Factorization

Finally, we shall take a look at this very elegant method introduced by Sargent. It consists of the following steps:

1. Write the model in terms of lead – and lag-polynomials in expectations.

2. Factor the polynomials, into one-order polynomials, deriving the roots.

3. Invert the factored one-order polynomials into the directions of converging forward polynomials of expectations.

Again, we use the simplifying definition

zt = bp2Xt + £pt:

so the model is again

Apt = bp1EtApt+i + bpiApt-i + zt.

Note that the forward, or lead, operator, F, and lag operator, L, only work on the variables and not expectations, so:

LEtzt = Etzt-i

FEtZt = Etzt+i L-i = F.

The model can then be written in terms of expectations as:

-bpiEtApt+i + EtApt – bpEApt-i = EtZt, and using the lead – and lag-operators:

(-bpiF + 1 — bpiL)EtApt = Etzt,

Подпись: F2 Подпись: 1 bf °pi image349 Подпись: 1 bf bpi Подпись: Etzt.

or, as a second-order polynomial in the lead operator:

Подпись: [(F - a-) (F - 02)] LEt Apt (F - a-) LEt Apt (1 - a-L) Apt (1 - a-L) Apt image353 Подпись: Etzt.

The polynomial in brackets is exactly the same as the one in (A.12), so we know it can be factored into the roots (A.13):

However, we know that (1/1-(1/a2)F) = ^=0(1/a2)iFi, since |1/a2| < 1,

Подпись: Apt Подпись: a-Apt-i + Подпись: bp2 bP-a2 image358 Подпись: 1 bP-a2 Подпись: £pt>

so we can write down the solution immediately:

where we have also substituted back for zt.

To derive the complete solution, we have to solve for

image361Etxt+i

given

(1 bXL)xt Єxt.

We can now appeal to the results of Sargent (1987, p. 304) that work as follows. If the model can be written in the form

yt = Etyt+- + xta(L)xt + et,

r

a(L) =1 -^2 aj L°

j=-

with the partial solution

yt = (A)i Et’xt+i,

i=o

1 – XL-[116]

r — 1

a(X)—1 1 + £ Y, Xk—jau I Lj

j=1 k=j+1

The solution therefore becomes Apt – a.1 Apt—1 = 1

o1^2 .

Подпись: is determined by
Подпись: yt = Zg(L)xt

Подпись: g(L) =

Подпись: 1 - Xa(X)—1a(L)L—1
Подпись: In our case
Подпись: z = bp2 bfp10.2 X = —, 0-2 a(L) 1 bxL:
Подпись: so g(L) will have the form
Подпись: g(L) = (1 - a1X) — 1 1
Подпись: 1 - bx(1/«2)'
Подпись: 1
Подпись: Apt = ouApt—1 +
image373

then the complete solution

as before.

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