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



(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,


a(L) =1 -^2 aj L°


with the partial solution

yt = (A)i Et'xt+i,


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 +

then the complete solution

as before.

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