Lectures on macroeconometric modelling
1 Macroeconomic models
Gunnar Bårdsen
Bologna, 12-13 June, 2014
Looking ahead I
From a discretized and linearized cointegrated VAR representation to a dynamic
Simultaneous Model (SEM) in three steps
1. Linearized and discretized approximation as a data-coherent
statistical system representation in the form of a cointegrated
VAR
u�
Δ𝐳u� = 𝝁 + Π𝐳u�−1 + ∑ Γu� Δ𝐳u�−u� + 𝐮u� ,
u�=1
2. Identify the steady state, by testing and imposing
overidentifying restrictions on the cointegration space:
′
u�
Δ𝐳u� = 𝝁 + 𝜶𝜷𝛼∗ 𝛽 ∗ 𝐳u�−1 + ∑ Γu� Δ𝐳u�−u� + 𝐮u� ,
u�=1
(1)
Looking ahead II
From a discretized and linearized cointegrated VAR representation to a dynamic
Simultaneous Model (SEM) in three steps
3. Identify the dynamics, by testing and imposing overidentifying
restrictions on the dynamics:
′
u�
𝐀0 Δ𝐳u� = 𝐀0 𝝁 + 𝐀0 𝛼∗ 𝛽 ∗ 𝐳u�−1 + ∑ 𝐀0 Γu�−u� Δ𝐳u�−u� + 𝐀0 𝐮u� .
u�=1
Plan of the lectures
To get there we will investigate
1. Characteristics of macroeconomic models
Plan of the lectures
To get there we will investigate
1. Characteristics of macroeconomic models
1.1 Steady state
Plan of the lectures
To get there we will investigate
1. Characteristics of macroeconomic models
1.1 Steady state
1.2 Dynamics with deviations from steady state: Equilibrium
correction mechanisms
Plan of the lectures
To get there we will investigate
1. Characteristics of macroeconomic models
1.1 Steady state
1.2 Dynamics with deviations from steady state: Equilibrium
correction mechanisms
2. A general derivation of dynamic systems with steady state
Plan of the lectures
To get there we will investigate
1. Characteristics of macroeconomic models
1.1 Steady state
1.2 Dynamics with deviations from steady state: Equilibrium
correction mechanisms
2. A general derivation of dynamic systems with steady state
2.1 Linearization
Plan of the lectures
To get there we will investigate
1. Characteristics of macroeconomic models
1.1 Steady state
1.2 Dynamics with deviations from steady state: Equilibrium
correction mechanisms
2. A general derivation of dynamic systems with steady state
2.1 Linearization
2.2 Discretization
Plan of the lectures
To get there we will investigate
1. Characteristics of macroeconomic models
1.1 Steady state
1.2 Dynamics with deviations from steady state: Equilibrium
correction mechanisms
2. A general derivation of dynamic systems with steady state
2.1 Linearization
2.2 Discretization
3. Examples of empirical modelling
Plan of the lectures
To get there we will investigate
1. Characteristics of macroeconomic models
1.1 Steady state
1.2 Dynamics with deviations from steady state: Equilibrium
correction mechanisms
2. A general derivation of dynamic systems with steady state
2.1 Linearization
2.2 Discretization
3. Examples of empirical modelling
3.1 Dynamic specification and interpretation
Plan of the lectures
To get there we will investigate
1. Characteristics of macroeconomic models
1.1 Steady state
1.2 Dynamics with deviations from steady state: Equilibrium
correction mechanisms
2. A general derivation of dynamic systems with steady state
2.1 Linearization
2.2 Discretization
3. Examples of empirical modelling
3.1 Dynamic specification and interpretation
3.2 Testing of model class restrictions
Macroeconomic models with steady state
We are interested in models with a steady state
They need not be long-run growth models, but they need to be
stable so
𝑓 (𝑍u� ) = 𝑓 (𝑍)̄ ,
u�→∞
where 𝑍 is a vector of suitably transformed stationary variables.
Otherwise we cannot do policy analysis. This is Samuelsons (?, ?)
correspondence principle—see ?) for a recent reappraisal.
To motivate, some classical and simple model examples follow.
Keynesian multiplier model
Foundation of policy and disequilibrium macro
No growth, discrete time:
𝑌u� = 𝐶u� + 𝐼 ̄
𝐶u� = 𝑐0 + 𝑐1 𝑌u�−1
with constant steady state
𝑌̄ =
𝑐0
1
+
𝐼.̄
1 − 𝑐1 1 − 𝑐1
if 𝑐1 < 1. The model is cast in Equilibrium Correction form
(EqCM) as:
𝑌u� − 𝑌u�−1 = 𝑐0 − (1 − 𝑐1 ) 𝑌u�−1 + 𝐼 ̄
⎛
⎞
⎡
⎤
1
⎜ 𝑐0
⎟
̄ ⎥
= − (1 − 𝑐1 ) ⎢𝑌u�−1 − ⎜
+
𝐼⎟
⎜
⎟
⎜⏟⏟⏟⏟⏟⏟⏟
⎢
1 − 𝑐1 1 − 𝑐1 ⎟⎥
⎣
⎝
⎠⎦
u�̄
̄
Δ𝑌u� = − (1 − 𝑐1 ) (𝑌u�−1 − 𝑌 ) .
Solow-Swan growth model I
Foundation of growth literature
In discrete time, with population growth and technological
progress, see f. ex. ?):
𝑌u� = 𝐶u� + 𝐼u�
𝑌u� = 𝐹 (𝐴u� 𝑁u� , 𝐾u�−1 )
𝐶u� = (1 − 𝑠) 𝑌u�
𝐼u� = 𝐾u� − 𝐾u�−1 + 𝛿𝐾u�−1
Δu�u�
u�u�−1
Δu�u�
u�u�−1
=𝑎
=𝑛
}
Δ (𝐴𝑁 )u�
(𝐴𝑁 )u�−1
= (1 + 𝑎) (1 + 𝑛) − 1 = 𝑔
(2)
(3)
(4)
(5)
(6)
Solow-Swan growth model II
Foundation of growth literature
If the production function satisfies the Inada conditions, the
stability condition (1−u�)
1+u� < 1 of
𝑠×𝑓(
𝐾u�−1
𝐾u�
𝐾
) = (1 + 𝑔) (
) − (1 − 𝛿) ( u�−1 )
(𝐴𝑁 )u�
(𝐴𝑁 )u�+1
(𝐴𝑁 )u�
(7)
gives per capita efficiency level of capital of
𝐾u�−1
𝑔+𝛿
=
.
u�→∞ (𝐴𝑁 )
𝑠
u�
Ramsey growth model in continous time
Foundation of the RBC literature
1. order Taylor expansion of model in per capita terms, no
technological progress or depreciation, see ?):
𝑐 ̇ (𝑡) = −𝛽 (𝑘 − 𝑘∗ )
𝑘̇ (𝑡) = − (𝑐 − 𝑐∗ ) + 𝜃 (𝑘 − 𝑘∗ )
The stability conditon is now
u�−√(4u�+u�2 )
2
< 0.
Phillips’ (?, ?) proportional control model
Foundation of policy control literature
Continous time with no growth, see ?, p. 322-323):
(
𝑌 ̇ (𝑡)
𝛼 (𝑐 − 1) 𝛼
𝑌 −𝑌∗
)
=
(
)
(
)
−𝛽𝛾
−𝛽
𝐺 − 𝐺∗
𝐺 ̇ (𝑡)
The stability conditions are
𝛼 (𝑐 − 1) − 𝛽 < 0
𝛼𝛽 (𝛾 + 1 − 𝑐) > 0,
New Keynesian Model with Taylor rule
Presently
Discrete time, no growth, see ?, p. 358-369):
Δ𝑝u� = 𝜇 + 𝛽𝐸u� Δ𝑝u�+1 + 𝛾𝑥u� + 𝑒u�u�
𝑥u� = 𝐸u� 𝑥u�+1 − 𝛼 (𝑅u� − 𝐸u� Δ𝑝u�+1 − 𝜃) + 𝑒u�u�
𝑅u� = 𝜃 + Δ𝑝∗ + 𝜇 (Δ𝑝u� − Δ𝑝∗ ) + 𝜈𝑥u� + 𝑒u�u�
where Δ𝑝 is inflation, 𝑥 is the output gap and 𝑅 is the interest
u�
rate and the 𝑒-s are shocks. If 1 > 1+u�(u�+u�u�)
> 0, the solutions
for inflation and output gap are
(1 + 𝛼𝜈) 𝑒u�u� + 𝛾 (𝑒u�u� − 𝛼𝑒u�u� )
1 + 𝛼 (𝜈 + 𝜇𝛾)
−𝛼𝜇𝑒u�u� − 𝑒u�u� + 𝛼𝑒u�u�
𝑥u� =
1 + 𝛼 (𝜈 + 𝜇𝛾)
Δ𝑝u� = Δ𝑝∗ +
so on average inflation Δ𝑝 is equal to the target Δ𝑝∗ and the
output gap 𝑥 is zero.
An example: the New Keynesian Phillips Curve
The model is
(8)
u�
u�
Δ𝑝u� = 𝑏u�1
𝐸u� Δ𝑝u�+1 + 𝑏u�1
Δ𝑝u�−1 + 𝑏u�2 𝑥u� + 𝜀u�u�
(9)
𝑥u� = 𝑏u� 𝑥u�−1 + 𝜀u�u�
where all coefficients are assumed to be between zero and one.
The solution (the statistical system) is
𝛼
Δ𝑝
(
) =( 1
𝑥
0
u�
(
𝑢u�
) =(
𝑢u�
u�
Derivation of the solution
1
u�u�u�1 u�2
0
u�u�2 u�u�
u�u�u�1 (u�2 −u�u� )
𝑏u�
)(
Δ𝑝
)
𝑥
+(
u�−1
u�u�2
u�u�u�1 (u�2 −u�u� )
1
)(
𝜀u�
)
𝜀u�
u�
𝑢u�
) ,
𝑢u�
u�
Hard to find I
Standard dynamic price-wage model
(
1
−𝛾2
Δ𝑝
1 𝛾1
Δ𝑝
)(
) =(
)(
)
−𝛿2
1
𝑥
𝛿1 1
𝑥
u�
u�−1
+(
𝑒
𝛾3 0
𝑔𝑎𝑝
)(
) + ( u� )
0 𝛿3
𝑢
𝑒u�
u�
u�
with statistical system (reduced form)
(
− u�2 u�1 +1
Δ𝑝
2 −1
) = ( u�u�21u�+u�
2
𝑥
−
u� u� −1
u�
2 2
(
𝑣u�
) =(
𝑣u�
u�
2
− u�u�1u�+u�−1
u�3
2 u�2 −1
−𝛾3 u� u�u�2−1
2 2
− u�
2 2
1 u�2 +1
− u�
u� u� −1
)(
2 2
u�3
2 u�2 −1
− u� u�u�3−1
2 2
−𝛾2 u�
Δ𝑝
)
𝑥
+(
u�−1
)(
𝑣u�
) ,
𝑣u�
𝑔𝑎𝑝
) +(
𝑢
u�
u�
u�u� +u�2 u�u�
u�2 u�2 −1
u�u� +u�2 u�u�
− u� u� −1
2 2
−
)
Hard to find II
could be observationally equivalent–same with conditional models.
▶
See ?) for testing with conditional models.
References I
Appendix: Solution of the New Keynesian Phillips Curve
Method of repeated substitution
All of the Rational expectations techniques rely on the law of
iterated expectations, saying
𝐸u� 𝐸u�+u� 𝑥u�+u� = 𝐸u� 𝑥u�+u� ,
that your average revisions of expectations given more information
will be zero. The method of repeated substitution is the brute force
solution, cumbersome and not very general. But since it’s also
instructive to see exactly what goes on, we use with this method.
See ?, Appendix A1) for alternative analytical methods and ?) and
?) for general methods based on matrix decompositions.
We start by using a trick to get rid of the lagged dependent
variable, following ?, p. 108-9), by defining
Δ𝑝u� = 𝜋u� + 𝛼Δ𝑝u�−1
(10)
where 𝛼 will turn out to be the backward stable root of the process
of Δ𝑝u� .
We take expectations one period ahead
𝐸u� Δ𝑝u�+1 = 𝐸u� 𝜋u�+1 + 𝛼𝐸u� Δ𝑝u�
𝐸u� Δ𝑝u�+1 = 𝐸u� 𝜋u�+1 + 𝛼𝜋u� + 𝛼2 Δ𝑝u�−1 .
Next, we substitute for 𝐸u� Δ𝑝u�+1 into original model:
u�
𝜋u� + 𝛼Δ𝑝u�−1 = 𝑏u�1
(𝐸u� 𝜋u�+1 + 𝛼𝜋u� + 𝛼2u�−1 Δ𝑝u�−1 )
u�
+𝑏u�1
Δ𝑝u�−1 + 𝑏u�2 𝑥u� + 𝜀u�u�
𝜋u� = (
u�
𝑏u�1
u�
1 − 𝑏u�1
𝛼
+(
) 𝐸u� 𝜋u�+1 + (
𝑏u�2
1−
u�
𝑏u�1
𝛼
u�
u�
𝑏u�1
𝛼2 − 𝛼 + 𝑏u�1
u�
1 − 𝑏u�1
𝛼
) 𝑥u� + (
1
u�
1 − 𝑏u�1
𝛼
) Δ𝑝u�−1
) 𝜀u� .
The parameter 𝛼 is defined by
u�
u�
𝑏u�1
𝛼2 − 𝛼 + 𝑏u�1
=0
or
𝛼2 −
1
u�
𝑏u�1
𝛼+
u�
𝑏u�1
u�
𝑏u�1
=0
(11)
with the solutions
u� u�
1 ± √1 − 4𝑏u�1
𝑏u�1
𝛼1
}=
.
𝛼2
2𝑏u�
(12)
u�1
The model will typically have a saddle point behaviour with one
root bigger than one and one smaller than one in absolute value. In
the following we will use the backward stable solution, defined by:
∣𝛼1 =
u� u�
1 − √1 − 4𝑏u�1
𝑏u�1
u�
2𝑏u�1
∣ < 1.
u�
u�
In passing might be noted that the restriction 𝑏u�1
= 1 − 𝑏u�1
often
imposed in the literature implies the roots
𝛼1 =
u�
1 − 𝑏u�1
u�
𝑏u�1
𝛼2 = 1,
as given in (12) as before. We choose |𝛼1 | < 1 in the following.
So we now have a pure forward-looking model
𝜋u� = (
u�
𝑏u�1
1−
u�
𝑏u�1
𝛼1
) 𝐸u� 𝜋u�+1 +(
𝑏u�2
1−
u�
𝑏u�1
𝛼1
) 𝑥u� +(
1
u�
1 − 𝑏u�1
𝛼1
) 𝜀u�u� .
Finally, using the relationship
𝛼1 + 𝛼 2 =
1
u�
𝑏u�1
between the roots, see f.ex. ?, p. 506), so:
u�
u�
1 − 𝑏u�1
𝛼1 = 𝑏u�1
𝛼2 ,
(13)
the model becomes
𝜋u� = (
𝑏u�2
1
1
) 𝐸u� 𝜋u�+1 + ( u�
) 𝑥u� + ( u�
) 𝜀u�u�
𝛼2
𝑏u�1 𝛼2
𝑏u�1 𝛼2
𝜋 = 𝛾𝐸u� 𝜋u�+1 + 𝛿𝑥u� + 𝑣u�u�
(14)
(15)
Following ?, p. 109–10), we now derive the solution in two steps:
1. Find 𝐸u� 𝜋u�+1
2. Solve for 𝜋u� .
Find 𝐸u� 𝜋u�+1 :
Define the expectational errors as:
𝜂u�+1 = 𝜋u�+1 − 𝐸u� 𝜋u�+1 .
We start by reducing the model to a single equation:
𝜋u� = 𝛾𝜋u�+1 + 𝛿𝑏u� 𝑥u�−1 + 𝛿𝜀u�u� + 𝑣u�u� − 𝛾𝜂u�+1 .
(16)
Solving forwards then produces:
𝜋u� = 𝛾 (𝛾𝜋u�+2 + 𝛿𝑏u� 𝑥u� + 𝛿𝜀u�u�+1 + 𝑣u�u�+1 − 𝛾𝜂u�+2 )
+ 𝛿𝑏u� 𝑥u�−1 + 𝛿𝜀u�u� + 𝑣u�u� − 𝛾𝜂u�+1
= (𝛿𝑏u� 𝑥u�−1 + 𝛿𝜀u�u� + 𝑣u�u� − 𝛾𝜂u�+1 )
+ 𝛾 (𝛿𝑏u� 𝑥u� + 𝛿𝜀u�u�+1 + 𝑣u�u�+1 − 𝛾𝜂u�+2 ) + (𝛾)2 𝜋u�+2
u�
= ∑ (𝛾)u� (𝛿𝑏u� 𝑥u�+u�−1 + 𝛿𝜀u�u�+u� + 𝑣u�u�+u� − 𝛾𝜂u�+u�+1 ) + (𝛾)u�+1 𝜋u�+u�+
u�=0
By imposing the transversality condition:
u�→∞
(𝛾)u�+1 𝜋u�+u�+1 = 0
and then taking expectations conditional at time 𝑡, we get the
“discounted solution”:
∞
𝐸u� 𝜋u�+1 = ∑ (𝛾)u� (𝛿𝑏u� 𝐸u� 𝑥u�+u� + 𝛿𝐸u� 𝜀u�u�+u�+1 + 𝐸u� 𝑣u�u�+u�+1 − 𝛾𝐸u� 𝜂u�+u�+
u�=0
∞
= ∑ (𝛾)u� (𝛿𝑏u� 𝐸u� 𝑥u�+u� ) .
u�=0
However, we know the process for the forcing variable, so:
𝐸u�−1 𝑥u� = 𝑏u� 𝑥u�−1
𝐸u� 𝑥u� = 𝑥u�
𝐸u� 𝑥u�+1 = 𝑏u� 𝑥u�
2
𝐸u� 𝑥u�+2 = 𝐸u� (𝐸u�+1 𝑥u�+2 ) = 𝐸u� 𝑏u� 𝑥u�+1 = 𝑏u�
𝑥u�
u�
𝐸u� 𝑥u�+u� = 𝑏u�
𝑥u� .
We can therefore substitute in:
∞
u�
𝐸u� 𝜋u�+1 = ∑ (𝛾)u� (𝛿𝑏u� 𝑏u�
𝑥u� )
u�=0
∞
= 𝛿𝑏u� ∑ (𝛾𝑏u� )u� 𝑥u�
u�=0
=(
𝛿𝑏u�
) 𝑥u� .
1 − 𝛾𝑏u�
and substitute back the expectation into the original equation:
𝜋u� = 𝛾𝐸u� 𝜋u�+1 + 𝛿𝑥u� + 𝑣u�u�
=𝛾(
𝛿𝑏u�
) 𝑥u� + 𝛿𝑥u� + 𝑣u�u� .
1 − 𝛾𝑏u�
Finally, using (10) and (15) we get the complete solution:
Δ𝑝u� − 𝛼1 Δ𝑝u�−1
+(
=(
u�
( u� u�2 ) 𝑏u� ⎞
⎛
u�u�1 u�2
⎜
⎟
⎟
= ( u�
)⎜
u�
⎜
⎟ 𝑥u�
u�
u�1
𝑏u�1 𝛼2
1 − ( u�
) 𝑏u�
u�u�1 u�2
⎝
⎠
u�
𝑏u�1
𝑏u�2
u�
𝑏u�1
𝛼2
) 𝑥u� + (
1
u�
𝑏u�1
𝛼2
(17)
) 𝜀u�u�
𝑏u�2 𝑏u�
𝑏u�2
1
1
) ( u�
) 𝑥u� + ( u�
) 𝑥u� + ( u�
) 𝜀u�u�
𝛼2
𝑏u�1 (𝛼2 − 𝑏u� )
𝑏u�1 𝛼2
𝑏u�1 𝛼2
Δ𝑝u� = 𝛼1 Δ𝑝u�−1 + (
Return
u�
𝑏u�1
𝑏u�2
(𝛼2 − 𝑏u� )
) 𝑥u� + (
1
u�
𝑏u�1
𝛼2
) 𝜀u�u� (18)