a question about ARMA(p,q)

[hardmann]

Dear Tom:

I allwasy have stupid question. I am now studying Beveridge-Nelson decomposion and related papers, such as Morley, Nelson & Zivot(2003), "Why Are the
Beveridge-Nelson and Unobserved-Components Decompositions of GDP So Different?
I encounter a question on representation of ARMA(p,q). In thier paper,Unobserved-component(UC) representation: (1): yt=τt+Ct ,(2): τt=τt-1 + μ + ηt, If Ct is a stationary: (3): Φp(L)Ct=Θq(L)εt.
However,yt first difference: Φp(L)(1-L)yt=Φp(1)μ + Φp(L)ηt + Θq(L)(1-L)εt. Though μ is a scalar but ηt is a vector, Why μ and ηt have different coef in (4) and the coef. of μ is Φp(1) rather than Φp(L) and what is Φp(1)?

Thanks
Hardmann

[TomDoan]

The manipulation of lag operators that leads to the final form for the change in y is given here:

bn_arma.gif (4.22 KiB) Viewed 7733 times

The last equality in the first line comes from pulling the inverse phi through the expression. PHI(1) is the result of applying the polynomial to the value 1. The second line gives delta y as an ARMA(p*,q*) where p*=p (same lag polynomial as for C's ARMA) and q*=max(q+1,p). For the latter result, see, for instance, Hamilton, pp 102-105 on sums of ARMA processes.

The following is a discussion of lag operators and the manipulation of lag polynomials:

Lag operators and filters (#2).pdf

Lag operators/Z transforms

(123.87 KiB) Downloaded 822 times

[hardmann]

Thanks Tom.
I understaned Lag operators and BN thought and formulas but phi(1). I understan your explanation, however, I still confused what is you said "PHI(1) is the result of applying the polynomial to the value 1"
I had relearned sums of arma processes and Lag operator and filters. My qustion reserved.


hardmann

[TomDoan]

Thanks Tom.
I understaned Lag operators and BN thought and formulas but phi(1). I understan your explanation, however, I still confused what is you said "PHI(1) is the result of applying the polynomial to the value 1"
I had relearned sums of arma processes and Lag operator and filters. My qustion reserved.

hardmann

PHI(z) is a polynomial---the Z transform of PHI(L). PHI(1) is that polynomial evaluated at z=1.