Modern Macroeconomics II - Osaka Universitytakii/macroII-slide.pdf · 2011. 2. 1. · macroeconomics, but also for understanding dynamic issues in any –elds of economics, such as

Modern Macroeconomics II

Katsuya Takii

OSIPP

Katsuya Takii (Institute) Modern Macroeconomics II 1 / 428

Introduction

Purpose: This lecture is aimed at providing students with standardmethods in modern macroeconomics. In particular, the lectureextends Solow model, which was taught in Modern MacroeconomicsI, into three directions. Firstly, I apply the dynamic optimizationtechniques to endogenize saving rate in Solow model. Secondly, Iintroduce stochastic shocks to analyze uncertainty in a dynamiccontext. Finally, I discuss how one can numerically analyze themodel. These methods are useful not only for understandingmacroeconomics, but also for understanding dynamic issues in anyelds of economics, such as public economics and nancialeconomics.

O¢ ce hour: Room 602, 9:30-10:20, 12:15-12:45 on Monday.Appointment is required for other time.

E-mail Address: [email protected]


Introduction

Grading Policy: 35% on assignments and 65% on a nal exam.1 I will give you X assignments. Students must hand them in at thefollowing lecture. If students turn an assignment in by the due date, Iwill give them 35/X points. If students turn an assignment in late, Iwill give them 14/X points. If students submit all assignments, youwill receive 35 points. Students must write their answers with a pen.I dont allow the typed answers for this assignment.

2 The full score of nal exam is 65 points. I guarantee that 40 pointsout of 65 points will come from the assignment and the slide of lecture.If you hand in all assignments and you perfectly answer the questionsappeared in assignments, you can certainly receive B.


Introduction

Remarks:1 I assume that students have already taken Decision Theory, ModernMicroeconomics 1, Modern Macroeconomics 1.

2 Because of the nature of the issues, the lectures are rather technical.Students are expected to prepare by themselves to understand thelecture. I highly recommend this course to the students who think ofeconomics as a major discipline and go to the doctoral program.

3 I will teach this course in English (supplemented by Japanese ifnecessary.) I encourage students to make comments and questions inEnglish. However, I will not prevent students from asking questions inJapanese. I can discuss your questions and comments in Japanese orEnglish at my o¢ ce hour.


Introduction

Course Outline1 Basic Dynamic Programming (2 lectures)2 Basic Growth Model (3 lectures)3 Continuous Time Growth Model (2 lectures)4 Stochastic Growth Model (2 lectures)5 Overlapping Generation Model (2 lectures)6 Discrete Choice and Unemployment (3 lectures)7 Final Exam (1 lecture)


Introduction to Dynamic Programming

Macroeconomics is a study to explain the behavior of aggregate datasuch as GDP per capita, ination rate, and unemployment rate.

For this purpose, we must infer the structure of the economy thatbrings the observable data.

In order to infer the structure, we need a theory. There are twocurrent consensuses among macroeconomists

1 Use the variants of dynamic stochastic general equilibrium models toanalyze economy.

2 Developing a quantitative theory is useful.

Theory can be used not only for conveying the idea, but also forextracting meaningful information from data.

The foundation of current macroeconomics that brings twoconsensuses is the neoclassical growth model. This is the mainsubject of this lecture.



Remember Kaldors Stylized Facts (1963)1 The growth rate of GDP per capita is nearly constant:2 The growth rate of capital per capita is nearly constant:.3 The growth rate of output per worker di¤ers substantially acrosscountries.

4 The rate of return to capital is nearly constant:5 The ratio of physical capital to output is nearly constant:6 The shares of labor and physical capital are nearly constant:

In order to meet this requirement, the growth model must exhibit thebalanced growth path.



The rst candidate is the extended Solow Model, which issummarized by the following 3 equations.

Kt+1 = F (Kt ,TtNt ) + (1 δ)Kt CtTt+1 = (1+ g)TtNt+1 = (1+ n)Nt

with an assumption.

Ct = (1 s) F (Kt ,TtNt )

But, As Lucas (1976) critique suggests, this constant saving rate maynot be a robust assumption.

We would like to develop a model which allows us to analyze theimpact of policy change on real economy through changes in thesaving rate.



In order to endogenously determine the saving rate, we must knowhow consumers decide their consumption.

One of a convenient assumption is that a representative consumermaximize the following utility function.

maxfctg∞

τ

∞

∑t=0

βtNtU (ct ) ,

Kt+1 = F (Kt ,TtNt ) + (1 δ)Kt ctNt , K0 is givenTt+1 = (1+ g)Tt , T0 is given

Nt+1 = (1+ n)Nt , N0 is given

where ct = CtNt.

This model is called the neoclassical growth model. In order to solvethis problem, we need to know how to solve a dynamic optimizationproblem.


A Finite Horizon Problem

Consider the following problem

maxfXtg

(T1∑t=τ

β(tτ)r (Xt ,St ) + β(Tτ)VT (ST )

),

s.t. St+1 = G (Xt ,St ) ,

Sτ is given.

The variable, fXtg, is called a control variable. It can be a vector.This is the variable which an agent tries to control in order tomaximize his objective function. The variable fCtg is the controlvariable in the neoclassical growth model.

The variable, fStg, is called a state variable. It can be a vector.The state variable summarizes the current state of the economy. Thevector f(Kt ,Tt ,Nt )g is a state vector in the neoclassical growthmodel.



The function r (Xt ,St+1) is called the one period return function.One period return function summarizes the reward from current statevariable and the current control variable. In the case of theneoclassical growth model, r (Xt ,St+1) = NtU (Ct ).

The function G (Xt ,St ) is called the transition function. It describesthe dynamic behavior of the state variables. In the case of theneoclassical growth model

G (Ct , (Kt ,Tt ,Nt )) =

8<:F (Kt ,TtNt ) + (1 δ)Kt Ct

(1+ g)Tt(1+ n)Nt

9=;The function, VT () is called the value function at the last period.Since the state variable summarizes the current state of economy.The value function indicates, how much value he expects to obtain inthe future when the current state is ST .



Given Sτ, when the agent chooses Xτ at date τ, Sτ+1 is automaticallydetermined. Given this Sτ+1, the agent can choose Xτ+1 at dateτ + 1, and it determines Sτ+2, and so on. In this way, the agent canrecursively decide his decisions.

Note that

maxfXtg

(T1∑t=τ


)

= maxfXtg

(....+ β(t1τ)r (Xt1,St1) + β(tτ)r (Xt ,St )+β(t+1τ)r (Xt+1,St+1) + ...β(

Tτ)VT (ST )

)

= maxfXsgs<t

8>>><>>>:....+ β(t1τ)r (Xt1,St1)

+maxfXx gxt

264 β(tτ)r (Xt ,St )+β(t+1τ)r (Xt+1,St+1) +

...β(Tτ)VT (ST )

3759>>>=>>>;



That is, once the agent knows, St , he does not need to worry aboutother past variables f(Xs ,Ss )gs<t because the state variablesummarizes every important information of the economy at that time.

Hence, the maximization problem at date T 1 can be simpliedas follows:

maxXT1

fr (XT1,ST1) + βVT (G (XT1,ST1))g ,

ST1 is given.

The solution to this problem depends on ST1. That is, we canderive a policy function xT1 () as a function of ST1:

xT1 (ST1) = arg maxXT1

fr (XT1,ST1) + βVT (G (XT1,ST1))g



Since the policy function is an optimal strategy, we can dene thevalue function at date T 1, VT1 (), as follows:

maxXT1

fr (XT1,ST1) + βVT (G (XT1,ST1))g

= r (xT1 (ST1) ,ST1) + βVT (G (xT1 (ST1) ,ST1))

VT1 (ST1) .

Using VT1 (ST1) we can rewrite the agents problem at datesT 2 as follows:

maxXT2

fr (XT2,ST2) + βVT1 (G (XT2,ST2))g ,

ST2 is given.

This is the similar problem as before. Therefore, we can deneVT2 (ST2) and using VT2 (ST2), we can describe the problemat date T 3 and so on.



In general, an original problem can be rewritten as the sequence of asimple static problem:

Vt (St ) maxXtfr (Xt ,St ) + βVt+1 (G (Xt ,St ))g ,

This equation is called the Bellman Equation.

Hence, a nite horizon problem can be solved by a sequence of policyfunctions and value functions, f(xt () , Vt ())gT1t=0 .

For any initial state variable, Sτ, the optimal policy function xτ (Sτ)determines Xτ and the transition function G (Xτ,Sτ) determinesSτ+1. Given this Sτ+1, the optimal policy function and the transitionfunction determines Xτ+1 and Sτ+2 and so on.



Assume that the rst order condition is valid. The rst ordercondition is

0 = r1 (xt (St ) ,St ) + βV 0t+1 (G (xt (St ) ,St ))G1 (xt (St ) ,St ) , (1)

for any t.

The rst term of the right hand side of equation (1) is the marginalreturn from changing Xt . However, when the agent changes Xt , ita¤ects not only the current return, but also the future returns bychanging the future state variable. When the agent slightly changesXt , it will change St+1 by G1 (Xt ,St ). If St+1slightly moves, itchanges the present value of the future reward at date t + 1 byV 0t+1 (St+1). Since the agent discount the future by β, the futureimpact of changing Xt is βV 0t+1 (G (Xt ,St ))G1 (Xt ,St ). If the agentoptimally chooses variables, these two e¤ects must be the same atany date t.



It is useful to derive the envelope condition:

V 0t (St )

= r2 (xt (St ) ,St ) + βV 0t+1 (G (xt (St ) ,St ))G2 (xt (St ) ,St ) +r1 (xt (St ) ,St ) + βV 0t+1 (G (xt (St ) ,St ))G1 (xt (St ) ,St )

x 0t (St )

= r2 (xt (St ) ,St ) + βV 0t+1 (G (xt (St ) ,St ))G2 (xt (St ) ,St )

for any t.



It shows the component of the marginal benet of changing the statevariable, St . When the state variable changes a little bit, it changesthe current return today by r2 (xt (St ) ,St ). But since changing thestate variable at date t will change the future state variable St+1 byG2 (Xt ,St ), it also has a dynamic e¤ect. Since changing St+1 a¤ectsthe present value of the future reword by V 0t+1 (St+1) and the agentdiscounts the future by β, the total e¤ect must beβV 0t+1 (G (Xt ,St ))G2 (Xt ,St ).



Example: Consider the following household problem,

maxfCtg

T1∑t=τ

β(tτ)U (ct ) + β(Tτ)VT (aT ) ,

s.t. at+1 = (1+ ρ) at + w ctgiven aτ

We can dene the Bellman equation as follows:

Vt (at ) = maxctfU (ct ) + βVt+1 (at+1)g ,

s.t. at+1 = (1+ ρ) at + w ct ,



The rst order condition is

U 0 (c (at )) = βV 0t+1 ((1+ ρ) at + w c (at ))

Envelope condition

V 0t (at ) = βV 0t+1 [(1+ ρ) at + w c (at )] [1+ ρ] .

Combining Two equations

U 0 (c (at1))β

= U 0 (c (at )) (1+ ρ)



Euler equation:U 0 (ct )

βU 0 (ct+1)= 1+ ρ

where ct+1 = c (at+1) and ct = c (at ). The left hand side is themarginal rate of substitution between consumption at date t andt + 1; the right hand side is the marginal gain from saving.



Euler Equation


An Innite Horizon Problem

Now let me consider the case, T goes innite. That is, an originalproblem is

U (Sτ) = maxfXtg

(∞

∑t=τ

β(tτ)r (Xt ,St )

),

s.t. St+1 = G (Xt ,St ) , Sτ is given.

If T goes innite, there is no last period and we cannot use theprevious method. However, we can explain it by analogy between thismodel and the previous one.

Assume that a policy function of this original problem is xu (Sτ).



Remember that once we control a current state, the past decisiondoes not inuence future maximization problem. Hence,

U (Sτ) = maxfXtg∞

τ

∞

∑t=τ

β(tτ)r (Xt ,St ) ,

= maxXτ

(r (Xτ,Sτ) + β max

fXtg∞τ+1

∞

∑t=τ+1

β(t(τ+1))r (Xt ,St )

),

= maxXτ


fXtg∞τ+1

U (Sτ+1)

)



Hence, we can consider the following recursive problem:

V (St ) = maxXtfr (Xt ,St ) + βV (St+1)g , (2)

St+1 = G (Xt ,St ) .

Assume that the policy function of this problem is x (St ).1 With certain mild conditions (r (X ,S) is bounded and continuous,f(St+1,Xt St ) jSt+1 G (Xt ,St )g is compact), we can show thatthere exists a unique value function V () and a policy function whichsatises equation (2) and that U (Sτ) = V (Sτ) and xu (Sτ) = x (Sτ).That is, Bellman equation (2) is equivalent to the original problem andthe value function and policy function is time invariant.

2 With the further assumptions (r (X ,S) is strict concave andf(St+1,Xt St ) jSt+1 G (Xt ,St )g is convex), V (S) is strictlyconcave and x (Sτ) is continuous and single valued.

3 With further mild condition ( r (X ,S) is continuously di¤erentiable), itis known that V is continuously di¤erentiable.



Example: Suppose that S 2 f1g. Therefore, St = G (Xt ,St ) = 1for any t. Suppose that.

x (1) = argmaxXfr (X , 1) + βV (G (X , 1))g

ThenTV (1) r (x (1) , 1) + βV (1) .



The Existence of a Unique Value Function



An economic interpretation of the Bellman equation (2) is that theagent maximizes the sum of current return, r (Xt ,St ), and thepresent value of the discounted future returns, βV (St+1). He isconcerned about the tradeo¤ between the current benets and thefuture benets when he chooses Xt . Since he lives forever, it doesnot matter when he makes his decisions. Therefore, the valuefunction and the policy function do not depend on time. His problemis stationary.



The original problem can be approximated by a nite horizon problem:

U (Sτ, τ) = limT!∞

maxfXtg

("T1∑t=τ


#),

s.t. St+1 = G (Xt ,St ) ,

Sτ is given.

It is shown that we can approximate the Bellman equation (2) by thecorresponding nite horizon problem.

V (S) = limt!∞

Vt (S) .

Vt (S) maxXfr (X ,S) + βVt+1 (G (X ,S))g (3)

Since there exist a unique value function V (), if we iterate equation(3) from an arbitrary initial value function VT (S) for any S , it mustconverge to V (S).



There are several methods to analyze the Bellman equation (2):1 Guess and Verify Method,2 Euler Equation3 Numerical method.


Guess and Verify Method

Guess and Verify method: One way to analyze the property ofequation (2) is guess and very method. The rst, guess what wouldbe the property of V () and assume that the property is true. Thenverify that your guess satisfy equation (2). Since we know the valuefunction is unique, if a property satises equation (2), the valuefunction must have the property.



Example: Assume the following investment problem:r (K , I ) = 1

1+ρ [pF (K , L) wL C (I ,K ) I ],G (K , I ) = (1 δ)K + I and β = 1

1+ρ . Assume that F (K , L) andC (I ,K ) are constant returns to scale.

The Bellman Equation of this problem can be written as

V (K ) =1

1+ ρmaxI ,L

pF (K , L) wL C (I ,K ) I + V

K 0

K 0 = I + (1 δ)K



Guess V (K ) = QK .

maxI ,L

pF (K , L) wL C (I ,K ) I +QK 0

= max

I ,L

(pF1, LK

w L

K

C IK , 1

I

K +QI+(1δ)K

K

)K

= maxg ,lfpφ (l) wl c (g) g +Q [g + (1 δ)]gK

where φ (l) = F (1, l), c (g) = C (g , 1), l = LK and g =

IK . If there

exists Q that satises

Q =1

1+ ρmaxg ,lfpφ (l) wl c (g) g +Q [g + (1 δ)]g .

Then our guess is veried. Since the value function is unique, thelinear must be the property of the value function.



Note that

Q = 1+ c 0 (g)

g = ψ [Q 1]

This is called the Q theory of investment. It says that the rmsgrowth rate is determined by Q. Note that Q can be estimated by

Q =V (K )K

.

This is the market value of a rm relative to the replacement cost ofcapital.



Assume that c (g) = 12A (g B)

2 and c 0 (g) = gBA . Hence

g BA

= Q 1g = B + A [Q 1]= B + AQ + ε

where B + ε = B A. Empirical researchers often estimate thisregression.



Since many applied economists restrict their attention to the case towhich we can apply the rst order condition, I explain this method byusing the rst order condition. The policy function x () must satisfythe rst order condition and the bellman equation for any S asfollows:

0 = r1 (x (S) ,S) + βV 0 (G (x (S) ,S))G1 (x (S) ,S) (4)

V (S) = r (x (S) ,S) + βV (G (x (S) ,S)) . (5)

Given the value function V (), equation (4) determines x (); givenx (), equation (5) determines V (). That is, these are simultaneousequations. We can analyze our model by examining these twoequations.

In general we cannot solve a closed form solution. However, there isa special case in which we can nd a closed form solution.



Example: Consider the following problem.

maxfctg

∞

∑t=0

βt ln ct ,

s.t. at+1 = (1+ ρ) at ct


V (at ) = maxctfln ct + βV (at+1)g ,

s.t. at+1 = (1+ ρ) at ct



Guess V (at ) = f + g log (at ). Then consider the problem

maxctfln ct + β [f + g ln (at+1)]g ,

s.t. at+1 = (1+ ρ) at ct ,First Order Condition implies

1c (at )

=βg

(1+ ρ) at c (at )Policy Function

c (at ) =(1+ ρ) at c (at )

βgβg + 1

βgc (at ) =

(1+ ρ) atβg

c (at ) =(1+ ρ) at

βg + 1



at+1

at+1 = (1+ ρ) at c (at )

= (1+ ρ) at (1+ ρ) at

βg + 1

=βg (1+ ρ) at

βg + 1



The Maximized Value

ln c (at ) + β [f + g ln (at+1)]

= ln(1+ ρ) at

βg + 1+ β

f + g ln

βg (1+ ρ) at

βg + 1

= (1+ βg) ln

(1+ ρ) at

βg + 1

+ β [f + g ln βg ]

= (1+ βg) ln(1+ ρ)

βg + 1+ β [f + g ln βg ] + (1+ βg) ln at

Hence, if we can nd f and g that satises

f = (1+ βg) ln(1+ ρ)

βg + 1+ β [f + g ln βg ]

g = 1+ βg

then we can verify our guess.



g

g =1

1 β

f

f = (1+ βg) ln(1+ ρ)

βg + 1+ β [f + g ln βg ]

=1

1 β

(1+ βg) ln

(1+ ρ)

βg + 1+ βg ln βg

=1

1 β

241+ β

1 β

ln(1+ ρ)

β1β + 1

+β

1 βln

β

1 β

35=

11 β

1+

β

1 β

ln (1+ ρ) (1 β) +

β

1 βln

β

1 β

=

11 β

ln (1+ ρ)

1 β+

β

1 βln β+ ln (1 β)



V (at )

V (at ) =1

1 β

ln (1+ ρ)

1 β+

β

1 βln β+ ln (1 β) + ln at



The guess and verify method demands many calculations. If we areonly interested in a policy function, but not a value function, there isone way to reduce calculation: the use of envelop condition.

The rst order condition and the envelop theorem of the Bellmanequation (2) imply that

0 = r1 (x (S) ,S) + βV 0 (G (x (S) ,S))G1 (x (S) ,S) (6)

V 0 (S) = r2 (x (S) ,S) + βV 0 (G (x (S) ,S))G2 (x (S) ,S) (7)

for any St . Again two unknown functions x () and V 0 () can besolved by two equations.



Note that the rst order condition (6) is the same as the rst ordercondition (4); the envelop condition (7) is not the same as theBellman equation (5).

Also note that the previous two equations determine x () and V ();the current two conditions determine x () and V 0 (). In general,there exists C such that

V (S) =ZV 0 (S) dS + C .

Since equation (6) and (7) cannot determine C , without a boundarycondition, equation (6) and (7) cannot solve the value function.However, if we are only interested in a policy function, equations (6)and (7) can solve it.



Example: Consider the previous problem again.

V (at ) = maxctfln ct + βV (at+1)g ,

s.t. at+1 = (1+ ρ) at ct

The rst order condition and envelop condition are

1c (at )

= βV 0 ((1+ ρ) at c (at ))

V 0 (at ) = β (1+ ρ)V 0 ((1+ ρ) at c (at )) ,



Guess V 0 (a) = ga . Then the rst order condition is

1c (at )

=gβ

(1+ ρ) at c (at )

Policy function

c (at ) =(1+ ρ) at c (at )

βgβg + 1

βgc (at ) =

(1+ ρ) atβg

c (at ) =(1+ ρ) at

βg + 1



Hence,

β (1+ ρ)V 0 ((1+ ρ) at c (at ))

=(1+ ρ)

c (at )

=(1+ ρ)(1+ρ)at

βg+1

=βg + 1at

Hence if we nd g that satises

g = βg + 1,

our guess is satised.



g

g =1

1 β.

V 0 (a)

V 0 (a) =1

(1 β) a



Assignment: Suppose that r (I ,K ) = zK pI A2 I2 and

G (I ,S) = I + (1 δ)K , where K is capital stock, I is investment, pis the investment price, z is the instantaneous return on capital and Ais the parameter of adjustment cost.

1 Dene the Bellman Equation.2 Find a Policy Function.


Euler Equation

Euler Equation: In general, it is di¢ cult to solve a closed formsolution. However, it is possible to analyze the property of solutionsfor more general class of functions. Remember that the rst ordercondition and envelop condition are

0 = r1 (x (S) ,S) + βV 0 (G (x (S) ,S))G1 (x (S) ,S) ,

V 0 (S) = r2 (x (S) ,S) + βV 0 (G (x (S) ,S))G2 (x (S) ,S) .

Using two equations, we can eliminate V 0 (S) and derive the rstorder nonlinear di¤erence equation. This is called Euler equation.

r1 (Xt ,St )βG1 (Xt ,St )

=r1 (Xt+1,St+1)G1 (Xt+1,St+1)

G2 (Xt+1,St+1) r2 (Xt+1,St+1) .(8)


Euler Equation

Although the rst order condition and the envelop condition canderive the Euler equation, the Euler equation cannot derive the rstorder condition and the envelop condition. That is, the elimination ofV 0 (St ) discards useful information. Therefore, the Euler equation isa necessary condition for the original problem, but not su¢ cient.

Note that we can derive the dynamics of Xt from Euler equation:

Xt = Φ (St+1,Xt+1,St )

We also have the transition equation for the dynamics of S .

St+1 = G (Xt ,St ) ,


Euler Equation

It means that

Xτ = Φ (Sτ+1,Xτ+1,Sτ)

..

XT1 = Φ (ST ,XT ,ST1)..

Sτ+1 = G (Xτ,Sτ)

..

ST = G (XT1,ST1)

..


Euler Equation

These equations imply that from period τ to period T 1, thenumber of equations are 2 (T τ) and the number of Xt and St is2 (T τ + 1). It means that we need 2 known variables to solve thesequence of Xt and St between τ and T 1.Now we have one initial condition Sτ. In a nite horizon case, sincethe value function at the end period is given, the rst order condition,

r1 (XT ,ST ) + βV 0T+1 (G (XT ,ST ))G1 (XT ,ST ) = 0,

determines XT as a function of ST . It gives us one more equation.This serves as a boundary condition.


Euler Equation

The di¢ cult question is what would be the appropriate boundarycondition in an innite horizon problem. It is known that thefollowing transversality condition is su¢ cient for an optimal solution:

limt!∞

βtV 0 (St ) St = limt!∞

βt1r1 (Xt1,St1)G1 (Xt1,St1)

St = 0. (9)

Since V 0 (St ) is the marginal value of St , βtV 0 (St ) St is the presentvalue of stock at t. The transversality condition implies that when tgoes innite, the present value of stock must be negligible. That is,we should not be concerned about an innite period later.

Although Euler equation is not su¢ cient for the original problem,given usual technical conditions such as concavity etc, it is knownthat the Euler equation together with the transversality condition issu¢ cient for the original problem.


Euler Equation

Example: Consider the following problem.

maxfctg

∞

∑t=0

βtU (ct ) ,

s.t. at+1 = (1+ ρ) at + w ct

Bellman Equation

V (at ) = maxctfU (ct ) + βV ((1+ ρ) at + w ct )g


Euler Equation

FOCU 0 (ct ) = βV 0 (at+1)

Envelop Condition

V 0 (at ) = β (1+ ρ)V 0 (at+1)


Euler Equation

Euler equation:

U 0 (c (at1))β

= U 0 (c (at )) (1+ ρ)

U 0 (ct )βU 0 (ct+1)

= 1+ ρ

Transversality Condition

limt!∞

βtV 0 (at ) at = limt!∞

βtU 0 (ct1)

βat = lim

t!∞βt1U 0 (ct1) at

Euler equation is the same as the nite horizon problem, but innitehorizon problem requires the transversality condition.


Euler Equation

Assignment: Assume the following investment problem:r (K , I ) = 1

1+ρ [pF (K , L) wL C (I ,K ) I ],G (K , I ) = (1 δ)K + I and β = 1

1+ρ . Assume that F (K , L) andC (I ,K ) are constant returns to scale.

1 Dene the Bellman Equation.2 Derive the Euler Equation.3 Derive the Transversality Condition.


The Neoclassical Growth Model

We apply Guess and Verify method and Euler Equation to theNeoclassical Growth Model.

We also explain the numerical method by using the NeoclassicalGrowth Model.



Consider the previous neoclassical growth model:

maxfCtg

∞

∑t=0

βtNtUCtNt

,

s.t. Kt+1 = F (Kt ,TtNt ) + (1 δ)Kt Ct , K0 is givenTt+1 = (1+ g)Tt , T0 is given


First, we reformulate the model so that the analysis becomes simpler.



Note that

Kt+1 =

"F

KtTtNt

, 1+ (1 δ) Kt

TtNt

CtTtNt

#TtNt ,

Kt+1Tt+1Nt+1

Tt+1Nt+1TtNt

= f (ket ) + (1 δ) ket cet ,

ket+1 (1+ g) (1+ n) = f (ket ) + (1 δ) ket cet ,

ket+1 =f (ket ) + (1 δ) ket cet

(1+ g) (1+ n)



Hence

maxfcetg

∞

∑t=0[β (1+ n)]t N0U (cetTt )

s.t. ket+1 =f (ket ) + (1 δ) ket cet

(1+ g) (1+ n)Tt+1 = (1+ g)Tt , given (k0,T0)



In order to satisfy the empirical regularity that economy must exhibita balanced growth, we need to assume that the coe¢ cient of therelative risk aversion is constant.

U00 (cetTt ) cetTtU 0 (ctTt )

= θ.

It is well-known that the following functions satisfy the abovecondition:

U (cetTt ) =(cetTt )

(1θ) 11 θ

This utility function is called the constant relative risk aversion utilityfunction.

Remark: When θ = 1, then it can be shown that

U (cetTt ) = ln (cetTt ) .



Assignment: Show that if utility function is

U (Ct ) =(Ct )

(1θ) 11 θ

,

Then

U00 (Ct )CtU 0 (Ct )

= θ.



It is also known that 1θ measures the elasticity of substitution betweenconsumption at any two points in time.To see this, consider the following two period problems. Supposethat U (Ct , Ct+1) = u (Ct ) + βu (Ct+1) and that the budgetconstraint is ptCt + pt+1Ct+1 = W , where pt and pt+1 are prices att and t + 1, and W is the wealth. Since the rst order condition is

pt+1pt

=U2 (Ct ,Ct+1)U1 (Ct ,Ct+1)

=βu0 (Ct+1)u0 (Ct )

,

If U (Ct ) =(Ct )

(1θ)11θ , then u0 (Ct+1) = Cθ

t . Hence,

pt+1pt

=βu0 (Ct+1)u0 (Ct )

=βCθ

t+1

Cθt



It means that the elasticity of substitution, d log

Ct+1Ct

d log(

pt+1pt )

, must satisfy

the following condition.

d log

Ct+1Ct

d log

pt+1pt

= d logCt+1Ct

d log

βCθ

t+1

Cθt

It is shown that

d log

Ct+1Ct

d log

βCθ

t+1

Cθt

= d [log (Ct+1) log (Ct )]dhlogCθt+1

log

Cθt

i= d [log (Ct+1) log (Ct )]

d fθ [log (Ct+1) log (Ct )]g

=1θ.



Suppose that θ 6= 1, then

maxfcetg

∞

∑t=0[β (1+ n)]t N0

(cetTt )1θ 1

1 θ

= maxfcetg

∞

∑t=0[β (1+ n)]t N0

cetT0 (1+ g)

t1θ 11 θ

= maxfcetg

(∑∞t=0 [β (1+ n)]

t N0 (1+ g)(1θ)t (cetT0)

1θ

1θ

∑∞t=τ [β (1+ n)]

t N01θ

)

= N0T0maxfcetg

(∞

∑t=0

hβ (1+ n) (1+ g)(1θ)

it (cet )1θ

1 θ

)

∞

∑t=τ

[β (1+ n)]tN01 θ



Hence

argmaxfcetg

∞

∑t=0[β (1+ n)]t N0

(cetTt )1θ 1

1 θ

= argmaxfcetg

(∞

∑t=0

hβ (1+ n) (1+ g)(1θ)

it (cet )1θ

1 θ

)



Suppose that θ = 1, then U (cetTt ) = ln cetTt . So

maxfcetg

∞

∑t=0[β (1+ n)]t N0 ln cetTt

= maxfcetg

∞

∑t=0[β (1+ n)]t N0

ln cet + ln (1+ g)

t + lnT0

= N0

maxfcetg ∑∞

t=0 [β (1+ n)]t ln cet

+∑∞t=0 [β (1+ n)]

t [t ln (1+ g) + lnT0]

Hence

argmaxfcetg

∞

∑t=0[β (1+ n)]t N0 ln cetTt = argmax

fcetg

∞

∑t=0[β (1+ n)]t ln cet



Hence, the original problem is equivalent to

maxfcetg

(∞

∑t=0(β)t u (cet )

)

s.t. ket+1 =f (ket ) + (1 δ) ket cet

(1+ g) (1+ n), given kτ

where u (cet ) =(cet )

1θ

1θ if θ 6= 1 and log (cet ) , if θ = 1 and

β = β (1+ n) (1+ g)(1θ).

We assume that β = β (1+ n) (1+ g)(1θ) < 1.




V (ket ) = maxcetfu [cet ] + βV (ket+1)g


(1+ g) (1+ n)


1θ

1θ if θ 6= 1 and log (cet ) , if θ = 1 and

β = β (1+ n) (1+ g)(1θ).



First Order Condition and Envelop Condition

u0 [c (ket )] =βV 0 (κ (ket ))(1+ g) (1+ n)

V 0 (ket ) =βV 0 (κ (ket )) [f 0 (ket ) + (1 δ)]

(1+ g) (1+ n)

where

κ (ket ) =f (ket ) + (1 δ) ket c (ket )

(1+ g) (1+ n)



Assignment: Suppose that θ = 1 and therefore u (c) = ln c , δ = 1and f (ke ) = kα

e . Show that

c (ket ) = (1 αβ) kαet

κ (ket ) =αβkα

et

(1+ g)θ


Guess and Verify Method for Growth Model



Euler Equation for Growth Model

We cannot analyze guess and verify method for more general case. Inthis case, we can derive Euler equation.

Remember that the rst order condition and envelop condition are

u0 [c (ket )] =βV 0 (κ (ket ))(1+ g) (1+ n)


(1+ g) (1+ n)



Hence, the Euler equation is

u0 (cet1) (1+ g) (1+ n)β

= u0 (cet )f 0 (ket ) + (1 δ)

.

(1+ g) (1+ n) u0 (cet )

β (1+ n) (1+ g)(1θ) u0 (cet+1)= f 0 (ket+1) + (1 δ)

u0 (cet )βu0 (cet+1)

=f 0 (ket+1) + (1 δ)

(1+ g)θ



Euler Equation 2



Note that u (cet ) =(cet )

1θ

1θ means u0 (cet ) = (cet )θ. Hence the

growth rate of the consumption is

(cet )θ

(cet+1)θ

=β [f 0 (ket+1) + (1 δ)]

(1+ g)θcet+1cet

θ

=β [f 0 (ket+1) + (1 δ)]

(1+ g)θ

cet+1cet

=

"β [f 0 (ket+1) + (1 δ)]

(1+ g)θ

# 1θ



One additional condition is the transversality condition:

0 = limt!∞

(β)t V 0 (ket ) ket

0 = limt!∞

[β]t1 [cet1]θ ket .

An economic interpretation of this transversality condition is that it isnot optimal to keep capital stock at the nal date when the marginalvalue of consumption is positive. If so, the agent can always lowercapital stock and consume more.



Optimal Growth

cet+1cet

=

"[f 0 (ket+1) + (1 δ)]

(1+ $) (1+ g)θ

# 1θ


(1+ g) (1+ n), k0 is given

0 = limt!∞

"(1+ g)1θ

1+ $

#t1[cet ]

θ ket

where β = 11+$ and $ is the discount rate.


Steady State Analysis

Steady State Analysis: the steady state is a point such that

ce = cet+1 = cet , ke = ket+1 = ket

On the steady state

f 0 (ke ) + (1 δ) = (1+ g)θ (1+ $)

f 0 (ke ) = (1+ g)θ (1+ $) (1 δ)

and

ke =f (ke ) + (1 δ) ke ce

(1+ g) (1+ n)ce = f (ke ) + (1 δ) ke (1+ g) (1+ n) ke

= f (ke ) (δ+ g + n+ ng) ke



Note that if an economy reaches the steady state, the transversalitycondition is satised.

0 = limt!∞

"(1+ g)1θ

1+ $

#t1[ce ]

θ ke

0 = limt!∞

"(1+ g)1θ

1+ $

#t1.



Saving rate:

s =Yt CtYt

=[f (ke ) ce ]TtNtf (ke )TtNt

=f (ke ) [f (ke ) (δ+ g + n+ ng) ke ]

f (ke )

=(δ+ g + n+ ng) ke

f (ke )

Hence the constant saving rate is supported around the steady sate.



Assume f (k) = kα. Then f 0 (k) = αkα1. Hence,

s =(δ+ g + n+ ng) ke

(ke )α

(ke )1α =

sδ+ g + n+ ng

ke =

s

δ+ g + n+ ng

11α

.

This condition is the same as that in Solow model.



Now we can endogenize s. Note that on the steady state,

f 0 (ke ) = (1+ g)θ (1+ $) (1 δ)

α (ke )α1 = (1+ g)θ (1+ $) (1 δ) ,

(ke )1α =

α

(1+ g)θ (1+ $) (1 δ),

Hence

s = (δ+ g + n+ ng) (ke )1α

=α (δ+ g + n+ ng)

(1+ g)θ (1+ $) (1 δ)

=α [(1+ g) (1+ n) (1 δ)]

(1+ g)θ (1+ $) (1 δ)



Note that saving rate is larger when we have more population growth.When you expect to have more children in future, you will have moveincentive to save.

When $ is large, the agent largely discounts his future. Since today ismore important than tomorrow, the agent saves less.

Similarly, the larger the marginal rate of substitution 1θ , the saving

rate is. When θ is small, the marginal rate of substitution is large.The agent is willing to change his consumption in response to thechanges in the return. When the economy is growing, the return ishigh. Therefore, this behavior implies that the agent saves more.



The steady state value of capital stop per unit of e¤ective labor is.

ke =

"1

δ+ g + n+ ngα (δ+ g + n+ ng)

(1+ g)θ (1+ $) (1 δ)

# 11α

,

=

"α

(1+ g)θ (1+ $) (1 δ)

# 11α

,

Using that y e = (ke )

α

YtNt=

"α

(1+ g)θ (1+ $) (1 δ)

# α1α

Tt .



Di¤erent from Solow model, population growth has no impact on theGDP per capita. Because the saving rate is an increasing function ofpopulation growth, the negative impacts of population growth onGDP per capita is canceled out.

The more the agent discount the future, (large $), the lower thesaving rate and, therefore, the lower the steady state value of percapita GDP.

Hence, the larger the marginal rate of substitution 1θ , the larger per

capita GDP is.



Assignment: Derive the consumption per capita on the steady state.


Numerical Analysis

In order to obtain a numerical solution, many economists use Matlab.

Although this may not be the fastest program, It is instructive todemonstrate how we can write Matlab code to obtain a numericalpolicy function and numerical value function of the neoclassicalgrowth model.

I rst explain the basic code of Matlab and later I use Matlab toderive a numerical policy function and numerical value function.


Numerical Analysis

Turning on Matlab: To access Matlab, you can go to the networkcenter.

Getting Help in Matlab: If you have questions about the use of aparticular command, you just type

help [command]

Example: the command "plot" plots variables. If you want help onhow to use it, you would typehelp plot


Numerical Analysis

Basic Math in Matlab:

Symbols for basic operations are "+" for addition, "-" for subtraction,"*" for multiplication and "/" for division. "^" is "to the power of."If you want Matlab to perform a computation, just type it.

Example: if you want Matlab to compute "2+2," just type into thecommand line2+2

If you want to assign a value to a variable, just type it out.

Example: if you want the variable "apple" to have the value 3 and thevariable orange to have the value 5, then typeapple=3orange=5


Numerical Analysis

Basic Math in Matlab:

Matlab will remember these values unless you change them.Computations with variables are just as simple as with numbers.

Example: if you want to nd out what apple+orange is, you just typeapple+orange

You will notice that when you type something like "apple=3," thecomputer repeats it. To stop that from happening, just put asemi-colon at the end.

Example: If you dont want the computer to repeat what you havedone, typeapple=3;


Numerical Analysis

Vectors and Matrices: For the most part, in Matlab you will beprogramming with vectors and matrices, not just numbers. This is nobig stretch, however. After all, a variable is just a 1 1 matrix.Matlab uses the notation X (i , j) to refer to the element in the ithrow and the jth column of a matrix X .

Example: if you have a matrix called "apple" and you want the 3rdelement of the second row to equal 4.5, you would typeapple(2,3)=4.5;If you havent told Matlab that you want a matrix called apple that isat least 2x3, however, it may give you an error. Therefore, it is a goodidea to initialize all matrices at the beginning of the program beforeyou actually use them. This is called "declaring" a variable. One wayto do this is to writeapple=zeros(2,3);Matlab creates the 2x3 matrix apple, and sets all its elements to equalzero.


Numerical Analysis

Loops in Matlab: All programming languages of which I am awaremake extensive use of something called "loops". This is a way ofrepeating a certain task many times or under certain conditions.

for... end: This is the classic loop.

Example: Suppose you wanted to create a 101x1 vector called"orange", where the elements are the numbers from zero to 1 inincrements of 0.01. You would do the following:

for i=1:101orange(i)=(i-1)/100;end

Matlab begins by setting i=1, then proceeds until it hits the word"end". Then it returns to the "for" line, sets i=2, and repeats thisprocess until i=101.


Numerical Analysis

You can have multiple concentric loops.

Example: suppose that for some reason you want to have a 100x100matrix X where Xij = i2 + j3. Then you would use:

X=zeros(100,100);for i=1:100for j=1:100

X(i,j)=i^2+j^3;end

end

Notice that what is inside each loop is indented. When you have a lotof loops it can be hard to keep track of things, and indentation is away of telling where you are.


Numerical Analysis

if... end: Another loop-like structure involves conditional statements.

Example: Suppose you have generated a vector of 100 numbers calledX . You want to see whether they are positive or negative, and storethat information in another vector called Y . You could proceed asfollows.

Y=zeros(100,1);for i=1:100if X(i)>0Y(i)=1;

endend

When it reads "if", Matlab checks whether the condition is met(X(i)>0) and if so performs what comes afterwards until it hits "end."If the condition is not met, it just ignores everything until "end".


Numerical Analysis

An alternative is

Y=zeros(100,1);for i=1:100if X(i)>0Y(i)=1;

elseY(i)=0;

endend

Here Matlab does the same as before, except that if the condition isnot met it does not ignore everything, rather it executes thecommands that lie after "else" instead. In this example, of course,that is redundant (since the default value of Y is zero).


Numerical Analysis

Writing your own Programs and Functions in Matlab:Programs: A "program" is nothing but a list of commands (such asthose above) that you save so that you can use them over and overagain. How do you write programs? Just nd any text editor, and typeout the sequence of commands that you desire. Then, save it to a disk

Example: You can save your program to a le such as "macroII.m" Ifyou want to run macroII.m, just typemacroIIThen Matlab will perform all the commands in the le, in order.

There is a function in Matlab called "save." However, "save" does notsave programmes, rather it saves the values of all the variables in thememory.

Example: if you type:save katsuyaMatlab will create a le called katsuya.mat. If for some reason youwanted to retrieve the values of the variables in that le, you wouldtype:load katsuya


Numerical Analysis

Functions: Sometimes you may nd it convenient to write down afunction that does not exist in Matlab. Then you can use acommand function.

Example: This would be a le takii.m:

function [z] = takii(x)z = (x+1)^2;After saving this le, if you input takii(2) into Matlab you should get9 as output: takii.m is now a function. In that program, x representsthe input and z represents the output.


Numerical Analysis

Solving equations: Matlab can also help us to solve non-linearequations.

Example: The following line embedded in another program nds thezeroes of takii, or fx : takii(x) = 0g:fsolve(takii, 10)Matlab starts at x = 10 and uses gradient methods to nd the value ofx such that takii(x) is zero.Why would you want to do this? Many economic problems are solved bynding values of some variable x such that f (x) = 0 for some functionf . For example, the deterministic steady state in the neoclassicalgrowth model is dened in this manner. Fsolve is one way to nd it..


Numerical Analysis

Comments: Good programming style follows certain conventions sothat others can easily pick up your code and see what you are doing.For this purpose, it is important to write comments on your program.

Example: If you want to make a comment, "This is where I computethe steady state." you would type:% Here I compute the steady stateMatlab ignores everything in a line after it sees "%", so you can writeanything there that you wish.


Numerical Analysis

We are now ready to write a program to derive a policy function anda value function of the growth model expressed by the followingBellman equation.

V (ket ) = maxcetfu [cet ] + βV (ket+1)g

ket+1 =kαet + (1 δ) ket cet(1+ g) (1+ n)


1θ

1θ if θ 6= 1, log (cet ) if θ = 1, and

β = (1+n)(1+g )(1θ)

1+$ .


Numerical Analysis

I explain the process with the help of the LECTURE TWO OFRECURSIVE MACROECONOMICS by MARTIN ELLISON. Thefollowing argument is based on his lecture note.

There are four main parts to the code:1 parameter value initialization,2 state space denitions,3 value function iterations4 output.


Numerical Analysis

Parameter value initialization: Start new program by clearing thework space. Parameters are initialized with theta the coe¢ cient ofrelative risk aversion, discount the discount rate, alpha the exponenton capital in the production function, g the growth rate ofproductivity, n the population growth rate, beta discount factor anddelta the depreciation rate.

clear alltheta=1.2;discount=.05;alpha=.3;g=0.02;n=0.02;beta=((1+n)*((1+g)^(1-theta)))/(1+discount);delta=.05;


Numerical Analysis

State space denitions: We now construct a grid of possible capitalstocks. The grid is centered on the steady-state capital stock per unitof e¤ective labor, with maximum ( ) and minimum ( ) values 105%and 95% of the steady-state value. There are N discrete points in thegrid. Note that

ke =

"α

(1+ g)θ (1+ $) (1 δ)

# 11α

.

N=1000;Kstar=(alpha/(((1+g)^(theta))*(1+discount)-(1-delta)))^(1/(1-alpha));Klo=Kstar*0.95;Khi=Kstar*1.05;step=(Khi-Klo)/N;K=Klo:step:Khi;l=length(K);


Numerical Analysis

K is the vector.

K = [Klo,Klo + step,Klo + 2step, ...,Khi ]| z l


Numerical Analysis

Value function iterations: Value function iteration requires 3 steps1 Set initial guess for the value function.2 Express one period return function as a function of current statevariable and future state variable.

In this way, the Bellman equation can be expressed only by currentstate variable and future state variable.

3 Iterations

Add one period return function to the discounted initial guess and ndthe optimal value. This value serves the initial guess in the next round.Iterate it until the initial guess meets the optimal value.


Numerical Analysis

Initial Guess for the value function: initial guess is taken byassuming that all capital is always consumed immediately. In thiscase,

v0 = ln (ytot) if θ = 1

v0 =(ytot)1θ

1 θ, if θ 6= 1.

ytot = kαet + (1 δ) ket .


Numerical Analysis

Hence, our initial guess is written as

ytot = K.âlpha+(1-delta)*K;ytot=ytot*ones(1,l);if theta==1v=log(ytot);elsev=ytot.^(1-theta)/(1-theta);endwhere ones(1,l).are 1 l vector of 1 and the syntax .^ deneselement-by-element operation, so each element of a vector or matrix isrisen to the power of the exponent.


Numerical Analysis

Note that v is l l matrix with every column being the same. Forexample, if θ 6= 1,

v =

2664[(Klo)α+(1δ)Klo]

1θ

1θ[(Klo)α+(1δ)Klo]

1θ

1θ[(Klo)α+(1δ)Klo]

1θ

1θ

.. .. ..

[(Khi )α+(1δ)Khi ]1θ

1θ[(Khi )α+(1δ)Khi ]

1θ

1θ[(Khi )α+(1δ)Khi ]

1θ

1θ

3775


Numerical Analysis

Express one period return function as a function of currentstate variable and future state variable. For use in the valuefunction iterations, we dene an l l matrix C

C =

24 ce (1, 1) ... ce (1, l).. .. ..

ce (l , 1) .. ce (l , l)

35ce (i , j) = kα

et (j) + (1 δ) ket (j) (1+ g) (1+ n) ket+1 (i)ket (j) = Klo + (j 1) step,

ket+1 (i) = Klo + (i 1) step


Numerical Analysis

The program for this operation is

for i=1:lfor j=1:lC(i,j) = K(j).âlpha + (1-delta)*K(j) - (1+g)*(1+n)*K(i);

endend


Numerical Analysis

Next, we need to map consumption levels into utility levels byapplying the suitable utility function. The syntax ./ deneselement-by-element operation rather than matrix division implied byusing the / operator.

if theta==1U=log(C);elseU=(C.^(1-theta))/(1-theta);end


Numerical Analysis

Iteration: Now we can use our rst guess of the value function to getthe second estimate v1, which will be the start of our next Titerations.

T=100for j=1:Tw=U+beta*v;v1=max(w);v=v1*ones(1,l);

end


Numerical Analysis

Note that w is the l l matrix. Suppose θ 6= 1. Then

w =

264[ce (1,1)]

1θ

1θ + βv (1, 1) ... [ce (1,l)]1θ

1θ + βv (l , l).. .. ..

[ce (l ,1)]1θ

1θ + βv (l , 1) .. [ce (l ,l)]1θ

1θ + βv (l , l)

375ce (i , j) = kα

et (j) + (1 δ) ket (j) (1+ g) (1+ n) ket+1 (i)

v (i , j) =

[ket+1 (i)]

α + (1 δ) ket+1 (i)1θ

1 θket (j) = Klo + (j 1) step,

ket+1 (i) = Klo + (i 1) step


Numerical Analysis

Hence, v1 nd the maximum value for each column.

v1 =h[ce (i 1 ,1)]

1θ

1θ + βv (i1 , 1) ..[ce (i l ,l)]

1θ

1θ + βv (il , l)i

where ij is the maximum indices of the j th column.

Hence, v is again, l l matrix with every column being the same.For example, if θ 6= 1,

v =

264[ce (i 1 ,1)]

1θ

1θ + βv (i1 , 1) ..[ce (i 1 ,1)]

1θ

1θ + βv (i1 , 1).. .. ..

[ce (i l ,l)]1θ

1θ + βv (il , l) ..[ce (i l ,l)]

1θ

1θ + βv (il , l)

375


Numerical Analysis

Output: Value function iterations are now complete. The nal valuefunction is stored in Val and the indices of the future ket+1 are storedin ind. These latter indices are converted into ket+1 values in optk.Using optk, we can create the optimal value of consumption, optc.

[val,ind]=max(w);optk = K(ind);optc = K.âlpha + (1-delta)*K - (1+g)*(1+n)*optk;


Numerical Analysis

To illustrate the results, I rst plot the value function.

gure(1)plot(K,v1,LineWidth,2)xlabel(K);ylabel(Value);legend(Value function,4);


Numerical Analysis


Numerical Analysis

Next I plot the policy function

gure(2)plot(K,optc,LineWidth,2)xlabel(K);ylabel(C);legend(Policy function,4);


Numerical Analysis


Numerical Analysis

It is also useful to plot how ket+1 varies as a function of ket . It showsthat an economy converges to the steady state.

gure(3)plot(K,optk,LineWidth,2)hold onplot(K,K,r,LineWidth,2)xlabel(K);ylabel(K);legend(Policy Function 2,45 degree line,4);


Numerical Analysis


Numerical Analysis

Another useful gure is showing net investment ket+1 ket as afunction of ket . The gure conrms that net investment is positivewhen ket < ke and negative when ket > k

e .

gure(4)plot(K,(optk-K),LineWidth,2)xlabel(K);ylabel(K- K);


Numerical Analysis


Numerical Analysis

To gain an insight into the dynamics of the adjustment process, it isuseful to calculate the path of convergence of capital if initial capitalis not equal to its steady-state value. p is the number of periods overwhich to follow convergence. The indices of the capital choices arestored in the p 1 vector mi and the capital stocks themselves arestored in the p 1 vector m. The initial index is taken as 1 so theinitial capital stocks are at its maximum value.

p=50;mi=zeros(p,1);m=zeros(p,1);mi(1)=N;m(1)=K(mi(1));for i=2:pmi(i)=ind(mi(i-1));m(i) = K(mi(i));end


Numerical Analysis

We can plot this dynamics by the following program.

t=1:1:50gure(5)plot(t,m)xlabel(t);ylabel(k(t));


Numerical Analysis


Numerical Analysis

Assignment: go to the following website of MARTIN ELLISON anddownload Matlab program for the deterministic growth model.Modify the program and replicate what I have done in this lecture.

www.economics.ox.ac.uk/members/martin.ellison/recursive/growdisc1.m


Recursive Competitive Equilibrium

Market Economy: So far, I discussed a command economy: thesocial planner maximizes the representative consumers utility functionsubject to the resource constraint. But how does this commandeconomy relate to the market economy? In order to answer thisquestion, remember that the rst and second welfare theorem. Therst welfare theorem says that the market economy is Pareto optimal.The second welfare theorem says that the Pareto optimum allocationcan be supported by a market economy with the income transfer.Note that our command economy is Pareto optimum and since everyagent is identical, there is no reason to talk income transfer in ourmodel. Hence, there must have prices which support our equilibrium.



In this section, we show that our economy can be supported by amarket economy, which we call recursive competitive equilibrium. Inparticular, we show that there is a price mechanism that canreproduce a following policy function and a value function that satisfythe rst order condition and envelope condition of our problem.

u0 [c (ket )] =βV 0 (κ (ket ))(1+ g) (1+ n)

,


(1+ g) (1+ n),

where

κ (ket ) =f (ket ) + (1 δ) ket c (ket )

(1+ g) (1+ n).



Let us rst consider the following representative household problem

maxfct ,ltg∞

t=0

∞

∑t=0

βtNtU(ct )

s.t. At+1 = (1+ ρ (ket ))At + w (ket )TtNt ctNt , A0 = ke0T0N0

ket+1 =f (ket ) + (1 δ) ket Cm (ket )

(1+ g) (1+ n), ke0 is given

Tt+1 = (1+ g)Tt , T0 is given


where ct is consumption per capita, At is the asset of this household,ket , Tt and Nt are aggregate state variables.Note that the interest rate ρ, and the wage per unit of e¤ective labor,w are function of the aggregate capital per unit of e¤ective labor, ket .

We also assume that an initial individual asset is the same as theaggregate capital, A0 = ke0T0N0.



Assignment: Suppose that


(1θ) 11 θ

,

Show that the original problem is equivalent to

maxfcetg

∞

∑t=0(β)t u (cet )

aet+1 =(1+ ρ (ket )) aet + w (ket ) cet

(1+ g) (1+ n), a0 = ke0

s.t. ket+1 =f (ket ) + (1 δ) ket Cm (ket )

(1+ g) (1+ n), ke0 is given


1θ

1θ if θ 6= 1, log (cet ) if θ = 1, and

β = β (1+ n) (1+ g)(1θ).



Bellman Equation

V h (aet , ket ) = maxcet

nu (cet ) + βV h (aet+1, ket+1)

oaet+1 =

(1+ ρ (ket )) aet + w (ket ) cet(1+ g) (1+ n)


(1+ g) (1+ n)



First order conditions

u0(c (aet , ket )) =βV h1 (aet+1, ket+1)(1+ g) (1+ n)

where c (aet , ket ) is a policy function

Envelop Condition

V h1 (aet , ket ) =βV h1 (aet+1, ket+1) [1+ ρ (ket )]

(1+ g) (1+ n)



Assume that a representative rm maximizes capital and labor given afactor price functions

maxKt ,Ht

nFK ft ,TtLt

r (ket )K ft w (ket )TtLt

o,

= Tt maxket ,Lt

nFk fet , 1

r (ket ) k fet w (ket )

oLt

= Tt maxket ,Lt

nfk fet r (ket ) k fet w (ket )

oLt

where k fet =K ftTtLt

. The rental price r is also the function of theaggregate capital per unit of e¤ective labor.The rst order conditions are

r (ket ) = f 0k fe (ket )

,

w (ket ) = fk fe (ket )

r (ket ) k fe (ket )



Arbitrage conditionρ (ket ) = r (ket ) δ

Labor and Capital Market

Lt = Nt ,

k fe (ket )TtLt = aetTtNt

We assume that the initial value of individual asset is the same as thatof the aggregate asset.

aetTtNt = ketTtNt

Two market condition with this initial condition imply

k fe (ket ) = ket .



Denition: Recursive competitive equilibrium consists of a value functionV h1 (aet , ket ), a policy function c (aet , ket ), a corresponding set ofaggregate decision Cm (ket ), production plans k fe (ket ), set of factor pricefunctions, fr (ket ) ,w (ket ) , ρ (ket )g such that these function satisfy

Household

u0(c (aet , ket )) =βV h1 (aet+1, ket+1)(1+ g) (1+ n)

V h1 (aet , ket ) =βV h1 (aet+1, ket+1) [1+ ρ (ket )]

(1+ g) (1+ n)

where

aet+1 =(1+ ρ (ket )) aet + w (ket ) c (aet , ket )

(1+ g) (1+ n)


(1+ g) (1+ n)



Firm

r (ket ) = f 0k fe (ket )

,

w (ket ) = fk fe (ket )

r (ket ) k fe (ket )

Arbitrage conditionρ (ket ) = r (ket ) δ

Capital and Labor Market with initial condition

k fe (ket ) = ket

the consistency of individual and aggregate decisions

Cm (ket ) = c (ket , ket )



Note that there exist k fe (ket ), r (ket ), ρ (ket ) and w (ket ) that satisfythe denition of RCE and

k fe (ket ) = ketr (ket ) = f 0 (ket )

ρ (ket ) = f 0 (ket ) δ,

w (ket ) = f (ket ) f 0 (ket ) ket .



From the denition of the recursive competitive equilibrium

aet+1 =(1+ r (ket ) δ) aet + w (ket ) c (ket , ket )

(1+ g) (1+ n)

=(1+ r (ket ) δ) ket + f (ket ) r (ket ) ket c (ket , ket )

(1+ g) (1+ n)

=fk fe (ket )

+ (1 δ) ket c (ket , ket )(1+ g) (1+ n)

=f (ket ) + (1 δ) ket Cm (ket )

(1+ g) (1+ n)= ket+1



Therefore, householdsrst order condition is

u0(Cm (ket )) =βV h1 (ket+1, ket+1)(1+ g) (1+ n)

Envelop condition is

V h1 (ket , ket ) =βV h1 (ket+1, ket+1) [f

0 (ket ) + 1 δ]

(1+ g) (1+ n)



Dene Vm0 (ket ) V h1 (ket , ket )

u0(Cm (ket )) =βVm0 (ket )(1+ g) (1+ n)

Vm0 (ket ) =βVm0 (ket+1) [f 0 (ket ) + 1 δ]

(1+ g) (1+ n)



Remember that Social Planner Problem is summarized by

u0 (C (ket )) =βV 0 (ket+1)(1+ g) (1+ n)

V 0 (ket ) =βV 0 (ket+1) [f 0 (ket ) + (1 δ)]

(1+ g) (1+ n)

We know that there exists the set of policy functions and the marginalvalue function fC (ket ) ,V 0 (ket )g that satises these equations existsand unique. It means that fCm (ket ) ,Vm0 (ket )g exists and

Cm (ket ) = C (ket ) ,Vm0 (ket ) = V 0 (ket ) .



Given the existence of the aggregate consumption function Cm (ket )and the price functions ρ (ket ) and w (ket ), there existsc (aet , ket ) ,V h1 (aet , ket )

that satises the denition of RCE

because these are the solutions to the standard DynamicProgramming given the price functions and the aggregate decisions.



We have proved that there is a price mechanism that can reproducesocial optimal allocation.

Obviously, we are interested in a decentralized economy, but notplanner problem. But, because the social planner problem is easier todeal with, we mainly discuss an optimal growth model in this lecture.



Assignment: Consider the following social planners problem.

maxfCtg

∞

∑t=0

βtU (Ct ) ,

s.t. Kt+1 = F (Kt ,N) + (1 δ)Kt Ct ,K0 is given.

where Ct is consumption, N is population, Kt is capital stock,β 2 (0, 1) is a discount factor, δ is the depreciation rate of capitalstock. Population is assumed to be constant. Assume that theaggregate production function, F (, ), satises usual assumptions: itexhibits the constant return to scale, the rst derivative is positive,the second is negative and it satises Inada condition. No populationgrowth and no technological growth.

1 Dene the recursive competitive equilibrium that can support theresource allocation of this problem.

2 Show that your equilibrium allocation is the same as the resourceallocation of this problem.


Continuous Time Problem

We discusses how the discrete dynamic programming can be appliedfor the optimal growth model.

However, this is not the only tool to analyze a dynamic optimizationproblem. Depending on the model, it is easier to analyze continuoustime model.

In particular, it is easier to analyze a transition dynamics usingcontinuous time model.

We will discuss the continuous time model as a limit of the discretetime model.



In the previous model, time goes like t, t + 1, t + 2.

In this section, I assume that time goes like t, t + ∆, t + 2∆....Then Idescribe the continuous time model by taking a limit of ∆: ∆ ! 0.

I assume that one period return between t and t + ∆ is constant,∆r (Xt ,St ). Therefore, the sum of the returns is

∆r (Xτ,Sτ) + β∆r (Xτ+∆,Sτ+∆) + β2∆r (Xτ+2∆,Sτ+3∆) + ...,

where τ is an initial period. We can dene at time t by tj = τ + j∆.Hence, j = tjτ

∆ .

Therefore, the above sum can be rewritten as

∆r (Xτ,Sτ) + βt1τ

∆ ∆r (Xt1 ,St1) + βt2τ

∆ ∆r (Xt2 ,St2) ...



Assume thatβ =

11+ ∆$

,

where $ is the discount rate.

Then the continuous version of the original model can be expressed as

U (Sτ) = maxfXtg

8><>: lim∆!0

∞

∑j=0

1

1+ ∆$

(tjτ)∆

∆rXtj ,Stj

9>=>; ,s.t. St+∆ = G (Xt ,St ) = ∆G c (Xt ,St ) + St ,

Sτ is given.



Since

e$t = lim∆!0

1

1+ ∆$

t∆

,

the continuous version of the previous model is written as

U (Sτ) = maxXt

Z ∞

τe$(tτ)r (Xt ,St ) dt

s.t. St = G c (Xt ,St )

Sτ is given.



I would like to derive the continuous version of the Bellman equationas the limit of the previous discrete model. Since β = 1

1+∆ρ

V (St ) = maxXt

∆r (Xt ,St ) +

11+ ∆$

V (St+∆)

,

St+∆ St = ∆G c (Xt ,St ) .

We can modify the value function as follows.

(1+ ∆$)V (St ) = maxXtf(1+ ∆$)∆r (Xt ,St ) + V (St+∆)g

∆$V (St ) = maxXt

∆r (Xt ,St ) + ∆2$r (Xt ,St )+ [V (St+∆) V (St )]

$V (St ) = max

Xt

(r (Xt ,St ) + ∆$r (Xt ,St )

+V (St+∆)V (St )∆

)



When ∆ goes 0, the continuous version of the Bellman equation isderived as follows.

$V (St ) = maxXt

r (Xt ,St ) + V 0 (St ) St

= max

Xt

r (Xt ,St ) + V 0 (St )G c (Xt ,St )

Bellman Equation:

$V (St ) = maxXt

r (Xt ,St ) + V 0 (St )G c (Xt ,St )



The rst order condition of this Bellman equation is

0 = r1 (x (St ) ,St ) + V 0 (St )G c1 (x (St ) ,St ) . (10)

Substitute the policy function into the Bellman equation,

$V (St ) = r (x (St ) ,St ) + V 0 (St )G c (x (St ) ,St ) . (11)

Again, the rst order condition (10) and the Bellman equation (11)solves the value function V () and x (). We can apply the Guessand veried method to solve two equations.



Assignment: Consider the following investment problem:

maxLt ,Kt

Z ∞

0[PF (Kt , Lt ) wLt C (It ,Kt ) It ] e$tdt

s.t. Kt = It δKt

where F (K , L) and C (I ,K ) are constant returns to scale in eachfactors.

1 Dene the Bellman equation.2 Show that the value function is linear in K .



The envelop condition can be also derived.

$V 0 (St ) = r2 (Xt ,St ) + V 00 (St )G c (Xt ,St ) + V 0 (St )G c2 (Xt ,St ) .

Note that V 00 (St )G c (Xt ,St ) = V 00 (St ) St =dV 0(St )dt . Hence the

envelop condition can be rewritten as

dV 0 (St )dt

= $V 0 (St )r2 (Xt ,St ) + V

0 (St )G c2 (Xt ,St ). (12)

This equation describes the dynamics of the marginal value of statevariable V 0 (St ).



Optimal Control: Dening the costate variable, λt , is equal to themarginal value of state variables,

λt V 0 (St ) ,

we can derive the result of alternative method to solve dynamicoptimization, optimal control.

Using λt , the rst order condition (10) and the envelop theorem (12)and the transition equation can be rewritten as

0 = r1 (x (St ,λt ) ,St ) + λtG c1 (x (St ,λt ) ,St ) , (13)

λt = $λt [r2 (x (St ,λt ) ,St ) + λtG c2 (x (St ,λt ) ,St )] , (14)

St = G c (x (St ,λt ) ,St ) , Sτ is given. (15)



The rst order condition (13) determines a policy function x (St )given λt . Given the policy function x (St ), equations (14) and (15)determine the path of the costate variable, λt and the state variable,St .

Note that since we have one boundary condition, Sτ, we need anadditional boundary condition to solve the equations (14) and (15).For this purposes, we need the following transversality condition:

limt!∞

e$tV 0 (St ) St = limt!∞

e$tλtSt = 0. (16)

Similar to the discrete time problem, given usual technicalassumptions, concavity etc, it is shown that equations (13), (14),(12) and (16) are su¢ cient conditions for the original problem.



The above conditions are nicely summarized by dening HamiltonianH (Xt ,St ,λt ):

H (Xt ,St ,λt ) = r (Xt ,St ) + λtG c (Xt ,St ) .

Then above conditions are expressed as

0 = HX (Xt ,St ,λt )

λt = $λt HS (Xt ,St ,λt )St = Hλ (Xt ,St ,λt ) , Sτ is given.

0 = limt!∞

e$tλtSt

These are the useful conditions for analyzing the continuous versionof the dynamic optimization problem.



Hamiltonian can be nicely interpreted. When we choose a dynamicoptimization problem, we know that current decision a¤ects not onlythe current return, but also the future returns. The second term ofHamiltonian summarizes the impact on the future returns.Remember, λt = V 0 (St ). That is, the costate variable can beinterpreted as the marginal impact of the state variable on the presentvalue of the discounted future returns. Since St = G c (Xt ,St ), thesecond term is interpreted as the impact of a change in St on thefuture returns. Therefore, Hamiltonian summarizes the importanttrade o¤ of the dynamic optimization problem: the impact on thecurrent return and the future returns. The rst order condition,0 = HX , implies that the agent chooses the control variable Xt suchas he maximizes Hamiltonian.




maxLt ,Kt

Z ∞

0[PF (Kt , Lt ) wLt C (It ,Kt ) It ] e$tdt

s.t. Kt = It δKt

where F (K , L) and C (I ,K ) are constant returns to scale in eachfactors.

1 Dene Hamiltonian of this problem.2 Derive the necessary conditions and transversality condition for thisproblem.


Continuous Time Growth Model

In this section I apply optimal control to the neoclassical growthmodel and analyze the neoclassical growth model. Consider thefollowing problem.

maxct

Z ∞

0e$tNtU (ct ) dt

Kt = F (Kt ,TtNt ) δKt ctNt ,Tt = gTt ,

Nt = nNt ,

where K0, T0 and N0 are given.



Note thatgket = g K

TN= gK (gT + gN )

Hence

ketket

=F (Kt ,TtNt ) δKt ctNt

KtTtTt+NtNt

,

=[F (ket , 1) δket cet ]TtNt

Kt (g + n) ,

=[f (ket ) δket cet ]

ket (g + n) ,

ket = f (ket ) cet (g + n+ δ) ket ,

where f (kt ) = F (kt , 1). From the second equation to the third, Iuse the assumption on the production function, the constant return toscale.



Lemma

Suppose that Xt = gXt . Then Xt = X0egt .

Proof.

Xt = gXt

0 = Xtegt gXtegt =d (Xtegt )

dt

=Z T

0

d (Xtegt )dt

dt = XT egT X0

XT = X0egT



Using Tt , kt and ct , I can rewrite the original problem asZ ∞

0e($n)tN0U (cetTt ) dt,

ket = f (ket ) cet (g + n+ δ) ket ,

Tt = gTt ,

In order to meet the requirement of the Balanced Growth, assumethat


(1θ) 11 θ



Suppose that θ 6= 1. ThenZ ∞

0e($n)tN0

(cetTt )(1θ) 1

1 θdt

=Z ∞

0e($n)tN0T

(1θ)t

(cet )(1θ)

1 θdt

Z ∞

0

e($n)tN01 θ

dt

=Z ∞

0e($n)tN0

T0egt

(1θ) (cet )(1θ)

1 θdt

Z ∞

0

e($n)tN01 θ

dt

= N0 (T0)(1θ)

Z ∞

0e[$n(1θ)g ]t (cet )

(1θ)

1 θdt

Z ∞

0

e($n)tN01 θ

dt



Hence

argmaxcet

Z ∞

0e($n)tN0

(cetTt )(1θ) 1

1 θdt

= argmaxcet

Z ∞

0e[$n(1θ)g ]t (cet )

(1θ)

1 θdt



Suppose that θ = 1. Then U (cetTt ) = ln cetTt . Hence,Z ∞

0e($n)tN0 ln cetTtdt

=Z ∞

0e($n)tN0 [ln cet + lnTt ] dt

=Z ∞

0e($n)tN0 ln cetdt +

Z ∞

0e($n)tN0 lnT0egtdt

= N0

Z ∞

0e($n)t ln cetdt + lnT0

Z ∞

0e($n)tgtdt

Hence

argmaxcet

Z ∞

0e($n)tN0 ln cetTtdt = argmax

cet

Z ∞

0e($n)t ln cetdt



The original problem becomes

maxcet

Z ∞

0e$t u (cet ) dt,

ket = f (ket ) cet (g + n+ δ) ket , ke0 is given


(1θ)

1θ and $ = $ n (1 θ) g . We assumethat $ > 0.

Dene Hamiltonian of this problem:

H (ct , kt ,λt ) =(cet )

(1θ)

1 θ+ λt [f (ket ) cet (g + n+ δ) ket ]



The rst order conditions are

λt = cθet

λt = $λt λtf 0 (ket ) (g + n+ δ)

, 0 = lim

t!∞λtketeρt

ket = f (ket ) ct (g + n+ δ) ket , ke0 is given



Note thatgλ = gcθ

e= θgce

Hence

θcetcet

=λtλt= $

f 0 (ket ) (g + n+ δ)

= [$ n (1 θ) g ]

f 0 (ket ) (g + n+ δ)

= ($+ δ+ θg) f 0 (ket )

ctct

=1θ

f 0 (ket ) ($+ δ+ θg)

(17)

Equation (17) is the continuous version of Euler equation.



The corresponding transversality condition is

0 = limt!∞

λtkete$t

0 = limt!∞

(cet )θ kete[$n(1θ)g ]t



Together with the transition equation of kt , the following twodi¤erential equations and two boundary conditions solve the optimalgrowth path:

cetcet

=1θ

f 0 (ket ) ($+ δ+ θg)

, (18)

ket = f (ket ) ct (g + n+ δ) ket , ke0 is given. (19)

0 = limt!∞

(cet )θ kete[$n(1θ)g ]t , (20)


Phase Diagram

The Phase Diagram: One of the merits of working with acontinuous model is that we can use the phase diagram. Let medescribe what the phase diagram is. The rst, let me dene thesteady state.

DenitionOn the steady state the path of (ce , k

e ) satises

ce = ke = 0.

Therefore, the following equation must be satised on the steady sate:

ce = 0 : f 0 (ke ) = $+ δ+ θg (21)

ke = 0 : ce = f (ke ) (g + n+ δ) ke (22)


Phase Diagram

ce = 0: Note that ke is uniquely determined by equation (21).1 When ket = ke , c

e = 0.

2 When ket < ke , f0 (ket ) > $+ δ+ θg . ) cet > 0.

3 When ket > ke , f0 (ket ) < $+ δ+ θg . ) cet < 0.


Phase Diagram

Phase Diagram 1


Phase Diagram

ke= 0:1 When ce = f (k

et ) (g + n+ δ) ket , ket = 0

2 When cet > f (ket ) (g + n+ δ) ket , ket < 0.3 When cet < f (ket ) (g + n+ δ) ket , ket > 0.

Equation (22) implies that

dcedke

= f 0 (ke ) (g + n+ δ) ,d2ced (ke )

2 = f00 (ke ) < 0

Hence, ce has the maximum value at

f 0kGRe

= g + n+ δ. (23)

Capital stock, kGRe is called the golden rule level of the capital stock.If you want to maximize ce when ke = 0 , equation (23) must besatised.


Phase Diagram

Phase Diagram 2


Phase Diagram

Compare

f 0 (ke ) = $+ δ+ θg , and f 0kGRe

= g + n+ δ.

Note that

g + n+ δ < $+ δ+ θg i¤ ($ n) > (1 θ) g .

Because we have assumed ρ n > (1 θ) g , f 0 (ke ) > f0 kGRe

,ke < k

GRe .

Since the agent discount future, it is not optimal to reduce currentconsumption to reach the golden rule level of the capital stock. Thesteady state value of capital stock, ke , is called the modied goldenrule level of capital stock.Note that given the assumption of ρ n > (1 θ) g ,

0 = limt!∞

(ce )θ ke e

[$n(1θ)g ]t .


Phase Diagram

Phase Diagram 3


Phase Diagram

The saddle point path: Note that given ke0, there exists c (ke0)which converges to the steady state. If ce0 = c (ke0), economy willeventually reach the steady state. Since the steady state satises thetransversality condition (20), this path is optimal. This path is calledthe saddle point path.


Phase Diagram

Phase Diagram 4


Phase Diagram

If ce0 > c (ke0), it is known that it hits ket = 0 line in a nite time.But when ket = 0, cet must jump to 0 from equation (19).Otherwise, ket becomes negative. However, the Euler equation (18)does not allow the jump. Hence this path violates the Euler equation(18).

If ce0 < c (ke0), the phase diagram says that this path eventuallyconverges to the point cet = 0 and ket < ∞. We show that this pathviolate the transversality condition (20).


Phase Diagram

Proof: Because for large t, ket > kGRet , f 0 (ket ) < f 0kGRe

. Hence

cetcet

=1θ

f 0 (ket ) ($+ δ+ θg)

<

1θ

hf 0kGRe

($+ δ+ θg)

i=

1θ[g + n+ δ ($+ δ+ θg)]

=1θ[(1 θ) g + n $] = $

θ


Phase Diagram

Proof: Hence, for large t

cet < ce0e$θ t .

This implies that

(cet )θ kete$t >

ce0e

$θ tθ

kete$t = ce0ket > 0

Hence,0 < lim

t!∞(cet )

θ kete.

It proves that this path violates the transversality condition.


Phase Diagram

In sum, the unique globally stable optimal path is characterized by thesaddle point path. On this path, for any given ke0, the agentoptimally chooses c (ke0) with his expectation that economy willconverge to the steady state.


Phase Diagram

Assignment: Consider the following social planners problem givengovernment expenditure G :

maxct

Z ∞

0e$tU (ct ) dt

kt = f (kt ) δkt ct g ,

where ct is consumption, g is constant government expenditure, kt iscapital stock, $ is the discount rate, δ is the depreciation rate ofcapital stock. Assume that the aggregate production function, f (),satises usual assumptions: f 0 () > 0, f 00 () < 0, f (0) = 0,f 0 (0) = ∞, f 0 (∞) = 0.

1 Draw and explain the Phase Diagram which summarizes the dynamicsof kt and ct

2 Discuss how government expenditure g inuences the steady statevalue of kt and ct .


Stochastic Dynamic Optimization

So far, we dont have any uncertainty in the model.

Introducing uncertainty in our model greatly increases theapplicability of our method to several other issues such as a nancialmarket, business cycle and labor search.

In this section, we simply add uncertainty to the basic dynamicprogramming.



We introduce random factors into the optimization problem.


τ

Eτ

(∞

∑t=τ

β(tτ)r (Xt ,St )

),

s.t. St+1 = G (Xt ,St , εt+1) ,Sτ is given,

where εt+1 has a distribution function Q (εt+1).



Note that


τ

Eτ

(∞

∑t=τ

β(tτ)r (Xt ,St )

)

= maxXτ


fXtg∞τ+1

Eτ

∞

∑t=τ+1

β(t(τ+1))r (Xt ,St )

)

= maxXτ

(r (Xτ,Sτ) +

βmaxfXtg∞τ+1EτEτ+1 ∑∞

t=τ+1 β(t(τ+1))r (Xt ,St )

)

= maxXτ


fXtg∞τ+1

EτU (Sτ)

)



The Bellman equation can be written as

V (St ) = maxXτ

r (Xt ,St ) + β

ZV (G (Xt ,St , εt+1)) dQ (εt+1)

where Z

V (G (Xt ,St , εt+1)) dQ (εt+1)

=ZV (G (Xt ,St , εt+1))Q 0 (εt+1) dεt or

=n

∑i=1V (G (Xt ,St , εi ))Pr (εt+1 = εi )

andPr (εt+1 = εi ) = Q (εi )Q (εi1)

We can analyze stochastic dynamic optimization problem using themethod similar to the deterministic case.



Example: Consider the following Lucas (1978) model of asset prices:1 Consider an economy consisting of a large number of identical agents.2 Each agent is born with one tree. Each period, each tree yieldsfruits (=dividends) in the amount dt into its owner at the beginning ofperiod t.

3 Let p (dt ) is the price of a tree in period t, which is measured in unitsof consumption goods per tree.



Household

maxfctg

Eτ

∞

∑t=τ

β(tτ)U (ct ) ,

s.t. ct + p (dt ) st+1 = [p (dt ) + dt ] stdt+1 = G (dt , εt+1) , εt+1~Q (εt+1)

where st is the number of trees owned at the beginning of period, t.

Capital market clearing condition: Because everything is identical,nobody sells and buys in an equilibrium.

st = 1



Bellman Equation

V (st , dt ) = maxct

U (ct ) +

βRV (st+1, dt+1) dQ (εt+1)

s.t. st+1 =

[p (dt ) + dt ] st ctp (dt )

dt+1 = G (dt , εt+1) , εt+1~Q (εt+1)



First Order Conditions:

U 0 (ct ) = βZ V1 (st+1, dt+1)

p (dt )dQ (εt+1)

Envelop Condition:

V1 (st , dt ) = βZ[p (dt ) + dt ]p (dt )

V1 (st+1, dt+1) dQ (εt+1)



Euler Equation:1 Combining the rst order condition and the envelop condition

[p (dt ) + dt ]U 0 (ct ) = V1 (st , dt )

2 Using envelop condition, Euler equation can be derived as

[p (dt ) + dt ]U 0 (ct )

= βZ[p (dt ) + dt ]p (dt )

[p (dt+1) + dt+1 ]U0 (ct+1) dQ (εt+1)

Hence

U 0 (ct ) = βZ[p (dt+1) + dt+1 ]

p (dt )U 0 (ct+1) dQ (εt+1)



TVC:

0 = limt!∞

βtE [V1 (st , dt ) st jdτ]

= limt!∞

βtEτ

[p (dt ) + dt ]U 0 (ct ) st

The operator Et [] is the conditional expectation given the informationavailable at date t. Hence, stochastic version of the transversalitycondition demands an expectation conditioning on the informationavailable at date τ.



Equilibrium Condition implies

st = 1) ct = dt

Euler equation and TVC are

U 0 (dt ) = βZ[p (dt+1) + dt+1]

p (dt )U 0 (dt+1) dQ (εt+1)

0 = limt!∞

βtEτ

[p (dt ) + dt ]U 0 (dt )



Euler Equation can be rewritten as

p (dt ) =ZMt+1 [p (dt+1) + dt+1] dQ (εt+1)

= Et [Mt+1 [p (dt+1) + dt+1]]

= E [Mt+1 [p (dt+1) + dt+1] jdt ]

where Mt+1 = βU 0(ct+1)U 0(ct )

= βU 0(dt+1)U 0(dt )

is the intertemporal marginal rateof substitution also known as the stochastic discount factor or pricingkernel. Because information needed to predict future sequences ofdt+i is summarized by dt , the last equality must be satised.



Note that

p (dt )

= Et [Mt+1 [p (dt+1) + dt+1]]

= Et [Mt+1dt+1] + Et [Mt+1p (dt+1)]

= Et [Mt+1dt+1] + Et [Mt+1Et+1 [Mt+2 [p (dt+2) + dt+2]]]

= Et [Mt+1dt+1] + Et [Mt+1Mt+2dt+2] + Et [Mt+1Mt+2p (dt+2)]

= Et [Mt+1dt+1] + Et [Mt+1Mt+2dt+2]

+Et [Mt+1Mt+2Et+2 [Mt+3 [p (dt+3) + dt+3]]]

= Et [Mt+1dt+1] + Et [Mt+1Mt+2dt+2]

+Et [Mt+1Mt+2Mt+3 [p (dt+3) + dt+3]]

=∞

∑i=1EtΠix=1Mt+xdt+i

+ limi!∞

EtΠix=1Mt+x [p (dt+i ) + dt+i ]



Note that

Πix=1Mt+x = β

U 0 (ct+1)U 0 (ct )

βU 0 (ct+2)U 0 (ct+1)

βU 0 (ct+3)U 0 (ct+2)

..βU 0 (ct+i )U 0 (ct+i1)

= βiU 0 (ct+i )U 0 (ct )

Hence

limi!∞

EΠix=1Mt+x [p (dt+i ) + dt+i ] jdt

= lim

i!∞E

βiU 0 (ct+i )U 0 (ct )

[p (dt+i ) + dt+i ] jdt

=limi!∞ βiE [U 0 (ct+i ) [p (dt+i ) + dt+i ] jdt ]

U 0 (ct )= 0



The share price is an expected discounted stream of dividends with astochastic discount factor. .

p (dt ) =∞

∑i=1EΠix=1Mt+xdt+i jdt

where

Πix=1Mt+x = βi

U 0 (dt+i )U 0 (dt )



Euler Equation can be also rewritten as

1 =ZMt (1+ ρrt ) dQ (εt ) = E [Mt (1+ ρrt )]

where ρrt =dt+p(dt )p(dt1)

p(dt1)is the return from this asset.



Lemma

Cov (Xt ,Yt ) = E [XtYt ] E [Xt ]E [Yt ]

Proof.

Cov (Xt ,Yt )

= E [(Xt E [Xt ]) (Yt E [Yt ])]= E [XtYt E [Xt ]Yt XtE [Yt ] + E [Xt ]E [Yt ]]= E [XtYt ] E [Xt ]E [Yt ] E [Xt ]E [Yt ] + E [Xt ]E [Yt ]= E [XtYt ] E [Xt ]E [Yt ]



Using lemma,

1 = E [Mt (1+ ρrt )] = E [Mt ]E [1+ ρrt ] + Cov(Mt , 1+ ρrt )

Hence we can get

E [1+ ρrt ] =1

E [Mt ][1 Cov(Mt , 1+ ρrt )]

E [ρrt ] =1

E [Mt ][1 Cov(Mt , 1+ ρrt )] 1 (24)



Risk free rate of return: If there is a risk free asset, the return onthis asset ρ must satisfy

Cov(Mt , 1+ ρ) = 0

Hence, equation (24) implies

1+ ρ =1

E [Mt ]

Return on the Risky Asset: Substituting the risk free rate intoequation (24),

E [1+ ρrt ] = (1+ ρ) [1 Cov(Mt , 1+ ρrt )]

E [ρrt ] = ρ (1+ ρ)Cov(Mt , 1+ ρrt )



A stock return must be equal to risk free rate of return +Riskpremium. The risk premium is now decreasing function of covariancebetween a return and Mt .

Why? Note that because Mt = βU 0(ct )U 0(ct1)

,

ct "= U 0 (ct ) #= Mt #

So when covariance is low, we will expect a high return when Mt islow, that is, your consumption is high, and a low return when yourconsumption is low. So you cannot smooth your consumption withthis asset much. This model is called Consumption Capital AssetPricing Model (Consumption CAPM).




maxIt ,Lt

E0∞

∑t=0

Πtx=0Mx [ztF (Kt , Lt ) wLt C (It ,Kt ) It ]

Kt = It + (1 δ)Ktzt+1 = zt + εt+1, εt+1~Q (εt+1)

where F (Kt , Lt ) and C (It ,Kt ) are constant returns to scale,Mt < β 1 is the stochastic discount factor. (For this exercise, thesequence of Mt is taken as given. )

1 Dene the Bellman Equation of this problem.2 Show that the value function is linear in Kt .3 Derive Euler Equation of this problem.


Stochastic Growth Model

In this section, we introduce a stochastic shock into the neoclassicalgrowth model.

In addition to the shock, we allow that household can value leisure.

These framework plays the benchmark for studying business cycle.



The general problem is as follows:

maxfct ,ltg∞

t=0

E0∞

∑t=0

βtNtU(ct , 1 lt )

s.t. Kt+1 = ztF (Kt ,Tt ltNt ) + (1 δ)Kt ctNt , K0 is givenTt+1 = (1+ g)Tt , T0 is given


ln zt+1 = ζ ln zt + εt+1, z0 is given, εt+1~Q (εt+1)



Assignment: Assume that the utility exhibits CRRA:

U (cetTt , lt ) =[cetTtv (1 lt )]1θ 1

1 θ

Show that the original problem can be rewritten as

maxfcet ,ltg

E0∞

∑t=0(β)t u (cet , 1 lt )

s.t. ket+1 =ztF (ket , lt ) + (1 δ) ket cet

(1+ g) (1+ n). ke0 is given


where u (cet , 1 lt ) = [cetv (lt )]1θ

1θ if θ 6= 1,u (cet , 1 lt ) = ln cet + ln v (1 lt ) if θ = 1 andβ = β (1+ n) (1+ g)(1θ).



Assuming that β < 1, the following Bellman Equation can bedened.

V (ket , zt ) = maxfcet ,ltg

u(cet , 1 lt ) + β

ZV (ket+1, zt+1) dQ (εt+1)

s.t. ket+1 =

ztF (ket , lt ) + (1 δ) ket cet(1+ g) (1+ n)

ln zt+1 = ζ ln zt + εt+1,



First Order Conditions

u1(c (ket , zt ) , 1 l (ket , zt )) =βRV1 (κ (ket , zt ) , zt+1) dQ (εt+1)

(1+ g) (1+ n)

u2(c (ket , zt ) , 1 l (ket , zt ))

=βRV1 (κ (ket , zt ) , zt+1) dQ (εt+1) ztF2 (ket , l (ket , zt ))

(1+ g) (1+ n)

where κ (ket , zt ) =ztF (ket ,l(ket ,zt ))+(1δ)ketc (ket ,zt )

(1+g )(1+n) .

Hence, consumption and leisure relationship is inuenced by

u2(cet , 1 lt )u1(cet , 1 lt )

= ztF2 (ket , lt )





Envelop condition

V1 (ket , zt ) =

βRV1 (κ (ket , zt ) , zt+1) dQ (εt+1)

[ztF1 (ket , l (ket , zt )) + 1 δ]

(1+ g) (1+ n)



Euler Equation1 Combining the rst order condition and envelop condition,

V1 (ket , zt ) = u1(cet , 1 lt ) [ztF1 (ket , lt ) + 1 δ]

2 Using envelop condition

u1(cet , 1 lt ) [ztF1 (ket , lt ) + 1 δ]

=

βRu1(cet+1, 1 lt+1) [zt+1F1 (ket+1, lt+1) + 1 δ] dQ (εt+1)

[ztF1 (ket , lt ) + 1 δ]

(1+ g) (1+ n)

Hence,

u1(cet , 1 lt )

=Z

βu1(cet+1, 1 lt+1)[zt+1F1 (ket+1, lt+1) + 1 δ]

(1+ g) (1+ n)dQ (εt+1)

=Z


(1+ g)θdQ (εt+1)



Euler Equation

u1(cet , 1 lt )

=Z


(1+ g)θdQ (εt+1)

TVC

0 = limt!∞

(β)t E0 [V1 (ket , zt ) ket ]

= limt!∞

(β)t E0 [u1(cet , 1 lt ) [ztF1 (ket , lt ) + 1 δ] ket ]



Summary Dynamics (cet , lt , ket , zt )Intratemporal substitution between leisure and consumption


= ztF2 (ket , lt )

Euler Equation and TVC

u1(cet , 1 lt )

=Z


(1+ g)θdQ (εt+1)

0 = limt!∞

(β)t E0 [u1(cet , lt ) [ztF1 (ket , 1 lt ) + 1 δ] ket ]

where β = β (1+ n) (1+ g)(1θ).



Summary Dynamics (cet , lt , ket , zt )Capital accumulation and the dynamics of productivity shocks.

ket+1 =ztF (ket , lt ) + (1 δ) ket cet

(1+ g) (1+ n). ke0 is given



Numerical Analysis

With their seminal analysis of business cycle, Kidland and Prescott(1982) develops a new method of empirical work, calibration, inmacroeconomics. In order to understand its spirits, Let Kidland andPrescott (1996) speak how computational experiment can beimplemented, in particular with a stochastic growth model.

According to Kidland and Prescott (1996), an economiccomputational experiment has four steps.

1 Pose a Question2 Use a well-tested theory3 Construct a model economy4 Calibrate the model economy5 Run the experiment


Numerical Analysis

Pose a question: The purpose of a computational experiment is toderive a quantitative answer to some well-posed question. Thus, therst step in carrying out a computational experiment is to pose such aquestion.Kidland and Prescott (1996) Question may involve

1 Assessments of the theoretical implications of changes in policy.2 the assessments of the ability of a model mimic features of the actualeconomy.

In case of a basic real business cycle model, Kidland and Prescott(1996) poses

Does a model designed to be consistent with long-term economicgrowth produce the sort of uctuation that we associate with thebusiness cycle?How much would the U.S. postwar economy have uctuated iftechnology shocks had been the only source of uctuations?


Numerical Analysis

Use a well-tested theory: With a particular question in mind, aresearcher needs some strong theory to carry out a computationalexperiment: that is, a researcher needs a theory that has been testedthrough use and found to provide reliable answers to a class ofquestions. Here, by theory we do not mean a set of assertions aboutthe actual economy. Rather, following Lucas (1980), economictheory is dened to be an explicit set of instructions for building... amechanical imitation system to answer a questionKidland andPrescott (1996)

The basic theory used in the modern study of business cycles is theneoclassical growth model. Kidland and Prescott (1996)


Numerical Analysis

Construct a Model Economy: With a particular theoreticalframework in mind, the third step in conducting a computationalexperiment is to construct a model economy. Here, key issues are theamount of detail and the feasibility of computing the equilibriumprocess. Kidland and Prescott (1996)

..an abstraction can be judged only relative to some given question.To criticize or reject a model because it is an abstraction is foolish: allmodels are necessarily abstractions. A model environment must beselected based on the question being addressed. Kidland and Prescott(1996)The selection and construction of a particular model economy shouldnot depend on th answer provided. In fact, searching within someparametric class of economies for the one that best ts a set ofaggregate time series makes little sense, because it isnt likely toanswer an interesting question. Kidland and Prescott (1996)


Numerical Analysis

In case of a basic real business cycle model, they add a productivityshock and the choice of leisure to the neoclassical growth modelbecause two-thirds of aggregate output are attributable touctuations of labor.


Numerical Analysis

Calibrate the Model Economy: Now that a model has beenconstructed, the fourth step in carrying out a computationalexperiment is to calibrate that model. Originally, in physical sciences,calibration referred to the graduation of measuring instrument.Kidland and Prescott (1996)

Generally, some economic questions have known answers, and themodel should give an approximately correct answer to them if we are tohave any condence in the answer given to the question with unknownanswer. Thus, data are used to calibrate the model economy so that itmimics the world as closely as possible along a limited, but clearlyspecied, number of dimensions. Kidland and Prescott (1996)


Numerical Analysis

Calibrate the Model Economy: In case of a standard business cyclemodel,

1 the parameters must meet the long run averages of certain statisticssuch as the share of output paid to labor and the fraction of availablehours worked per household, both of which have been remarkablystable over time.

2 In addition, empirical results obtained in micro studies conducted atthe individual or household level are often used as a means of pinningdown certain parameters.


Numerical Analysis

Calibrate the Model Economy: The di¤erence between estimationand calibration is also emphasized.

Note that calibration is not an attempt at assessing the size ofsomething: it is not estimation, It is important to emphasize thatthe parameter values selected are not the ones that provide the best tin some statistical sense. In some cases, the presence of a particulardiscrepancy between the data and the model economy is a test of thetheory being used. Kidland and Prescott (1996)Prescott (1986) also claims that The models constructed within thistheoretical framework are necessarily highly abstract. Consequently,they are necessarily false, and statistical hypothesis testing will rejectthem. This does not imply, however, that nothing can be learned fromsuch quantitative theoretical exercises.


Numerical Analysis

Run the Experiment: The fth and nal step in conducting acomputational experiment is to run the experiment.Kidland andPrescott (1996)

If the model economy has no aggregate uncertainty, then it is simply amatter of comparing the equilibrium path of the model economy withthe path of the actual economy.If the model economy has aggregate uncertainty, rst a set ofstatistics summarize relevant aspects of the behavior of the actualeconomy is selected. Then the computational experiment is used togenerate many independent realizations of the equilibrium process forthe model economy. In this way, the sampling distribution of this setof statistics can be determined to any degree of accuracy for the modeleconomy with the value of the set of statistics for the actual economy.Kidland and Prescott (1996)

A standard business cycle exercise typically compares the secondmoment such as the variance and the covariance of variables.


Numerical Analysis

Cooley and Prescott (1995) describes the process of calibration moreprecisely. They identify three steps.

1 The rst step is to restrict a model to a parametric class.2 to construct a set of measurements that are consistent with theparametric class of models.

3 The third step is to assign values to the parameters of our models.

Let us conduct these steps using our models.


Numerical Analysis

Restricting the model: we need to restrict technology andpreference.

We have already restricting a part of our model to meet the balanced

growth path: u (cet , 1 lt ) = [cetv (1lt )]1θ

1θ if θ 6= 1,u (cet , lt ) = ln cet + ln v (1 lt ) if θ = 1. Cooley and Prescott (1995)assumes that θ = 1 and ln v (1 l) = η ln (1 l). Therefore,u (cet , 1 lt ) = ln cet + η ln (1 lt ).Knowing that labor share and capital share is stable, it is common toassume F (ke , l) = kα

e l1α. Because it is known that

α = rKY = 1 wL

Y , even if the economy is on the transition path, theshare must be stable.We assume that Q (εt+1) is normally distributed with the mean 0 andthe standard deviation of σε.


Numerical Analysis

Note that u (cet , 1 lt ) = ln cet + η ln (1 lt ) andyet = ztF (ke , l) = ztkα

e l1α implies u1 (cet , 1 lt ) = 1

cet,

u2 (cet , 1 lt ) = η1lt , ztF1 (ket , lt ) = αztkα1

et l1αt = α yetket and

ztF1 (ket , lt ) = (1 α) ztkαet lαt = (1 α) yetlt . Therefore, our

economy is summarized by the following equations

Intratemporal substitution between leisure and consumption


= ztF2 (ket , lt )

ηcet1 lt

= (1 α)yetlt

where yet = ztkαet l1αt .


Numerical Analysis

Euler Equation and TVC

u1(cet , 1 lt )

=Z


(1+ g)θdQ (εt+1)

1cet

=Z

β

cet+1

hα yet+1ket+1

+ 1 δi

1+ gdQ (εt+1)

0 = limt!∞

(β)t E0 [u1(cet , lt ) [ztF1 (ket , 1 lt ) + 1 δ] ket ]

= limt!∞

[β (1+ n)]t E0

24hα yetket + 1 δ

iket

cet

35where yet = ztkα

et l1αt .


Numerical Analysis

Capital accumulation and the dynamics of productivity shocks.

ket+1 =yet + (1 δ) ket cet(1+ g) (1+ n)

. ke0 is given

ln zt+1 = ζ ln zt + εt+1, z0 is given, εt+1~N0, σ2ε

where yet = ztkα

et l1αt .


Numerical Analysis

Dening Consistent Measurements: Cooley and Prescott (1995)was a quite careful about the measurement of capital stock, andoutput. Because the model is abstract, there is no treatment ofgovernment sector, household production, foreign sector andinventory. Therefore, the model economys capital K includes capitalused in all these sectors plus the stock of inventory. Similarly,output includes the output productivity by all of this capital. Theydid some imputation. I discuss how they impute the values.

1 Estimate the return on asset ρt from private capital stock.2 They estimate depreciation rate δi separately for consumer durablesand government sector.

3 Using the estimated ρ and δi , they estimate the service ow forconsumer durables and government sector, riK it = (ρ+ δi )K it


Numerical Analysis

They estimate the return on asset ρt from private capital stock.

Note that ρt =rKpδKpKp

where Kp is private capital stock.

Kp is estimated by (1) the net stock of xed reproducible privatecapital (not including the stock of consumer durables) + (2) the stockof inventories and (3) the stock of land where (1) is from Musgrave(1992), (2) is from the NIPA and (3) is from the Flow of FundsAccounts.rKp is estimated by unambiguous capital income + αpambiguouscapital income +δK where αp is estimated to satisfy

αp =the estimated rKp

GDP , .δK is estimated by GNP-NNP (consumption ofxed capital), unambiguous capital income is the sum of rental income,corporate prots and net interest, and ambiguous capital income is sumof proprietors income + net national product - national income.


Numerical Analysis

Next they estimate depreciation rate for government sector andconsumer durable δi by the following equation

K it+1Yt+1

Yt+1Yt

= (1 δi )K itYt+I itYt

where i is consumer durables or government sector and Yt is GNP.For consumer durables,

I itis consumption of consumer durable as

reported in the NIPA. For government investmentI itis also from

NIPA. The capital stockK itfor consumer durable and government

investment are taken from Musgrave (1992).

Finally, they estimate the service ows of capital from the followingequation.

riK it = (ρt + δi )K it


Numerical Analysis

Calibrating a Specic Model Economy: Remember that oureconomy is summarized by 4 equations with suitable boundaryconditions

ηcet1 lt

= (1 α)yetlt

1cet

=Z

β

cet+1

hα yet+1ket+1

+ 1 δi

1+ gdQ (εt+1)


, given ke0

ln zt+1 = ζ ln zt + εt , given z0 εt+1~N0, σ2ε

where yet = ztkα

et l1αt and 0 = limt!∞ [β (1+ n)]

t E0

[α yetket +1δ]ket

cet

.

Note that yet , cet and ket are interpreted as the detrended GNP,Consumption and Capital per capita.We need to calibrate α, β, δ, η, n, g , ζ and σε.


Numerical Analysis

g = 0.0156...measured by the time series average of rate of growth ofreal GNP per capita (On the balanced growth, g Y

N= g).

n = 0.012...average population growth.

α = 0.4...capital share is calibrated by the time series average of

α =rtKtYt

=rpK

pt + rcK

ct + rgK

gt

GNP + rcK ct + rgKgt


Numerical Analysis

ζ = 0.95 and σε = 0.0007: Note that

ln zt = lnYt a lnKt (1 α) ln ltTtNt

It means that

ln zt+1 ln zt= (lnYt+1 lnYt ) a (lnKt+1 Kt )

(1 α) (ln lt+1Nt+1 ln ltNt ) (1 α) (lnTt+1 lnTt )= (lnYt+1 lnYt ) a (lnKt+1 Kt )

(1 α) (ln lt+1Nt+1 ln ltNt ) (1 α) g

So given α and g we can generate a series of ln zt+1 ln zt andtherefore ln zt given ln z0. (Cooley and Prescott (1995) actuallymeasures it assuming g = 0. Note that it does not change theresults because ln z0 is arbitrary.) These sequences are used tocalibrate ζ and σε of

ln zt+1 = ζ ln zt + εt+1, εt+1~N0, σ2ε


Numerical Analysis

In order to calibrate the remaining parameters β, δ, and η. We usemore theory and stylized facts on the balanced growth.

δ = 0.012 : On the steady state, capital accumulation equationimplies

ke =y e + (1 δ) ke ce(1+ g) (1+ n)

(1+ g) (1+ n) ke = (1 δ) ke + ie

(1 δ) ke = (1+ g) (1+ n) ke ie1 δ = (1+ g) (1+ n) ie

ke

δ =IK+ 1 (1+ g) (1+ n)

Given the measured values of g and n, Cooley and Prescott (1995)calibrate δ using this equation.


Numerical Analysis

β = 0.987: If there is no uncertainty, Euler equation becomes

1cet

=β

cet+1

hα yet+1ket+1

+ 1 δi

1+ g

On the steady state,

1β

= αy eke+ 1 δ

=αYK + 1 δ

1+ g

Given the measured value of δ, Cooley and Prescott (1995) calibrateβ using this equation.


Numerical Analysis

η = 1.78: Intratemporal substitution between consumption andleisure implies

ηcet1 lt

= (1 α)yetlt

η = (1 α)yetcet

1 ltlt

.

Given the knowledge of α, Cooley and Prescott (1995) calibrate ηusing this equation. For lt , they use micro evidence from timeallocation studies by Ghez and Becker (1975) and Juster and Sta¤ord(1991) that shows one third of their discretionary time is allocated tomarket activities.


Numerical Analysis

Final task is to run experiment. For this purpose, we need tocompute an equilibrium. Note that our economy is summarized by 4equations with suitable boundary conditions

ηcet1 lt

= (1 α)yetlt

1cet

=Z

β

cet+1

hα yet+1ket+1

+ 1 δi

1+ gdQ (εt+1)


, given ke0


where yet = ztkα

et l1αt and 0 = limt!∞ [β (1+ n)]

t E0

[α yetket +1δ]ket

cet

.


Numerical Analysis

If the set of possible states of nature were small, one could calculatean approximate value function and policy function. The advantagesof this method would be

1 It would preserve the curvature of the model.2 It would allow study of the model in an equilibrium that is far from thesteady sate.

If the set of possible states of nature were large, we need to rely on alinear approximation of the model.


Numerical Analysis

One way to approximate model is to conduct the log-linearapproximation of the derived dynamics around the steady state. Thismethod has four step processes.

1 To nd the steady states without stochastic shocks.2 To construct a log-linear approximation of the model around steadysate.

3 To derive the solution of the resulting log-linear approximation of themodel.

4 Analyze the solution.

You can write your program to conduct this process using Matlab.


Dynare

Dynare can greatly help writing this program.

What is Dynare?

Dynare is a powerful and highly customizable engine, with an intuitivefront-end interface, to solve, simulate and estimate DSGE models.(Dynare User Guide)..., it is a pre-processor, and a collection of Matlab routines that hasthe great advantages of reading DSGE model equations written almostas in an academic paper. (Dynare User Guide)Basically, the model and its related attributes, like a shock structurefor instance, is written equation by equation in an editor of your choice.The resulting le will be called the .mod le. This le is the calledfrom Matlab. This initiates for the Matlab routines used to eithersolve or estimate the model. (Dynare User Guide)


Dynare

Installing Dynare1 Go to

http://www.dynare.org/download

2 Download Dynare 4.1 to your holder at the network center.3 It is also useful to download the user guide from

http://www.dynare.org/documentation-and-support/user-guide


Dynare

The Procedure:1 You write a lename.mod le, which contains the essential informationon your model. Then, using this information, Dynare constructs alename.m le that Matlab can then run.

2 Save your lename.mod le somewhere that Matlab can nd it.3 In Matlab,

1 Set a path to Dynare folder including sub-folder.2 Change the directory to the folder that contains lename.mod.3 Type

dynare lename


Dynare

Basic Instructions for the program by Dynare.1 Each instruction must be terminated by a semicolon, ;.2 You can comment out any line by starting the line with two forwardslashes, //, or comment out an entire section by starting with /* andending with */.


Dynare

The program contains the following 4 blocks.1 Preamble: lists variables and parameters.2 Declaration of the model: spells out the model.3 Set initial conditions: Specify initial conditions and gives indicationto nd the steady state of a model.

4 Specifying Shocks: denes the shocks to the system.5 Computation: instructs Dynare to undertake specic operations (e.g.forecasting, estimating impulse response functions, and providingvarious descriptive statistics.)


Dynare

Preamble consists of the some setup information: the endogenousand exogenous variables, the parameters and assign values to theseparameters.

1 var species the lists of endogenous variables, to be separated bycommas. In our example, endogenous variables are detrendedconsumption per capita ce , detrended capital per capita ke , detrendedGNP per capita (ye ), hours of work (l), and the technology (z).

2 varexo species the list of exogenous variables that will be chocked.In our example, an exogenous variable is a stochastic term, ε.

3 parameters species the list of parameters of the model and assignsvalues:


Dynare

var c, k, l, y, lnz;varexo epsilon;parameters beta, alpha, eta, delta, g, n zeta;beta=0.987; alpha=0.4; eta=1.78; delta=0.012; g=0.0156; n=0.012;zeta=0.95;


Dynare

Declaration of the model: Now it is time to specify the modelitself. You must write

model;describing your model;end;


Dynare

There are several rules for describing your model.1 There need to be as many equation as you declared endogenousvariables.

2 Time subscripts: "x" means xt , "x(-n)" means xtn and "x(+n)"means xt+n .

3 In dynare, the timing of each variable reects when that variable isdecided. Because state variables are in general determined at past, atransition equation should be written as St = G (Xt1,St1, εt ), butnot St+1 = G (Xt ,St , εt+1). On the other hand, a the variable shownas x (+1) tells Dynare to count that variable as a forward-lookingvariable. These distinctions are important to insure saddle pointstability.

4 You can ignore the expectation terms when you write a program.


Dynare

Remember that our economy is summarized by the following 5equations.

ηcet1 lt

= (1 α)yetlt

1cet

=Z

β

cet+1

hα yet+1ket+1

+ 1 δi

1+ gdQ (εt+1)


, given ke0


yet = ztkα

et l1αt

and 0 = limt!∞ [β (1+ n)]t E0

[α yetket +1δ]ket

cet

.


Dynare

model;(eta*c)/(1-l)=(1-alpha)*(y/l);1/c=(beta/c(+1))*((alpha*(y(+1)/k)+1-delta)/(1+g));k=(y+(1-delta)*k(-1)-c)/((1+g)*(1+n));lnz=zeta*lnz(-1)+epsilon;y=exp(lnz)*(k(-1)âlpha)*(l^(1-alpha));end;


Dynare

Set initial conditions: Next, you must tell Dynare initial conditions.

initval;Listing the initial conditionsend;

Dynare can help nding your models steady sate by calling theappropriate Matlab functions. But it is usually only successful if theinitial values you entered are closed to the true steady state. Henceit is good idea that you can solve your steady state and nd thesteady state values by yourself.


Dynare

Assignment: Compute the steady state value of (ce , ye , ke , l) whenβ = 0.987, α = 0.4, η = 1.78, δ = 0.012, g = 0.0156, n = 0.012,and ζ = 0.95 using Matlab. You have Two ways of doing so.Choose one of them.

1 Derive an analytical solution of (ce , ye , ke , l) on the steady state whenσ2ε = 0. Then compute (ce , ye , ke , l) using Matlab or Excel.

2 Use the command fsolveand functionin Matlab to nd the steadystate value of (ce , ye , ke , l) when σ2ε = 0.


Dynare

In addition, if you start your analysis from your steady state, you canadd

steady;

Then Dynare recognizes that your initial conditions are justapproximations and you wish to start your analysis from your steadysate.


Dynare

initval;k=15.84207; c=0.992578; l=0.35531; y=1.622889; lnz=0;epsilon=0;end;steady;


Dynare

Adding Shocks: We then specify the innovations and their matrix ofvariancecovariance. This is done by writing:

shocks;var [the name of stochastic term] =[number];end;

In our program,

shocks;var epsilon=0.0007^2;end;


Dynare

Computation: Finally, you tell Dynare that you are done by typing

stoch_simul;

This commands instructs Dynare to compute a Taylor approximationof the decision and transition function for the model, impulseresponse functions, dxt+jdut

and various descriptive statistics (moments,variance decomposition, correlation and autocorrelation coe¢ cients.)

There are several options which you can learn later. For example, bydefault Dynare drops the rst 100 values, but you can change thatusing the appropriate option.


Dynare


Dynare


Dynare


Dynare


Dynare

Implus Response Functions


Dynare

Assignment: Reproduce what I did by using Dynare.


Dynare

Log-linearization: Let me briey illustrate how dynare conductslog-linearization. We just discuss the rst order approximation,though Dynare can conduct the second order approximation. Thespirits of derivation for the second order approximation is the same asthe rst order one.

Note that DSGE model is in general summarized by the collection ofthe rst order equilibrium conditions that take the general form:

Et [f (xt+1, xt , xt1, ut )] = 0

where xt is the vector of endogenous variables and ut is the vector ofstochastic shocks.

In our example, xt = [ct , kt , lt , yt , zt ] and ut = [0, 0, 0, 0, εt ]


Dynare

LemmaSuppose that Γ (x) is n 1 vector and x is m 1 vector. Then

Γ (x) Γ (x) +rΓ (x) (x x)

where

rΓ (x) =

264∂Γ1(x )

∂x1..

∂Γ1(x )∂xm

.. ..∂Γn(x )

∂x1..

∂Γn(x )∂xm

375and Γi (x) and xj are ith element of Γ (x) and jth element of xrespectively.


Dynare

Corollary

Γ (x) Γ (x) +Mx

where

M =

264∂Γ1(x )

∂x1x1 ..

∂Γ1(x )∂xm

xm.. ..

∂Γn(x )∂x1

x1 ..∂Γn(x )

∂xmxm

375where x = ln x ln x.


Dynare

Proof: Note that

Γ (x) = Γ (exp y) = Γ (y)

where y = ln x . Applying lemma

Γ (exp y) Γ (exp y ) +rΓ (y ) (y y )

Note that

rΓ (y ) =

264∂Γ1(y )

∂y1..

∂Γ1(y )∂ym

.. ..∂Γn(y )

∂y1..

∂Γn(y )∂ym

375and

∂Γi (y )∂yj

=∂Γi (exp yj )

∂ exp yjexp yj

Knowing that exp y = x, the desired result is immediate.


Dynare

Applying the corollary,

f (xt+1, xt , xt1, ut )

= f (x, x, x, 0) + [M+,M0,M,Mu ] [xt+1, xt , xt , ut ]T

= f (x, x, x, 0) +M+xt+1 +M0xt +Mxt +Mu ut

where M+, M0, M and Mu are the Matrix with the element of∂Γ(y )

∂y y where y is xt+1, xt , xt1 and ut evaluated at the steady state.Note that on the steady sate

f (x, x, x, 0) = 0

and by denition

Et [f (xt+1, xt , xt1, ut )] = 0

Hence, after the rst order approximation, the dynamics of DSGEmodel around the steady state is summarized by

0 = Et [M+xt+1 +M0xt +Mxt1 +Mu ut ]


Dynare

Solution to the log-linearized dynamics: Because the dynamics ofDSGE model is linear around steady state, we seek to obtain thefollowing solution.

xt = Px xt1 + Pu ut

and the dynamics is stable.

Note that this solution implies

xt+1 = Px xt + Pu ut+1= Px [Px xt1 + Pu ut ] + Pu ut+1


Dynare

Hence,

0 = Et [M+xt+1 +M0xt +Mxt1 +Mu ut ]

= Et

M+ [Px [Px xt1 + Pu ut ] + Pu ut+1]

+M0 [Px xt1 + Pu ut ] +Mxt1 +Mu ut

=

M+P2x +M0Px +M

xt1

+ f[M+Px +M0]Pu +Mug ut +M+PuEt [ut+1]=

M+P2x +M0Px +M

xt1 + f[M+Px +M0]Pu +Mug ut

for any xt1 and ut .


Dynare

Hence, Px and Pu must be chosen to satisfy

0 = M+P2x +M0Px +M0 = [M+Px +M0]Pu +Mu

Dynare compute Px and Pu to satisfy this relationship and to insurethe stability of dynamics.


Overlapping Generation Model

So far, we analyze an economy based on a representative agentmodel. In this model, we assume that agents behave as if they neverdie. Some of you may think that this assumption is extreme.

There is another popular model to analyze economic growth. It iscalled an overlapping generation model (OGM). The main di¤erencefrom the representative agent model is that there is turnover in thepopulation. New generation comes to society; old generation leaves.This model is especially useful when you are interested in interactionsacross generations.

The main purpose of this section is to show that (1) when we haveturnover in the population, economy may not be Pareto optimal and(2) a representative agent model can be viewed as the model in whichparents care about their children.



Basic Model:

For the simplest case, each agent lives in two periods; young and old.In each period, new young enters the economy and old leaves.Therefore there is turnover in the population in each period.The agents are identical in their generation, but di¤er acrossgeneration.





Household: The agent with a constant risk aversion solves thefollowing problem:

maxCyt ,Cot+1

(c (1θ)yt 11 θ

+1

1+ $

c (1θ)ot+1 11 θ

)s.t.snt = wt cyt (25)

cot+1 =1+ ρt+1

snt (26)

where snt is net saving, cyt and cot are consumption when he is youngand old, respectively. When the agent is young, he earns wage wt .He decides whether he consumes today or save it for the next period.When he gets age, he retires and receives income from his saving.Since he does not have the next period, he consumes all his wealthwhen he is old.



This model can be simplied by

maxsnt

([wt snt ](

1θ) 11 θ

+1

1+ $

1+ ρt+1

snt(1θ) 1

1 θ

)The rst order condition is

[cyt ]θ =

1+ ρt+1

[cot+1]

θ

1+ $

Hence,cot+1cyt

=

1+ ρt+11+ $

1θ

(27)

This is OGM version of the Euler equation. The agent compares theinterest rate and the discount rate. When the interest rate is higherthan the discount rate, he saves more and increases his consumptiontomorrow. When the discount rate is larger, he increases hisconsumption today.



Combing two budget constraints (25) and (26), I can derive theintertemporal budget constraint:

cyt +cot+11+ ρt+1

= wt . (28)

It shows that the present value of the stream of lifetime consumptionis equal to the present value of lifetime income, which, in this case,wages during young.Substituting the Euler equation (27) into (28),2641+

1+ρt+11+$

1θ

1+ ρt+1

375 cyt = wt .Therefore, consumption is linear in wages.

cyt =(1+ $)

1θ

(1+ $)1θ +

1+ ρt+1

1θθ

wt



We can derive the saving rate, sρt+1

, as the function of the

interest rate:

sρt+1

=

cyt wtwt

=

1+ ρt+1

1θθ

(1+ $)1θ +

1+ ρt+1

1θθ

As you can see, the saving rate can be an increasing or decreasingfunction of the interest rate:

s 0ρt+1

> 0, if 1 > θ

s 0ρt+1

< 0, if 1 < θ

s 0ρt+1

, if 1 = θ



This is the result of a usual substitution e¤ect and income e¤ect.When the interest rate increases, the return to saving increases.Therefore the agent saves more (substitution e¤ect). On the otherhand an increase in the interest rate implies an increases in lifetimeincome. Therefore, the agent can consumes more (income e¤ect).As I discussed in the representative agent model, the parameter 1θ canbe interpreted as the elasticity of substitution between consumptiontoday and tomorrow. It measures the sensitivity of consumption tothe change in price. When 1

θ is larger than 1, θ is smaller than 1, theagent is sensitive to the change in the return to saving. Therefore,substitution e¤ect dominates income e¤ect. The saving rate is anincreasing function of the interest rate.

Using this saving rate, the net saving is

snt = sρt+1

wt



Firm: Firmsmaximization conditions are the same as before. Thefollowing prot maximization condition characterizes the rmsbehavior:

rt = f 0 (ket )

wt =f (ket ) f 0 (ket ) ket

Tt

As usual, I express the rst order conditions by the e¢ ciency unitterm.

Intermediation: The following arbitrage condition is also the same asbefore:

ρt = rt δ



Labor Market Clearing Conditions: Since only the young works in thiseconomy, the demand for labor must be equal to the population ofthe young.

Lt = Nyt

where Nyt is the population of the young.



Capital Market Clearing Conditions: We assume that current netsaving by young increases the asset at the next period. But when theagent becomes old, he consumes every asset he has, which reducestotal asset in the economy in the next period. Therefore, thefollowing condition must be satised.

At+1 At = snt Nyt snt1Nyt1

Because this relationship must be satised for any t,

At+1 = snt Nyt

Since the demand for capital must be equal to the supply of asset,

Kt+1 = At+1 = snt Nyt .



We assume that the growth rate of productivity and population is gand n, respectively:

Tt+1 = (1+ g)TtNyt+1 = (1+ n)Nyt

Using e¢ ciency unit, both market clearing conditions are summarizedby

Kt+1Tt+1Nyt+1

Tt+1Tt

Nyt+1Nyt

Tt = snt

ket+1 (1+ g) (1+ n)Tt = snt .

where ket+1 =Kt+1

Tt+1Nyt+1. Note that the denominator is not total

population, but the number of young population because only youngcan work in this economy.



Equilibrium: The market equilibrium consists of the sequence ofsnt , ρt+1,wt , rt , kt

∞t=0, which satises

Consumer maximizes his utility:

snt = sρt+1

wt (29)

where sρt+1

=

1+ ρt+1

1θθ

(1+ $)1θ +

1+ ρt+1

1θθ

Firm maximizes its prots:

rt = f 0 (ket ) (30)

wt =f (ket ) f 0 (ket ) ket

Tt (31)



The arbitrage condition:

ρt+1 = rt+1 δ (32)

Market clearing condition:

ket+1 (1+ g) (1+ n)Tt = snt . (33)



Combining equilibrium conditions, we can derive

ket+1 =snt

(1+ g) (1+ n)Tt

=sρt+1

wt

(1+ g) (1+ n)Tt

=s (rt+1 δ) [f (ket ) f 0 (ket ) ket ]Tt

(1+ g) (1+ n)Tt

=s (f 0 (ket+1) δ) [f (ket ) f 0 (ket ) ket ]

(1+ g) (1+ n)

ket+1 =(f 0 (ket+1) + 1 δ)

1θθ

(1+ $)1θ + (f 0 (ket+1) + 1 δ)

1θθ

[f (ket ) f 0 (ket ) ket ](1+ g) (1+ n)

(34)



As you can see, this is a nonlinear rst order di¤erence equation.Hence, potentially, we can solve it given an initial condition k0.Unfortunately, we cannot say much about the property of thisdynamic equation. It is well known that the OGM can produce thevariety of dynamics. It is possible to have multiple steady states inthe OGM. In another case, we cannot determine the dynamics ofOGM. Moreover, OGM can yield a chaotic uctuation. Thefollowing gures are some examples.





The dynamics of the OGM depends on the parameters of theproduction function and the utility function. This can be seen as theadvantages and disadvantages of OGM. On one hand, it gives uspossible explanations for several puzzling evidence including businesscycle. On the other hand, it is di¢ cult to tell where we stand andwhat would be policy implication. Anything goes.



Two additional assumptions simplify the dynamics of our solution.Assume that θ = 1 and the production function is Cobb-Douglasf (k) = kα.. Then equation (34) is

ket+1 =(1 α) (ket )

α

(2+ $) (1+ g) (1+ n)

A (ket )α

where A = (1α)(2+$)(1+g )(1+n)





In this case, there is a unique steady state and economy globally andmonotonically converges to the steady state. Let me derive thesteady state value of ket . On the steady state kt = kt+1 = k. Then

k =

1 α

(2+ $) (1+ g) (1+ n)

11α

(35)


Dynamic Ine¢ ciency

Dynamic Ine¢ ciency: Let me discuss the potential ine¢ ciency ofthe OGM. To do so, I examine the level of capital stock whichmaximizes consumption per capita on the steady state, which is calledthe golden rule level of capital stock. Then I compare this capitalstock and the steady state capital stock which the OGM can reach.

Note that if the steady state capital stock level is larger than thegolden rule level of capital stock, we can easily nd new allocation inwhich everybody is better o¤. To see this, lower the capital stock onthe steady state to the golden rule level during period t. Since theagent can enjoy more consumption during period t, the agent isbetter o¤. From the next period, economy reaches the golden levelof capital stock. Since the golden rule level of capital stockmaximizes consumption per capita on the steady state, the followinggenerations can enjoy higher consumption and they are better o¤. Inother word, the steady state is not e¢ cient.



Because the representative agent model is always Pareto Optimal, thesteady state capital stock, ke , is always smaller than the golden rulelevel of capital stock, kGR . Remember that if ke > k

GR , the TVCcannot be satised. Note that because OGM model does not need toconsider an innite period later, there is no TVC.

Phase Diagram 3



Firstly, note that maximizing consumption per capita is the same asmaximizing consumption per workers:

CtNyt +Not

=Ct

Nyt +Nyt1+n

,

=1+ n2+ n

CtNyt

,

where Ct = cytNyt + cotNot . Hence, I nd the level of capital stockwhich maximizes consumption per workers given any level oftechnology.



Social planner faces resource constraint any time. The resourceconstraint is

Kt+1 = F (Kt ,TtNyt ) Ct + (1 δ)Kt

This means

Kt+1Tt+1Nyt+1

Tt+1Nyt+1TtNyt

= F

KtTtNyt

, 1 CtTtNyt

+ (1 δ)KtTtNyt

ket+1 (1+ g) (1+ n) = f (ket ) cet + (1 δ) ket

where cet = CtTtNyt

and ket = KtTtNyt

.



On the steady state ket = ket+1 = ke and cet = ce . Hence,

ce = f (ke ) + [(1 δ) (1+ g) (1+ n)] ke= f (ke ) (δ+ g + n+ ng) ke

Hence, the level of capital stock which maximizes consumption perworkers given the level of productivity is

dc

dk= f 0

kGR

(δ+ g + n+ ng) = 0

Note that if ng 0, this golden rule is the same as the golden rule inthe representative agent model.



Assume that f (k) = kα, the golden rule level of capital stock perunit of e¤ective labor is

αkGR

α1= (δ+ g + n+ ng)

kGR =

α

δ+ g + n+ ng

11α

(36)



Because

k =

1 α

(2+ $) (1+ g) (1+ n)

11α

I get

kGR < k, ifα

δ+ g + n+ ng<

1 α

(2+ $) (1+ g) (1+ n).



Hence, there exists parameter values with which the steady state isnot Pareto optimal. This result is contrasted with the one of therepresentative agent model in which the allocation is always Paretooptimal. This result sounds surprising since market is competitiveand no externality in Overlapping generation model. The mainreason is that we have the innite number of agents in our economy.The social planner can transfer resources from young to old without amarket. Obviously, the current old is better o¤ and the currentyoung is worse o¤. However, the social planner can compensate thecurrent young when he becomes old by transferring resources fromthe next generation. Since we have the innite number ofgenerations, nobody may be worse o¤. This is the reason forine¢ ciency. This is called the dynamic ine¢ ciency.





Assignment: A household with a constant risk aversion solves thefollowing problem:

maxcyt ,cot+1

log cyt +

11+ $

log cot+1

s.t.snt = wt cyt τ, cot+1 =

1+ ρt+1

snt + B

Government levies a lump sum tax, τ and uses the proceeds to paybenets to old individual by B . Suppose that Government balancesbudget constrained: B = (1+ n) τ. Assume that productionfunction is K α

t (TtLt )(1α) and the maintain other assumptions in our

lecture note.1 Derive the di¤erence equation of ket = Kt

TNyt. How is the equation

inuenced by τ?2 Derive the steady state level of capital stock per unit of e¤ectiveworkers, ke . How is it inuenced by τ?

3 Compare the derived ke and kGR . Discuss how τ a¤ects the welfare.


Altruism

The above reasoning suggests that if one generation cares about thenext generation, we may be able to recover e¢ ciency.

This motivate us to modify the model to include altruism.


Altruism

Suppose that the agents cares about their children. More concretely,the young during period t maximizes the following utility function:

Ut = u (Cyt ) +1

1+ $

u (cot+1) +

1+ n1+ ϕ

Ut+1

,

where ϕ is the measure of selshness. If ϕ = 0, parents treat theirchildren like themselves. The variable Ut is the total sum ofdiscounted utility of the young during period t. During period t + 1,an agent expects to have 1+ n children. Since the parents are selshin that they care about themselves more than their children, thebenets from their children are discounted by 1

1+ϕ . Hence their

utility from their children is 1+n1+ϕUt+1.


Altruism

Assume that0 = lim

s!∞βsUt+s .

where β = 1+n(1+$)(1+ϕ)

. Then utility of generation t is

Ut = u (Cyt ) +1

1+ $u (cot+1) + βUt+1

= u (Cyt ) +1

1+ $u (cot+1) + β

u (Cyt+1) + 1

1+$u (cot+2)+βUt+2

= u (Cyt ) +

11+ $

u (cot+1)

+β

u (Cyt+1) +

11+ $

u (cot+2)+ β2Ut+2

=∞

∑s=0

βsu (cyt+s ) +

11+ $

u (cot+s+1)+ lims!∞

βsUt+s


Altruism

Ut

Ut =∞

∑s=0

βsu (cyt+s ) +

11+ $

u (cot+s+1)

Assume that parents can transfer their income to their children.Then the budget constraint of each generation must change:

cyt + st = wt +mt (37)

cot+1 + (1+ n)mt+1 =1+ ρt+1

st (38)

where mt is the transfer from the parents to their children. Hence,

(1+ n)mt+1 =1+ ρt+1

[wt +mt cyt ] cot+1

mt+1 =1+ ρt+11+ n

[wt +mt cyt ]cot+11+ n


Altruism

This problem is equivalent to

maxfCyt+s ,Cot+s+1g

∞

∑s=0

βsu (cyt+s ) +

11+ $

u (cot+s+1)

s.t mt+1 =1+ ρt+11+ n


where β = 1+n(1+$)(1+ϕ)

.

This is essentially the problem of representative agent. In therepresentative agent model, the market economy is Pareto optimal.(Remember that a social planner model can be attained by RCE.)The overlapping generation model with altruism can also attainPareto optimal. Because parents care about childrens utility andbudget constraint is connected by bequests, any transfer from youngto old cannot improve olds utility.


Altruism

Note that this analysis implicitly assume that income transfer alwaysoccurs. When the parents are selsh enough, they do not want toleave any bequests. To see this, we can dene the Bellman equationof this problem as

V (mt , t) = maxCyt ,Cot

u (cyt ) +

11+ $

u (cot+1)+ βV (mt+1, t + 1)

s.t. mt+1 =

1+ ρt+11+ n


where β = 1+n(1+$)(1+ϕ)

.


Altruism

The FOCs are

u0 (cyt ) = βV1 (mt+1, t + 1)1+ $t+11+ n

=1+ $t+1

(1+ $) (1+ ϕ)V1 (mt+1, t + 1)

u0 (cot+1) = βV1 (mt+1, t + 1)1+ $

1+ n

=1

(1+ ϕ)V1 (mt+1, t + 1)


Altruism

Although the equation presumes the FOCs are hold with equality, if ϕis large enough, the rst order conditions may not be satised withequality. It is optimal for the parents to consume their all income. Inthis case, the intergenerational linkages are broken and the transfersfrom the next generation can improve the utility of the currentgeneration. Hence, the equivalence of the representative agent modeland the OGM occurs if the parentsaltruism is strong enough.


Discrete Choice and Labor Search

So far, we presume that the rst order conditions are valid. Thisassumption may not be suitable when the choice set is discrete.

There are several discrete decisions in our life: to marry withsomebody, to accept a job, to move a di¤erent region and so on.

In particular, unemployment is likely to be inuenced by the behaviorof job search.

We discuss how discrete choice model can be analyzed in a frameworkof a job search model.



Original Problem:

maxEτ

∞

∑t=τ

β(tτ) [I (st = e)wt + I (st = u) b]

s.t. st+1 = G [st , ot , dt ] ,

wt+1 = Ω [st , ot , dt ,wt , εt ]

where w is wage b is the unemployment benets or/and the value ofleisure, I (st = x) = 1 if st = x I (st = x) = 0 otherwise, and

1 st 2 fe, ug, e and u indicate employment and unemployment,respectively.

2 ot 2 fm, ng, m and n indicate meeting and not meeting with a job,respectively

3 dt 2 fa, rg, a and r indicates accepting or rejecting an o¤er,respectively.



We assume the following probability distribution

Qεo (εt jot = m) = Qε (εt ) , Qεo (εt jot = n) = 0

Qos (mjst = e) = 0, Qos (njst = e) = 1,Qos (mjst = u) = λ, Qos (njst = u) = 1 λ

We also assume the following properties of the transition functions:

e = G [e, n, dt ] , u = G [u, n, dt ] = G [u,m, r ] , e = G [u,m, a] ,

w = Ω [e, ot , dt ,w , εt ] , 0 = Ω [u, n, dt , 0, εt ] = Ω [u,m, r , 0, εt ] ,εt = Ω [u,m, a, 0, εt ]



Bellman Equation

W (e,w) = w + βW (G [e, n, a] ,w)

W (u, 0) = b+ β

24 λRmax

W (G [u,m, a] , ε) ,W (G [u,m, r ] , 0)

dQ (ε)

+ (1 λ)W (G [u, n, r ] , 0)

35



DeneU W (u, 0) ,W (w) = W (e,w)

We can rewrite Bellman equation by

W (w) = w + βW (w)

and

U = b+ β

λZmax fW (ε) ,Ug dQ (ε) + (1 λ)U

= b+ β

λZmax fW (ε) U, 0g dQ (ε) + U

(1 β)U = b+ βλ

Zmax fW (ε) U, 0g dQ (ε)



Note that W (w) can be solved as

W (w) =w

1 β

HenceW 0 (w) > 0



Reservation wage: Strictly increasing function of W () means thatthere exists a unique reservation wage wR that has a property

W (wR ) = U

W (w) > U, 8w > wRW (w) < U, 8w < wR

Optimal Decision: an optimal decision is

at = a, w wR= r , w < wR



Value Function: Dene W (w) = max [W (w) ,U ]. Then

W (w) = max [W (w) ,W (wR )]

=w

1 βif w wR

=wR1 β

if w < wR





Reservation wage:

(1 β)W (wR ) = b+ βλZmax fW (ε)W (wR ) , 0g dQ (ε)

wR = b+ βλZmax

ε

1 β wR1 β

, 0dQε

wR = b+βλ

1 β

Zmax fε wR , 0g dQε > b

where βλ1β

RWR[ε wR ] dQ (ε) is the option value to wait.



LemmaIf w > b, there exist a unique wR that satises

wR = b+βλ

1 β

ZwR[ε wR ] dQ (ε)



Proof: Dene

T (wR : b) : T (wR : b) = wR b+

βλ

1 β

ZwR[ε wR ] dQ (ε)

Note that T function has a property:

T1 (wR : b) = 1 βλ

1 β

[wR wR ]

ZwRdQ (ε)

= 1+

βλ

1 β

ZwRdQ (ε)

> 0

T (0 : b) = b+

βλ

1 β

Z0

εdQ (ε)< 0, T (w : b) = w b > 0

Because T (wR : b) is continuous in wR , there exist a unique wR thatsatises T (wR : b) = 0.



dwRdb : Note that

T2 (wR : b) < 0

Because T (wR : b) = 0 for any pair of wR and b,

0 = T1 (wR : b) dwR + T2 (wR : b) dbdwRdb

= T2 (wR : b)T1 (wR : b)

> 0

That is, increases in unemployment benets raise workersreservationwages.



Hazard Rate: the probability of an employed worker accepting a jobat date t is called the hazard date.

H = P (st = ejss = u, 8s < t) = λ [1Q (wR )]

Using hazard rate, the probability of being unemployed for exactly mperiods can be expressed as follows.

P (m) = (1H)m1 H

Therefore, the expected duration of unemployment is

E [m] =∞

∑m=1

m (1H)m1 H



LemmaThe expected duration of unemployment is the inverse of hazard rate:

E [m] =1H



Proof: Note that for all H

1 =∞

∑m=1

(1H)m1 H

Hence

0 =d ∑∞

m=1 (1H)m1 H

dH

=∞

∑m=1

(1H)m1 ∞

∑m=1

(m 1) (1H)m2 H

= (1H)∞

∑m=1

(1H)m1 ∞

∑m=1

(m 1) (1H)m1 H

=∞

∑m=1

(1H)m1 ∞

∑m=1

m (1H)m1 H



Proof:∞

∑m=1

m (1H)m1 H =∞

∑m=1

(1H)m1

=∑∞m=1 (1H)

m1 HH

=1H



dE [m]db

E [m] =1

λ [1Q (wR )]

Because dwRdb > 0,

dE [m]db > 0.

Because an increase in unemployment benets raise the reservationvalue of workers, workers are more likely to reject o¤ers. Therefore,the expected duration of unemployment is longer.


Stochastic continuous DP and Labor Search

In general, it is not easy to analyze stochastic process undercontinuous time model.

However, if we assume a particular process such as a poisson processor Brownian motion, we can derive a handy solution.

Researchers often analyze a continuous search model under theassumption that the arrivals of a job o¤er follow a poisson process.

Using the similar model, we discuss how we can analyze continuousmodel.



Consider a possibility of multiple o¤ers

U = b+ β

∑∞n=1 λ (n)

Rmax fW (ωn) ,Ug dQ (ωn : n)

+ [1∑∞n=1 λ (n)]U

= b+ β

"∞

∑n=1

λ (n)Zmax fW (ωn) U, 0g dQ (ωn : n) + U

#

(1 β)U = b+ β∞

∑n=1

λ (n)Zmax fW (ωn) U, 0g dQ (ωn : n)

where ωn = max fε1, ε2, ..εng.



Assume that β = 11+∆$ and replace w , b and λ (n) by ∆w , ∆b and

λ (n,∆). We can rewrite the discrete problem as follows.

W (w) = ∆w +1

1+ ∆$W (w)

1 1

1+ ∆$

U

= ∆b+1

1+ ∆$

"∞

∑n=1

λ (n,∆)Zmax fW (ωn) U, 0g dQ (ωn : n)

#



W (w)

W (w) = ∆w +1

1+ ∆$W (w)

(1+ ∆$)W (w) = (1+ ∆$)∆w +W (w)

∆$W (w) = ∆w + ∆2$w$W (w) = w + ∆$w

HenceW (w) =

w$



U 1 1

1+ ∆$

U

= ∆b+∑∞n=1 λ (n,∆)

Rmax fW (ωn) U, 0g dQ (ωn : n)

1+ ∆$

∆$U = (1+ ∆$)∆b

+∞

∑n=1


= ∆b+ ∆2$b

+∞

∑n=1


$U = b+ ∆$b+∞

∑n=1

λ (n,∆)∆

Zmax fW (ωn) U, 0g dQ (ωn : n)



Assume that λ (n,∆) is given by a Poisson density function withparameter λ∆:

λ (n,∆) =(λ∆)n eλ∆

n!Note that

lim∆!0

λ (n,∆)∆

= lim∆!0

λn∆n1eλ∆

n!= λ if n = 1

= 0 if n 2

This limiting case is called a Poisson process.



Hence, the continuous version of Bellman equation is

$U = b+ λZmax fW (ω1) U, 0g dQ (ω1 : 1)

= b+ λZmax fW (ε) U, 0g dQ (ε)



Assignment:1 Suppose that W (w) = max fW (w) ,Ug. Draw the picture of thevalue function W (w).



Because W (w) = w$ ,

$U = b+ λZmax

ε

$ U, 0

dQ (ε)

Hence, there exist a reservation wage wR

wR$= U.

and it must satisfy

wR = b+ λZmax

ε

$ wR

$, 0dQ (ε)

= b+λ

$

ZwR(ε wR ) dQ (ε) > b



Assignment:1 Show that If w > b, there exist a unique wR that satises

wR = b+λ

$

ZwR[ε wR ] dQ (ε)

2 Show that dwRdb > 0.



Hazard Rate: Similar to the discrete time model, hazard rate can beobtained as follows.

H = P (st = ejss = u, 8s < t) = λ [1Q (wR )]

Let us assume that F (t) is the probability to accept before at date t.Then the hazard rate can be written as

P (st = ejss = u, 8s < t) =F 0 (t)1 F (t)

It means that

F 0 (t) = H [1 F (t)]F 0 (t) +HF (t) = H



LemmaSuppose that xt follows a di¤erential equation:

dxtdt= at + btxt

1 Suppose that xτ is given. Then the solution to this di¤erentialequation is

xT = eR T

τ bsdsxτ +

Z T

τate

R tτ bsdsdt

2 Suppose that 0 = limT!∞ xT e

R Tt bsds . Then the solution to this

di¤erential equation is

xτ = Z ∞

τate

R tτ bsdsdt



Proof: Note that the di¤erential equation can be rewritten as

ateR t

τ bsds =dxtdte

R tτ bsds btxte

R tτ bsds

=dhxte

R tτ bsds

idtZ T

τate

R tτ bsdsdt =

Z T

τ

dhxte

R tτ bsds

idt

dt

= xT eR T

τ bsds xτ



.

Proof:

Suppose that xτ is given.

xT = eR T

τ bsdsxτ +

Z Tτate

R tτ bsdsdt

Suppose that 0 = limT!∞ xT e

R Tt bsds .

xτ = Z ∞

τate

R tτ bsdsdt



.

Applying the lemma to this context

H = F 0 (t) +HF (t)

HeHt = F 0 (t) eHt +HF (t) eHt

=dF (t) eHt

dtZ T

0HeHtdt =

hF (t) eHt

iT0h

eHtiT0

= F (T ) eHT F (0)

eHT 1 = F (T ) eHT

F (T ) = eHtheHT 1

i= 1 eHt



.

Hence, a probability that a worker stay unemployment pool exactly tperiod is

F 0 (t) = HeHt



Lemma

E [t] =Z ∞

0tHeHtdt =

1H

Proof.Note that for any H

1 =Z ∞

0HeHtdt

0 =Z ∞

0

dHeHt

dHdt =

Z ∞

0

heHt tHeHt

idt

Hence Z ∞

0tHeHtdt =

Z ∞

0eHtdt =

e

Ht

H

∞

0=1H


Equilibrium Unemployment

So far, we demonstrate an individual search decision problem.

In this section, we extend our analysis to the labor market equilibriummodel and derive equilibrium unemployment.

For our simple analysis, we construct a model that has a uniquemarket wage and everybody accepts the o¤er when they meet a jobo¤er.

This model is the standard unemployment model that can beintegrated into the stochastic growth model we have studied. This isthe current benchmark model for the study of unemployment.



Search and Labor Wedge: Recently Chari, Kehoe and McGattan(2007) argue that the e¢ ciency and labor wedges together account foressentially all of uctuations. Because e¢ ciency wedges is just like aproductivity, zt , labor wedges might be the main source of distortion.

What is the labor wedge? Note that our social planner problemsuggests that the marginal substitution between leisure andconsumption must be equal to the marginal productivity of labor.


= ztF2 (ket , lt )

Hence any deviation of the marginal productivity of labor from themarginal substitution of leisure and consumption is considered to belabor wedge.

Because a search model explicitly considers labor market friction, itcan naturally derive this deviation.



There are three key factors in a model:1 The importance of specic investment: As we discuss in modernmacroeconomics 1, specic investment is necessary to derive anunemployment. We reemphasize our points in a dynamic marketequilibrium model. In this way, we can construct a model that cananalyze what inuences the natural rate of unemployment throughmeeting rate and separation rate.

2 Whether or not a rm can commit its wage payment: Because ofthe existence of specic investment, when they make a production,they both have rent to share. How the sharing rule inuences thenature of unemployment.

3 Trading Externality: When there is something to share, searchingsomebody is useful not only for themselves but also for your potentialpartners. This suggests that searching behavior naturally containsexternality. We will discuss what kind of externality exists and how itinuences economy.



Matching Function: One of standard assumption is the existence ofthe matching function:

mN = m (uN, vN)

where N is the number of labor force, u is unemployment rate, v isthe vacancy rate, m is the meeting rate, which shows the degree ofjob match per unit of time. We assume that m (uN, vN) is constantreturns to scale.Given this assumption, the poisson arrival rate of the vacant jobsnding a worker is dened as

m (uN, vN)vN

= muv, 1= m

1vu, 1 p (θ)

where θ = vu is the measure of labor tightness. and the poisson

arrival rate of an unemployed workers nding a job is

m (uN, vN)uN

=m (uN, vN)

vNvNuN

= p (θ) θ



We assume that

p0 (θ) < 0, p00 (θ) > 0,dp (θ) θ

dθ> 0,

d2p (θ) θ

d2θ< 0

limθ!0

p0 (θ) = ∞, limθ!∞

p0 (θ) = 0, limθ!0

dp (θ) θ

dθ= ∞, lim

θ!∞

dp (θ) θ

dθ= 0

limθ!0

p (θ) = ∞, limθ!∞

p (θ) = 0, limθ!0

p (θ) θ = 0, limθ!∞

p (θ) θ = ∞





Dynamics of unemployment:

ut = s [1 ut ] p (θ) θut

where s is the separation rate, under which a worker and a job areseparated.

Natural Rate of Unemployment:

s [1 u] = p (θ) θu

(p (θ) θ + s) u = s

un =s

p (θ) θ + s

Hence, the dynamics of unemployment rate and the natural rate ofunemployment is entirely inuenced by labor tightness, θ. We needto develop a model that can determine θ.



Assignment: Assume that θ is constant over time. Solve thedi¤erential equation


and shows that the unemployment rate eventually converges to thenatural rate of unemployment un.



Bellman equations for workers

$W = w + s [U W ]$U = b+ p (θ) θ [W U ]

where W and U are value functions when the worker is employed orunemployed, respectively.

Di¤erence from the individual decisions:1 The wage, w , is a unique wage determined in a equilibrium. Hence,everybody accepts the wage when the worker meets the job.

2 There is a possibility of separation, and the probability of separation isassumed to follow poisson process with a parameter s.



Bellman equations for rms,

$J = mpl w s [J V ]$V = ψs + p (θ) [J V ψe ]

where J and V are the value functions when the job is occupied andvacant, respectively. The parameters, mpl = maxk ff (k) rkg, ψe ,ψs , denote the marginal productivity of labor, training costs (= costfor education at a rm) and search cost, respectively.

We assume r as given.Note that a job must incur search costs ψs , before it nds workers, butit pays training cost ψe after it nds workers. This di¤erence is turnout to be important.



Free Entry Condition:V = 0

J: Free entry condition implies that J must be solved by : followingequation:

$J = mpl w sJ

J =mpl w

$+ s

This condition implies that the rms values are discounted sum of thedi¤erences between marginal productivity of labor and wage payment.It is shown that specic investment actually guarantees this deviation.So even if household optimally chooses optimal leisure give the marketwage, w , the marginal productivity of labor can deviate from themarginal rate of substitution between consumption and leisure- laborwedge.



θ: Free entry condition also implies that

ψs = p (θ) [J ψe ]

ψsp (θ)

= J ψe

Note that the expected duration of searching workers is 1p(θ) . Hence,

ψsp(θ) is the expected cost of search. The expected cost of search mustbe equal to the expected prots, J ψe under 0 prot conditions,which determines the tightness of labor market, θ.



Equilibrium given wage, w : Given wage w ,

Firms value function J and labor tightness θ are determined by

J =mpl w

$+ s.

ψsp (θ)

= J ψe .

Workers value functions, W and U are determined by

$W = w + s [U W ] .$U = b+ p (θ) θ [W U ] .

unemployment rate is determined by

u = s [1 u] p (θ) θu.



Assignment: Derive U and W as a function of w and θ.


Wage Without Commitment

Wage bargaining (Pissarides 1985): Because a rm sank a searchcost or/and a training cost before making production, changingpartner is costly. Therefore, there are some rents for them to share.

W + J U + V = U

If a rm is unable to commit a particular wage, there is a room tobargain over. Pissarides (1985) assumes that wage can bedetermined by a bargaining.



Generalized Nash Bargaining: It is standard to use generalizedNash bargaining to solve this problem. Let me briey describe whatNash bargaining is.Bargaining: Consider two players 1 and 2. They are negotiatingsomething, say wage, w 2 W . Dene the set of all utility pairs:

U =u (w) 2 R2 : u (w) = (u1 (w) , u2 (w)) ,w 2 W

,

where ui is the ith players utility. Dene reservation utilityd = (d1, d2) 2 D is utility pairs which an agent can get when theycould not reach an agreement. Bargaining is dened by a functionf : U D ! W .Generalized Nash bargaining outcome is the solution of thefollowing problem

w (u, d) = argmaxw(u1 (w) d1)γ (u2 (w) d2)(1γ) .

s.t. u1 (w) d1, u2 (w) d2, γ 2 (0, 1) .Katsuya Takii (Institute) Modern Macroeconomics II 367 / 428




The solution of generalized Nash bargaining, w is a unique solutionwhich satises following 3 axioms.

1 Invariance to equivalent utility representations: Ifu0i (w) = αui (w) + β and d 0i = αdi + β where i = 1 or 2, and α > 0,then u0i (w

(u0, d 0)) = αui (w (u, d)) + β. (This means that even ifwe change the unit of measurent, the result does not change. )

2 Independence of irrelevant outcome: Suppose that U U 0 andu (w (u0, d)) 2 U, where u0 2 U 0. Thenu (w (u0, d)) = u (w (u, d)). (This means that even if we addedirrelevant alternative choices in the choice set, the result does notchange.)

3 Pareto e¢ ciency: If u (w), u (w 0) 2 U and u (w 0) > u (w), thenu (w ) 6= u (w).

Rubinstein (1982) shows that alternating o¤ers bargaining model hasthe same solution as that of the generalized Nash bargaining.



Note that when γ ! 1 then problem is equivalent to

w (u, d) = argmaxw(u1 (w) d1) .

s.t. u2 (w) d2.

This is the case that a player 1 is a principal and player 2 is an agent,and that a principal makes a take-or-leave-it o¤er to the agent.

On the other hand, γ ! 0, then

w (u, d) = argmaxw(u2 (w) d2) .

s.t. u1 (w) d1.

Now, a player 2 is a principal and a player 1 is an agent. A principalmakes a take-or-leave-it o¤er.

Hence, we will sometimes call γ the bargaining power of a player 1.



Let me apply generalized Nash bargaining to our problem:

maxw(Wi U)γ J1γ

i

s.t. Wi =wi + sU

$+ s, J =

mpl wi$+ s

where i refers to ith pair.FOC:

γ (Wi U)γ1 Ji 1γ

$+ s=

(1 γ) (Wi U)γ Jiγ

$+ s(1 γ) (W U) = γJ

The parameter γ shows that labor shares of the total surplus thatoccupied job created:

W U = γ [J +W U ]



w : Substituting the Denition of J and W , we obtain the followingequation.

w + sU$+ s

U = γ

mpl w

$+ s+w + sU$+ s

U

w + sU = γ [mpl w + w + sU ($+ s)U ] + ($+ s)Uw = γ [mpl $U ] + $U

It shows that the wage is the reservation ow utility, $U, plus the partof a ow surplus of this match, γ [mpl $U ]:



$U: Note that

$U = b+ p (θ) θ [W U ]= b+ p (θ) θ

γ

1 γJ

Moreover, free entry condition shows the relationship between labortightness and the value of a rm:

ψsp (θ)

= J ψe ) J =ψsp (θ)

+ ψe

This relationship indicates that there is a relationship between thereservation ow utility of a worker, $U, and labor tightness, θ:



1 If more jobs enter in the market, the reservation utility would be largebecause they can easily nd new jobs (Positive trading externality).

2 In this economy there is the rent to share. Because unemployedworkers expect to obtain this rent in future, this rent inuences thereservation utility of workers, $U. But note that if the training costand the average search cost ψs

p(θ) , which is inuenced by labor tightnessθ, are large, the amount of rent must be also large. Otherwise, therm cannot enter the market. In this way, the rent inuenced by θa¤ects $U.



This means

$U = b+p (θ) θγ

hψsp(θ) + ψe

i1 γ

= b+γ [θψs + p (θ) θψe ]

1 γ= b+

γhvNψsuN +

mNψeuN

i1 γ



The opportunity cost ρU, is inuenced not only by unemploymentbenet b but also by aggregate search costs per unemployed workers,θψs =

vNψsuN and aggregate training cost per unemployed workers,

p (θ) θψe =mNψeuN .

Because increases in search costs and training costs reduce thereservation value of the rm and increase rms surplus J V (= J)and, therefore, the rent that a worker can exploit,W + J U V (= W + J U). Because unemployed workersexpect that they can enjoy this rent when they nd their jobs, theirreservation utility would be larger. In this way, search costs andtraining costs inuence workersreservation utility.



w : Substituting $U into the wage equation, we obtain

w = γ [mpl $U ] + $U

w = γ

mpl b

γ [θψs + p (θ) θψe ]

1 γ

+ b+


1 γ

= γmpl + (1 γ)

b+


1 γ

= γ [mpl b+ θψs + p (θ) θψe ] + b

wage, w , is inuenced by marginal product of labor, mpl ,unemployment benet b plus aggregate search costs per unemployedworkers, θψs , aggregate training cost per unemployed workers,p (θ) θψe .



θ: Using the derived wage, we can determine θ, and therefore, u.Note that the free entry condition provides an information on therelationship between θ and w .

ψsp (θ)

= J ψe

1p (θ)

=1

ψs

mpl w

$+ s ψe

This equation shows that the high w reduces the labor tightness, θ,and therefore the average duration to nd workers, 1

p(θ) . When thewage, w , is high, the expected prot, J, is low. Therefore, a fewvacancy, v , is posted. Hence, the labor tightness, θ, would be low.



Hence, the following two equations solves w and θ.

w = γ [mpl b+ θψs + p (θ) θψe ] + b

Ω (θ, b) , Ω1 (θ, b) > 0,Ω2 (θ, b) > 01

p (θ)=

1ψs

mpl w

$+ s ψe

) θ = JC (w) , JC 0 (w) < 0

Assume the steady state. Then we know the natural rate ofunemployment is determined by θ:

un =s

p (θ) θ + s Υ (θ) , Υ0 (θ) < 0





dudb : an increase in b pushes up an equilibrium wage, w , and therefore,reduces the expected prots and the number of vacancy. Hence,labor tightness becomes lower θ. Because the unemployment rate isstock variable, it does not change immediately. However, the lowerlabor tightness implies that it is more di¢ cult nd a job. Followingthe dynamics of the unemployment rate,


the unemployment rate gradually goes up and reaches a new highernatural rate of unemployment.





Assignment: Using the similar graph, discuss the economic logic ofthe following e¤ects.

1 Discuss the temporal and long run impacts of the productivity shockmpl on the wage and the unemployment rate.

2 Discuss the temporal and long run impacts of search costs, ψc andtraining costs, ψe , on the wage and the unemployment rate.



Optimality of unemployment rate: If a part of unemployment isinevitable in a frictional economy, what is the optimal unemploymentrate? Can market economy attain optimal unemployment rate?

Because we have unemployed workers and employed workers in aneconomy, there are potentially many Pareto optimal allocations.

We focus on those that maximize the sum of agents utility, orequivalently, that maximize the present discounted value of outputnet of the disutility of work and search and training cost.



Social Planner Problem: Let us dene a social planner model asfollows.

maxθt

Z ∞

0

mpl (1 ut )

+ [b (ψsθt + p (θt ) θtψe )] ut

e$tdt

s.t. ut = s [1 ut ] p (θt ) θtut , u0 is given

where1 mpl (1 ut ) is the aggregate output per capita at date t.2 but is the aggregate unemployment benets (or the benets fromleisure) per capita at date t.

3 [p (θt ) θtψe + ψs θt ] ut is the sum of aggregate search cost andtraining cost per capita at date t.



Bellman equation

$V (ut ) = maxθ

mpl (1 ut ) + [b (ψsθt + p (θt ) θtψe )] ut

+V 0 (ut ) [s [1 ut ] p (θt ) θtut ]

TheoremThe socially optimal labor tightness θo must satisfy the following equation

dp (θo ) θo

dθo[mpl b+ p (θo ) θoψe + ψsθ

o ]

=

ψs +

dp (θo ) θo

dθoψe

(ρ+ s + p (θo ) θo )



Proof: Guess V (ut ) = α0 + α1ut

maxθ

mpl (1 ut ) + [b (ψsθt + p (θt ) θtψe )] ut

+α1 [s [1 ut ] p (θt ) θtut ]

= max

θ

mpl + α1s +

mpl + b (ψsθt + p (θt ) θtψe )

α1 (s + p (θt ) θt )

ut

= mpl + α1s min

θt

mpl b+ (ψsθt + p (θt ) θtψe )

+α1 (s + p (θt ) θt )

ut

= mpl + α1s

mpl b+ α1s

+minθt [ψsθt + (ψe + α1) p (θt ) θt ]

ut



Proof: Hence, if there exists α0 and α1 that satisfy the followingequations, our guess is satised.

$α0 = mpl + α1s

$α1 = mpl b+ α1s +min

θt[ψsθt + (ψe + α1) p (θt ) θt ]

FOC

0 = ψs + (ψe + α1)dp (θt ) θtdθt

This equation means that θ is constant.SOC

(ψe + α1)d2p (θt ) θt

dθ2t> 0) α1 < ψe



Proof: From FOC, α1 must satisfy

dp (θ) θ

dθα1 =

ψs +

dp (θ) θ

dθψe

α1 =

"ψs

dp(θ)θd θ

+ ψe

#< ψe



Proof: Using the equation that α1 must satisfy,

$α1 =

mpl b+ α1s+minθ [ψsθ + (ψe + α1) p (θ) θ]

$α1 =

mpl b+ α1s

+ [ψsθ + (ψe + α1) p (θ) θ]

($+ s + p (θ) θ) α1 = [mpl b+ (ψsθ + ψep (θ) θ)]



Proof: Substituting α1 to the equation, we can derive

mpl b+ ψsθ + p (θ) θψe =

"ψs

dp(θ)θd θ

+ ψe

#($+ s + p (θ) θ)

Hence, optimal labor tightness must satisfy the following condition.

dp (θ) θ

dθ[mpl b+ p (θ) θψe + ψsθ]

=

ψs +

dp (θ) θ

dθψe

($+ s + p (θ) θ)



Labor tightness in the Search Equilibrium: Let us analyticallyderive the market labor tightness at the market economy. Rememberthat the market equilibrium must satisfy two equations.

w = γ [mpl b+ θψs + p (θ) θψe ] + b

1p (θ)

=1

ψs

mpl w

$+ s ψe

TheoremLabor tightness in the search equilibrium, θ, must satisfy the followingequation.

(1 γ) p (θ) [mpl b+ θψs + p (θ) θψe ]

= (ψs + p (θ)ψe ) (ρ+ s + p (θ

) θ)



Proof: Note that

1p (θ)

=1

ψs

mpl w

$+ s ψe

ψsp (θ)

+ ψe =mpl w

$+ s

w = mpl ($+ s)

ψsp (θ)

+ ψe



Proof: Hence θ must satisfy the following equation.

mpl ($+ s)

ψsp (θ)

+ ψe

= γ [mpl b+ θψs + p (θ) θψe ] + b

It means that

mpl b γ [mpl b+ θψs + p (θ) θψe ]

= ($+ s)

ψsp (θ)

+ ψe



Proof: Rearranging equation we can derive

(1 γ) [mpl b+ θψs + p (θ) θψe ]

=($+ s)p (θ) θ

[θψs + p (θ) θψe ] + θψs + p (θ) θψe

=

1+

($+ s)p (θ) θ

[θψs + p (θ) θψe ]

=[θψs + p (θ) θψe ]

p (θ) θ(ρ+ s + p (θ) θ)


= (ψs + p (θ)ψe ) (ρ+ s + p (θ) θ)



Comparison

Social Optimal:

dp (θo ) θo

dθo[mpl b+ p (θo ) θoψe + ψs θ

o ]

=

ψs +

dp (θo ) θo

dθoψe

(ρ+ s + p (θo ) θo )

Market Economy:


= (ψs + p (θ)ψe ) (ρ+ s + p (θ

) θ)



TheoremSuppose that ψe = 0. Then the market economy attains social optimal ifthe bargaining power of workers γ, equal to the elasticity of matchingfunction with respect to u.

γ = p0 (θ) θ

p (θ).

Note thatp (θ) = m

uv, 1.

Hence we can show that the right hand side is the elasticity ofmatching function with respect to u.

This condition is called Hosios Condition.



Proof: Suppose ψe = 0. Then

Social Optimal

dp (θo ) θo

dθo[mpl b+ θoψs ] = ψs (ρ+ s + p (θ

o ) θo )

Market Economy

(1 γ) p (θ) [mpl b+ θψs ] = ψs (ρ+ s + p (θ) θ)

Hence, the market economy attains social optimal if

dp (θ) θ

dθ= (1 γ) p (θ)



Proof: Note that

dp (θ) θ

dθ= p0 (θ) θ + p (θ)

Hence, the market economy attains social optimal if

p0 (θ) θ + p (θ) = (1 γ) p (θ)

γp (θ) = p0 (θ) θ

γ = p0 (θ) θ

p (θ)



There is several reasons that a market economy is ine¢ cient.

Hold-up problem (I will discuss it later): Intuitively, when a rmmakes investment specic to a particular workers before production, arm must expect to receive enough return from the investment. But ifwe cannot write an explicit contract on the investment, after theinvestment, workers can threaten rms to leave. Expecting thispossibility, a rm hesitates to makes enough investment. Hence,output is too small. This is called hold-up problem.Trading externality:

1 Posting vacancy makes it easy for unemployed workers to nd new jobs(positive externality). Because individual rms do not take intoaccount this e¤ect, the number of employed workers and, therefore,output is too small.

2 Posting vacancy makes it di¢ cult for other job to nd new workers(negative externality). Because individual rms do not take intoaccount this e¤ect, too many rms enter and society must incur toomany search costs.



Because of several e¤ects in our model, labor tightness in the marketcan be too large or too small. Therefore, the unemployment rate inthe market equilibrium is also too many or too few. Hosios conditionnicely balances these e¤ects.



TheoremSuppose that ψs = 0. Then socially optimal unemployment rate is 0,though a market economy maintains a positive unemployment rate. Thatis, the entry is too few under a market economy.



Proof: Suppose that ψs = 0. Social optimal labor market tightnessmust satisfy

dp (θ) θ

dθ[mpl b+ p (θ) θψe ] =

dp (θ) θ

dθψe (ρ+ s + p (θ) θ)

0 =dp (θ) θ

dθ[mpl b ψe (ρ+ s)]

As we implicitly assume mpl > b+ ψe (ρ+ s) (otherwise nobody

search), dp(θ)θd θ = 0. That is, θ = ∞. Hence, the socially optimalunemployment rate is

ut = s [1 ut ] limθ!∞

p (θt ) θtut = ∞

Hence, ut = 0



Proof: On the other hand, under a market economy, a labor markettightness, θ must satisfy

(1 γ) p (θ) [mpl b+ p (θ) θψe ] = p (θ)ψe (ρ+ s + p (θ) θ)

Hence,

0 = p (θ) f(1 γ) [mpl b+ p (θ) θψe ] ψe (ρ+ s + p (θ) θ)g= p (θ) fmpl b ψe (ρ+ s) γ (mpl b+ ψep (θ) θ)g

Hence, there exist θ < ∞ that satisfy

mpl b ψe (ρ+ s) = γ (mpl b+ ψep (θ) θ)

It means that the unemployment eventually converges to the naturalrate of unemployment:

ut =s

p (θ) θ + s> 0



Although the model still contains the hold-up problem and searchexternality, because there is no search cost, negative externality doesnot have any impacts on the society.

Although posting vacancy makes it di¢ cult for other jobs to nd newworkers, without any search cost (=sunk cost a rm must incur beforending workers), there is no social loss from posting new jobs.Therefore, the impact of positive externality and hold-up problemalways dominates negative externality and the number of marketeconomy in the market economy is too few. (Note that given theassumption of mpl > b+ ψe (ρ+ s), creating new job always increasesnet output. )



TheoremSuppose that ψs = 0. The market economy can attain social optimal,therefore, there is no unemployment, if a worker does not have anybargaining power: γ = 0.

Proof.When γ = 0,

0 = p (θ) [mpl b ψe (ρ+ s)]

As we assume mpl > b+ ψe (ρ+ s) to insure positive entries, p (θ) = 0.Therefore, θ = ∞.

If workers do not have any bargaining power, they cannot threatenrms. Hence, there is no hold-up problem.Note also that the impact of positive externality is 0 when θ = ∞:limθ!∞

dp(θ)θd θ = 0.


A Short Detour to Discussions on Hold Up Problem

Consider the following two period model: t = 1 or 2.

A supplier and a buyer try to trade a good which is suitable to aspecic demand of a buyer. You can interpret a supplier as a worker,and a buyer as a rm. Both parties are risk neutral.At the rst period, a worker invests to improve human capital, h withcost C (h) where C 0 (h) > 0 and C 00 (h) > 0. Since human capital isuseful only for this rm, if they re the worker, the worker does not getanything.At the second period, human capital yields output and the rm can sellthe output at a price 1 and pays wage w .



The key assumption is that two parties cannot make a contract at therst period. That is because

1 both parties may not be able to foresee contingent future, or2 even though they can foresee, they may not be able to describe thecontingent future, or

3 even though they can describe, writing every contingency is quite costly.

That is, both party can observe the level of investment at the rstperiod, but it is not veriable.



The rst best: The rst best investment maximizes expected netbenet. If we assume that an agent does not discount future, then

maxIfh C (h)g ,

FOC1 = C 0

hbest

.



Bargaining Problem:

The second period : At the second period, everything becomes clear.A rm and a worker may be able to negotiate the wage w . Assumethat if workers quite a job worker can get U and if a rm res theworker, a rm can get 0. Then the wage bargaining determines

w = U + γ (h U)

where γ 2 (0, 1) is the bargaining power of workers.The rst period : A worker knows that the wage will be determined asabove. Given this knowledge his problem is

W = maxh[w C (h)] ,

s.t. w = U + γ (h U)



FOC:γ = C 0 (h)

Because γ < 1, and C 00 > 0

h < hbest .



That is, a worker does underinvestment. The reason is that sinceinvestment is specic to the rm, after making investment, theinvestment is sunk. If the rm re the worker, investment is useless.Bargaining over wage reduces the marginal benet from theinvestment. Since he knows it will happen, it discourages hisinvestment. He will optimally reduce his investment. This is calledHold-up problem.

The above analysis suggest that an increase in γ increases humancapital accumulation. If γ = 1, h = hbest . Because a worker is aperson to make investment decision, a larger bargaining power ofworker encourages human capital accumulation. Because h is lowerthan social optimal value, larger human capital is welfare improving.So it is good to provide more power on workers.



Let us illustrate the importance of an explicit contract at the rstperiod.

Suppose that a rm can o¤er an a¢ ne contract based on output:that is, w = α0 + α1h.

V = maxα0,α1

fh (α0 + α1h)g

s.t. h = argmax fα0 + α1h C (h)g (IC )

U α0 + α1h C (h) (IR)



h

α1 = c 0 (h)) h = h (α1)

Hence, a rm can reduce α0 without changing the incentive ofworkers until IR condition is bound.

U = α0 + α1h (α1) C (h (α1))α0 + α1h (α1) = U + C (h (α1))



Hence, the rms problem can be rewritten as follows

V = maxα1,h

fh (α1) C (h (α1)) Ug

FOC

0 =1 C 0 (h (α1))

h0 (α1)

1 = C 0 (h (α1))

Note that 1 = C 0hbest

. Hence,

h (α1) = hbest



In this case,

α1 = c 0 (h (α1)) = 1

α0 = U + C (h (α1)) h (α1)V = α0

That is, it is optimal for the rm to sell a job by price α0 and let aworker receive all outcome from his own e¤ort.


Wage With Commitment

As we show that an market economy is not in general optimal in thatit does not maximizes total sum of utilities. Moreover, the bargainingpower of workers is the key parameter to insure the optimality. Thisindicates that ine¢ cient outcome might be related to the way thatwage is determined.

We also discuss that a rm can avoid hold-up problem if it can writean explicit contract on the investment.

It motivates us to analyze how an explicit contract helps attaininge¢ cient outcome.

We will next analyze an alternative popular method to determinewage, wage posting.



Moen (1997) provides a convenient model of wage posting, whichattains social optimal outcome.

In his model, jobs post their wages before they enter the market.Unemployed workers direct their search to the most attractive jobs.

Because high posted wage, w , attracts more applicants, which reducesworkerscontact rate p (θ) θ and raises the jobscontact rate p (θ).

Knowing this relationship between w and θ, unemployed workersdecide which jobs for them to apply. Hence, all unemployed workerschoose jobs that maximize their total sum of discounted utility, U.

In other words, only jobs which post w that maximizes U survive inthe equilibrium.



Hence, wage posting problem can be formally rewritten as

maxwU,

s.t. V = 0) ψs = p (θ) [J ψe ]

where

the jobsvalue function is determined by

J =mpl w

$+ s.

Workers value functions, W and U are determined by

$W = w + s [U W ] .$U = b+ p (θ) θ [W U ] .



W U: the value function of employed workers and unemployedworkers imply

$ (W U) = w b+ (s + p (θ) θ) (U W )

W U =w b

$+ s + p (θ) θ

U

$U = b+p (θ) θ (w b)$+ s + p (θ) θ

=($+ s) b+ p (θ) θw

$+ s + p (θ) θ

U =($+ s) b+ p (θ) θw

$ [$+ s + p (θ) θ]



The relationship between w and θ. The value function of a job, J,and prot 0 condition imply

ψs = p (θ)mpl w

$+ s. ψe

ψsp (θ)

+ ψe =mpl w

$+ s

w = mpl ($+ s)

ψsp (θ)

+ ψe



Wage posting problem can be rewritten as

maxw

($+ s) b+ p (θ) θw$ [$+ s + p (θ) θ]

s.t. w = mpl ($+ s)

ψsp (θ)

+ ψe

This is equivalent to

maxθ

(ρ+ s) b+ p (θ) θnmpl (ρ+ s)

hψsp(θ) + ψe

ioρ (ρ+ s + p (θ) θ)

= maxθ

(ρ+ s) (b ψsθ) + p (θ) θ [mpl (ρ+ s)ψe ]

ρ (ρ+ s + p (θ) θ)



TheoremLabor tightness under competitive search equilibrium, θ, must satisfy

dp (θ) θ

dθ[mpl b+ p (θ) θψe + ψsθ

]

=

ψs +

dp (θ) θ

dθψe

(ρ+ s + p (θ) θ)

Hence, the competitive search equilibrium attains an social optimalallocation.



Proof: FOC implies

0 =

ndp(θ)θd θ [mpl (ρ+ s)ψe ] (ρ+ s)ψs

o(ρ+ s + p (θ) θ)

[(ρ+ s) (b ψsθ) + p (θ) θ [mpl (ρ+ s)ψe ]]dp(θ)θd θ

ρ (ρ+ s + p (θ) θ)2

=

dp(θ)θd θ

[mpl (ρ+ s)ψe ] (ρ+ s + p (θ) θ)

(ρ+ s) (b ψsθ) + p (θ) θ [mpl (ρ+ s)ψe ]

(ρ+ s)ψs (ρ+ s + p (θ) θ)

ρ (ρ+ s + p (θ) θ)2



Proof: Hence,

0 =

dp(θ)θd θ f[mpl (ρ+ s)ψe ] (ρ+ s) (ρ+ s) (b ψsθ)g

(ρ+ s)ψs (ρ+ s + p (θ) θ)

ρ (ρ+ s + p (θ) θ)2

=

(ρ+ s)

(dp(θ)θd θ fmpl (ρ+ s)ψe (b ψsθ)g

ψs (ρ+ s + p (θ) θ)

)ρ (ρ+ s + p (θ) θ)2



Proof: Rearranging equation, we can derivedp (θ) θ

dθfmpl (ρ+ s)ψe (b ψsθ)g = ψs (ρ+ s + p (θ) θ)

dp (θ) θ

dθ[mpl b+ ψsθ] = ψs (ρ+ s + p (θ) θ)

+ (ρ+ s)ψedp (θ) θ

dθ

dp (θ) θ

dθ[mpl b+ θψs + p (θ) θψe ]

= ψs (ρ+ s + p (θ) θ) + (ρ+ s + p (θ) θ)ψedp (θ) θ

dθ

dp (θ) θ

dθ[mpl b+ p (θ) θψe + ψsθ]

=

ψs +

dp (θ) θ

dθψe

(ρ+ s + p (θ) θ)



Note that we have two sources of ine¢ ciency: holdup problem andtrading externality.

If a job can commit a wage policy, there is no room for bargaining.Therefore, there is no hold-up problem.Because of prot 0 condition, workers obtain all benets and costs ofexternality. But, when unemployed workers apply particular jobs, theytake into account not only wage payment, w , but also the expectedtime to wait before nding the job, 1

p(θ)θ . In this way, the externalityis internalized.

Therefore, wage posting model actually eliminate all possible sourceof ine¢ ciency.


Final Words

Congratulation! You nish all subjects in this lecture.

Topics in macroeconomics are broad. So there are many subjects thislecture did not cover such as monetary economy, internationalmacroeconomics and endogenous growth, heterogeneity amongagents, incomplete market and so on.

However, I guarantee that you have already studied the mostimportant fundamental methods to analyze these problems. So youwill be able to read many textbooks and papers in macroeconomicsby yourself. Try it.


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Modern Macroeconomics II - Osaka Universitytakii/macroII-slide.pdf · 2011. 2. 1. · macroeconomics, but also for understanding dynamic issues in any –elds of economics, such as