Solution techniquesMartin Ellison

University of Warwick and CEPR

Bank of England, December 2005

State-space form

10110 tttt vBXAXEA

Generalised state-space form

Many techniques available to solve this class of models

We use industry standard: Blanchard-Kahn

Alternative state-space form

10101

101

tttt vBAXAAXE

11 tttt BvAXXE

A B

Partitioning of model

11

1

t

t

t

tt

t Bvyw

AyEw

t

tt y

wX

backward-looking variables

predetermined variables

forward-looking variablescontrol variables

Jordan decomposition of A

11

1

t

t

t

tt

t Bvyw

AyEw

1 PPA

eigenvectors 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

tw

ty

Too many stable roots

0w

multiple solutions

equilibrium path not unique

need alternative techniques

tw

ty

Too many unstable roots

0w

no solution

all paths are explosive

transversality conditions violated

tw

ty

Blanchard-Kahn satisfied

0w

one solution

equilibrium path is unique

system has saddle path stability

Rearrangement of Jordan form

111

1

11

t

t

t

tt

t BvPyw

PyEw

P

11

1

1

t

t

t

tt

t Bvyw

PPyEw

R

Partition of model

2

1

00

*22

*21

*12

*111

PPPP

P

11

1

11

t

t

t

tt

t Rvyw

PyEw

P

2

1

RR

R

stable

unstable

Transformed problem

1

1~~

t

tt yw

E

t

t

yw~~

ttt

ttt

yyPwP

wyPwP~

~*22

*21

*12

*11

12

1*22

*21

*12

*11

2

1

1

1*22

*21

*12

*11

00

t

t

t

tt

t vRR

yw

PPPP

yEw

PPPP

12

1

2

1

1

1~~

00

~~

t

t

t

tt

t vRR

yw

yEw

Decoupled equations

12

1

2

1

1

1~~

00

~~

t

t

t

tt

t vRR

yw

yEw

1111~~

ttt vRww

1221~~

tttt vRyyE

Decoupled equations can be solved separately

stable

unstable

Solution strategy

Solve unstable transformed equation ty~

Translate back into original problem

tw~

t

t

yw

Solve stable transformed equation

Solution of unstable equation

As , only stable solution is12 tyt 0~

0~ *22

*21 ttt yPwPy

tj

jtt yyE ~~2

tt wPPy *21

1*22

Solve unstable equation forward to time t+j

Forward-looking (control) variables are function of backward-looking (predetermined) variables

Solution of stable equation

tt

ttt

wPPy

yPwPw*21

1*22

*12

*11

~

tj

jtt wwE ~~1

tt wPPPPw )(~ *21

1*22

*12

*11

Solve stable equation forward to time t+j

As , no problems with instability11

Solution of stable equation

111*

211*

22*12

*11

*21

1*22

*12

*111

1*21

1*22

*12

*111

)(

)()(

t

tt

vRPPPP

wPPPPPPPPw

1111~~

ttt vRww

tt wPPPPw )(~ *21

1*22

*12

*11

11*21

1*22

*12

*11

~)( tt wwPPPP

Future backward-looking (predetermined) variables are function of current backward-looking (predetermined) variables

Full solution

111*

211*

22*12

*11

*21

1*22

*12

*111

1*21

1*22

*12

*111

*21

1*22

)(

)()(

t

tt

tt

vRPPPP

wPPPPPPPPw

wPPy

All variables are function of backward-looking (predetermined) variables: recursive structure

Baseline DSGE model

tt

t

tt

tt vx

ExE

0ˆˆ

11

ˆˆ

01 11

1

11

11 ttt vv

State space form

To make model more interesting, assume policy shocks vt follow an AR(1) process

New state-space form

tv

111

1

1

11

001

ˆˆ

101

00

ˆˆ

0010

001

t

t

t

t

tt

tt

t

xv

ExEv

ttx ,ˆ

One backward-looking variable

Two forward-looking variables

Blanchard-Khan conditions

t

tt

tt

xy

vw

ˆ

Require one stable root and two unstable roots

Partition model according to

Next steps

Exercise to check Blanchard-Kahn conditions numerically in MATLAB

Numerical solution of model

Simulation techniques