We assume below that Zt ∼ W N (0, σ 2 ), B is a backshift operator.
6.1 For the model (1 − B)(1 − 0.2B)Xt = (1 − 0.5B)Zt :
a) Classify the model as an ARIMA(p, d, q) process (i.e. find p, d, q). ARIMA(1,1,1)
b) 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 θ(z), z = 2, lies outside the unit circle, the process is invertible .
c) Evaluate the first three ψ-weights of the model whenP
expressed as an AR(∞) model.
∞
Xt can be expressed as an AR(∞) process of the form Xt = k=0 ψk Zt−k
∞
X
θ(B)
Zt = (1 − 0.5B)
(1.25 − 0.25(0.2)k )B k
Xt =
ϕ(B)
!
Zt
k=0
= Zt +
∞
X
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 whenP
expressed as an MA(∞) model.
∞
Zt can be expressed as a MA(∞) process of the form Zt = k=0 πk Xt−k
∞
X
ϕ(B)
Xt = (1 − 1.2B + 0.2B 2 )
0.5k B k
Zt =
θ(B)
!
Xt
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 the properties of the model found in (b).
P∞
An ARMA model is only causal if k=0 |ψk | < ∞, where ψk are the coefficients from the AR(∞) process found in
(c). Note that in out case, this infinite sum is given by
1+
∞
X
|0.625 + 0.075(0.2)k−1 |
k=1
which
P∞ does not converge, confirming that the given process is not causal. Similarly, an 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
1
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.
6.2 Show that the AR(2) process Xt = Xt−1 + cXt−2 + Zt is stationary provided −1 < c < 0. Show that the
AR(3) process Xt = Xt−1 + cXt−2 + cXt−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
√
1+4c
was plotted for
In order to determine where these roots lie in relation to the unit circle, the norm of z = 1±−2c
−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 ϕ(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 non-stationary for all c.
6.3 Write
a model
P
P for the following SARIMA(p, d, q) × (P, D, Q)s processes in an explicit form:
ai Xt−i =
bj Zt−j .
a) (0, 1, 0) × (1, 0, 1)12 ;
A SARIMA(p, d, q) × (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
ϕ0 (z) = 1
Φ(B 12 )(∇Xt ) = Θ(B 12 )Zt
Φ1 (z) = 1 − αz
∇1 ∇012 = ∇ = (1 − B)
(1 − αB 12 )(1 − B)Xt = (1 + βB 12 )Zt
=⇒
(1 − B − αB 12 + αB 13 )Xt = (1 + βB 12 )Zt
θ0 (z) = 1
Xt − Xt−1 − αXt−12 + αXt−13 = Zt + βZt−12
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
2
ϕ2 (z) = 1 − ϕ1 z − ϕ2 z 2
ϕ(B)Xt = θ(B)Zt
Φ0 (z) = 1
∇
0
∇01
=1
(1 − ϕ1 B − ϕ2 B 2 )Xt = (1 + θ1 B + θ2 B 2 )Zt
=⇒
θ2 (z) = 1 + θ1 z + θ2 z 2
Xt − ϕ1 Xt−1 − ϕ2 Xt−2 = Zt + θ1 Zt−1 + θ2 Zt−2
Θ0 (z) = 1
ϕ1 (z) = 1 − ϕz
Φ1 (z) = 1 − αz
∇
1
∇14
4
= (1 − B)(1 − B )
ϕ(B)Φ(B 4 )(∇∇4 Xt ) = θ(B)Θ(B 4 )Zt
=⇒
(1 − ϕB)(1 − αB 4 )(1 − B)(1 − B 4 )Xt = (1 + θB)(1 + βB 4 )Zt
θ1 (z) = 1 + θz
Θ1 (z) = 1 + βz
Expanding the lefthand side of the equation yields:
LHS = (1 − ϕB)(1 − αB 4 )(1 − B)(1 − B 4 )Xt
= [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 10 ]Xt
= Xt − (ϕ + 1)Xt−1 + ϕXt−2 − (α + 1)Xt−4 + (ϕ + 1)(α + 1)Xt−5 − ϕ(α + 1)Xt−6 + αXt−8 − · · ·
α(ϕ + 1)Xt−9 + ϕαXt−10
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
6.4 For each of the following ARIMA models specify their order (p, d, q) and find out whether Xt is
stationary, causal, invertible.
a)
b)
c)
d)
e)
f)
g)
Xt + 0.2Xt−1 − 0.48Xt−2 = Zt ;
Xt + 1.9Xt−1 + 0.88Xt−2 = Zt + 0.2Zt−1 + 0.7Zt−2 ;
Xt + 0.5Xt−2 = Zt−2 ;
Xt − 2Xt−1 + Xt−2 = Zt − 0.3Zt−1 ;
Xt − 0.3Xt−1 = Zt − 0.3Zt−1 ;
Xt = Xt−1 + Zt ;
Xt − 2Xt−1 + Xt−2 = Zt + 0.5Zt−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.
3
(p, d, q)
a)
(2,0,0)
ϕ(z), θ(z)
roots
ϕ(z) = 1 + 0.2z − 0.48z
2
z=
property*
− 45 , 43
θ(z) = 1
b)
(2,0,2)
I
ϕ(z) = 1 + 1.9z + 0.88z 2
d)
e)
f)
(2,0,2)
(0,2,1)
(1,0,1)
(0,1,0)
z = − 54 , − 10
11
ϕ(z) = 1 + 0.5z 2
θ(z) = z 2
z = ±1
ϕ(z) = 1 − 2z + z 2
z = 1, 1
(0,2,1)
I
S, C
θ(z) = 1 − 0.3z
z=
10
3
I
ϕ(z) = 1 − 0.3z
z=
10
3
S, C
θ(z) = 1 − 0.3z
z=
10
3
I
ϕ(z) = 1 − z
z=1
θ(z) = 1
g)
S
√
z = − 71 ± 69i
7
√
z=± 2
θ(z) = 1 + 0.2z + 0.7z 2
c)
S, C
I
ϕ(z) = 1 − 2z + z
z = 1, 1
θ(z) = 1 + 0.5z
z = −2
*S = Stationary, C = Causal, I = Invertible
4
I