Second Order Approximation Methods forDSGE Models

The University of YorkDr. Nicola Branzoli

11-12 June 2009

Contents

1 Introduction 2

2 Starting Examples 32.1 A Simple Economic Example . . . . . . . . . . . . . . . . . . . . 32.2 A Simple Numerical Example . . . . . . . . . . . . . . . . . . . . 4

3 Algebraic and Economic Preliminary De�nitions 53.1 Algebraic Tools . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53.2 Economic Tools . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

4 First Order Approximation 7

5 Higher Order Approximations 95.1 A General Approach . . . . . . . . . . . . . . . . . . . . . . . . . 95.2 Using �rst order methods to solve up to the second order . . . . 115.3 A Simple RBC application . . . . . . . . . . . . . . . . . . . . . . 13

6 Normative Analysis 156.1 Policies not a¤ecting the steady state . . . . . . . . . . . . . . . . 15

6.1.1 An RBC application . . . . . . . . . . . . . . . . . . . . . 156.2 Policies a¤ecting the steady state . . . . . . . . . . . . . . . . . . 16

6.2.1 An RBC application . . . . . . . . . . . . . . . . . . . . . 166.3 Exercises policy not a¤ecting the steady state: an RBC model

with Money and Optimal Policy . . . . . . . . . . . . . . . . . . 186.4 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 evolu-tion 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 understudy) are used for normative analysis

Economic Research

Identify interesting empirical facts#

Write down and solve (derive �rst order conditions) the model that describethese 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 LectureWe will study numerical methods to implement the last three steps and

(some) theory behind these methodsOur 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 decom-position)

4. Higher Order Taylor Approximations (the Schmitt-Grohé-Uribe and Lom-bardo 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 ofMoneySuppose that in�ation and aggregate demand can be described:

�t = �Et�t+1 + �4yt "Phillips Curve"4mt = �t +4yt Agg. Demand Function (quantity Theory of Money)

This is a system of two equations in three variables (describing dynamicbehavior rather than "static" values)4mt is a policy variable.

IF this model were true, what are the implications?A simple economic example: solve the systemThis system implies:

�t = �Et�t+1 + � [4mt � �t]) �t =�

1 + �Et�t+1 +

�

1 + �4mt )

�t =�

1 + �Et

��

1 + �Et+1�t+2 +

�

1 + �4mt+1

�+

�

1 + �4mt:::)

�t =�

1 + �

1Xi=0

Et

"��

1 + �

�i4mt+i

#4yt = 4mt � �tA simple economic example: implication of the model

1. Constant Money Supply: 4mt+i = m 8i

�t =�

1 + �

1Xi=0

Et

"��

1 + �

�i4mt+i

#=

�

1 + �

1Xi=0

"��

1 + �

�im

#=

�

1 + �� �m

4yt = 4mt � �t =1� �

1 + �� �m

2. AR(1) Money Supply 4mt+1 = �4mt + "t+1

( Et"t+1 = 0;�unpredictable, from the private sector, component of mon-etary policy):

�t =�

1 + �

1Xi=0

Et

"���

1 + �

�i4mt

#=

�

1 + �� ��m

4mt+1 = �4mt + "t+1 ) �t+1 = ��t +1 + �� ��

�"t+1

What you "feed in" as a description of the policy is also the descriptionof the evolution of in�ation. Match evidence?!

3

2.2 A Simple Numerical Example

A simple numerical example: a monetary model with two period stickinessConsider the following system of equations (this model comes from the over-

lapping contract model à la Taylor 1979 1980):

Optimal Price Equation : xt =1

2xt�1 +

1

2Et�1xt+1 + :::

:::+�

2Et�1yt +

�

2Et�1yt+1

The price Level : pt =1

2xt +

1

2xt�1

Agg. Demand : yt = mt � ptPolicy Rule : 4mt = g4pt + "t

A simple numerical example: the solutionNote that the policy implies:

mt = gpt � gpt�1 +mt�1 + "t

Substitute into the Agg Demand to get Et�1yt and Et�1yt+1(note the im-plicit assumption of rational expectations):

Et�1yt = � (1� g)Et�1pt � gpt�1 +mt�1

Substituting in the Optimal Price Equation:

�� (mt�1 � gpt�1) =(1� h)2

xt�1 � (1 + h)Et�1xt +(1� h)2

Et�1xt+1

h =(1� g)�

2

The Key Question of solving dynamic equations: for what law of motion thisequation is always true (whatever is the value of mt�1, pt�1 and xt�1)? Solvinga system of dynamic equations "boils down" to answer this questionA simple numerical example: the solution I"Guess and Verify" MethodNote 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 equationwe obtain:

(1� h)2

�d2xt�1 + (d+ 1) b (mt�1 � gpt�1)

�+ :::

:::� (1 + h) [dxt�1 + b (mt�1 � gpt�1)] +(1� h)2

xt�1 = �� (mt�1 � gpt�1)

A simple numerical example: the solution II

4

"Guess and Verify" MethodFor this equation to be true:

(1� h)2

d2 � (1 + h) d+ (1� h)2

= 0

(1� h) (d+ 1) b2

� (1 + h) b = ��

Which gives us the solution:

d =1�

ph

1 +ph

; d =1� d1 + g

This method is clearly unfeasible and/or ine¢ cient in general! However themethods we see/saw in this course rely on the same ideas: solving backward orforward some equations (�rst example) to �nd a low of motion for endogenousvariable (second example).

3 Algebraic and Economic Preliminary De�ni-tions

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 (�) =lXi=0

�iAi

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 �0s s.t. det (A0; A1) = 0

Algebraic De�nitions

De�nition 4 Unitary Matrix: A square complex matrix Q is said to be unitaryif it satis�es:

QQ� = In

where Q� is its conjugate transpose (take the transpose and than the complexconjugate 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 orthonormalvectors)

Algebraic De�nitions

De�nition 7 Generalized Schur Decomposition (GSD): let P (�) = (A;B) bea regular pencil matrix. Then 9 unitary matrices Q and Z such that:

QAZ = S is upper triangular

QBZ = T is upper triangular

� (A;B) =

�tiisii: sii 6= 0

�Corollary 8 A GSD always exists (given regularity)

Corollary 9 If tii = sii = 0 than 9 a linear combination of lines (rows) in eachmatrix 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 vectorobtained stacking all the columns of X one under each other.

Corollary 11 Let A;B;C matrices such that the product ABC exists, thenvec (ABC) = (C 0 A) vec (B)

De�nition 12 Let X be a general square nxn matrix. Than vech(X) is an(n+1)

2 x1 vector obtained stacking by columns the lower-triangular part of X.

Corollary 13 For any symmetric square matrix X there exist a unique matrixL such that vec (X) = Lvech (X)Furthermore Lhvec (X) = vech (X) where L+ = (L0L)�1 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 mar-tingale di¤erence process if Et ("t) = 0 8t

De�nition 15 Backward-looking (state) variables: a variable yt is said to be abackward-looking (state) variable if:

� yt+1 � Et [yt+1] is martingale di¤erence process;

� y0 is given

6

Economic Tools: obtaining a VAR(1) representationWe are interested in studying a system of expectational di¤erence equation

of the form:AnxnEt[xt+1]

nx1= Bxt + �

nxm�tmx1

where xt is a vector of correlated endogenous variables and �t is a set ofexogenous 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 lagIf the model has more than one lag:

Et [xt+1] = Bxt + Cxt�1

introduce an auxiliary variable to write:

Et [xt+1] = Bxt + C~xt

~xt+1 = xt

to obtain an equivalent VAR(1) system.More than one expectationIf 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~xt

xt+1 = ~xt + "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: TheGeneralized Schur DecompositionThe ProblemWith a small loss of generality we are interested in studying a system like:

AnxnEt[xt+1]

nx1= Bxt + �

nxmztmx1

It�s not uncommon that the matrixA is not invertible, so that pre-multiplyingby A�1 and apply the Jordan decomposition is unfeasibleThe Generalized Schur Decomposition: a quick reviewTransformation of Variables

7

Re-order the variables in the system, states must come �rst:

xt ��x1tx2t

�label all non-states variables controls.Consider the following transformation of variables:�

stut

�| {z }

yt

= ZH�x1tx2t

�

where ZH is the conjugate transpose of the Z matrix in the GSD.The Generalized Schur Decomposition: a quick reviewTransformation of VariablesNote that:

AEt [xt+1] = Bxt + ��t ) Q�1SZ�1Et [xt+1] = Q�1TZ�1xt + �zt

) Q�1SZHEt [xt+1] = Q�1TZHxt + �zt

) Q�1SEt [yt+1] = Q�1Tyt + �zt

) SEt [yt+1] = Tyt +Q�zt

Partitioned as:�S11 S210 S22

�Et

�st+1ut+1

�=

�T11 T210 T22

� �stut

�+

�Q1Q2

��zt

This system is clearly equivalent to the original one.The Generalized Schur Decomposition: a quick reviewSolving for controlsAll controls are solved forward:

S22Et [ut+1] = T22ut +Q2�zt

The solution is:

ut = �T221Xk=0

�T�122 S22

�kQ2�Et [zt+k]

that can be easily solved depending on the structure of the endogenousprocesses.The Generalized Schur Decomposition: a quick reviewSolving for statesSolved for controls, we can treat present controls and their expectations as

given and solve backward for states:

S22Et [ut+1] = T22ut +Q2�zt

8

The solution is:

ut = �T221Xk=0

�T�122 S22

�kQ2�Et

��t+k

�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 re-

sponse functions (dynamic response to exogenous shocks)However they have been proved to be ill suited to develop welfare analy-

sis: second order terms (related to the variances of endogenous variables) aregenerally important for the equilibrium welfare functions (Kim et Kim (JIE2005), see Woodford (2002) for a discussion of the su¢ ciency of the �rst orderapproximation 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, y0 controls at period t+1.Clearly we can write the system characterizing the model as:

Et [f (y0; y; x0; x)]nx1

= 0 (1)

higher order approximations are based on this simple representationA general approach to approximation methodsDe�ne implicitly:

y = g (x; �) (2)

x0 = h (x; �) + '�"t (3)

where � is a variance (scalar) parameter.Than an n-th order Taylor approximation of these functions around the non-

stochastic steady-state are:

y = g (x; �) = g (x; 0) +nXi=1

1

i!

24@igx�x1; 0

�@ix

(x� x)i + @ig� (x; 0)

@ix�i

35x0 = h (x; �) + '�"t = h (x; 0) +

nXi=1

1

i!

�@ihx (x; 0)

@ix(x� x)i + @

ih� (x; 0)

@ix�i�

9

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 (y0; y; x0; x)]nx1

= 0 for ANY value of y0; y; x0; x and �

SO@f

@ix@jy@s�= 0 for 8i; j; s; t; x; y

A direct Taylor approximation: state variables

� Substitute eqs.(2)-(3) in (1):

Etf (g (h (x; �) + '�"t; �) ; g (x; �) ; h (x; �) + '�"t; x)nx1

= 0 (4)

� and take the total derivative w.r.t. x :@f

@x= 0)

@f

@y0 j(�x;0)nxny

@g

@x0 j(�x;0)nyxnx

@h

@x j(�x;0)nxxnx

+@f

@y j(�x;0)nxny

@g

@x j(�x;0)nyxnx

+ :::

+@f

@x0 j(�x;0)nxnx

@h

@x j(�x;0)nxxnx

+@f

@x j(�x;0)nxnx

= 0nxnx

� System of nxnx equations in nxnx variables.

A direct Taylor approximation: stochastic parameter

� Take the total derivative w.r.t. � :@f

@�= 0)

@f

@y0 j(�x;0)

@g

@x0 j(�x;0)

@h

@� j(�x;0)+@f

@y0 j(�x;0)

@g

@� j(�x;0)+ :::

+@f

@y j(�x;0)

@g

@� j(�x;0)+@f

@x0 j(�x;0)

@h

@� j(�x;0)= 0

� this is a linear homogenous equation in @h@� j(�x;0) and

@g@� j(�x;0) so if a unique

solution exist it must be @h@� j(�x;0) = [0] and

@g@� j(�x;0) = [0]

10

� 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] = fy0y0gxhx + fy0ygx + fy0x0hx + fy0xgxhx + fy0gxxhxhx + fy0gxhxx + ::: =

(fyy0gxhx + fyygx + fyx0hx + fyx) gx + fygxx + (fx0y0gxhx + fx0ygx + fx0x0hx + fx0x)hx + fx0hxx

fxy0gxhx + fxygx + fxx0hx + 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 independentof the volatility of the shocks

A second order approximation to the policy function of a stochastic modeldi¤er from the non-stochastic version only by a constant parameter of thecontrol vector.

5.2 Using �rst order methods to solve up to the secondorder

Finding second order solutions using �rst order methodsA second-order approximation of model can be written as:

A1nxnEt[xt+1]

nx1= A2

nxnxtnx1

+ A3nxm

ztmx1

+ A4nx

n(n+1)2

�tn(n+1)

2 x1

+ A5nx

n(n+1)2

Et [�t+1]n(n+1)

2 x1

where:�t = vech (~xt~x

0t)

where:

~xt �

24 ztx1tx2t

35Finding second order solutions using �rst order methods

11

Suppose we have applied the Schur decomposition to solve the �rst orderapproximation obtaining the state-space form:

fx1t+1 = F1zt + F2fx

1t

fx2t = P1fx

1t + P2fxt

and we can compactly write the solution as:24 ztfx

1t

fx2t

35 =

�ztfx

1t

��ztfx

1t

�| {z }

st

= �

�zt�1fx

1t�1

�+ �"t

where fx1t is the �rst order approximation of the states, and � are appro-priate matrices constructed using F1; F2; P1; P2 and IFinding second order solutions using �rst order methodsSince we are already dealing with an approximation, we can substitute x1t

with its �rst order approximation.Note that:

�t = vech (~xt~x0t) = Lvec

�ztx1t

� �ztx1t

�00

!

= L ( ) vec �

ztx1t

� �ztx1t

�0!

= L ( )Lhvech �

ztfx

1t

� �ztfx

1t

�0!= RVt

Finding second order solutions using �rst order methodsSecond:

Vt = vech (sts0t) = vech

�[�st�1 + �"t] [�st�1 + �"t]

0�= vech

��st�1s

0t�1�

0 +�st�1�0"0t + �"ts

0t�1�

0 + �"t�0"0t�

= Lvec��st�1s

0t�1�

0�+ Lvec (�st�1"0t�0) + Lvec ��"ts0t�1�0�+ Lvec (�"t"0t�0)= L (� �)Lhvech

�st�1s

0t�1�+ L (� �)Lhvech ("t"0t) +

+L (� �)Lhvech (st�1"0t) + L (� �)Lhvech�"ts

0t�1�

= ~�Vt�1 + ~�~"t + ~~�t

where ~ = L (� �)Lh + L (� �)LhP and P is such that Pvech (X 0) =vech (X)

12

Therefore the second order elements of the system are themselves an inde-pendent system.Finding second order solutions using �rst order methodsHence we have now an augmented system:

A1Et [xt+1] = A2xt +A3zt +G�t +H�

Vt = ~�Vt�1 + ~�~"t + ~~�t

zt = Nzt�1 + "t

where G = A4R+A5R~� H = A5R~� and � = Et~"t:We can than solve backward for Vt and than treating it andH� as exogenous

variables in the main representation.

5.3 A Simple RBC application

Finding second order solutions: A Simple RBC applicationWe now consider a simple application. The Ramsey model is:

maxE0

1Xt=0

C1� t

1�

s:t:Kt+1 = (1� �)Kt + It

It + Ct = AtK�t

logAt+1 = � logAt + "t+1

we assume "t � N (0; 1) and, for simplicity, � = 1 and � = 0:Finding second order solutions: A Simple RBC applicationFOC�s imply:

C� t = ��Et�C� t+1At+1K

��1t+1

�Kt+1 = AtK

�t � Ct

logAt+1 = "t+1

which implies:

� logCt = log (��)� Et [logCt+1] + Et [logAt+1] + (�� 1)Et [logKt+1]

log (Kt+1 + Ct) = log [AtK�t ]

logAt+1 = "t+1

So:

Et�f�x2t+1; x

2t ; x

1t+1; x

1t

��nx1

� Et

� logCt + log (��)� Et [logCt+1] + Et [logAt+1] + (�� 1)Et [logKt+1]

log (Kt+1 + Ct)� log [AtK�t ]

�= 0

13

x1t = Kt; x2t = Ct and zt = at

Finding second order solutions: taking the �rst order approximationRecall that in this case we have one state and one control.For the SGU Method:

fy0gh+ fyg + fx0h+ fx = [0]�� 0

�gh+

� Css

Kss+Css

�g +

�(�� 1)Kss

Kss+Css

�h+

�0�

�= [0]

Note that we are looking for the roots of this system of equations (see ex-ample at the beginning of the class).For the GSD:

� Etct+1 + (�� 1) kt+1 = � ctKss

Kss + Csskt+1 = � Css

Kss + Cssct + �kt + at

In Matrix notation:�(�� 1) � Kss

Kss+Css0

� �kt+1Etct+1

�+

�0 �� Css

Kss+Css

� �ktct

�+

�0�1

�at = [0]

Finding second order solutions: the second order approximation for SGUMethodFor the second order we have:

[0] = fy0y0gxhx + fy0ygx + fy0x0hx + fy0xgxhx + fy0gxxhxhx + fy0gxhxx + ::: =

(fyy0gxhx + fyygx + fyx0hx + fyx) gx + fygxx + (fx0y0gxhx + fx0ygx + fx0x0hx + fx0x)hx + fx0hxx

fxy0gxhx + fxygx + fxx0hx + fxx

which is a system of two equations in two unknowns.Finding second order solutions: the second order approximation for the Lom-

bardo Sutherland MethodFor the second order we have:

� Etct+1 + (�� 1) kt+1 +1

2Et

��at+1 + Etct+1 + (�� 1) kt+1

�2�= � ct +

2

2c2t

� CssKss + Css

ct + �kt + at �1

2

CssKss + Css

c2t +�2

2k2t +

1

2a2t + �ktat =

Kss

Kss + Csskt+1 +

1

2

Kss

Kss + Cssk2t+1

In matrix notation:

A1

�kt+1Etct+1

�= A2

�ktct

�+A3at +A4�t +A5Et [�t+1]

14

where:

A1 =

�(1� �) � Kss

Kss+Css0

�; A2 =

�0 � � Css

Kss+Css

�; A3 =

�01

�; A4 =

"0 0 0 0 0 � 2

212 � �2

2 0 0 � 12

CssKss+Css

#

A5 =

"12 (�� 1) (��1)2

2 (�� 1) 2

2

0 0 � 12

Kss

Kss+Css0 0 0

#

trough which we can construct the auxiliary system.

6 Normative Analysis

Normative AnalysisWe 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 alwayseasy to implement) way.

6.1 Policies not a¤ecting the steady state

6.1.1 An RBC application

Policies not a¤ecting the steady stateThe standard Ramsey model augmented with consumption taxes is:

maxct;kt+1;it

E0

" 1Xt=0

�tc1� t

1�

#

s:t:kt+1 = (1� �) kt + it(1 + � t) ct + ii = atk

�t

yt = atk�t

at+1 = �at + "t

Consider an exogenous set of policies:

� t = �kEt [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

FOC�s imply the following system of 3 equations in 3 variables:

c� t = �Et�c� t+1

��at+1k

��1t+1 + (1� �)

�kt+1 = (1� �) kt + atk�t � (1 + � t) ct� t = �kEt [yt+1]

Plus:

c� t = (1 + � t)�t

ii = atk�t � (1 + � t) ct

Policies not a¤ecting the steady stateThis 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 betterto 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 Policies a¤ecting the steady state

6.2.1 An RBC application

Policies a¤ecting the steady state: the problemConsider the following problem:

maxct;kt+1;it;�t

1Xt=0

�t

"c1� t

1�

#

s:t:kt+1 = (1� � � � t) kt + itct + it = atk

�t x

1����t

xt = � tkt

log at+1 = � log at

That is the government can impose wealth taxes to provide services thatenter in the production function.In this problem the level of taxes a¤ects the steady state.Policies a¤ecting the steady state: analytical solution

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Write the lagragian:

L =1Xt=0

�t

8<: c1� t

1� + �t1x2

24 (1� � � � t) kt + atk1��t (� t)1���� � ct � kt+1

atk1��t (� t)

1���� � ct � itlog at+1 = � log at

359=;FOC�s imply:

c� t = �1t

�1t = �n�1t+1

h(1� �) at+1 (kt+1)�� (� t+1)1���� + 1� � � � t+1

iokt = (1� �� �) atk1��t (� t)

����

+ the constraints

Policies a¤ecting the steady state: analytical solutionIn steady state the FOC�s imply:

c� = �1

k =

�1� � (1� � � �)� (1� �) �1����

�� 1�

1 = (1� �� �) k�� (�)����

c = (�� � �) k + k1�� (�)1����

which implies:

1 = (1� �� �) 1� � (1� � � �)� (1� �) �

which implies:

� =(1� �� �) [1� � (1� �)]

[��]

Policies a¤ecting the steady state: numerical solutionThis problem is simple, so we can use it to test a numerical approach.Consider the problem written in matrix notation:

L =1Xt=0

�t�Ot + �t

1xncLtncx1

�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 stateFirst order conditions w.r.t endogenous variables imply:

@Ot@xt

+ �t@Lt@xt

= [0]

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Note that this is a system of n variables in nc unknowns (the lagrange mul-tipliers) and therefore can not be solved exactly.Furthermore @Lt

@xtis not a square matrix and therefore can not be inverted.

However we can, given a value of the policy parameter that determine thevalue of endogenous variables, solve this system by OLS (i.e. the values of thelagrange 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: anRBC model with Money and Optimal Policy

maxct;kt+1;it;mt;bt

1Xt=0

�t��c1��t + (1� �)m1��

t

� 1� 1��

1�

s:t:kt+1 = (1� �) kt + it

ct + it +BtPt+Mt

Pt= atk

�t + (1 +Rt)

Bt�1Pt

+Mt�1Pt

log at+1 = � log at

which implies:

ct + kt+1 + bt +mt = atk�t + (1� �) kt + (1 +Rt)

bt�1�t

+mt�1�t

Lagrangian and First Order Conditions are:

maxct;kt+1;mt;bt

1Xt=0

�t

8<: [�c1��t +(1��)m1��t ]

1� 1��

1� +

+�t

hatk

�t + (1� �) kt + (1 +Rt)

bt�1�t

+ mt�1�t

� ct � kt+1 � bt �mt

i9=;

��c1��t + (1� �)m1��

t

� 1� 1���1 � (1� �) c��t = �t

�t = ��t+1��at+1k

��1t+1 + (1� �)

��t = ��t+1

�1 +Rt+1�t+1

���c1��t + (1� �)m1��

t

� 1� 1���1 (1� �) (1� �)m��

t = �t � ��t+1�t+1

Couple of notes:

This is a system of 7 variables ct;mt; kt; bt; �t; at; Rt in 4 equations, thegovernment budget constraint, the exogenous low of motion and the policy rule:��c1��t + (1� �)m1��

t

� 1� 1���1 c��t = �

��c1��t+1 + (1� �)m1��

t+1

� 1� 1���1 c��t+1

��at+1k

��1t+1 + (1� �)

�18

��c1��t + (1� �)m1��

t

� 1� 1���1 c��t = �

��c1��t+1 + (1� �)m1��

t+1

� 1� 1���1 c��t+1

�1 +Rt+1�t

�

��c1��t + (1� �)m1��

t

� 1� 1���1 (1� �) (1� �)m��

t =��c1��t + (1� �)m1��

t

� 1� 1���1 � (1� �) c��t +

����c1��t+1 + (1� �)m1��

t+1

� 1� 1���1 (1� �) c��t+1

�t+1

ct + kt+1 + bt +mt = atk�t + (1� �) kt + (1 +Rt)

bt�1�t

+mt�1�t

log at+1 = � log at

Bt �Bt�1 +Mt �Mt�1 = Rt�1Bt�1

Ptbt + Ptmt � Pt�1mt�1 = (1 +Rt�1)Pt�1bt�1

�tbt + �tmt �mt�1 = (1 +Rt�1) bt�1

Steady state: a = 1 than

k =

�1� � (1� �)

��

� 1��1

R =�

�� 1

(1� �)m��t = �c��t � �

�c��t+1�t+1

m =

"�� ���(1� �)

#� 1�

| {z }AA

c

b =� � 1

1 +R� �m

b =

�� � 1

1 +R� �

� ��� � ��(1� �)

�� 1�

| {z }BB

c

19

c+BBc+AAc = k�t � �k + (1 +R)BB

�c+

AA

�c�

1 +AA

�1� 1

�

�+

�1� 1 +R

�

�BB

�c = k�t � �k

c =k� � �k�

1 +A�1� 1

�

�+�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 RBCmodel withMoney

Consider the same model with money, but with instantaneous utility function:

log ct +mt exp (� mt)

Lagrangian and First Order Conditions are:

maxct;kt+1;mt;bt

1Xt=0

�t

(log ct +mt exp (� mt)+

+�t

hatk

�t + (1� �) kt + (1 +Rt)

bt�1�t

+ mt�1�t

� ct � kt+1 � bt �mt

i )

1

ct= �t

�t = ��t+1��at+1k

��1t+1 + (1� �)

��t = ��t+1

�1 +Rt+1�t+1

�(1� mt) exp (� mt) =

1

ct+

�

�t+1

1

ct+1

The last four equation constitute our model.In steady state:

k =

�1� � (1� �)

��

� 1��1

R =�

�� 1

c (1� m) exp (� m) =�1� �

�

�

20

b =� � 1

1 +R� �m

b =� � 1�� � �

m

b =� � 1��1���

�| {z }

BB

m

b =� (� � 1)� (1� �)| {z }

BB

m

k� � �k + (1 +R) b�+m

�= c+ b+m

k� � �k = c+

�1� (1 +R)

�

�b+

�1� 1

�

�m

k� � �k = c+

�1� 1

�

�b+

�1� 1

�

�m

k� � �k = c+

�1� 1

�

�� (� � 1)� (1� �)m+

�1� 1

�

�m

k� � �k = c+

�� � 1�

�� (� � 1)� (1� �)m+

�1� 1

�

�m

k� � �k = c� (� � 1)�

m+

�1� 1

�

�m

k� � �k = c

Since steady state consumption is independent of money, we need to set anin�ation rate to maximize utility derived by money holdings. Than maximiza-tion 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.

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