STAT 758, Fall 2016
Home Work 6
SARIMA
We assume below that Zt is a white noise with mean 0 and variance σZ2 .
Problem 1
For the model (1 − B)(1 − 0.2 B)Xt = (1 − 0.5 B)Zt :
a) Classify the model as an ARIMA(p, d, q) process (i.e. find p, d, q).
ARIM A(1, 1, 1)
textbfb) Determine whether the process is stationary, causal, invertible.
• The process is stationary if all roots of φ(z) are off of the unit circle.
φ(z) = (1 − z)(1 − 0.2z) = 0
→
z = 1, 5
Because φ(z) has root z = 1, which lies on the unit circle, the process is
not stationary .
• The process is causal if all roots of φ(z) are outside of the unit circle. Because
φ(z) has one root z = 1, which lies on the unit circle, the process is not causal .
• The process is invertible if all roots of θ(z) are outside of the unit circle.
θ(B) = (1 − 0.5z) = 0
→
B=z
Because the only root of theta(z), z = 2, lies outside the unit circle, the process
is invertible .
c) Evaluate the first three ψ-weights of the model when expressed as an
MA(∞) model.
P
Xt can be expressed as an MA(∞) process of the form Xt = ∞
k=0 ψk Zt−k
∞
X
θ(B)
Xt =
Zt = (1 − 0.5B)( (1.25 − 0.25(0.2)k )B k )Zt
φ(B)
k=0
∞
X
= Zt +
(0.625 + 0.075(0.2)k−1 )Zt−k
k=1
= Zt + 0.7Zt−1 + 0.64Zt−2 + 0.628Zt−3 + ...
d) Evaluate the first four π-weights of the model when expressed as an
AR(∞) model.
Zt can be expressed as a AR(∞) process of the form Zt =
P∞
k=0
πk Xt−k
∞
X
φ(B)
2
Zt =
Xt = (1 − 1.2B + 0.2B )( (0.5)k B k )Xt
θ(B)
k=0
= Xt − 0.7Xt−1 +
∞
X
−0.15(0.5)k−2 Xt−k
k=2
= Xt − 0.7Xt−1 − 0.15Xt−2 − 0.075Xt−3 − 0.0375Xt−4 + ...
Discuss how the behavior of the weights is related to
Pthe properties of the
model found in b). An ARIMA model is only causal if ∞
k=0 |ψk | < ∞, where ψk
are the coefficients from the AR(∞) process found in (c). Note that this infinite sum
is given by
1+
∞
X
|0.625 + 0.075(0.2)k−1 |
k=1
which does not converge, confirming
the given process is not causal. Similarly, an
Pthat
∞
ARMA model is only invertible if k=0 |πk | < ∞, where πk are the coefficients from
the MA(∞) process found in (d). In this case, the corresponding infinite sum is given
by
1.7 +
∞
X
| − 0.15(0.5)k−2 |
k=2
which does converge, confirming that the given process is invertible.
Problem 2
Show that the AR(2) process Xt = Xt−1 + c Xt−2 + Zt is stationary provided
−1 < c < 0. Show that the AR(3) process Xt = Xt−1 + c Xt−2 − c Xt−3 + Zt is
non-stationary for all values of c.
The AR(2) process Xt = Xt−1 + cXt−2 + Zt is stationary if and only if all roots of φ(z)
are off the unit circle. The roots of φ(z) are given by
√
1 ± 1 + 4c
2
φ(z) = −cz − z + 1 = 0
→
z=
−2c
In order
√ to determine where these roots lie in relation to the unit circle, the norm of
1+4c
z = 1±−2c
was plotted for −1 < c < 0, as shown in Figure 1. It can be seen that the
roots (real or complex) always have norm greater than 1 for this range of c (i.e., the
roots are always outside of the unit circle). Thus, the given AR(2) process is stationary
for −1 < c < 0.
Figure 1
Similarly, the AR(3) process Xt = Xt−1 + cXt−2 − cXt−3 + Zt is stationary if and
only if all roots of φ(z) are off the unit circle. The roots of phi(z) are given by
(
1,
if c = 0,
3
2
2
φ(z) = cz − cz − z + 1 = (z − 1)(cz − 1) = 0
→
z=
−0.5
1, ±c , if c 6= 0
Thus the roots of φ(z) always include z = 1 and the given AR(3) process is nonstationary for all c. Problem 3
Write a model P
for the following
SARIMA (p, d, q) × (P, D, Q)s processes in an
P
explicit form:
ai Xt−i = bj Zt−j .
a) (0, 1, 0) × (1, 0, 1)12 A SARIMA(p, d, q)X(P,D,Q)s process is typically described
in the following form
s
φp (B)ΦP (B s )(∇d ∇D
s Xt ) = θq (B)ΘQ (B )Zt
In this case, this yields the following explicit process
∇1 ∇012
φ0 (z) = 1
Φ(B 12 )(∇Xt ) = Θ(B 12 )Zt
Φ1 (z) = 1 − αz
(1 − α ∗ B 12 )(1 − B)Xt = (1 + βB 12 )Zt
= ∇ = (1 − B)
→
(1 − B − α ∗ B 12 + α ∗ B 13 )Xt = (1 + β ∗ B 12 )Zt
Xt − Xt−1 − α ∗ Xt−12 + α ∗ Xt−13 = Zt + β ∗ Zt−12
θ0 (z) = 1
Θ1 (z) = 1 + βz
b) (2, 0, 2) × (0, 0, 0)1 In this case, P = D = Q = 0, which yields the following
explicit process.
c) (1, 1, 1) × (1, 1, 1)4 In this case, except for s = 4, all other parameters are set to
1, which yields the following explicit process
∇0 ∇01 = 1
θ2 (z) = 1 + θ1 z + θ2 z
2
→
φ2 (z) = 1 − φ1 z − φ2 z 2
Φ0 (z) = 1
φ(B)Xt = θ(B)Zt
2
(1 − φ1 B − φ2 B )Xt = (1 + θ1 B + θ2 B 2 )Zt
Xt − φ1 Xt−1 − φ2 Xt−2 = Zt + θ1 Zt−1 + θ2 Zt−2
Θ0 (z) = 1
φ1 (z) = 1 − φz
Φ(z) = 1 − αz
∇1 ∇14 = (1 − B)(1 − B 4 )
→
φ(B)Φ(B 4 )(∇∇4 Xt ) = θ(B)Θ(B 4 )Zt
θ1 (z) = 1 + θz
(1 − φB)(1 − αB 4 )(1 − B)(1 − B 4 )Xt = (1 + θB)(1 + βB 4 )Zt
T heta1 (z) = 1 + βz
Expanding the lefthand side of the equation yields:
LHS = (1 − φB)(1 − αB 4 )(1 − B)(1 − B
= [1 − (φ + 1)B + φB 2 − (α + 1)B 4 + (φ + 1)(α + 1)B 5 − φ(α + 1)B 6 + αB 8 − α(φ + 1)B 9 + φαB
= Xt − (φ + 1)Xt−1 + φXt−2 − (α + 1)Xt−4 + (φ + 1)(α + 1)Xt−5 − φ(α + 1)Xt−6 + αXt−
α(φ + 1)Xt−9 + φα
Expanding the righthand side of the equation yields:
RHS = (1 + θB)(1 + βB 4 )Zt
= (1 + θB + βB 4 + θβB 5 )Zt
= Zt + θZt−1 + βZt−4 + θβZt−5
Problem 4
For each of the following ARIMA models specify their order (p, d, q) and find out
whether Xt is stationary, causal, invertible.
a) Xt + 0.2 Xt−1 − 0.48 Xt−2 = Zt
b) Xt + 1.9 Xt−1 + 0.88 Xt−2 = Zt + 0.2 Zt−1 + 0.7 Zt−2
c) Xt + 0.5 Xt−2 = Zt−2
d) Xt − 2 Xt−1 + Xt−2 = Zt − 0.3 Zt−1
e) Xt − 0.3 Xt−1 = Zt − 0.3 Zt−1
f) Xt = Xt−1 + Zt
g) Xt − 2 Xt−1 + Xt−2 = Zt + 0.5 Zt−1
Blue text denotes roots inside the unit circle, while red text denotes roots on the
unit circle, and black text denotes roots outside the unit circle.
(p, d, q)
φ(z), θ(z)
a)
(2, 0, 0)
b)
(2, 0, 2)
c)
(2, 0, 2)
d)
(0, 2, 1)
e)
(1, 0, 1)
f)
(0, 1, 0)
φ(z) = 1 + 0.2z − 0.48z 2
θ(z) = 1
φ(z) = 1 + 1.9z + 0.88z 2
θ(z) = 1 + 0.2z + 0.7z 2
φ(z) = 1 + 0.5z 2
θ(z) = z 2
φ(z) = 1 − 2z + z 2
θ(z) = 1 − 0.3z
φ(z) = 1 − 0.3z
θ(z) = 1 − 0.3z
φ(z) = 1 − z
θ(z) = 1
φ(z) = 1 − 2z + z
θ(z) = 1 + 0.5z
g)
(0, 2, 1)
roots
z=
−5 4
,
4 3
z = −5
, −10
4
11
√
69i
z = −1
±
7
√ 7
z=± 2
z = ±1
z = 1, 1
z = 10
3
z = 10
3
z = 10
3
z=1
property*
S,C
I
S
I
S,C
I
S,C
I
I
z = 1, 1
z = −2
Table 1: *S = Stationary, C = Causal, I = Invertible
I