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19 Pages·2008·0.21 MB·English

An Alternative Method to Obtain the Blanchard and
Kahn Solutions of Rational Expectations Models
∗
Santiago Acosta Ormaechea Alejo Macaya
† ‡
August 8, 2007
Abstract
In this paper we show that the method of variation of parameters can be used as an
alternativeprocedureforsolvinglarge-scaleRationalExpectationsModels. Considering
thesamedynamicalsystemasBlanchardandKahn(1980)weexplain, inparticular, the
following issues: (i) how to apply the method of variation of parameters to obtain the
particular solution of the system (ii) that the solutions under the Rational Expectations
or the Perfect Foresight hypothesis can be obtained after considering a boundary value
problem and (iii) that some conditions must be satisfied for these solutions to be finite
or bounded. We also include a brief comparison between the Rational Expectations
and the Perfect Foresight solutions, where we show that the existence of a time-varying
information set affects the form in which each problem is solved and interpreted. For
robustness, we show that this method yields the solutions proposed in Obstfeld and
Rogoff (1996), based in either recursive substitution or factorization of polynomials
(considering the lag and forward operators L and L 1, respectively).
−
JEL: C61, E00
We would like to thank Daniel Heymann and Marcelo Sánchez for helpful comments and suggestions.
∗
Any remaining error or omission is the responsibility of the authors.
Department of Economics, University of Warwick, CV4 7AL, Coventry, UK. E-mail: S.L.E.Acosta-
†
[email protected].
Centro de Investigación en Métodos Cuantitativos Aplicados a la Economía y la Gestión, Facultad
‡
de Ciencias Económicas, Universidad de Buenos Aires. Córdoba 2122, Buenos Aires, Argentina. E-mail:
[email protected].
1
1 Introduction
The objective of this note is to present the method of variation of parameters1 (MVP) as
an alternative procedure to obtain the solutions of Blanchard and Kahn (1980) for solving
models under the Rational Expectations or the Perfect Foresight hypothesis. By applying
this method we are able to obtain the solution of the system of linear difference equations
without relying in "recursive substitutions" as it is done by these authors. This aspect of the
method is particularly useful, since it facilitates the computations needed to obtain the final
solution of the problem.
We emphasize in this note the fact that the system under the Rational Expectations or
the Perfect Foresight hypothesis can be solved as a boundary value problem. We also make
explicittheconditionsunderwhichthe"fundamentalsolution"isfiniteorbounded. Although
mentioned in Blanchard and Kahn, these conditions are not explicitly applied in their paper.
For robustness, we compare the solutions obtained through this method with those given
by Obstfeld and Rogoff (1996); that involve either recursive substitutions or factorization of
polynomials. We show that all these methods yield the same results.
The rest of the paper is organized as follows. Section 2 explains the method of variation of
parameters. Section 3 applies this method to the problem considered in Blanchard and Kahn
(1980), which originally assumes Rational Expectations. We then illustrate this method
with a simple stochastic version of the well-known Dornbusch (1976) overshooting model.
Section 4 considers a natural extension: the Perfect Foresight problem. Section 5 discusses
the differences in the solutions under the two behavioral assumptions and Section 6 presents
concluding remarks.
2 The method
In this section we develop the method of variation of parameters. This method is widely used
to solve systems of differential or difference equations in many math textbooks that deal with
dynamic systems (see for differential equations Nagel et al (2004) and for difference equations
Aiub (1985) or Elaydi (2005) to name just a few references). Its application to solve dynamic
economicmodelsismuchlessextended. Oneofthefewexampleswherethismethodisapplied
is given in Turnovsky (2000).
To facilitate comparability with those solutions obtained in Blanchard and Kahn we will
1Themethodofvariationofparameters(alsoknownasthemethodofvariationofconstants)wasinvented
by J. Lagrange in 1774.
2
followthenotationusedintheirpaper. Letusconsiderthefollowingsystemoflineardifference
equations with constant coefficients (with a slight abuse of notation regarding the subindex
t) :
X = AX +γZ , t t , t +1,..., t , (1)
t+1 t t 0 0
∈ { }
where X is a (n x 1) vector of endogenous variables, A is a (n x n) matrix of coefficients,
t
γ is a (n x k) matrix of coefficients and Z is a (k x 1) matrix of exogenous or fundamental
t
variables. From here onwards we assume that A is a full rank matrix. We can transform A
as follows,
A = C 1JC,
−
where J is the Jordan canonical form of matrix A 2. Let us introduce the following change
of basis: U = CX . The system stated in Eq. 1 thus yields,
t t
U = JU +CγZ . (2)
t+1 t t
The homogeneous solution can be expressed as,
Uh = Jt t0K. (3)
t −
The method of variation of parameters consists in proposing a solution for the complete
system taking as a benchmark the homogeneous solution previously obtained. Since we are
interested in spanning the space of solutions for the complete system, however, we have to
propose a solution which is linearly independent of Eq. 3. We thus postulate,
U = Jt t0K , (4)
t − t
where K is a (n x 1) vector, which is now a function of time and has to be determined.
t
Since this solution must satisfy Eq. 2 we have,
Jt+1 t0(K K ) = CγZ .
− t+1 t t
−
Let ∆K K K denote the difference operator on K between periods t+1 and t.
t t+1 t t
≡ −
The above system can be written as,
2To preserve comparability with Blanchard and Kahn, C 1 denotes a matrix with the associated eigen-
−
vectors of A in its columns.
3
Jt+1 t0∆K = CγZ ,
− t t
or3
∆K = J (t+1 t0)CγZ ,
t − − t
Applying the sum operator on both sides of the previous expression between periods t
0
and t 1 gives,
−
t 1 t 1
− −
∆K = J (s+1 t0)CγZ .
s − − s
sX=t0 sX=t0
Notice that the sum and the finite difference are inverse operators. It is easy to see then
that t 1 ∆K = K K 4. The above equation, therefore, reduces to:
s−=t0 s t − t0
P t 1
−
K = K + J (s+1 t0)CγZ ,
t t0 − − s
sX=t0
expression that determines the solution of the vector K . From Eq. 4 the solution of the
t
transformed system takes the form,
t 1
−
U = Jt t0K + J(t s 1)CγZ .
t − t0 − − s
sX=t0
Considering again the transformation U = CX we can obtain the solution of the original
t t
system:
t 1
−
X = C 1Jt t0K + C 1J(t s 1)CγZ . (5)
t − − t0 − − − s
sX=t0
Eq. 5 is the general solution of the system of difference equations stated in Eq. 1. If the
problem is one of initial values where X = X is given, then the particular solution of the
t0 0
problem takes the form5,
t 1
−
X = C 1Jt t0CX + C 1J(t s 1)CγZ . (6)
t − − 0 − − − s
sX=t0
3Note that since A is a full rank matrix (i.e., all eigenvalues are different from zero), J 1 always exists.
−
4This expression is also known as "telescopic sum".
5Forthesum ts0=−t01C−1J−(s+1−t0)CγZs weadopttheconventionthatisthesumofelementsofanempty
set. Therefore, we take the neutral value for the sum operator ( 0). Also notice that we adopt t as the
0
P ≡
initial period, so as to have the possibility of studying the behavior of the solution whenever t is set to any
0
arbitrary large value in the past.
4
The solution can also be expressed in terms of the original matrix A :
t 1
−
X = At t0X + A(t s 1)γZ . (7)
t − 0 − − s
sX=t0
Note that the above solution indicates that X is defined by the sum of the capitalized
t
value of X and the weighted sum of the past values of the exogenous variables.
0
3 The Rational Expectations Solution
In this section we consider an application of the method presented in the previous section
undertheassumptionthatagentshaveRationalExpectations(RE).Wewillfollow,essentially,
the case analysed in Blanchard and Kahn (1980). Let us define the following system of linear
difference equations with constant coefficients:
X X
t+1 = A t +γZ , t t , t +1,..., t, ..., T , (8)
Pte+1 Pt t ∈ { 0 0 }
· ¸ · ¸
where X is a (n x 1) vector of predetermined or state variables; P is a (m x 1) vector
t t
of non-predetermined or jump variables; Z is a (k x 1) vector of exogenous variables; A is
t
a ((n+m) x (n+m)) matrix of coefficients and γ is a ((n+m) x k) matrix of parameters
associated with the exogenous variables of the system. Following Muth (1961), the agent’s
expectation of P (=Pe ) will equate what the theory would predict conditioning on the
t+1 t+1
information set available at time t, E(P /Ω ); where E( /Ω ) denotes the expectation of any
t+1 t t
·
given variable conditional on Ω .
t
Note that the information set available at time t (i.e., Ω ) not only includes the past and
t
currentrealisationsoftheexogenousvariables,butalsotheirassociateddistributionfunctions;
that will allow agents to predict rationally the mean of the future sequence of these variables.
Since E( /Ω ) is a linear operator we can express the system stated in Eq. 8 as,
t
·
E(X /Ω ) E(X /Ω )
t+1 t = A t t +γE(Z /Ω ), (9)
E(P /Ω ) E(P /Ω ) t t
t+1 t t t
· ¸ · ¸
Recalling that E( /Ω ) of any vector of variables at period t is equal to the same vector
t
·
of variables, from Eq. 5 the solution of the system conditional on the information set Ω is
t
given by,
t 1
X −
t = C 1Jt t0K + C 1J(t s 1)CγZ . (10)
P − − t0 − − − s
t
· ¸ sX=t0
5
X
To solve for K we set t = t , thus obtaining: K = C t0 . Therefore, conditional
t0 0 t0 P
· t0 ¸
on the information set available at period t (= Ω ) Eq. 10 can be written as,
0 t0
t 1
E(X /Ω ) X −
t t0 = C 1Jt t0C t0 + C 1J(t s 1)CγE(Z /Ω ).
E(P /Ω ) − − P − − − s t0
· t t0 ¸ · t0 ¸ sX=t0
Letting t = T gives,
T 1
E(X /Ω ) X −
T t0 = C 1JT t0C t0 + C 1J(T s 1)CγE(Z /Ω ). (11)
E(P /Ω ) − − P − − − s t0
· T t0 ¸ · t0 ¸ sX=t0
In order to facilitate obtaining the solution of the system, we will partition the following
matrices as indicated below:
C C B B J 0
11 12 11 12 1
C = (nx n) (nx m) , C 1 = (nx n) (nx m) , J = (nx n) (nx m) and
 C21 C22  −  B21 B22   0 J2 
(mx n) (mx m) (mx n) (mx m) (mx n) (mx m)
   γ   
1
γ = (nx k) .
 γ 
2
(mx k)
 
 
With this partition, it will be possible to decouple the system depending on the whether
the roots of the matrix A are inside or outside the unit circle. Moreover, to guarantee the
existence and uniqueness of the RE solution we will follow Blanchard and Kahn assuming
that there are n roots inside and m roots outside the unit circle in the matrices J and J ,
1 2
respectively, ordered from the lowest to the highest absolute values6.
ToobtaintheparticularsolutionoftheREproblemwewillassumethefollowingboundary
or side conditions for every period t t , t +1,..., t, ..., T :
0 0
∈ { }
X given, (12)
t
lim J (T t)B 1E(P /Ω ) = 0. (13)
2− − 2−2 T t
T +
→ ∞
Eq. 12 indicates that at period t, the initial or inherited value of the vector of prede-
termined variables is given. Eq. 13 is often called "transversality condition" when solving
problems of intertemporal optimisation. It requires that as T + the expected value of
→ ∞
the vector of jump variables conditional on the information set Ω , discounted back to the
t
6For a proof of this proposition see Appendix A.
6
current period, is zero. This condition will imply the absence of "bubbles" as is often called
in the literature.
Forthisparticularsolutiontobefinitewefurtherassumeforeveryperiodt t ,t +1,...,
0 0
∈ {
t, ..., T that,
}
T 1
lim J (T t)B 1B − J(T s 1)(C γ +C γ )E(Z /Ω ) = 0, (14)
2− − 2−2 21 1 − − 11 1 12 2 s t
T +
→ ∞ s=t
X
Eq. 14 requires that the expected value of the vector of exogenous or "fundamental"
variables, conditional on the information set Ω (i.e., E(Z /Ω )), does not grow "too fast" as
t s t
T + .
→ ∞
Notice that we are interested in obtaining the solutions of the endogenous variables con-
ditioning on the information set Ω . Solving for E(P /Ω ) Eq. 11 gives:
t T t
E(P /Ω ) = (B JT tC +B JT tC )X +(B JT tC +B JT tC )P (15)
T t 21 1− 11 22 2− 21 t 21 1− 12 22 2− 22 t
T 1
+ − B J(T s 1)(C γ +C γ )+B J (s+1 T)(C γ +C γ ) E(Z /Ω ).
{ 21 1 − − 11 1 12 2 22 2− − 21 1 22 2 } s t
s=t
X
From this equation we can obtain the solution of P (see Appendix B for details):
t
P = C 1C X C 1 ∞ J (s+1 t)(C γ +C γ )E(Z /Ω ), (16)
t − 2−2 21 t − 2−2 2− − 21 1 22 2 s t
s=t
X
expression that gives a relation between P , X and the future sequences of Z conditional
t t s
on Ω . Since the path of P is conditional on the information set available at each period
t t
t, the complete solution of the system must take this fact into account. This aspect of the
RE solution is a key difference with respect to the Perfect Foresight solution, as it will be
discussed in the next section.
A A
11 12
Partitioning the matrix A as A = (nx n) (nx m) and substituting the expression for
 A21 A22 
(mx n) (mx m)
P derived in Eq. 16 into Eq. 9 yields, 
t
X = B J B 1X A C 1 ∞ J (s+1 t)(C γ +C γ )E(Z /Ω )+γ Z . (17)
t+1 11 1 1−1 t − 12 2−2 2− − 21 1 22 2 s t 1 t
s=t
X
Observe that since X is a predetermined variable we have that E(X /Ω ) = X . We
t t+1 t t+1
have also made use of the following relation: A A C 1C = B J B 1. Note that Eq.
11 − 12 2−2 21 11 1 1−1
17 represents a (non-homogeneous) system of linear difference equations in X , that can be
t
7
solved through the method of variation of parameters. The solution of this system can be
obtained from Eq. 6 and is given by the following expression7,
t 1
Xt = B11J1t−t0B1−11X0 +B11 − J1(t−s−1)B1−11γ1Zs (18)
sX=t0
t 1 +
B − J(t s 1)B 1A C 1 ∞ J (v+1 s)(C γ +C γ )E(Z /Ω ).
− 11 1 − − 1−1 12 2−2 2− − 21 1 22 2 v s
sX=t0 Xv=s
Observing that A C 1 = B B J B 1B J 1 J , it can be seen that Eq. 18 is the
12 2−2 12 − 11 1 1−1 12 2− 2
same as Eq. (4) in Blanchard and Kahn (1980, p. 1308) for the case in which t = 0. The
£ ¤ 0
solution of P is obtained by substituting Eq. 18 in Eq. 16 and considering the fact that
t
C 1C = B B 1 :
2−2 21 − 21 1−1
t 1
Pt = B21J1t−t0B1−11X0 +B21 − J1(t−s−1)B1−11γ1Zs (19)
sX=t0
t 1
B − J(t s 1)B 1A C 1 ∞ J (v+1 s)(C γ +C γ )E(Z /Ω )
− 21 1 − − 1−1 12 2−2 2− − 21 1 22 2 v s
sX=t0 Xv=s
C 1 ∞ J (s+1 t)(C γ +C γ )E(Z /Ω ).
− 2−2 2− − 21 1 22 2 s t
s=t
X
This expression is the same as Eq. (5) in Blanchard and Kahn (1980, p. 1308) whenever
t = 0. Having obtained the solutions of X and P , it is worth giving an interpretation of
0 t t
why there are double sums appearing in those expressions. This is a direct consequence of
the presence of a time-varying information set. Notice that the sum over v is the discounted
value, from any period s, of the expected path of the exogenous or forcing variables. The
sum in s, that goes up to t 1, gives an average over those discounted values at different
−
points in the past (i.e., starting at a different s, from t to t 1). The whole expression is,
0
−
therefore, a weighted average seen from s = t up to t 1 (i.e., in the past), of the discounted
0
−
value of the expected path of the exogenous variables. This particular aspect where the
past "matters" in the solution of current variables is a consequence of the fact that how was
perceived the expected evolution of Z, conditional on each information set, affects the value
of the predetermined vector of variables in previous periods; and through it implicitly affects
the vector of predetermined variables in the current period.
7To obtain this solution we can associate the matrix A in Eq. 1 with B J B 1 and γZ with
11 1 1−1 t
−foAllo1w2Cin2−g21rela∞s=titoJn2s−:(sC+1=−tB)(C12,1γJ1=+JC2a2nγd2)EC(Z1s=/ΩBt)+. Tγh1Zets.oSluinticoenwtheuksnfoowllotwhsa.t A = C−1JC we have the
1−1 1 − 11
P
8
3.1 An example: the Exchange Rate Overshooting
Toillustratethismethodweconsiderasimplifiedversionofthewell-knownDornbusch(1976)’s
model developed in Taylor (1986). It is worth noting that Taylor considers the method of
undetermined coefficients to solve it. The model is summarized by the following equations:
m p = αr (20)
t t t
− −
r = E(e /Ω ) e (21)
t t+1 t t
−
p p = β(e p ), (22)
t t 1 t t
− − −
where m is the nominal quantity of money, p is the price level, r is the domestic interest
t t t
rate and e denotes the nominal exchange rate, measured as the domestic price of foreign
t
exchange, α > 0 and β > 0 are associated parameters. All variables, except the interest rate,
are in logs. Eq. 20 defines money market equilibrium, Eq. 21 is the UIP condition where we
have assumed that r = 0 and Eq. 22 states that domestic inflation increases in the excess
∗
demand for domestic goods (i.e., a positive function of the real exchange rate). The system
can be written as:
p 1 β p 0
t = (1+β) 1 t 1 + m
E(e /Ω ) − α 1 1+β(1+α 1) e − α 1 t
t+1 t − − t −
· ¸ · ¸· ¸ · − ¸
Observe that here p is a predetermined variable while E(e /Ω ) is a non-predetermined
t t+1 t
orjumpvariable. Hence,thissystemhasthesameformoftheonestatedinEq. 9. AsinTaylor
(1986) we assume that the nominal quantity of money follows the following autorregresive
process:
∞
m = θ ε .
t i t i
−
i=0
X
The boundary conditions are given in this case by:
p given
t
lim λ (T t)b 1E(e /Ω ) = 0,
2− − −22 T t
T +
→ ∞
9
and, to obtain a finite fundamental solution we assume:
T 1
lim b 1α 1b λ (T t) − λT s 1E(m /Ω ) = 0.
T + − −22 − 21 2− − 1− − s t
→ ∞ s=t
X
From Eqs. 18 and 19 we have the following solutions for the price level and the exchange
rate:
t 1 +
p = λtp +βα 1(1+β) 1 − λt s 1 ∞ λ (v+1 s)E(m /Ω ),
t 1 1 1 − − 1− − 2− − v s
− −
s=0 v=s
X X
t 1 + +
e = b b 1λtp +b b 1βα 1(1+β) 1 − λt s 1 ∞λ (v+1 s)E(m /Ω )+α 1 ∞λ (s+1 t)E(m /Ω )
t 21 −11 1 1 21 −11 − − 1− − 2− − v s − −2 − s t
−
s=0 v=s s=t
X X X
Following Taylor we assume that p = 0 and that the money supply is equal to zero
1
−
before a temporary and unexpected monetary shock hits the economy at t = 0. This shock
implies that the sequence of the exogenous variable is given by E(m ) = ε and E(m ) = 0
0 0 t
t > 0.
∀
The solution for the price level for t > 0 takes the form:
p = βα 1(1+β) 1λ 1λt 1ε .
t 1 − − −2 1− 0
−
The solution for the exchange rate at t = 0 is given by:
e = α 1λ 1ε .
0 − −2 0
For t > 0 its solution is:
e = b b 1βα 1(1+β) 1λt 1λ 1ε .
t 21 −11 − − 1− −2 0
Letting α = β = 1, it can be seen that the eigenvalues of the coefficient matrix are
given by: λ = 1 √2/2. Hence, the matrix of eigenvectors takes the form: C 1 = B =
1,2 −
∓
1 1
. Replacing in the above solutions yields:
1 √2 1+√2
· − ¸
1 t 1
p = (1/2) 1+√2/2 − 1 √2 − ε
t 1 0
− −
1
e0 = 1+√2¡/2 − ε0 ¢ ¡ ¢
t 1 1
et = (¡1/2) 1 ¢√2 1 √2/2 − 1+√2/2 − ε0.
− −
Letting ε ¡= 1, we¢¡can constr¢uct T¡able 1, w¢hich summarises the results obtained here.
0
This table coincides with that presented in Taylor (1986, p. 2024).
10