Second Order Approximation Methods for
DSGE Models
The University of York
Dr. Nicola Branzoli
11-12 June 2009
Contents
1 Introduction
2
2 Starting Examples
2.1 A Simple Economic Example . . . . . . . . . . . . . . . . . . . .
2.2 A Simple Numerical Example . . . . . . . . . . . . . . . . . . . .
3
3
4
3 Algebraic and Economic Preliminary De…nitions
3.1 Algebraic Tools . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.2 Economic Tools . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5
5
6
4 First Order Approximation
7
5 Higher Order Approximations
5.1 A General Approach . . . . . . . . . . . . . . . . . . . . . . . . .
5.2 Using …rst order methods to solve up to the second order . . . .
5.3 A Simple RBC application . . . . . . . . . . . . . . . . . . . . . .
9
9
11
13
6 Normative Analysis
15
6.1 Policies not a¤ecting the steady state . . . . . . . . . . . . . . . . 15
6.1.1 An RBC application . . . . . . . . . . . . . . . . . . . . . 15
6.2 Policies a¤ecting the steady state . . . . . . . . . . . . . . . . . . 16
6.2.1 An RBC application . . . . . . . . . . . . . . . . . . . . . 16
6.3 Exercises policy not a¤ecting the steady state: an RBC model
with Money and Optimal Policy . . . . . . . . . . . . . . . . . . 18
6.4 Exercise with Money policy a¤ecting the steady state: optimal
level of in‡ation in an RBC model with Money . . . . . . . . . . 20
1
1
Introduction
Numerical Methods and Systems of Partial Di¤erential Equations
A model is a system of partial di¤erential equations describing the evolution of (economic) variables over time (dynamic approach)
Di¤erent models are judged by their implications (variances, covariances,
impulse responses etc.)
"Good" models (models whose implications match empirical facts under
study) are used for normative analysis
Economic Research
Identify interesting empirical facts
#
Write down and solve (derive …rst order conditions) the model that describe
these empirical facts
#
Solve the system of di¤erential equations (global solutions or local solutions
like Taylor expansion)
#
Study the implications of the model (variances and variance decomposition,
impulse response functions)
#
Perform Normative Analysis
Today’s Lecture
We will study numerical methods to implement the last three steps and
(some) theory behind these methods
Our road map for today is:
1. Present two simple examples to get basic ideas
2. Algebraic and economic preliminary de…nitions
3. First Order Taylor Approximation (recap of the generalized Schur decomposition)
4. Higher Order Taylor Approximations (the Schmitt-Grohé-Uribe and Lombardo Sutherland approaches)
5. RBC application
6. Normative Analysis
2
2
Starting Examples
2.1
A Simple Economic Example
A simple economic example: A Phillips Curve and the Quantity Theory of
Money
Suppose that in‡ation and aggregate demand can be described:
=
=
t
4mt
Et t+1 + 4yt "Phillips Curve"
t + 4yt Agg. Demand Function (quantity Theory of Money)
This is a system of two equations in three variables (describing dynamic
behavior rather than "static" values)
4mt is a policy variable.
IF this model were true, what are the implications?
A simple economic example: solve the system
This system implies:
t
=
t
=
t
=
4yt
Et
1+
t+1
+ [4mt
Et
1+
= 4mt
1
X
1+
"
Et
i=0
Et+1
t]
)
t+2
+
i
1+
t
=
1+
1+
#
Et
4mt+1
t+1
+
+
1+
1+
4mt )
4mt ::: )
4mt+i
t
A simple economic example: implication of the model
1. Constant Money Supply: 4mt+i = m 8i
"
#
1
i
X
Et
4mt+i =
t =
1 + i=0
1+
1+
4yt
= 4mt
t
=
1
1+
1
X
i=0
"
i
1+
#
m =
m
2. AR(1) Money Supply 4mt+1 = 4mt + "t+1
( Et "t+1 = 0; unpredictable, from the private sector, component of monetary policy):
"
#
1
i
X
Et
4mt =
m
t =
1 + i=0
1+
1+
4mt+1
=
4mt + "t+1 )
t+1
=
t
+
1+
"t+1
What you "feed in" as a description of the policy is also the description
of the evolution of in‡ation. Match evidence?!
3
1+
m
2.2
A Simple Numerical Example
A simple numerical example: a monetary model with two period stickiness
Consider the following system of equations (this model comes from the overlapping contract model à la Taylor 1979 1980):
Optimal Price Equation
:
xt =
1
xt
2
1
1
+ Et
2
1 xt+1
::: +
The price Level
:
Agg. Demand
Policy Rule
:
:
E t 1 y t + Et
2
2
1
1
pt = xt + xt 1
2
2
y t = mt pt
4mt = g4pt + "t
+ :::
1 yt+1
A simple numerical example: the solution
Note that the policy implies:
mt = gpt
gpt
1
+ mt
Substitute into the Agg Demand to get Et
plicit assumption of rational expectations):
Et
1 yt
=
(1
g) Et
1 pt
1
+ "t
1 yt
gpt
and Et
1
+ mt
1 yt+1 (note
the im-
1
Substituting in the Optimal Price Equation:
(mt
1
gpt
1)
=
h
=
(1
h)
2
(1
xt
1
(1 + h) Et
1 xt
+
(1
h)
2
Et
1 xt+1
g)
2
The Key Question of solving dynamic equations: for what law of motion this
equation is always true (whatever is the value of mt 1 , pt 1 and xt 1 )? Solving
a system of dynamic equations "boils down" to answer this question
A simple numerical example: the solution I
"Guess and Verify" Method
Note that we can (wisely) guess that the solution has the form:
xt = dxt
1
+ b (mt
1
gpt
1)
where d and b are coe¢ cients to be determined. Substituting this equation
we obtain:
(1 h) 2
d xt 1 + (d + 1) b (mt 1 gpt 1 ) + :::
2
(1 h)
::: (1 + h) [dxt 1 + b (mt 1 gpt 1 )] +
xt 1 =
(mt 1 gpt 1 )
2
A simple numerical example: the solution II
4
"Guess and Verify" Method
For this equation to be true:
(1
h)
2
d2
(1
(1 + h) d +
h) (d + 1) b
2
(1
h)
2
(1 + h) b
=
0
=
Which gives us the solution:
p
h
1
p
d=
1+ h
; d=
1 d
1+g
This method is clearly unfeasible and/or ine¢ cient in general! However the
methods we see/saw in this course rely on the same ideas: solving backward or
forward some equations (…rst example) to …nd a low of motion for endogenous
variable (second example).
3
Algebraic and Economic Preliminary De…nitions
3.1
Algebraic Tools
Algebraic De…nitions
De…nition 1 Matrix Pencil: let A0 ; A1 ; :::; Al be nxn complex matrices s.t.
Al 6= [0] than the matrix valued function:
P( )=
l
X
i
Ai
i=0
is called a matrix pencil, written as (A0 ; A1 ; :::; Al ) :
When l=1 we have a linear matrix pencil.
De…nition 2 A linear matrix pencil is said to be regular: if 9 s.t. det (A0 ; A1 ) 6=
0
Corollary 3 The eigenvalues of P ( ) are the set of
0
s s.t. det (A0 ; A1 ) = 0
Algebraic De…nitions
De…nition 4 Unitary Matrix: A square complex matrix Q is said to be unitary
if it satis…es:
QQ = In
where Q is its conjugate transpose (take the transpose and than the complex
conjugate of each element).
When l=1 we have a linear matrix pencil.
5
Corollary 5 Q = Q
1
, i.e. (roughly speaking Q is made of independent lines)
Corollary 6 If Q is real, it is said to be orthogonal (made of orthonormal
vectors)
Algebraic De…nitions
De…nition 7 Generalized Schur Decomposition (GSD): let P ( ) = (A; B) be
a regular pencil matrix. Then 9 unitary matrices Q and Z such that:
QAZ
QBZ
= S is upper triangular
= T is upper triangular
tii
: sii 6= 0
(A; B) =
sii
Corollary 8 A GSD always exists (given regularity)
Corollary 9 If tii = sii = 0 than 9 a linear combination of lines (rows) in each
matrix that gives 0 at the same line (row).
Algebraic Operators
De…nition 10 Let X be a general nxm matrix. Than vec(X) is a mnx1 vector
obtained stacking all the columns of X one under each other.
Corollary 11 Let A; B; C matrices such that the product ABC exists, then
vec (ABC) = (C 0 A) vec (B)
De…nition 12 Let X be a general square nxn matrix. Than vech(X) is a
n(n+1)
x1 vector obtained stacking by columns the lower-triangular part of X.
2
Corollary 13 For any symmetric square matrix X there exist a unique matrix
L such that vec (X) = Lvech (X)
1
Furthermore Lh vec (X) = vech (X) where L+ = (L0 L) L0 and L+ L = I
3.2
Economic Tools
Economic De…nition
De…nition 14 Martingale Di¤ erence Process: A process "t is said to be a martingale di¤ erence process if Et ("t ) = 0 8t
De…nition 15 Backward-looking (state) variables: a variable yt is said to be a
backward-looking (state) variable if:
yt+1
Et [yt+1 ] is martingale di¤ erence process;
y0 is given
6
Economic Tools: obtaining a VAR(1) representation
We are interested in studying a system of expectational di¤erence equation
of the form:
A Et [xt+1 ] = Bxt +
t
nxn
nxmmx1
nx1
where xt is a vector of correlated endogenous variables and t is a set of
exogenous variables.
Most, if not all, models can be written into this form. We now brie‡y discuss
how to accomplish this task.
More than one lag
If the model has more than one lag:
Et [xt+1 ] = Bxt + Cxt
1
introduce an auxiliary variable to write:
Et [xt+1 ] = Bxt + C x
~t
x
~t+1 = xt
to obtain an equivalent VAR(1) system.
More than one expectation
If the model has more than one lag:
Et [xt+1 ] = Bxt + CEt
1
[xt ]
introduce an auxiliary variable to write:
Et [xt+1 ] = Bxt + C x
~t
xt+1 = x
~ t + "t
where E ("t ) = 0 to accommodate the hypotesis of rational expectation.
This two cases are enough to handle probably the 99% of macro models
nowadays used...
4
First Order Approximation
Solving a System of Partial Di¤erential Equations up to the …rst order: The
Generalized Schur Decomposition
The Problem
With a small loss of generality we are interested in studying a system like:
A Et [xt+1 ] = Bxt +
nxn
nx1
zt
nxmmx1
It’s not uncommon that the matrix A is not invertible, so that pre-multiplying
by A 1 and apply the Jordan decomposition is unfeasible
The Generalized Schur Decomposition: a quick review
Transformation of Variables
7
Re-order the variables in the system, states must come …rst:
x1t
x2t
xt
label all non-states variables controls.
Consider the following transformation of variables:
st
= ZH
ut
| {z }
x1t
x2t
yt
where Z H is the conjugate transpose of the Z matrix in the GSD.
The Generalized Schur Decomposition: a quick review
Transformation of Variables
Note that:
AEt [xt+1 ]
=
)
)
)
Bxt + t ) Q 1 SZ 1 Et [xt+1 ] = Q 1 T Z
Q 1 SZ H Et [xt+1 ] = Q 1 T Z H xt + zt
Q 1 SEt [yt+1 ] = Q 1 T yt + zt
SEt [yt+1 ] = T yt + Q zt
1
xt + zt
Partitioned as:
S11
0
S21
S22
st+1
ut+1
Et
=
T11
0
T21
T22
st
ut
+
Q1
Q2
zt
This system is clearly equivalent to the original one.
The Generalized Schur Decomposition: a quick review
Solving for controls
All controls are solved forward:
S22 Et [ut+1 ] = T22 ut + Q2 zt
The solution is:
ut =
T22
1
X
T221 S22
k
Q2 Et [zt+k ]
k=0
that can be easily solved depending on the structure of the endogenous
processes.
The Generalized Schur Decomposition: a quick review
Solving for states
Solved for controls, we can treat present controls and their expectations as
given and solve backward for states:
S22 Et [ut+1 ] = T22 ut + Q2 zt
8
The solution is:
ut =
T22
1
X
T221 S22
k
Q2 Et
t+k
k=0
that can be easily solved depending on the structure of the endogenous
processes.
5
Higher Order Approximations
5.1
A General Approach
Higher order approximations: why we need them?
First order approximation techniques are useful when studying impulse response functions (dynamic response to exogenous shocks)
However they have been proved to be ill suited to develop welfare analysis: second order terms (related to the variances of endogenous variables) are
generally important for the equilibrium welfare functions (Kim et Kim (JIE
2005), see Woodford (2002) for a discussion of the su¢ ciency of the …rst order
approximation for welfare analysis)
A general approach to approximation methods.
For simplicity let me change the notation: x are states in period t, x0 states
in period t+1, y are controls, y 0 controls at period t+1.
Clearly we can write the system characterizing the model as:
Et [f (y 0 ; y; x0 ; x)] = 0
(1)
nx1
higher order approximations are based on this simple representation
A general approach to approximation methods
De…ne implicitly:
y
x0
= g (x; )
= h (x; ) + ' "t
(2)
(3)
where is a variance (scalar) parameter.
Than an n-th order Taylor approximation of these functions around the nonstochastic steady-state are:
2
3
i
1
n
i
X
1 4 @ gx x ; 0
@
g
(x;
0)
i
i5
(x x) +
y = g (x; ) = g (x; 0) +
ix
ix
i!
@
@
i=1
x0
= h (x; ) + ' "t = h (x; 0) +
n
X
1 @ i hx (x; 0)
(x
i!
@ix
i=1
9
i
x) +
@ i h (x; 0)
@ix
i
Objective: …nd higher order approximations of g and h around x, y and
= 0;
Observations:
By de…nition
y = g (x; 0)
x = h (x; 0)
and:
Et [f (y 0 ; y; x0 ; x)] =
0 for ANY value of y 0 ; y; x0 ; x and
nx1
SO
@f
i
@ x@ j y@ s
=
0 for 8i; j; s; t; x; y
A direct Taylor approximation: state variables
Substitute eqs.(2)-(3) in (1):
Et f (g (h (x; ) + ' "t ; ) ; g (x; ) ; h (x; ) + ' "t ; x) = 0
(4)
nx1
and take the total derivative w.r.t. x :
@f
@x
=
0)
@f
@g
@h
@f
@g
+
+ :::
@y 0 j(x;0) @x0 j(x;0) @x j(x;0) @y j(x;0) @x j(x;0)
ny xnx
nxny
+
nx xnx
nxny
ny xnx
@f
@h
@f
+
= 0
0
@x j(x;0) @x j(x;0) @x j(x;0) nxnx
nxnx
nx xnx
nxnx
System of nxnx equations in nxnx variables.
A direct Taylor approximation: stochastic parameter
Take the total derivative w.r.t.
@f
@
=
:
0)
@f
@g
@h
@f
@g
+
+ :::
@y 0 j(x;0) @x0 j(x;0) @ j(x;0) @y 0 j(x;0) @ j(x;0)
+
@f
@g
@f
@h
+ 0
=0
@y j(x;0) @ j(x;0) @x j(x;0) @ j(x;0)
@g
@h
@ j(x;0) and @ j(x;0)
and @@g j(x;0) = [0]
this is a linear homogenous equation in
solution exist it must be
@h
@ j(x;0)
= [0]
10
so if a unique
the same method shows that
@h
@ @x1t j(x;0)
= [0] and
@g
@ @x1t j(x;0)
= [0]
A direct Second Order Taylor approximation: state variables
[0]
= fy0 y0 gx hx + fy0 y gx + fy0 x0 hx + fy0 x gx hx + fy0 gxx hx hx + fy0 gx hxx + ::: =
(fyy0 gx hx + fyy gx + fyx0 hx + fyx ) gx + fy gxx + (fx0 y0 gx hx + fx0 y gx + fx0 x0 hx + fx0 x ) hx + fx0 hxx
fxy0 gx hx + fxy gx + fxx0 hx + fxx
General Results
Theorem 16 Consider the model whose equations can be written as (1), then:
@h
@ j(x;0)
= [0]
@g
@ j(x;0)
= [0]
@h
@ @x j(x;0)
= [0]
@g
@ @x j(x;0)
= [0]
Corollary 17 The low of motion of the linearized system is independent
of the volatility of the shocks
A second order approximation to the policy function of a stochastic model
di¤ er from the non-stochastic version only by a constant parameter of the
control vector.
5.2
Using …rst order methods to solve up to the second
order
Finding second order solutions using …rst order methods
A second-order approximation of model can be written as:
A1 Et [xt+1 ]
nxn
nx1
= A2 xt + A3 zt +
nxnnx1
nxmmx1
A4
nx
t
n(n+1) n(n+1)
x1
2
2
+
A5
nx
where:
t
= vech (~
xt x
~0t )
where:
x
~t
3
zt
4 x1t 5
x2t
2
Finding second order solutions using …rst order methods
11
Et [
t+1 ]
n(n+1) n(n+1)
x1
2
2
Suppose we have applied the Schur decomposition to solve the …rst order
approximation obtaining the state-space form:
1
f xt+1
2
f xt
= F1 zt + F2f x1t
= P1f x1t + P2f xt
and we can compactly write the solution as:
2
3
zt
zt
4 f x1t 5 =
1
f xt
2
f xt
zt 1
1
f xt 1
zt
=
1
f xt
| {z }
+ "t
st
where f x1t is the …rst order approximation of the states, and are appropriate matrices constructed using F1 ; F2 ; P1 ; P2 and I
Finding second order solutions using …rst order methods
Since we are already dealing with an approximation, we can substitute x1t
with its …rst order approximation.
Note that:
!
0
z
z
t
t
0
xt x
~0t ) = Lvec
t = vech (~
x1t
x1t
!
0
zt
zt
= L(
) vec
x1t
x1t
!
0
z
z
t
t
= L(
) Lh vech
1
1
f xt
f xt
= RVt
Finding second order solutions using …rst order methods
Second:
Vt
0
1 + "t ]
st 1 s0t 1 0 + st 1 0 "0t + "t s0t 1 0 + "t 0 "0t
st 1 s0t 1 0 + Lvec ( st 1 "0t 0 ) + Lvec "t s0t 1 0
) Lh vech st 1 s0t 1 + L (
) Lh vech ("t "0t ) +
) Lh vech (st 1 "0t ) + L (
) Lh vech "t s0t 1
= vech (st s0t ) = vech [ st
= vech
= Lvec
= L(
1
+ "t ] [ st
+ Lvec ( "t "0t 0 )
+L (
= ~ Vt 1 + ~ ~"t + ~ ~t
where ~ = L (
vech (X)
) Lh + L (
) Lh P and P is such that P vech (X 0 ) =
12
Therefore the second order elements of the system are themselves an independent system.
Finding second order solutions using …rst order methods
Hence we have now an augmented system:
A1 Et [xt+1 ] = A2 xt + A3 zt + G t + H
Vt = ~ Vt 1 + ~ ~"t + ~ ~t
zt = N z t 1 + " t
where G = A4 R + A5 R ~ H = A5 R ~ and = Et ~"t :
We can than solve backward for Vt and than treating it and H
variables in the main representation.
5.3
as exogenous
A Simple RBC application
Finding second order solutions: A Simple RBC application
We now consider a simple application. The Ramsey model is:
max E0
1
X
C1
t
t=0
s:t:Kt+1
It + Ct
log At+1
1
= (1
) Kt + It
= At Kt
=
log At + "t+1
we assume "t N (0; 1) and, for simplicity, = 1 and = 0:
Finding second order solutions: A Simple RBC application
FOC’s imply:
Ct
Kt+1
log At+1
=
Et Ct+1 At+1 Kt+11
= At Kt
Ct
= "t+1
which implies:
log Ct = log ( )
Et [log Ct+1 ] + Et [log At+1 ] + (
log (Kt+1 + Ct ) = log [At Kt ]
log At+1 = "t+1
1) Et [log Kt+1 ]
So:
Et f x2t+1 ; x2t ; x1t+1 ; x1t
Et
log Ct + log (
nx1
=
0
13
)
Et [log Ct+1 ] + Et [log At+1 ] + (
log (Kt+1 + Ct ) log [At Kt ]
1) Et [log Kt+1 ]
x1t = Kt ; x2t = Ct and zt = at
Finding second order solutions: taking the …rst order approximation
Recall that in this case we have one state and one control.
For the SGU Method:
fy0 gh + fy g + fx0 h + fx
0
gh +
(
g+
Css
Kss +Css
1)
Kss
Kss +Css
h+
0
=
[0]
=
[0]
Note that we are looking for the roots of this system of equations (see example at the beginning of the class).
For the GSD:
Et c^t+1 + (
1) k^t+1
Kss
k^t+1
Kss + Css
=
=
c^t
Css
c^t + k^t + a
^t
Kss + Css
In Matrix notation:
(
1)
Kss
Kss +Css
0
k^t+1
Et c^t+1
k^t
c^t
0
+
Css
Kss +Css
+
0
1
a
^t = [0]
Finding second order solutions: the second order approximation for SGU
Method
For the second order we have:
[0]
= fy0 y0 gx hx + fy0 y gx + fy0 x0 hx + fy0 x gx hx + fy0 gxx hx hx + fy0 gx hxx + ::: =
(fyy0 gx hx + fyy gx + fyx0 hx + fyx ) gx + fy gxx + (fx0 y0 gx hx + fx0 y gx + fx0 x0 hx + fx0 x ) hx + fx0 hxx
fxy0 gx hx + fxy gx + fxx0 hx + fxx
which is a system of two equations in two unknowns.
Finding second order solutions: the second order approximation for the Lombardo Sutherland Method
For the second order we have:
2
1
^t+1 + Et c^t+1 + (
1) k^t+1
1) k^t+1 + Et a
2
2
Css
1
Css
1 2
c^t + k^t + a
^t
c^2t +
k^2 + a
^ + k^t a
^t
Kss + Css
2 Kss + Css
2 t
2 t
Et c^t+1 + (
In matrix notation:
A1
k^t+1
Et c^t+1
= A2
k^t
c^t
+ A3 a
^t + A4
14
t
+ A5 Et [
t+1 ]
2
=
=
c^t +
2
c^2t
Kss
1
Kss
k^t+1 +
k^2
Kss + Css
2 Kss + Css t+1
where:
A1
(1
=
A5
=
"
)
Kss
Kss +Css
1
2
0
(
0
1)
0
; A2 =
0
1)2
2
Kss
1
2 Kss +Css
Css
Kss +Css
(
(
0
1)
0
; A3 =
2
2
0
0
1
; A4 =
#
"
0
1
2
trough which we can construct the auxiliary system.
6
Normative Analysis
Normative Analysis
We distinguish between two cases:
policies NOT a¤ecting the steady state, stochastic environment
policies a¤ecting the steady state, non stochastic environment
The two cases can be combined in a conceptually easy (tough not always
easy to implement) way.
6.1
6.1.1
Policies not a¤ecting the steady state
An RBC application
Policies not a¤ecting the steady state
The standard Ramsey model augmented with consumption taxes is:
"1
#
X c1
t t
max E0
ct ;kt+1 ;it
1
t=0
s:t:kt+1
(1 + t ) ct + ii
yt
at+1
= (1
) kt + it
= at kt
= at kt
=
at + "t
Consider an exogenous set of policies:
^t =
k Et
[^
yt+1 ]
where the hat denote deviations from equilibrium level.
Taxes on consumption are set as a (linear) function function of expected
level of production.
Policies not a¤ecting the steady state
15
0
0
2
2
0 0
0 0
2
2
Css
1
2 Kss +Css
#
FOC’s imply the following system of 3 equations in 3 variables:
ct
kt+1
^t
)
Et ct+1 at+1 kt+11 + (1
(1
) kt + at kt
(1 + t ) ct
yt+1 ]
k Et [^
=
=
=
Plus:
ct
ii
= (1 +
= at kt
t)
t
(1 +
t ) ct
Policies not a¤ecting the steady state
This is the basic RBC model with an additional variable and an additional
equation.
Type of questions we can address:
What is the optimal taxation policy in this class?
What is the optimal taxation within linear policies? i.e. would it be better
to respond to past or current level of production?
What are the implication of the policy that maximizes private welfare?
Tomorrow we will discuss how to code and solve this problem.
6.2
6.2.1
Policies a¤ecting the steady state
An RBC application
Policies a¤ecting the steady state: the problem
Consider the following problem:
"
1
X
c1t
t
max
ct ;kt+1 ;it ; t
1
t=0
s:t:kt+1
ct + it
xt
log at+1
= (1
= at kt x1t
=
t kt
=
log at
#
t ) kt
+ it
That is the government can impose wealth taxes to provide services that
enter in the production function.
In this problem the level of taxes a¤ects the steady state.
Policies a¤ecting the steady state: analytical solution
16
Write the lagragian:
8
2
1
(1
< c1
X
t
t
L=
+ t4
:1
1x2
t=0
1
+ at kt1 ( t )
1
( t)
ct
log at+1 = log at
t ) kt
at kt1
ct
kt+1
it
FOC’s imply:
ct
=
1
t
=
kt
=
1
t
n
1
t+1
h
(1
) at+1 (kt+1 )
(
1
t+1 )
+1
(1
) at kt1 ( t )
+ the constraints
t+1
39
=
5
;
io
Policies a¤ecting the steady state: analytical solution
In steady state the FOC’s imply:
c
1
=
k
=
1
=
1
(1
(1
)
(1
c =
1
)k
(
1
)
( )
)k + k
1
1
( )
which implies:
1 = (1
)
1
(1
(1
)
)
which implies:
=
(1
) [1
[ ]
(1
)]
Policies a¤ecting the steady state: numerical solution
This problem is simple, so we can use it to test a numerical approach.
Consider the problem written in matrix notation:
L=
1
X
t
Ot +
t=0
t Lt
1xnc nc x1
we have two sets of …rst order conditions, those w.r.t. endogenous variables
and those w.r.t. the lagrange multipliers.
The latter are the constraints, that we can use to solve for the steady state,
given a value of the policy variable.
Policies a¤ecting the steady state
First order conditions w.r.t endogenous variables imply:
@Ot
+
@xt
t
@Lt
= [0]
@xt
17
Note that this is a system of n variables in nc unknowns (the lagrange multipliers) and therefore can not be solved exactly.
t
Furthermore @L
@xt is not a square matrix and therefore can not be inverted.
However we can, given a value of the policy parameter that determine the
value of endogenous variables, solve this system by OLS (i.e. the values of the
lagrange multipliers that minimize the sum of squared residuals).
Than we can check if this is a solution.
Tomorrow we will see how to implement this.
6.3
Exercises policy not a¤ecting the steady state: an
RBC model with Money and Optimal Policy
1
X
max
ct ;kt+1 ;it ;mt ;bt
c1t
t
+ (1
1
t=0
s:t:kt+1
Bt
Mt
ct + it +
+
Pt
Pt
log at+1
=
(1
) kt + it
= at kt + (1 + Rt )
=
1
1
) m1t
Bt 1
Mt 1
+
Pt
Pt
log at
which implies:
ct + kt+1 + bt + mt = at kt + (1
) kt + (1 + Rt )
bt
1
+
mt
t
Lagrangian and First Order Conditions are:
8
1
1
<
[ c1t +(1 )m1t ] 1
X
+
t
1
h
max
mt 1
bt 1
: +
ct ;kt+1 ;mt ;bt
) kt + (1 + Rt ) t + t
t=0
t at kt + (1
c1t
t
1
1
) m1t
+ (1
=
(1
at+1 kt+11 + (1
t+1
t
1
) ct
1
t
=
ct
kt+1
bt
mt
i
9
=
;
t
)
1 + Rt+1
=
t+1
t+1
c1t
+ (1
1
1
) m1t
1
(1
) (1
) mt
=
t+1
t
t+1
Couple of notes:
This is a system of 7 variables ct ; mt ; kt ; bt ; t ; at ; Rt in 4 equations, the
government budget constraint, the exogenous low of motion and the policy rule:
c1t
+ (1
) m1t
1
1
1
ct
=
c1t+1 + (1
18
) m1t+1
1
1
1
ct+1
at+1 kt+11 + (1
)
c1t
+ (1
c1t
+ (1
1
1
) m1t
1
1
1
) m1t
1
ct
c1t+1 + (1
=
(1
) (1
1
1
) m1t+1
) mt
c1t
=
1
ct+1
1 + Rt+1
t
) m1t
+ (1
c1t+1 + (1
1
1
) m1t+1
t+1
ct + kt+1 + bt + mt = at kt + (1
) kt + (1 + Rt )
bt
1
+
t
log at+1 = log at
B t B t 1 + Mt Mt
Pt bt + Pt mt Pt 1 mt
mt
t b t + t mt
= Rt 1 B t
= (1 + Rt
= (1 + Rt
1
1
1
1
1 ) Pt 1 bt 1
1 ) bt 1
Steady state: a = 1 than
1
k=
(1
(1
R=
1
=
ct
) mt
"
m =
|
b
=
b
=
1
1
)
1
ct+1
(1
{z
)
#
AA
t+1
1
c
}
m
1+R
1
1
|
1+R
{z
BB
19
(1
)
c
}
mt
1
t
1
1
1
(1
1
(1
) ct +
) ct+1
c + BBc + AAc = kt
1 + AA 1
1
1+R
+ 1
k + (1 + R)
BB c = kt
c+
AA
c
k
k
c =
BB
1+A 1
1
k
+ 1
1+R
B
Than we are ready to solve the model.
6.4
Exercise with Money policy a¤ecting the steady state:
optimal level of in‡ation in an RBC model with Money
Consider the same model with money, but with instantaneous utility function:
log ct + mt exp (
mt )
Lagrangian and First Order Conditions are:
(
1
X
log ct + mt exp ( mt ) +
h
t
max
+ t at kt + (1
) kt + (1 + Rt ) bt t1 + mt t 1 ct
ct ;kt+1 ;mt ;bt
t=0
1
=
ct
t
=
at+1 kt+11 + (1
t+1
t
t
)
1 + Rt+1
=
t+1
t+1
(1
mt ) exp (
mt ) =
1
+
ct
The last four equation constitute our model.
In steady state:
1
(1
)
k=
c (1
m) exp (
20
t+1
1
1
R=
1
m) = 1
1
ct+1
kt+1
bt
mt
i
)
b
=
b
=
b
=
b
=
1
1+R
1
m
m
1
m
1
|
{z
BB
}
(
1)
m
(1
)
| {z }
BB
k
k + (1 + R)
b
+
m
= c+b+m
k
k
= c+ 1
k
k
= c+ 1
k
k
= c+ 1
k
k
= c+
k
k
= c
k
k
= c
(
(1 + R)
1
1
b+ 1
(
(1
(
(1
1
b+ 1
1
m
m
1)
m+ 1
)
1)
1
m+ 1
)
1)
1
m+ 1
m
1
1
m
m
Since steady state consumption is independent of money, we need to set an
in‡ation rate to maximize utility derived by money holdings. Than maximization of mt exp ( mt ) w.r.t mt implies m = 1 ;which is true if = :
We are ready to solve the model and check the numerical solution.
21