THE ECONOMETRICS OF MACROECONOMIC MODELLING

Undetermined coefficients

This method is more practical. It consists of the following steps:

1. Make a guess at the solution.

2. Derive the expectations variable.

3. Substitute back into the guessing solution.

4. Match coefficients.

We will first use the technique, following the excellent exposition of Blanchard and Fisher (1989: ch. 5), to derive the solution conditional upon the expected path of the forcing variable, as in Gall et al. (2001), so we will ignore any information about the process of the forcing variable.

In the following we will define

zt — bp2xt + &pt-

Since the solution must depend on the future, a guess would be that the solution will consist of the lagged dependent variable and the expected values of the forcing value:

Ж

Apt — aAp—i + 53 PiEtZt+i■ (A.19)

i=0

We now take the expectation of the solution of the next period, using the law of iterated expectations, to find the expected outcome

Ж

EtApt+i — aApt + 53 PiEtZt+i+i, i=0

which we substitute in the guessing solution Apt — bfp1 ^aApt + ^ fiiEtZt+i+^j + bpiApt-i + zt,
Apt —	Etzt+i+i■
(A.20)

Finally, the undetermined coefficients are now found by matching the coefficients of the variables between (A.19) and (A.20).

We start by matching the coefficients of Apt-i:

bpi

1 - abfpi

This gives, as above, the second-order polynomial in a:

with the solutions given in (A.13).

Using ai, we may now match the remaining undetermined coefficients of Etzt+i, giving

1

bf

bpi

1 - bp1ai

so, using (A.14), the coefficients can therefore be written as

declining as time move forwards.

Substituting back for

zt = bp2xt + £pt,

the solution can therefore be written

which is the same as in Gall et al. (2001), except the error term which they ignore.

To derive the complete solution, we need to substitute in for the forcing process xt. We can either do this already in the guessing solution, or by sub­stituting in for the expected terms Etxt+i. Here we choose the latter solution. The expectations, conditional on information at time t, are:

Etxt = xt,

Etxt+i = bxxt,

Etxt+2 = Et(Et+ixt+2) = Etbx xt+i = bXxt,

Etxt+j = bX xt,

where we again have used the law of iterated expectations. So the solution becomes