This is obviously a theoretically incorrect way of dealing with the problem. However, it is
very likely that the terminal condition will not have a big impact on the solution at the
beginning of the simulation. Therefore, if it is possible to run simulations over a long enough
horizon it is likely that the solution reaches a steady state at some point before it then departs
from the steady state as the (incorrect) terminal condition exerts its influence on the solution.
If one ignores the solution after the model has reached the steady state, then nothing would
be lost by applying this method. In that regard a comparison of this solution method with the
other, theoretically more adequate methods discussed below is instructive since it will
illustrate whether, and over which time horizon, this solution might be close to the correct
solution. The big advantage of this method is of course its simplicity. However, the
computational costs can be excessive, since applying this method may require very long
stacks in order to obtain reliable results over the adjustment period.
0HWKRG 7HUPLQDO &RQGLWLRQ Lq /HYHOV 7&/ - Lh use the correct terminal condition as
calculated from the “static” or “equilibrium” counterpart of the dynamic model.
The correct way of dealing with the problem of the ‘unknown’ solution in period T+1 is to
construct, prior to simulating the dynamic model (1), the equilibrium counterpart to the
dynamic model, Lh set up a system which gives the long run solution of \t to any vector [* of
exogenous variables, where [* denotes the long run level of the exogenous variables. Let this
system be given by
fs(∖*,[*) = 0
and let \* be the long run solution for \t. To give an example, suppose the vector ∖t can be
split into O predetermined state variables and m jumping variables contained in the vectors ∖st
and \ t and let the dynamic model be represented by
(3)
\ = $ \ Li + %\ + Y1[r
(, \+ι = &\ Li+ ' \ + γ 2 [,
(4)
Define Π =
$%
&'
then the steady state version of this model would be given by
\ *
\J *
= (I - ∏)-iγX*
(5)
Using this detour it is possible to calculate the long run solution implied by a given dynamic
model by running a simple simulation of the model fs(..) in order to calculate \* for the long
run values [* of the exogenous variables. The dynamic simulation can then be computed by
imposing the solution values ∖j* from this static model on the terminal conditions ∖t+T+1. This
method certainly has the advantage that the solution procedure is theoretically correct.
However, it has the disadvantage that the solution method becomes very cumbersome and
requires the maintenance of two models which should be identical up to the dynamic
specification.
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