Y 0.4Y 0.45Y Y to a proper ARMA specification.
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1 HG Jan 04 ECON 50 Exercises II - 0 Feb 04 (wih answers Exercise. Read secion 8 in lecure noes 3 (LN3 on he common facor problem in ARMA-processes. Consider he following process Y 0.4Y 0.45Y 0.5 ( where ~ WN(0,. This is an ARMA process of he form ( L Y ( L wihou consan erm. A. Invesigae if he wo lag-polynomials have a common facor. If so, reformulae he Y o a proper ARMA specificaion. difference equaion specificaion for B. Is here a causal saionary soluion for Y? If yes, wrie up he soluion as a MA( process. Is he MA specificaion inverible? If so, wrie up he AR( soluion for. ANSWER: A: The companion polynomials: Roos ( z 0.4z 0.45z ( z z 0.5z - - Facorized: ( z ( 0.5 z( 0.9z (z ( 0.5z
2 To have a proper ARMA we should cancel he common facor, giving new lag polynomials ( L 0.9L ( L 0.5L The proper ARMA(, specificaion Y 0.9Y 0.5 (. B: The soluion of (. is causal saionary since 0.9. Y 0.5L 0.9L 's We need o find he in 0.5z 0.9z zz (. 0.5 z ( 0.9 z z z or Muliplying ou he righ side (puing 0.9, 0.5 z z z z 3 3 z z z 3 ( z ( z ( z 3 3 Hence (using he uniqueness of power series coefficiens (see appendix o he exercises II ( j j j j j j j ( (0.9 (.4 for,3,
3 3 The process is inverible since 0.5. The AR( soluion is 0.9 L Y L L Y 0.5L ( Puing 0.9, 0.5, and using he companion forms, we ge z ( z( z z z z z 3 3 z z z 3 ( z ( z ( z 3 3 Hence (.4 j j j ( ( ( ( 0.5 (.4 for j,3, j Some values: j j j E-06.55E- Exercise. A. Suppose ha Y ~ ARMA( p, q and causal, saisfying ( L Y 0 ( L, where q ( L L L, ( L ( L L. Inroduce he cenered series, p p y Y, leading o ( L y ( L wihou he consan 0, and where Ey ( 0 (see exercise 3 of seminar. Explain why saemen (7 on page 5 of LN3, saying ( L ( h 0 and ( L ( h 0 for h max( p, q q, is rue, where ( h, ( h are he auocovariance funcion and acf respecively. [Hin. Assume h max( p, q and wrie
4 4 ( h E( y y E y y y y y y ec. ] h h p h p h q hq ANSWER: Coninuing he hin we ge (* ( h E( y y ( h ( h p E( y E( y h p h q hq (i If h p Suppose, e.g., h p. Then, remembering ha ( (, (* becomes, he answer I gave on he seminar was no good. The poin is raher: ( h E( y y ( h ( (0 ( E( y h p p p h ( h ( (0 ( E( y p p p h ( h ( ( (0 E( y p p p h and we see ha he difference equaion par for ( h has changed (he coefficiens are no longer he same.. Therefore, we mus have h p, for he difference equaion for ( h hold. (ii Since he (causal soluion for y depends on,,,, we mus have h q (or h q in order ha all he las -erms become zero. Hence h max( p, q. o B. Inroducion. If Y ~ AR ( wih y y and, we ge from page 3 in LN ha he auocovariance funcion, h ( h ( h which implies h ha he acf is ( h ( h for h 0,,, We are ineresed in his exercise o find ou wha he effec is on he AR( acf, ( h by adding a MA-erm, o y y y. Assume herefore ha,, and ~ WN(0, y, i.e., an ARMA(,, where. We need he auocovariance, ( h ( h for y in his new siuaion. Firs ( h and acf : Using (8 on page 5 in LN3 and A., we have
5 5 (i ( h ( h 0 for h,3, wih iniial condiions (ii (iii, i.e., (0 ( ( 0 ( (0 ( 0 where 0, are found from z z 0 zz and (0 ( ( 0 Quesion. Show ha 0 [Hin. Wrie he lag-equaion (in erms of companion series z ( z( 0 z z. Then muliply ou he righ side,. ] sufficienly o deermine 0 ANSWER: This was done in exercise B C. Show firs from ii. and iii. ha (0 ( ( ( ( ( h and hen ha ( h for h. ANSWER: 0,, ( h ( h for h,3, gives h ( h ( for h,3, From ii., iii. we ge
6 6 (* (0 ( ( ( ( (0 ( ( ( ( ( ( ( ( ( ( ( Subsiuing in (* gives ( ( (0 ( ( ( ( ( ( ( ( ( Hence (** ( ( (0, ( ( ( h ( h for h,3,, and ( ( ( ( h ( D. Show ha he acf of y can be expressed as h where ( h is he acf of he AR( process.,
7 7 ANSWER: From (** we ge ( ( ( h ( h for h,,3, Considering, ( h h ( ( ( ( h (, we ge h. E. The consan facor in fron of ( h, characerize he effec has on he AR( acf, ( h. Look a his effec (facor in he wo special cases, 0.9 and 0., i.e., for which values of will he AR( acf increase and for which values will i decrease? [Hin. The facor in fron of ( h is no so easy o discuss analyically. The bes way, in my view, is simply o plo he facor as a funcion of for in he wo cases. The easies is probably o plo i wih a compuer. This is, for example, easy in Saa. The following command, for example, woway funcion y=*x^-, range(-.5 plos he funcion y x for x.5 ] ANSWER: The facor in fron of ( h of AR(. Wriing x for, he facor is he funcion ( x( x gx ( ( xx represens he effec he MA-erm has on he acf Case 0.9 : Saa-plo : woway funcion y=(x+.9*(+.9*x/(.9*(+*.9*x+x^, range(-
8 y 0..4 y Phi = x g g( (.056, g g( ( max (We see ha if increases from 0 (e.g., ( h will increase also from ( h much. Case 0.: min, bu no Phi = x
9 9, is here any value of ha urns F. For any wih so, for which value? y ino whie noise? If ANSWER: From he figures we see ha here is a ha makes ( h 0 for h 0. From he formula we see ha his happens for. We also see from he soluion series y L L ha here is a common facor when, so ha y is whie noise for his value of. (For he appendix on power series see he original exercise ex
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