Due on Wednesday, 11 April, 2012

Zaliapin, 1:00pm

Tracy Backes

1

Tracy Backes

STAT 758 (Zaliapin): HW #6

Problem #1

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 when expressed as an AR(∞) model. ∞ Xt can be expressed as an AR(∞) process of the form Xt = k=0 ψk Zt−k Xt = θ(B) Zt = (1 − 0.5B) ϕ(B) ∞ ∞

(1.25 − 0.25(0.2)k )B k

k=0

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 π-weights of the model when expressed as an MA(∞) model. ∞ Zt can be expressed as a MA(∞) process of the form Zt = k=0 πk Xt−k Zt = ϕ(B) Xt = (1 − 1.2B + 0.2B 2 ) θ(B) ∞ ∞

0.5k B k

k=0

Xt

= Xt − 0.7Xt−1 +

k=2

−0.15(0.5)k−2 Xt−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 |πk | < ∞, where πk are the coeﬃcients from the MA(∞) process found in (d). In this case, the corresponding

Problem #1 continued on next page. . .

Page 2 of 7

Tracy Backes inﬁnite sum is given by

STAT 758 (Zaliapin): HW #6

Problem #1 (cont’d)

∞

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 √ 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. 3.5

√

3

2.5

|z|

2

1.5

1

0.5 −1

−0.8

−0.6 c

−0.4

−0.2

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) = cz 3 − cz 2 − z + 1 = (z − 1)(cz 2 − 1) = 0 =⇒ z= 1 1, ±c−0.5 if c = 0, if c = 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: 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 ϕp (B)ΦP (B s )( d D s Xt )

= θq (B)ΘQ (B s )Zt

In this case, this...