 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 в€ј W In (0, Пѓ 2 ), B is actually a backshift user. 6. you For the model (1 в€’ B)(1 в€’ zero. 2B)Xt sama dengan (1 в€’ 0. 5B)Zt: a) Classify the version as a great ARIMA(p, g, q) process (i. elizabeth. п¬Ѓnd p, d, q). ARIMA(1, you, 1) b) Determine whether the process is stationary, origin, invertible. вЂў The process is definitely stationary in the event that all origins of П•(z) are oп¬Ђ of the product circle. П•(z) = (1 в€’ z)(1 в€’ zero. 2z) = 0 =в‡’ z = 1, 5

Because П•(z) has main z = 1, which will lies on the unit ring, the process is not fixed. вЂў The procedure is causal if almost all roots of П•(z) are outside of the machine circle. Mainly because П•(z) features one basic z sama dengan 1, which usually lies around the unit group, the process is definitely not causal. вЂў The process is invertible if most roots of Оё(z) happen to be outside of the machine circle. Оё(B) = (1 в€’ 0. 5z) sama dengan 0 =в‡’ 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 в€’ zero. 7Xtв€’1 в€’ 0. 15Xtв€’2 в€’ 0. 075Xtв€’3 в€’ 0. 0375Xtв€’4 + В· В· В· Discuss how the behavior of the weights relates to the real estate of the style found in (b). в€ћ A great ARMA unit is only origin if k=0 |П€k | < в€ћ, where П€k are the coeп¬ѓcients from the AR(в€ћ) process found in (c). Remember that in out case, this kind of inп¬Ѓnite sum is given by в€ћ

1+

k=1

|0. 625 & 0. 075(0. 2)kв€’1

which will not converge, conп¬Ѓrming that the provided process is definitely not causal. Similarly, a great ARMA version is only invertible в€ћ if perhaps k=0 |ПЂk | < в€ћ, wherever ПЂk would be the coeп¬ѓcients through the MA(в€ћ) process found in (d). In this case, the related

Problem #1 continued on next page...

Site 2 of 7

Tracy Backes inп¬Ѓnite 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, conп¬Ѓrming which the given procedure is invertible.

6. two Show the AR(2) process Xt sama dengan Xtв€’1 & cXtв€’2 + Zt is definitely stationary presented в€’1 < c < 0. Show that the AR(3) process Xt = Xtв€’1 + cXtв€’2 + cXtв€’3 + Zt is non-stationary for all beliefs of c. The AR(2) process Xt = Xtв€’1 + cXtв€’2 + Zt is standing if and only if all roots of П•(z) are oп¬Ђ the system circle. The roots of П•(z) are given by в€љ 1 В± 1 + 4c two П•(z) = в€’cz в€’ z + 1 = 0 =в‡’ z= в€’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 = 1В±в€’2c в€’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 в€’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 = Xtв€’1 + cXtв€’2 в€’ cXtв€’3 + Zt is fixed if in support of if most roots of П•(z) happen to be oп¬Ђ the unit circle. The roots of П•(z) receive by П•(z) = cz 3 в€’ cz two в€’ z + 1 = (z в€’ 1)(cz 2 в€’ 1) sama dengan 0 =в‡’ z= 1 1, В±cв€’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 )

= Оёq (B)ОQ (B s i9000 )Zt

In such a case, this...

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