMA(1,1) Process, � = 0.9, ✓ = 0.25
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h (lag)
IX–9
ACVF for ARMA(1,1) Process, � = 0.9, ✓ = 0.5
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IX–10