Exercise Session
Advanced Macroeconomics II
Prof. Michal Kejak
TA: Petr Harasimovic
Spring 2007
Using the Uhlig Toolbox1
­ Handout
This handout derives the systems of linear equations for the examples
No. 0 and 1 in the Uhlig toolbox. These examples are contained in the
files exampl0.m and exampl1.m. Before you start using Uhlig toolbox type
readme in the Matlab command window.2
1 Example 0 ­ Neoclassical Stochastic Growth
Model
The model we want to solve is
max
{Ct,Kt}
t=0
E

t=0
t C1t
- 1
1 - 
,
s.t.
Ct + Kt = ZtK
t-1 + (1 - )Kt-1
log(Zt) = (1 - ) log( Z) +  log(Zt-1) + t,
t  i.i.d. N(0, 2
 )
First order conditions generate the following Euler Equation
Ct
= E C-
t+1(Zt+1K-1
t + (1 - ))
As we want to solve the model for five variables (C, K, Y, R, Z) we need
1
Available at http://www2.wiwi.hu-berlin.de/institute/wpol
2
Don't forget to put the folder with the toolbox on Matlab's search path either by the
command addpath ­ then put it on the top of the search path by the argument -begin or
by copying the toolbox in your working directory.
1
five equations3
, which are
Rt = ZtK-1
t-1 + (1 - ) (1.1)
Ct = Yt + (1 - )Kt-1 - Kt (1.2)
Yt = ZtK
t-1 (1.3)
1 = E
Ct
Ct+1

Rt+1 (1.4)
log Zt =  log Z + (1 - ) log Zt-1 +  (1.5)
Eliminating the time subscripts we can solve for steady state4
K =
 Z
R - 1 + 
1
1-
(1.6)
Y = Z K
(1.7)
C = Y -  K (1.8)
R =  Z K-1
+ (1 - ) (1.9)
1 =  R (1.10)
To use the Uhlig toolbox we have to linearize the system of equations
(1.1)-(1.5) first. Define the lowercase variables as ct  log Ct-log C, similarly
for all the other variables. By construction we have Ct = Cect
and using the
first order Taylor approximation5
it follows Cect
 C(1 + ct).
Using this substitution in equation (1.1) gives
R + Rrt   Z K-1
+ (1 - ) +  Z K-1
(zt + ( - 1)kt-1)
using (1.9)
Rrt  ( R - (1 - ))(zt + ( - 1)kt-1)
which after substituting (1.10) becomes
rt  (1 - (1 - ))zt - (1 - (1 - ))kt-1 (1.11)
3
We could alternatively use smaller number of equations and compute the other variables
later substituting the solution. However, the algorithm can handle more equations
easily and there is thus no reason to reduce the system. It is,in fact, convenient to solve
for all the variables simultaneously.
4 Z is an arbitrary parameter.
5
Recall that ex
 e0
+ e0
(x - 0).
2
The second equation is linearized as follows
C + Cct  Z K
+ Z K
(zt + kt-1) + (1 - ) K + (1 - ) Kkt-1 - Kkt - K
using (1.8)
ct 
Y
C
zt + 
Y
C
+ (1 - )
K
C
kt-1 -
K
C
kt
further, using the fact that
K
C

Y
C
+ (1 - ) =
K
C
R we get
ct  1 +
 K
C
zt +
K
C
kt-1 -
K
C
kt (1.12)
The third equation becomes
Y + Y yt  Z K
+ Z K
(zt + kt-1)
which reduces to
yt  zt + kt-1 (1.13)
The Euler Equation yields
1  E (
Cect
Cect+1
) R(1 + rt+1)
1  E  R(1 + (ct - ct+1) + rt+1 + (ct - ct+1)rt+1)
since the last term in the brackets is negligible and using again (1.10) we get
0  E [(ct - ct+1) + rt+1] (1.14)
The last equation can be transformed as
log( Zezt
) = (1 - ) log Z +  log( Zezt-1
) + t
log Z + zt = log Z + zt-1 + t
zt = zt-1 + t (1.15)
The last step before running the toolbox is to rewrite the system of linear
equations (1.11)-(1.15) in the form
0 = Ax(t) + Bx(t - 1) + Cy(t) + Dz(t)
0 = Et [Fx(t + 1) + Gx(t) + Hx(t - 1) + Jy(t + 1) + Ky(t) + Lz(t + 1) + Mz(t)]
z(t + 1) = Nz(t) + (t + 1)
where x is the vector of endogenous state variables, y is the vector of other
endogenous variables, and z is the vector of exogenous state variables. Obviously,
x = k, y = [c, r, y], and z = z. After feeding the toolbox with the
required matrices we are done.
3
2 Example 1 ­ Hansen's RBC Model
The model we want to solve reads
max
{Ct,Kt,Nt,It}
t=0
E

t=0
t
(log Ct - ANt) ,
s.t.
Ct + It = Yt
Yt = ZtK
t-1N1-
t
Kt = It + (1 - )Kt-1
log(Zt) = (1 - ) log( Z) +  log(Zt-1) + t,
t  i.i.d. N(0, 2
 )
The first order conditions yield
A =
1
Ct
(1 - )ZtK
t-1N-
t
1
Ct
= E
1
Ct+1
(Zt+1K-1
t N1t+1
+ (1 - ))
We have the following seven equations describing the model
Ct = Yt - It (2.1)
Kt = It + (1 - )Kt-1 (2.2)
Yt = ZtK
t-1N1t
(2.3)
A =
1
Ct
(1 - )ZtK
t-1Nt
(2.4)
Rt = ZtK-1
t-1 N1t
+ (1 - ) (2.5)
1 = E
Ct
Ct+1

Rt+1 (2.6)
log Zt =  log Z + (1 - ) log Zt-1 +  (2.7)
4
The steady state values are given by6
I =  K (2.8)
C = Y -  K (2.9)
Y = Z K N1-
(2.10)
R =
1

(2.11)
Y
K
=

R +  - 1
(from (2.10)) (2.12)
K =
Y
K
Z
1
-1
N (from (2.3)) (2.13)
A =
(1 - )Y
C N
(from (2.4) and using (2.10)) (2.14)
Now we have to linearize the system of equations (2.1)-(2.7). By the same
procedure as in the previous example the first three equations simply
Cct  Y yt - Iit (2.15)
Kkt  Iit + (1 - ) Kkt-1 (2.16)
yt  zt + kt-1 + (1 - )nt (2.17)
Using eq. (2.17) the eq. (2.4) becomes
ct  yt - nt (2.18)
The expression for interest rate (eq. (2.5)) can be approximated as
R + Rrt   Z K1- N1+
(1 - ) +  Z K1- N1(zt
+ ( - 1)kt-1 + (1 - )nt)
which after using eqs. (2.11) and (2.17) becomes
Rrt  
Y
K
yt - 
Y
K
kt-1 (2.19)
The last two equations are the same as in the previous example and they
read
0  E [(ct - ct+1) + rt+1] (2.20)
zt = zt-1 + t (2.21)
The vectors x, y and z now become x = k, y = [c, y, n, r, i], and z = z.
6
Following Uhlig's example we take N as a parameter and solve for the steady state
value of A. Usually, however, A is calibrated and N is derived. This would give
N = 1-
A
Y
K
Y
K
-
.
5