Solving Linear Rational Expectation Models
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1 Solving Linear Rational Expectation Models Dr. Tai-kuang Ho Associate Professor. Department of Quantitative Finance, National Tsing Hua University, No. 101, Section 2, Kuang-Fu Road, Hsinchu, Taiwan 30013, Tel: , ext , Fax: , This lecture note is based on Klein s paper and
2 1 The problem Approaches to solve linear rational expectation models include Sims (2002), Anderson and Moore (1985), Binder and Pesaran (1994), King and Watson (1998), Klein (2000), and Uhlig (1999). A recent view is Anderson (2008), who compares the accuracy and computational speed of alternative approaches to solving linear rational expectations models. Martin Uribe s Lectures in Open Economy Macroeconomics, Appendix of Chapter 4, provides a very clear explanation of the linear solution method to dynamic general equilibrium models. Fabrice Collard s lecture note. The latter contains many typos and I have tried my best to make they right in this document.
3 Martin Uribe s Lectures are available from his website. McCandless, George (2008), The ABCs of RBCs: An Introduction to Dynamic Macroeconomic Models, Harvard University Press. This book provides an detailing introduction to solving dynamic stochastic general equilibrium model. It is very practical for beginners as the author explains the deduction step by step, and the book includes many examples and solutions that facilitate learning. The author uses rst-order approximation to the model, and adopts Uhlig s toolkits and related computer programs to solve the log-linearized model.
4 Klein (2000) uses a complex generalized Schur decomposition to solve linear rational expectation models. Why generalized Schur decomposition? First, it treats in nite and nite unstable eigenvalues in a uni ed way. Second, Schur decomposition is computationally more preferable. Setting the stage
5 Measurement equation, which describe variables of interest, such as output or gross interest rate. N y Y t = N x X t + N z Z t (1) Endogenous variables M x0 E t X t+1 + M y0 E t Y t+1 + M z0 E t Z t+1 = M x1 X t + M y1 Y t + M z1 Z t (2)
6 Exogenous shocks (forcing variables) Z t = Z t 1 + t (3) Dimension of variables Y t : n y 1 X t : n x 1 Z t : n z 1
7 Predetermined variables X b t : n b Jump (control) variables X f t : n f n x = n b + n f X t = X b t X f t Dimension of matrices
8 N y (n y n y ) M x0 N x (n y n x ) M y0 (n x n x ) (n x n y ) M x1 M y1 (n x n x ) (n x n y ) (n z n z ) (n z n e ) N z (n y n z ) M z0 (n x n z ) M z1 (n x n z ) N y is invertible, which means that the variable of interest is uniquely de ned. All eigenvalues of lies within the unit circle
9 t N (0; ) Transforming the problem Y t = Ny 1 N x X t + Ny 1 N z Z t E t Z t+1 = Z t Substitute and rewrite equation (2) as:
10 AE t X t+1 = BX t + CZ t A = M x0 + M y0 Ny 1 N x B = M x1 + M y1 Ny 1 N x C = M z1 + M y1 Ny 1 N z Mz0 + M y0 Ny 1 N z
11 This system comes from the linearization of the individual optimization conditions and market clearing conditions in a dynamic equilibrium model. The matrix A is allowed to be singular. A singular matrix A implies that static (intra-temporal) equilibrium conditions are included among the dynamic relationships.
12 2 Generalized Schur decomposition The idea of Klein s approach is to use complex generalized Schur decomposition to reduce the system into an unstable and a stable block of equations. The stable solution is found by solving the unstable block forward and the stable block backward. De nition of predetermined or backward-looking variables: a process k is called backward-looking if the prediction error t+1 k t+1 E t k t+1 is an exogenous martingale di erence process (E t t+1 = 0) and k 0 is exogenous given. The dynamic equation
13 AE t X t+1 = BX t + CZ t Generalized Schur decomposition of the pencil (A; B) S = QAZ T = QBZ
14 QQ 0 = ZZ 0 = I See my handout for a description of generalized Schur decomposition The dynamic equation can be rewritten as AZZ 0 E t X t+1 = BZZ 0 X t + CZ t {z } I {z } I! t = Z 0 X t
15 AZE t! t+1 = BZ! t + CZ t QAZE t! t+1 = QBZ! t + QCZ t R QC SE t! t+1 = T! t + RZ t
16 Don t confuse Z with Z t. Remark Z = Z 11 Z 12 Z 21 Z 22!! t = Z 0 X t = " Z 0 11 Z 0 21 Z 0 12 Z0 22 # " X b t X f t # = 2 4 Z0 11 Xb t + Z0 21 Xf t Z 0 12 Xb t + Z0 22 Xf t 3 5 " #! b t! f t
17 ! b t = Z 0 11 Xb t + Z 0 21 Xf t! f t = Z0 12 Xb t + Z 0 22 Xf t The generalized eigenvalues of the system are T ii S ii
18 We sort the generalized eigenvalues in ascending order. n s stable eigenvalues n u unstable eigenvalues Blanchard and Kahn condition: if n b = n s (and n f = n u ) then the system admits a unique saddle path. There are as many predetermined variables as there are stable eigenvalues. How likely is that n b = n s?
19 In practice, very likely. If the system of equations is derived from a linear-quadratic dynamic optimization problem, we are almost guaranteed that n b = n s. Partition the system Z = Z 11 Z 12 Z 21 Z 22!
20 S = S 11 S 12 0 S 22! T = T 11 T 12 0 T 22! Z 11 : n s n s Z 12 : n s n f Z 21 : n f n s Z 22 : n f n f
21 Rewrite the system as S 11 S 12 0 S 22! E t! b t+1 E t! f t+1 1 A = T 11 T 12 0 T 22!! b t! f t! + R 1 R 2! Z t S 11 and T 22 are invertible by construction. 3 The forward part of the solution Look at the unstable part of the system
22 S 22 E t! f t+1 = T 22! f t + R 2Z t! f t = T 1 22 S 22E t! f t+1 T 1 22 R 2Z t! f t = lim T 1 k k!1 22 S 22 Et! f t+k 1X k=0 T 1 22 S k 22 T 1 22 R 2 k Z t E t! f t+k < 1
23 lim T 1 k k!1 22 S 22 Et! f t+k = 0! f t = 1 X k=0 T 1 22 S k 22 T 1 22 R 2 k Z t = Z t 1X k=0 T 1 22 S k 22 T 1 22 R 2 k Some matrix algebraic
24 vec (A + B) = vec (A) + vec (B) vec (ABC) = C 0 A vec (B) (AB CD) = (A C) (B D) S = 1X k=0 A k BC k = B + ASC
25 Back to the problem 1X T 1 22 S k 22 T 1 22 R 2 {z} k=0 {z } {z } k = T 1 22 R 2 + T 1 22 S 22 vec ( ) = vec T22 1 R 2 + vec T 1 22 S 22
26 vec ( ) = I T22 1 vec (R2 ) + 0 T22 1 S 22 vec ( ) vec ( ) = I T 1 22 vec (R2 ) + 0 T22 1 (I (S22 )) vec ( ) (I T 22 ) vec ( ) = vec (R 2 ) + 0 S 22 vec ( ) vec ( ) = 0 S 22 I T 22 1 vec (R2 )
27 Important! f t = Z t Recall that! f t = Z0 12 Xb t + Z 0 22 Xf t Z 0 12 Xb t + Z0 22 Xf t = Z t
28 Guess a solution for X f t X f t = Xb t + Z t Z 0 12 Xb t + Z 0 22 X b t + Z t = Zt Z Z0 22 = 0 (4)
29 Z 0 22 = (5) ;? Z 0 Z = I Z 0 11 Z0 21 Z 0 12 Z0 22! Z 11 Z 12 Z 21 Z 22! = I
30 " Z 0 11 Z 11 + Z 0 21 Z 21 Z 0 11 Z 12 + Z 0 21 Z 22 Z 0 12 Z 11 + Z 0 22 Z 21 Z 0 12 Z 12 + Z 0 22 Z 22 # = " # I 0 0 I Z 0 12 Z 11 + Z 0 22 Z 21 = 0 Z Z0 22 Z 21Z 1 11 = 0 Recall equation (4)
31 Z Z0 22 = 0 (4) = Z 21 Z 1 11 Let = ~ Recall equation (5) Z 0 22 = (5)
32 Z 0 22 = Z 0 22 ~ = Z 0 12 Z 11 + Z 0 22 Z 21 = 0 Z 0 12 = Z0 22 Z 21Z 1 11
33 Z 0 12 Z 12 + Z 0 22 Z 22 = I Z 0 22 Z22 Z 21 Z 1 11 Z 12 = I Z 0 22 Z22 Z 21 Z 1 11 Z 12 {z } = ~ = Z 22 Z 21 Z11 1 Z 12
34 = Z 22 Z 21 Z11 1 Z 12 We obtain the forward part of the solution X f t = Z 21Z 1 11 Xb t + Z 22 Z 21 Z 1 11 Z 12 Zt or express the solution as
35 X f t = F xx b t + F z Z t F x Z 21 Z 1 11 F z Z 22 Z 21 Z11 1 Z 12 vec ( ) = 0 S 22 I T 22 1 vec (R2 )
36 4 The backward part of the solution The upper part of the transformed system is S 11 E t! b t+1 + S 12E t! f t+1 = T 12! f t + T 11! b t + R 1 Z t E t! b t+1 = S 1 11 T 11! b t + S 1 11 T 12! f t S 1 11 S 12E t! f t+1 + S 1 11 R 1Z t Recall that
37 ! t = Z 0 X t (! b t = Z 0 11 Xb t + Z 0 21 Xf t X f t = Z 21Z 1 11 Xb t + Z 22 Z 21 Z 1 11 Z 12 Zt Substitute and rearrange,! b t = Z Z0 21 Z 21Z11 1 X b t + Z21 0 Z22 Z 21 Z11 1 Z 12 Zt
38 Z11 0 Z 11 + Z21 0 Z 21 = I ) Z Z0 21 Z 21Z11 1 = Z 11 1 Z 0 21 Z 22 + Z 0 11 Z 12 = 0 ) Z 0 21 Z 22 = Z 0 11 Z 12 Z 0 21 Z22 Z 21 Z11 1 Z 12 = Z 0 21 Z 22 Z21 0 Z 21Z 1 = Z Z21 0 Z 21Z11 1 = Z11 0 Z Z 12 Z12
39 ! b t = Z 1 11 Xb t Z 1 11 Z 12 Z t Z 1 11 Xb t =! b t + Z 1 11 Z 12 Z t X b t = Z 11! b t + Z 11Z 1 11 Z 12 Z t = Z 11! b t + Z 12 Z t z } { =! f t = Z 11! b t + Z 12! f t X b t are predetermined variables
40 X b t+1 E t X b t+1 = 0 Klein (2000) assumes X b t+1 E t X b t+1 = t+1 X b t = Z 11! b t + Z 12! f t X b t+1 E t X b t+1 = 0
41 Z 11! b t+1 E t! b f t+1 + Z12! t+1 E t! f t+1 = 0! b t+1 = E t! b t+1 Z11 1 Z f 12! t+1 E t! f t+1! f t+1 E t! f t+1 = t+1 *! f t = Z t! b t+1 = E t! b t+1 Z 1 11 Z 12 t+1
42 Recall that E t! b t+1 = S 1 11 T 11! b t + S 1 11 T 12! f t S 1 11 S 12E t! f t+1 + S 1 11 R 1Z t! b t+1 = S 1 11 T 11! b t + S 1 11 (T 12 S 12 + R 1 ) Z t Z 1 11 Z 12 t+1! b t = Z 1 11 Xb t Z 1 11 Z 12 Z t
43 ! b t+1 = Z 1 11 Xb t+1 Z 1 11 Z 12 Z t+1 Z11 1 Xb t+1 Z11 1 Z 12 Z t+1 = S11 1 T 11 Z 1 11 Xb t Z11 1 Z 12 Z t +S11 1 (T 12 S 12 + R 1 ) Z t Z11 1 Z 12 t+1 X b t+1 Z 12 Z t+1 = Z 11 S 1 11 T 11 Z 1 11 Xb t Z 1 11 Z 12 Z t +Z 11 S 1 11 (T 12 S 12 + R 1 ) Z t Z 12 t+1
44 Z t+1 = Z t + t+1 X b t+1 = Z 11S 1 11 T 11Z11 1 Xb t + h Z 11 S 1 11 T12 S 12 T 11 Z 1 11 Z 12 + R 1 + Z12 i Z t The dynamics of the predetermined variables is X b t+1 = M xx b t + M z Z t
45 M x = Z 11 S11 1 T 11Z11 1 M z = Z 11 S 1 11 T12 S 12 T 11 Z 1 11 Z 12 + R 1 + Z12 5 Computing variables of interest N y Y t = N x X t + N z Z t = h N x1 N x2 i " X b t X f t # + N z Z t = N x1 X b t + N x2 X f t + N zz t
46 N y Y t = N x X t + N z Z t = N x1 X b t + N x2 Fx X b t + F z Z t + Nz Z t Y t = Ny 1 (N x1 + N x2 F x ) Xt b + Ny 1 (N z + N x2 F z ) Z t Y t = P x X b t + P z Z t P x N 1 y (N x1 + N x2 F x )
47 P z N 1 y (N z + N x2 F z ) Express the solution of the whole system is state-space form X b t+1 = M xx b t + M z Z t Z t+1 = Z t + t+1
48 X f t = F xx b t + F z Z t Y t = P x X b t + P z Z t 6 When X b t+1 E t X b t+1 = t+1 Klein (2000) assumes X b t+1 E t X b t+1 = t+1
49 X b t = Z 11! b t + Z 12! f t X b t+1 E t X b t+1 = t+1 Z 11! b t+1 E t! b t+1 + Z12! f t+1 E t! f t+1 = t+1! b t+1 = E t! b t+1 Z11 1 Z f 12! t+1 E t! f t+1 + Z 1 11 t+1
50 ! b t+1 = E t! b t+1 Z 1 11 Z 12 t+1 + Z 1 11 t+1 Recall that E t! b t+1 = S 1 11 T 11! b t + S 1 11 T 12! f t S 1 11 S 12E t! f t+1 + S 1 11 R 1Z t! b t+1 = S 1 11 T 11! b t + S 1 11 (T 12 S 12 + R 1 ) Z t Z 1 11 Z 12 t+1 + Z 1 11 t+1
51 ! b t = Z 1 11 Xb t Z 1 11 Z 12 Z t! b t+1 = Z 1 11 Xb t+1 Z 1 11 Z 12 Z t+1 Z11 1 Xb t+1 Z11 1 Z 12 Z t+1 = S11 1 T 11 Z 1 11 Xb t Z11 1 Z 12 Z t +S11 1 (T 12 S 12 + R 1 ) Z t Z11 1 Z 12 t+1 + Z11 1 t+1
52 Xt+1 b Z 12 Z t+1 = Z 11 S11 1 T 11 Z 1 11 Xb t Z11 1 Z 12 Z t +Z 11 S11 1 (T 12 S 12 + R 1 ) Z t Z 12 t+1 + t+1 Z t+1 = Z t + t+1 X b t+1 = Z 11S 1 11 T 11Z11 1 Xb t + h Z 11 S 1 11 T12 S 12 T 11 Z 1 11 Z 12 + R 1 + Z12 i Z t + t+1
53 The dynamics of the predetermined variables is X b t+1 = M xx b t + M z Z t + t+1 M x = Z 11 S11 1 T 11Z11 1 M z = Z 11 S 1 11 T12 S 12 T 11 Z 1 11 Z 12 + R 1 + Z12
54 7 In practice In practice, we can treat exogenous shocks as part of the pre-determined variables. In other words, we rede ne the dynamic system as AE t X t+1 = BX t ; t+1 ; 3 7 5
55 X t = 2 X b t Z t 6 4 Xt f " X b t X f t # X b t " X b t Z t # The solution to the system becomes X b t+1 = M xx b t + t+1 (6)
56 M x = Z 11 S11 1 T 11Z11 1 t+1 = " # ; t+1 X f t = F xx b t (7) F x = Z 21 Z 1 11
57 This is the solution form employed in Klein s MATLAB code. To implement Paul Klein s method, you need 3 MATLAB m les: qzswitch.m; and qzdiv.m. solab.m; These MATLAB m les are available from Paul Klein s website. The 2 MATLAB m les, qzswitch.m and qzdiv.m, are originally written by C. Sims.
58 References [1] Anderson, Gary S. (2008), "Solving Linear Rational Expectations Models: A Horse Race," Computational Economics, 31, pp [2] Klein, Paul (2000), "Using the generalized Schur form to solve a multivariate linear rational expectations model," Journal of Economic Dynamics and Control, 24, pp
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