HW6 sol

Prévia do material em texto

We assume below that Zt ∼WN(0, σ2), 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 when expressed as an AR(∞) model.
Xt can be expressed as an AR(∞) process of the form Xt =
∑∞
k=0 ψkZt−k
Xt =
θ(B)
ϕ(B)
Zt = (1− 0.5B)
( ∞∑
k=0
(1.25− 0.25(0.2)k)Bk
)
Zt
= Zt +
∞∑
k=1
(
0.625 + 0.075(0.2)k−1
)
Zt−k
= Zt + 0.7Zt−1 + 0.64Zt−2 + 0.628Zt−3 + · · ·
d) Evaluate the first four pi-weights of the model when expressed as an MA(∞) model.
Zt can be expressed as a MA(∞) process of the form Zt =
∑∞
k=0 pikXt−k
Zt =
ϕ(B)
θ(B)
Xt = (1− 1.2B + 0.2B2)
( ∞∑
k=0
0.5kBk
)
Xt
= Xt − 0.7Xt−1 +
∞∑
k=2
−0.15(0.5)k−2Xt−k
= 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).
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 +
∞∑
k=1
|0.625 + 0.075(0.2)k−1|
which does not converge, confirming that the given process is not causal. Similarly, an ARMA model is only invertible
if
∑∞
k=0 |pik| <∞, where pik are the coefficients from the MA(∞) process found in (d). In this case, the corresponding
1
infinite sum is given by
1.7 +
∞∑
k=2
| − 0.15(0.5)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
ϕ(z) = −cz2 − z + 1 = 0 =⇒ z = 1±
√
1 + 4c
−2c
In order to determine where these roots lie in relation to the unit circle, the norm of z = 1±
√
1+4c
−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 ϕ(z) are given by
ϕ(z) = cz3 − cz2 − z + 1 = (z − 1)(cz2 − 1) = 0 =⇒ z =
{
1 if c = 0,
1, ±c−0.5 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 for the following SARIMA(p, d, q)× (P,D,Q)s processes in an explicit form:∑
aiXt−i =
∑
bjZt−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
ϕp(B)ΦP (B
s)(∇d∇Ds Xt) = θq(B)ΘQ(Bs)Zt
In this case, this yields the following explicit process
ϕ0(z) = 1
Φ1(z) = 1− αz
∇1∇012 = ∇ = (1−B) =⇒
θ0(z) = 1
Θ1(z) = 1 + βz
Φ(B12)(∇Xt) = Θ(B12)Zt
(1− αB12)(1−B)Xt = (1 + βB12)Zt
(1−B − αB12 + αB13)Xt = (1 + βB12)Zt
Xt −Xt−1 − αXt−12 + αXt−13 = Zt + βZt−12
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− ϕ1z − ϕ2z2
Φ0(z) = 1
∇0∇01 = 1 =⇒
θ2(z) = 1 + θ1z + θ2z
2
Θ0(z) = 1
ϕ(B)Xt = θ(B)Zt
(1− ϕ1B − ϕ2B2)Xt = (1 + θ1B + θ2B2)Zt
Xt − ϕ1Xt−1 − ϕ2Xt−2 = Zt + θ1Zt−1 + θ2Zt−2
ϕ1(z) = 1− ϕz
Φ1(z) = 1− αz
∇1∇14 = (1−B)(1−B4) =⇒
θ1(z) = 1 + θz
Θ1(z) = 1 + βz
ϕ(B)Φ(B4)(∇∇4Xt) = θ(B)Θ(B4)Zt
(1− ϕB)(1− αB4)(1−B)(1−B4)Xt = (1 + θB)(1 + βB4)Zt
Expanding the lefthand side of the equation yields:
LHS = (1− ϕB)(1− αB4)(1−B)(1−B4)Xt
= [1− (ϕ+ 1)B + ϕB2 − (α+ 1)B4 + (ϕ+ 1)(α+ 1)B5 − ϕ(α+ 1)B6 + αB8 − α(ϕ+ 1)B9 + ϕαB10]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 + βB4)Zt
= (1 + θB + βB4 + θβB5)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) Xt + 0.2Xt−1 − 0.48Xt−2 = Zt;
b) Xt + 1.9Xt−1 + 0.88Xt−2 = Zt + 0.2Zt−1 + 0.7Zt−2;
c) Xt + 0.5Xt−2 = Zt−2;
d) Xt − 2Xt−1 +Xt−2 = Zt − 0.3Zt−1;
e) Xt − 0.3Xt−1 = Zt − 0.3Zt−1;
f) Xt = Xt−1 + Zt;
g) 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.
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(p, d, q) ϕ(z), θ(z) roots property*
a) (2,0,0) ϕ(z) = 1 + 0.2z − 0.48z2 z = − 54 , 43 S, C
θ(z) = 1 I
b) (2,0,2) ϕ(z) = 1 + 1.9z + 0.88z2 z = − 54 ,− 1011 S
θ(z) = 1 + 0.2z + 0.7z2 z = − 17 ±
√
69i
7 I
c) (2,0,2) ϕ(z) = 1 + 0.5z2 z = ±√2 S, C
θ(z) = z2 z = ±1
d) (0,2,1) ϕ(z) = 1− 2z + z2 z = 1, 1
θ(z) = 1− 0.3z z = 103 I
e) (1,0,1) ϕ(z) = 1− 0.3z z = 103 S, C
θ(z) = 1− 0.3z z = 103 I
f) (0,1,0) ϕ(z) = 1− z z = 1
θ(z) = 1 I
g) (0,2,1) ϕ(z) = 1− 2z + z z = 1, 1
θ(z) = 1 + 0.5z z = −2 I
*S = Stationary, C = Causal, I = Invertible
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