Equilibrium correction; Solution of dynamic
equations; Dynamic systems
Ragnar Nymoen
Department of Economics, UiO
1 September 2009
ECON 3410/4410: Lecture 3 and 4
Lecture 3 and 4: Overview
The ECM transformation of the dynamic equation, and its
interpretation.
The formal solution of the …rst order dynamic model.
The formal analysis of a simple dynamic system.
Main reference is IDM, Ch 2.5-2.8.2 and 2.8.5.
Read 2.8.3 as background to one of the seminar exercises
where we apply the concepts that have been introduced to
Solow’s growth model. Chapter 2.8.4, on the RBC model, will
be lectured later.
ECON 3410/4410: Lecture 3 and 4
ECM: error-correction model, or
equilibrium correction model
The ECM is an 1-1 transformation of the ADL model.
But in many applications, the ECM form of the dynamic
equation is easier to interpret than the ADL.
We …rst go through the transformation of:
yt =
0
+
1 xt
+
2 xt 1
+ yt
1
+ "t :
to ECM,
and then explain the acronym and give some examples.
ECON 3410/4410: Lecture 3 and 4
(1)
The ECM transformation
1
Subtract yt
yt
2
1
on both sides of the ADL equation, which gives
=
Add and subtract
yt
3
yt
1
yt
1
=
0
+
1 xt
1 xt 1
0 + 1 (xt
+
2 xt 1
+(
1)yt
1
+ "t
on the right hand side:
xt
1 )+( 1 + 2 )xt 1 +(
1)yt
1 +"t
Use the di¤erence operator , meaning for example
yt = yt yt 1 , and write the ECM as:
yt =
0
+
1
xt + (
1
+
2 )xt 1
+(
1)yt
1
+ "t (2)
This is a 1-1 a re-parameterization: If the sequence of y
values y0 ; y1 ; y2 ; y3 .... satis…es (1), it also satis…es (2) for the
same sequence of x and " values.
ECON 3410/4410: Lecture 3 and 4
ECM and continuos time analogue
The ECM
yt =
0
+
1
xt + (
1
+
2 )xt 1
+(
1)yt
1
+ "t (2)
makes explicit that in a dynamic model, the growth of yt is
explained by the change in the explanatory variable and the
past levels of xt and yt .
The continuous time dynamic mode is formulated in close
analogue to (2).
With reference to Box 1.1 in IDM, we can extend the model
found there to:
y_ = a0 + b |{z}
x_ + c x(t) + ay (t) + "(t), a < 0,
|{z}
|{z}
|{z}
yt
xt
xt
1
yt
1
ECON 3410/4410: Lecture 3 and 4
(3)
ECM: correcting deviations from equilibrium (I)
yt =
Collect yt
yt =
1
0
+
and xt
0
+
xt + (
1
1
1
1
+
2 )xt 1
+(
1)yt
1
+ "t
inside a bracket:
xt
(1
) y
1
1
+
2
x
+ "t ;
(4)
t 1
and assume the following long-run relationship for a situation with
constant growth rates (possibly zero) in y and x:
y = k + x;
(5)
where y denotes the steady-state equilibrium for yt . We want
equation (4) to be consistent with this steady-state. The slope
coe¢ cient must therefore be equal to the long-run multiplier of
the ADL:
1+ 2
= long run
,
1< <1
(6)
1
ECON 3410/4410: Lecture 3 and 4
ECM: correcting deviations from equilibrium (II)
Since y = k + x, the expression inside the brackets in (4) can be
rewritten as
y
1
+
2
1
x =y
x =y
y + k.
(7)
Using (7) in (4) we obtain
yt =
0
(1
)k +
1
xt
(1
)fy y g + "t ;
| {z }
y
showing that
yt is explained by
1<
<1
t 1
x k
(8)
xt , and
the correction of the last period’s disequilibrium y
y .
So the E in the acronym ECM can either refer to equilibrium
correction model, or error correction model.
Error correction is perhaps most natural in applications where
y is a target that is derived from economic theory.
ECON 3410/4410: Lecture 3 and 4
Steady-state, with and without growth
Consider the steady-state situation with constant growth
xt = gx and yt = gy , and zero disturbance: "t = 0.
Imposing this in
yt =
0
(1
)k +
and noting that fy
steady-state, gives
gy =
y gt
0
xt
1
1
(1
) fy
y gt
1
+ "t
= 0 by de…nition of the
(1
)k +
1 gx ;
Mathematically, this shows that k in the model of
y : y = k + x depends on the two growth constants gy and
gx :
gy + 0 + 1 gx
k=
; if
1< <1
(9)
1
The case of a stationary steady-state (no growth), is a special
case: gx = gy = 0, (9) simpli…es to k = 0 =(1
).
ECON 3410/4410: Lecture 3 and 4
The homogenous ECM
If the long-run slope coe¢ cient is restricted to 1, we get the
Homogenous ECM, the last model in the model Typology
from Lecture 2.
= 1 ()
1
+
2
=
(
1)
The name refers to the property that a permanent unit
increase in the exogenous variable leads to a unit long-run
increase in y (as with homogeneity of degree one of
equilibrium market prices in a general equilibrium model).
The realism of the = 1 restriction cannot be taken as
granted, but in our example of the Norwegian consumption
function, it …ts to perfection:
ln Ct = 0:04 + 0:13 ln(INCt )
0:21 fln C
ln INC gt
1
implying that the steady-state savings rate (s) in independent
of income. The next slide gives the detailed argument.
ECON 3410/4410: Lecture 3 and 4
Long-run homogeneity and the savings rate
ln(1
ln(C ) = k + ln(INC ), in steady-state.
C
ln(
)=k +(
1) ln(INC )
INC
C
ln(1 1 +
)=k +(
1) ln(INC )
INC
INC C
) ( 1) s, (remember ln(1 + ( s))
| INC
{z }
s
Therefore the steady-state consumption function implies:
s = k + (1
@s
= (1
@ ln(INC )
) ln(INC )
) = 0 i¤
= 1:
ECON 3410/4410: Lecture 3 and 4
s).
The two interpretations of static models, revisited.
As noted in the …rst lecture, a static relationship has two distinct
interpretations in macroeconomics
1
As a model of actual behaviour when the adjustment speed is
“in…nitely great” (Frisch)
This corresponds to 2 = 0 and = 0 in the ADL.
2
As a long-run relationship, which applies to a steady-state
situation.
The validity of this interpretation only hinges on being “less
than one” ( 1 < < 1).
3
We advise to distinguish carefully between the two, to avoid
misunderstandings.
ECON 3410/4410: Lecture 3 and 4
Is Purchasing Power Parity (PPP) about the short-run or
the long-run?
PPP is one of the most used hypotheses in macroeconomic
models of the open economy.
It is used to make theoretical predictions about
the nominal exchange rate (if the monetary policy regime is
‡oating), or
the domestic price level (in a …xed exchange rate regime),
We start by de…ning the real-exchange rate:
EP
=
(10)
P
where P denotes the foreign price level, E is the nominal
exchange rate (Kroner/Euro) and P is the domestic price
level.
The purchasing power parity hypothesis, PPP, says that “the
real exchange rate is constant”.
But what is the time perspective of the PPP hypothesis?
ECON 3410/4410: Lecture 3 and 4
PPP and exchange rate pass-through
Assume a …xed exchange rate regime.
If PPP is applied to the short-run, then
(constant), and (10) gives the theory:
ln Pt =
ln
0
t
=
t 1
=
0
+ ln Et + ln Pt ,
which is a static price equation saying that the pass-through
of a change in Et on Pt is full and immediate.
If PPP is used as a long-run hypothesis, we have instead that
in a steady-state situation, and
t =
ln Pt = gE +
where gE and
denote the long-run constant growth rates of
Et and foreign prices. On this interpretation, PPP implies full
long-run pass-through, but the short-run pass-through can be
much lower, as in
ln Pt = (
1) ln Pt
1
ln Et
1
Pt
1
ECON 3410/4410: Lecture 3 and 4
+ "t
Solution of dynamic models
The solution of a single dynamic equation is a sequence of
values that satis…es the equation for given sequences of values
for the exogenous variables (including the stochastic shocks).
For models with 2 or more equations, the solution consists of
separate sequences for each endogenous variable in the model.
The existence and uniqueness of the solution depends on
initial condition , or on the terminal conditions.
In this lecture, and in most the course, we will work with
dynamic models that have unique solutions that only depend
on initial conditions. But example of the importance of
terminal conditions will be given later: they arise in dynamic
models that include the rational expectations hypothesis.
ECON 3410/4410: Lecture 3 and 4
Solving the ADL equation for y (I)
We now seek a solution to
yt =
0
+
1 xt
+
2 xt 1
+ yt
1
+ "t :
(1)
for a period that runs form t = 1 and forward in time:
t = 1; 2; 3:::::
For simplicity, and with no loss of generality, we assume the
following about the exogenous variables
"t = 0 for t = 0; 1; :::; and
xt = mx (constant x) for t = 0; 1:::
ECON 3410/4410: Lecture 3 and 4
Solving the ADL equation for y (II)
We write the simpli…ed model as
yt =
0
+ Bmx + yt
We regard the parameters
numbers.
1,
0,
where B =
1
1,
as known
2
and
+
2.
(11)
(11) holds for t = 0; 1; 2; :::. Remember that t = 0 is called
the initial period.
When y0 is a …xed and known number, there exists a unique
sequence of numbers y0 ; y1 ; y2 ; ::: which is the solution of (11).
The solution is found very easily by …rst solving (11) for y1 ,
then for y2 and so on.
This a recursive solution method, using the same logic that
we used when we obtained the dynamic multipliers.
ECON 3410/4410: Lecture 3 and 4
Forward recursion
1
y1 =
0
+ Bmx + y0
2
y2 =
0
+ Bmx + y1 =
0 (1
+ ) + (1 + )Bmx +
2
y0
and so on.....
Motivating, by induction:
yt = (
0
+ Bmx )
t 1
X
s
+
t
y0 , t = 1; 2; :::
s =0
ECON 3410/4410: Lecture 3 and 4
(12)
Uniqueness and stability of the solution
Uniqueness
The solution is obviously conditional on what we assume about
xt and "t , but given that, the only source of non-uniqueness of
the solution is variation in the initial condition, y0 :
But y0 cannot vary! Because it has been determined by
history— it is predetermined.
Uniqueness of the solution extends to all dynamic models and
systems that can be solved by forward recursion, no matter
how complicated dynamics.
The solution can be stable, unstable or explosive.
stable: often implicit in model based discussions about the
e¤ects of policy change (compare …scal policy packages)
unstable: some theories predict that the e¤ects of a shock
linger on after the shock itself has gone away, this is referred
to as hysteresis.
explosive: Example: “bubbles” in capital and credit markets.
ECON 3410/4410: Lecture 3 and 4
The stable solution
The condition
1<
<1
(13)
is the necessary and su¢ cient condition for the globally
asymptotically stable solution.
t 1
X
s ! 1 as t ! 1, we obtain the asymptotic solution:
Since
1
s =0
+ Bmx )
0
=
+ long run mx
(14)
1
1
where y is identical to the steady-state solution that we can
obtain from equation (1), if we assume from the outset that the
solution is stable.
Using y , the solution (12) above can also be expressed as (see
page 52 of IDM)
yt = y + t (y0 y ).
y =
(
0
showing that equilibrium correction is also embedded in the
solution.
ECON 3410/4410: Lecture 3 and 4
The unstable solution
In the stable solution, the in‡uence of the initial condition on
the solution is reduced as we move away from the initial
period. Asymptotically the in‡uence is completely removed, as
the expression
( + Bmx )
y = 0
1
shows.
In the unstable solution, the in‡uence of the initial condition
is not reduced: When = 1, we obtain from equation (12):
yt = (
0
+ Bmx )t + y0 , t = 1; 2; :::
(15)
showing that the solution also contains a deterministic time
trend, with coe¢ cient ( 0 + Bmx ):
ECON 3410/4410: Lecture 3 and 4
The explosive solution
When is greater than unity in absolute value the solution is
called explosive, for reasons that become obvious when we
again consult (12).
yt = (
0
+ Bmx )
t 1
X
s
+
t
y0 ; t = 1; 2; :::
s =0
The e¤ect of the initial condition is now “powered” up and
may dominate after a few solution periods.
If there is a change in mx somewhere in the solution period,
that change will also have explosive tendencies.
ECON 3410/4410: Lecture 3 and 4
Example of solutions paths
5.10
8
DFystable_ar.m8
DFystable_ar.8
6
5.05
4
5.00
2
200
205
DFyun stable
210
DFyun stable_2 01
200
205
210
DFyexplosiv e
150 00
= 0:8
b)
=
125 00
d) Explosive
100 00
750 0
3.60
500 0
200
205
0:8
c) Unstable
3.70
3.65
a)
200
210
220
230
ECON 3410/4410: Lecture 3 and 4
Deterministic trend and (in)stability of the solution
If B = 1 + 2 6= 0, the unstable solution implies a trend.
The converse implication does not hold. The presence of a
trend in the solution does not imply that the solution is
unstable.
Assume that the sequence of xt values is given by the process
xt = gx + xt
1,
t = 1; 2; :::
instead of
xt = gx , t = 0; 1; 2; :::
If 1 < < 1, the solution for yt is still stable since there is
no trace of y0 is the steady state solution, which however
contains a linear trend which is due to xt .
Solow’s growth model with technical change is an example of
a dynamic model where the stable solution contains a
deterministic trend for log GDP per capita.
ECON 3410/4410: Lecture 3 and 4
Finding the solution by simulation
The method with forward recursion is general
In practice, macro models contain tens (or hundreds) of
equations. Today’s’computer programs …nd the solution very
fast.
The solution method is called simulation.
Having found a solution for one set of assumptions about xt
and "t (t = 1; 2; ::), often called the baseline, it is easy to …nd
a solution for an alternative time path for xt . The di¤erence
between the two solutions give the dynamic multipliers.
Although the recursive method generalizes to any model, the
analysis of stability of the solution does not generalize. Often,
higher order lags of yt is part of the implied dynamics, and
then stability does not depend on one single autoregressive
parameter.
We shall however see some examples where stability of the
system can be answered with a simple analysis.
ECON 3410/4410: Lecture 3 and 4
Dynamic analysis of a dynamic macro model (I)
In the seminar exercises we do formal analysis of the
“cobweb-model”.
We now take a simple Keynesian model as our …rst
macroeconomic example.
The model consists of linear speci…cation of the consumption
function.
Ct = 0 + 1 INCt + Ct 1 + "t
(16)
together with the product market equilibrium condition
INCt = Ct + Jt
(17)
where Jt is endogenous and denotes autonomous expenditure,
and INC is now interpreted as GDP. Ct and INCt are the
endogenous variables.
We known that if we can …nd an ADL model for Ct that only
depends on exogenous and predetermined variables, we can
…nd the solution for Ct by recursion.
ECON 3410/4410: Lecture 3 and 4
Dynamic analysis of a dynamic macro model (II)
Substitute INC from (17) to give:
Ct = ~ 0 + ~ Ct
~ 0 and ~ are
~ 2 = 1 =(1
and
).
1
0
1
+ ~ 2 Jt + ~"t
divided by (1
1 ),
(18)
and
(18) is the …nal equation for Ct . It is an ADL of the type that
we have learnt how to solve.
If
1 < ~ < 1, the solution is stable.
The impact multiplier of consumption with respect to
autonomous expenditure is ~ 2 , and long run is ~ 2 =(1
Interim multiplier can be obtained from (18).
~ ).
We see that INCt inherits the stability of Ct , so we don’t have
to derive a …nal equation for INCt to check stability (only if
we want the interim multipliers)
ECON 3410/4410: Lecture 3 and 4
The method with a short-run and long-run model (I)
Often, the …nal equation is di¢ cult (impossible) to derive
manually.
In these cases we use the method that we motivated in the
…rst lecture.
We …rst analyze the short-run model, and then, as second step,
the long-run model, assuming that a stable steady-state exists.
This method can always be used to answer two questions:
1
What is the short-run e¤ect of a change in an exogenous
variable?
2
What is the long-run e¤ect of the change?
ECON 3410/4410: Lecture 3 and 4
In the simple Keynes model the short-run model is already given by
(16) and (17).
The long-run model is de…ned for the situation of
Ct = C ; INCt = I NC , Jt = J and "t = 0.
C=
0
1
I NC = C + J
+
1
1
I NC
(19)
(20)
The impact multiplier of a permanent (or temporary) rise in
Jt is obtained from the short-run model (16) and (17), while
subject to dynamic stability, the long-run multipliers can be
derived from the system (19)-(20).
It is a useful exercise to check that the long-run multipliers
from the long-run model are the same as the ones you can
obtain from the …nal equation for Ct and INCt .
ECON 3410/4410: Lecture 3 and 4
What is general, and what is special about this example?
It is general that in a dynamic system, the solution for a
variable, and the question of dynamic stability has to be found
from the …nal equation for that variable.
It is not general that the …nal equation has only …rst order
autoregressive dynamics.
With higher order dynamics in the …nal equation, stability is
depends on the roots of the characteristic equation.
For example, if we have a …nal equation of the form
yt =
0
+
1 xt
+
2 xt 1
+
1 yt 1
+
2 yt 2
+ "t
it is su¢ cient for a stable solution, but not necessary, that
j j j < 1 j = 1; 2:
The su¢ cient and necessary condition is that the associated
characteristic roots are less than one in magnitude. But this is
beyond the math that we use in this course.
In the notes to lecture 4 we will dig a little bit deeper into
these issues.
ECON 3410/4410: Lecture 3 and 4
NAM-‡owchart
ECON 3410/4410: Lecture 3 and 4