To avoid explosive behavior on the solution obtained in the previous slide:
... we have to impose the condition:
If
... where
Consider the following parameter values:
In the previous slide, we got the solution:
So, with those parameter values, eq. (1') can be rewritten as:
From eq. (2'), we can easily conclude:
Consider that the process is on its deterministic steady-state
What happens to the value of
Using the original equation
But
As there are no more shocks in this exercise, for period
Doing the same for
So, the solution will be:
The method used above is not very efficient. Suppose the shock occurred long ago, for example,
According to the previous method, we had to perform 50 operations to get the value of
There is a better way to obtain that value: use directly eq. (2'):
So, what is the value of
By the same way, what is the value of
Repeating the same exercise, we can collect the other results.
For example, what is the value of
So, the solution will be the same:
To avoid explosive behavior on the solution
... we have to impose the condition:
We get the following solution to this block at the
The solution to eq. (3), can be written as:
The expected-unconditional value of
Therefore, the solution to
In the previous slide we obtained that the solution to
Considering the information we have about the parameters:
So, we get:
Therefore, with unconditional expectations, the value of
As the expected-conditional value of
And as we have the information that
Then,
Therefore, the solution to
From eq.(7b) in the previous slide, we got:
This is a geometric sum, with a solution:
Therefore, it is easy yo see that:
If
As from a previous slide we know that
It is very simple to calculate the values 0f
So,
And
The static block is given by the equation:
From the previous slide, we know that:
From eq. (3), we know that:
Once we know the values of
Assuming that
Blanchard, O. and Kahn, C. M. (1980). The solution of linear difference models under rational expectations. Econometrica, 48(5), 1305-1311.
Write the model in state space form
Multiplying both sides of (9) by
Suppose we have a square matrix
The Jordan decomposition of
Our system was given by (10):
Apply the decomposition
Multiply both sides by
Let us assume that there are no shocks affecting the forward-looking block:
Next, we apply a partition to the matrices:
Our transformed model looks much easier now:
When solving these models, the most demanding task is to apply the correct partition to these three matrices.
Suppose a model with 1 backward-looking variable, one static, and the third is a forward-looking variable (as the simple model above).
The partitions should be as follows:
The original model:
To write the model in matrix form, put all variables expressed at
So, the model can be written as:
left hand-side: endogenous variables at
right hand-side: endogenous variables at
Detailed specification of the model:
Detailed specification of the model:
The model in state space representation:
A = zeros(3,3)
B = zeros(3,3)
C = zeros(3,3)
A[1,1] = 1.0
A[2,1] = -ϕ
A[2,2] = 1.0
A[2,3] = -μ
A[3,3] = β
B[1,1] = ρ
B[3,1] = -θ
B[3,3] = 1.0
C[1,1] = 1.0
D =[φ ; 0.0 ; -α]
Write the model in state space form:
Multiplying both sides of (4) by
The Jordan decomposition is given by:
Apply the decomposition to (A2):
Multiply both sides by
Let us assume that there are no shocks affecting the forward-looking block:
Next, we apply a partition to the matrices:
Our transformed model looks much easier now:
Transformed model written as a set of decoupled equations:
We can now apply our well known strategy. Iterate to:
Solve the predetermined transformed block and get the equilibrium levels of the predetermined (backward-looking) variables:
Solve the forward-looking transformed block and get the equilibrium values of the forward-looking variables:
Iterating forward this block, and as the shocks to this block are
If we assume
Then, the only stable solution will be
Now, from the partition of
From (A3)=(A4), the forward-looking block only depends on predetermined one:
Iterating forward this block, we get
If we assume that:
The process is stable, and from the partition of
Now, inserting eq. (A5) into (A6), we can obtain:
As from eq. (A7) we have
Then, for
But, as from eq. (Predetermined block) we have:
By mere substitution of (A8) and (A9) into (A10), we derive our final result:
Blanchard, O. and Kahn, C. M. (1980). The solution of linear difference models under rational expectations. Econometrica, 48(5), 1305-1311.