Transcript Practical DSGE modelling

Solution techniques
Martin Ellison
University of Warwick and CEPR
Bank of England, December 2005
State-space form
Generalised state-space form
A0 Et X t 1  A1 X t  B0vt 1
Many techniques available to solve this class of
models
We use industry standard: Blanchard-Kahn
Alternative state-space form
1
0
1
1
0
A
B
Et X t 1  A A X t  A B0vt 1
Et X t 1  AXt  Bvt 1
Partitioning of model
 wt 
Xt   
 yt 
backward-looking variables
predetermined variables
forward-looking variables
control variables
 wt 1 
wt 
 E y   A y   Bvt 1
 t t 1 
 t
Jordan decomposition of A
 wt 1 
wt 
 E y   A y   Bvt 1
 t t 1 
 t
A  PP
eigenvectors
1
diagonal matrix
of eigenvalues
Blanchard-Kahn condition
The solution of the rational expectations model
is unique if the number of unstable eigenvectors
of the system is exactly equal to the number of
forward-looking (control) variables.
i.e., number of eigenvalues in Λ greater than 1
in magnitude must be equal to number of
forward-looking variables
Too many stable roots
yt
multiple solutions
equilibrium path
not unique
w0
wt
need alternative
techniques
Too many unstable roots
yt
no solution
all paths are
explosive
w0
wt
transversality
conditions violated
Blanchard-Kahn satisfied
yt
one solution
equilibrium path
is unique
w0
wt
system has saddle
path stability
Rearrangement of Jordan form
w
 wt 1 
1  t 
 E y   PP  y   Bvt 1
 t t 1 
 t
w
 wt 1 
1  t 
1
P 
 P    P Bvt 1

 Et yt 1 
 yt 
1
R
Partition of model
w
 wt 1 
1  t 
P 
 P    Rvt 1

 Et yt 1 
 yt 
1
stable
 1 0 

  
 0 2 
unstable
*

P
P 1   11*
 P21
P12* 

* 
P22 
 R1 
R   
 R2 
Transformed problem
 P11*
 *
P
 21
*
w

0


P  t 1   1
 P11

 *






P  Et yt 1   0  2  P21
P
P
~ 
w
Et  ~t 1 
 yt 1 
~
w
t
 ~y 
 t
*
12
*
22
~
P11* wt  P12* yt  w
t
P* w  P* y  ~
y
21
t
22
t
t
*
12
*
22
wt   R1 
    vt 1
 y
 t   R2 
~  
~  R 
w
0
 w


t 1
t
1
1



vt 1


E ~



~



 t yt 1   0  2  yt   R2 
Decoupled equations
~  
~  R 
w
0
 w


t 1
t
1
1



vt 1


E ~



~



 t yt 1   0  2  yt   R2 
~  w
~ Rv
w
t 1
1 t
1 t 1
stable
Et ~
yt 1  2 ~
yt  R2vt 1
unstable
Decoupled equations can be solved separately
Solution strategy
Solve unstable
transformed equation
~
yt
Solve stable
transformed equation
~
wt
Translate back into
original problem
 wt 
y 
 t
Solution of unstable equation
Solve unstable equation forward to time t+j
j~
~
Et yt  j   2  yt
As
2  1 ,
only stable solution is ~yt  0 t
*
*
~
yt  P21wt  P22 yt  0
yt  P22*1P21* wt
Forward-looking (control) variables are function
of backward-looking (predetermined) variables
Solution of stable equation
Solve stable equation forward to time t+j
j ~
~
Et wt  j  1  wt
As
1  1 ,
no problems with instability
~  P* w  P* y
w
t
11 t
12 t
*1 *
22
21
yt   P P wt
~  (P*  P* P*1P* )w
w
t
11
12 22
21
t
Solution of stable equation
~  w
~ Rv
w
t 1
1 t
1 t 1
~
( P11*  P12* P22*1P21* )wt 1  w
t 1
~  (P*  P* P*1P* )w
w
t
11
12 22
21
t
wt 1  ( P11*  P12* P22*1P21* ) 1 1 ( P11*  P12* P22*1P21* ) wt
* *1 * 1
12 22
21
 ( P  P P P ) R1vt 1
*
11
Future backward-looking (predetermined)
variables are function of current backwardlooking (predetermined) variables
Full solution
yt   P22*1 P21* wt
wt 1  ( P11*  P12* P22*1 P21* ) 1 1 ( P11*  P12* P22*1 P21* ) wt
 ( P11*  P12* P22*1 P21* ) 1 R1vt 1
All variables are function of backward-looking
(predetermined) variables: recursive structure
Baseline DSGE model
State space form
 1  1  Et xˆt 1   1  1  xˆt   1 



   
vt



 0   E ˆ

 ˆ
 0 
1

 t t 1  
 t  

To make model more interesting, assume
policy shocks vt follow an AR(1) process
vt 1  vt   t 1
New state-space form
 1 0 0  vt 1   


  1
1
 0 1   Et xˆt 1    
 0 0   E ˆ   0

 t t 1  
0  vt   1 
   
1
1    xˆt    0  t 1

1  ˆ t   0 
0
One backward-looking variable
vt
Two forward-looking variables
xˆt , ˆt
Blanchard-Khan conditions
Require one stable root and two unstable roots
wt  vt
Partition model according to
 xˆt 
yt   
 ˆ t 
Next steps
Exercise to check Blanchard-Kahn conditions
numerically in MATLAB
Numerical solution of model
Simulation techniques