Solving a Dynamic Equilibrium Model, Exams of Stochastic Processes

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Solving a Dynamic Equilibrium Model

Jes´

us Fern´

andez-Villaverde

University of Pennsylvania

1

Basic RBC

•^

Social Planner’s problem:

max

E

∞X t=

t β

log

ct

ψ

log (

lt

ct

k

t+

αk t

ze t^ lt

− α^

δ

)^ k

,^ t

∀^

t >

zt

ρz

t−

1

ε

,^ t

εt

N

,^ σ

•^

This is a dynamic optimization problem.

2

Equilibrium ConditionsFrom the household problem+

firms’s problem+aggregate conditions:

(^1) ct

β

E

( t

ct

³^ 1 +

α

αk − 1 t^

(e

zt lt

− α^

δ

)´

ψ

ct 1

lt

α

)^ k

α t^

(e

zt^ lt

−α

−l

1 t

ct

k

t+

k

α t^

(e

zt^ l)t

1 −

α^

δ

)^ k

t

zt

ρ

zt

− 1

ε

t

4

Finding a Deterministic Solution

•^

We search for the

fi

rst component of the solution.

•^

If^

σ^

,^ the equilibrium conditions are:

(^1) ct

β

ct

³^ 1 +

α

αk − 1 t^

(^1) l −α t

δ

´

ψ

ct 1

lt

α

)^ k

αl t^

− α t

ct

k

t+

k

αl t^

1 −

α t^

δ )^ k

t

5

Solving the Steady StateSolution:

k^

μ Ω

ϕ

μ

l^

ϕ k

c^

k

y^

αk

(^1) l −α

where

ϕ

³ 1 α

³^1 β

δ ´´

1 1 −

α^ ,

ϕ

1 −

α^

δ and

μ

(^1) ψ

α

)^ ϕ

− α.

7

Linearization I

•^

Loglinearization or linearization?

-^

Advantages and disadvantages

-^

We can linearize and perform later a change of variables.

8

Linearization IIIWe get:^ −

1 c^

(c

−t

c) =

E

( t

1 ( c^

ct

c

α

α

)^ β

y zk^

t+

α (α

β

y 2 k

(k

t+

k

α

α

)^ β

y kl

(l

t+

l)

)

1 c^

(c

−t

c) +

l)

(l

−t

l) = (

α

)^ z

+t

α k

kt

k

)^ −

α l

lt^

l)

(c

−t

c

kt

k

 

y μ^ (

α

)^ z

+t

α k

kt

k

(

− α) l

(l

−t

l)

¶

δ

kt

k

 

zt

ρ

zt

− 1

ε

t

10

Rewriting the System IOr: α

ct

c

E

{t α

ct

c

α

z 2 t+

α

kt

k

α

lt+

l)

(c

−t

c

α

z 5

+t

α k

c^ (

kt

k

α

lt^

l)

(c

−t

c

kt

k

α

z 7

+t

α

kt

k

α

lt^

l)

zt

ρ

zt

− 1

ε

t

11

Rewriting the System IIIAfter some algebra the system is reduced to:

A

(k

t+

k

B

(k

−t

k

C

(l

−t

l) +

Dz

= 0t

E

(t

G

(k

t+

k

H

(k

−t

k

J

(l

t+

l) +

K

(l

−t

l) +

Lz

t+

Mz

) = 0t

E

zt t+

ρ

zt

13

Guess Policy FunctionsWe guess policy functions of the form (

kt

k

P

kt

k

Qz

t^ and

(l

−t

l) =

R

(k

−t

k

Sz

, plug them in and get:t

A

(P

kt

k

Qz

) +t

B

(k

−t

k

C

(R

(k

−t

k

Sz

) +t

Dz

= 0t

G

(P

kt

k

Qz

) +t

H

(k

−t

k

J

R

(P

kt

k

Qz

) +t

SNz

)t

K

(R

(k

−t

k

Sz

) + (t

LN

M

)^ z

= 0t

14

Solving the System II

•^

We have a system of four equations on four unknowns.

-^

To solve it note that

R

(^1) C

AP

B

(^1) C

AP

(^1) C

B

•^

Then:

P^

2

μ

B A

K J

GC JA

¶

P^

KB

HC

JA

a quadratic equation on

P

16

Solving the System III

•^

We have two solutions: P

 

B A

K J

GC JA

à μ

B A

K J

GC JA

¶^2

KB

HC

JA

!^0

.^5  

one stable and another unstable.

-^

If we pick the stable root and

fi

nd

R

(^1) C^

(AP

B

) we have to a

system of two linear equations on two unknowns with solution:

Q

D

(JN

K

CLN

CM

AJN

AK

CG

CJR

S^

ALN

AM

DG

DJR

AJN

AK

CG

CJR

17

General Structure of Linearized SystemGiven

m

states

x

, nt

controls

y

,^ t

and

k

exogenous stochastic processes

zt

, we have:

Ax

+t

Bx

t−

1

Cy

+t

Dz

= 0t

E

(t F x

t+

Gx

+t

Hx

t−

1

Jy

t+

Ky

+t

Lz

t+

Mz

) = 0t

E

zt t+

Nz

t

where

C

is of size

l×

n, l

n

and of rank

n,

that

F

is of size (

m

n

l)

×

n,

and that

N

has only stable eigenvalues.

19

Policy FunctionsWe guess policy functions of the form:

xt

P x

t−

1

Qz

t

yt

Rx

t−

1

Sz

t

where

P, Q, R,

and

S

are matrices such that the computed equilibrium is

stable.

20