Solution Dissertation

STAT 758: Groundwork #6

Thanks on Thursday, 11 Apr, 2012

Zaliapin, 1: 00pm

Tracy Backes

1

Tracy Backes

STAT 758 (Zaliapin): HW #6

Problem #1

We suppose below that Zt ∼ W In (0, σ 2 ), B is actually a backshift user. 6. you For the model (1 − B)(1 − zero. 2B)Xt sama dengan (1 − 0. 5B)Zt: a) Classify the version as a great ARIMA(p, g, q) process (i. elizabeth. find p, d, q). ARIMA(1, you, 1) b) Determine whether the process is stationary, origin, invertible. • The process is definitely stationary in the event that all origins of ϕ(z) are off of the product circle. ϕ(z) = (1 − z)(1 − zero. 2z) = 0 =⇒ z = 1, 5

Because ϕ(z) has main z = 1, which will lies on the unit ring, the process is not fixed. • The procedure is causal if almost all roots of ϕ(z) are outside of the machine circle. Mainly because ϕ(z) features one basic z sama dengan 1, which usually lies around the unit group, the process is definitely not causal. • The process is invertible if most roots of θ(z) happen to be outside of the machine circle. θ(B) = (1 − 0. 5z) sama dengan 0 =⇒ B=z

As the only reason for Оё(z), z = a couple of, lies outside of the unit ring, the process is invertible. c) Evaluate the п¬Ѓrst three П€-weights of the unit when indicated as an AR(в€ћ) style. в€ћ Xt can be expressed as a great AR(в€ћ) means of the form Xt = k=0 П€k Ztв€’k Xt = Оё(B) Zt = (1 в€’ 0. 5B) П•(B) в€ћ в€ћ

(1. 25 в€’ zero. 25(0. 2)k )B k

k=0

Zt

= Zt +

k=1

0. 625 + zero. 075(0. 2)kв€’1 Ztв€’k

sama dengan Zt + 0. 7Ztв€’1 + zero. 64Ztв€’2 & 0. 628Ztв€’3 + В· В· В· d) Measure the п¬Ѓrst 4 ПЂ-weights with the model once expressed because an MA(в€ћ) model. в€ћ Zt could be expressed like a MA(в€ћ) means of the form Zt = k=0 ПЂk Xtв€’k Zt = П•(B) Xt = (1 в€’ 1 . 2B & 0. 2B 2 ) Оё(B) в€ћ в€ћ

0. 5k N k

k=0

Xt

sama dengan Xt в€’ 0. 7Xtв€’1 +

k=2

в€’0. 15(0. 5)kв€’2 Xtв€’k

= Xt − zero. 7Xt−1 − 0. 15Xt−2 − 0. 075Xt−3 − 0. 0375Xt−4 + · · · Discuss how the behavior of the weights relates to the real estate of the style found in (b). ∞ A great ARMA unit is only origin if k=0 |ψk | < ∞, where ψk are the coefficients from the AR(∞) process found in (c). Remember that in out case, this kind of infinite sum is given by ∞

1+

k=1

|0. 625 & 0. 075(0. 2)kв€’1

which will not converge, confirming that the provided process is definitely not causal. Similarly, a great ARMA version is only invertible ∞ if perhaps k=0 |πk | < ∞, wherever πk would be the coefficients through the MA(∞) process found in (d). In this case, the related

Problem #1 continued on next page...

Site 2 of 7

Tracy Backes infinite sum is given simply by

STAT 758 (Zaliapin): HW #6

Trouble #1 (cont'd)

в€ћ

1 ) 7 &

k=2

| в€’ 0. 15(0. 5)kв€’2

which does are staying, confirming which the given procedure is invertible.

6. two Show the AR(2) process Xt sama dengan Xt−1 & cXt−2 + Zt is definitely stationary presented −1 < c < 0. Show that the AR(3) process Xt = Xt−1 + cXt−2 + cXt−3 + Zt is non-stationary for all beliefs of c. The AR(2) process Xt = Xt−1 + cXt−2 + Zt is standing if and only if all roots of ϕ(z) are off the system circle. The roots of ϕ(z) are given by √ 1 ± 1 + 4c two ϕ(z) = −cz − z + 1 = 0 =⇒ z= −2c 1+4c was plotted to get In order to identify where these roots lie in relation to the system circle, the norm of unces = 1±−2c −1 < c < 0, since shown in Figure 1 ) It can be seen that the origins (real or complex) have always norm higher than 1 for this range of c (i. at the., the roots are always outside of the unit circle). Thus, the given AR(2) process is usually stationary for −1 < c < 0. three or more. 5

в€љ

3

2 . 5

|z

2

1 . 5

one particular

0. 5 в€’1

в€’0. 8

в€’0. 6 c

в€’0. some

в€’0. 2

0

Figure 1 Similarly, the AR(3) process Xt = Xt−1 + cXt−2 − cXt−3 + Zt is fixed if in support of if most roots of ϕ(z) happen to be off the unit circle. The roots of ϕ(z) receive by ϕ(z) = cz 3 − cz two − z + 1 = (z − 1)(cz 2 − 1) sama dengan 0 =⇒ z= 1 1, ±c−0. 5 in the event c = 0, if c sama dengan 0

Therefore the beginnings of П•(z) always include z sama dengan 1 and the given AR(3) process is usually non-stationary for any c. six. 3 Create a model for the following SARIMA(p, d, q) Г— (P, D, Q)s processes within an explicit kind: ai Xtв€’i = bj Ztв€’j. a) (0, one particular, 0) Г— (1, zero, 1)12; A SARIMA(p, deb, q) Г— (P, G, Q)s process is typically referred to in the following form П•p (B)О¦P (B s )( d M s Xt )

= θq (B)ΘQ (B s i9000 )Zt

In such a case, this...



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