ramsey model chapter 3 - doug hanleydoughanley.com/grad_macro/lectures/chapter_3.pdf · 2/16/2020...

2/16/2020 Econ 2130

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Chapter 3Chapter 3Ramsey ModelRamsey Model Econ 2130: MacroeconomicsEcon 2130: Macroeconomics

University of Pittsburgh, 2020University of Pittsburgh, 2020

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MotivationMotivationSolow model assumes constant exogenous saving rate

Questions:

what determines the saving rate?when agents decide saving in a competitive equilibriumenvironment, do they choose a Pareto optimal saving rate?

Objective of this chapter: construct a general equilibrium model to analyzeaggregate saving

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AssumptionsAssumptionsModeling choices:

number of agent single representativetime horizon finite / infinitedecision maker planner / decentralized

Simplest consumption-saving model:

Single agent, finite horizon, no depreciation

→→

→

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Ramsey-Cass-Koopmans ModelRamsey-Cass-Koopmans Model

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Frank Plumpton Ramsey (1903-1930)Frank Plumpton Ramsey (1903-1930)

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Tjalling Koopmans (1910-1986)Tjalling Koopmans (1910-1986)

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David Cass (1937-2008)David Cass (1937-2008)

General of the Army

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AgentsAgentsPopulation grows at rate and so that

Objective function of each agent at

Utility from in evaluated at time is

Assumption

n L(0) = 1 L(t) = exp(nt)

t = 0

U(0) = exp(−ρt)u(c(t))L(t)dt∫0

∞

= exp(−(ρ − n)t)u(c(t))dt∫0

∞

c t t 0 exp(−ρt)u(c )t

→ ρ > n

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DiscountingDiscountingWhere does the exponential form of discounting come from?

intuition: calculate the value of $1 in periods

Divide interval into equally-sized subintervals

Discount rate in each subinterval is

Value of 1 util, periods from now

t

[0, t] t/Δ

Δρ ⋅ r

t

v(t ∣ Δ) = (1 − Δ ⋅ ρ)t/Δ

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Continuous limitContinuous limitLetting

Use L'Hopital's rule:

Δ → 0

v(t) = v(t∣Δ) ≡ (1 − Δ ⋅ ρ)Δ→0lim

Δ→0lim t/Δ

v(t) = exp log(1 − Δ ⋅ ρ)[Δ→0lim t/Δ]

= exp [Δ→0lim

Δ/tlog(1 − Δ ⋅ ρ)

]

= exp [Δ→0lim

1/t−ρ/(1 − Δρ)

]

= exp(−ρt)

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Budget constraintBudget constraintAgents earn a wage at every instant and can borrow and save assets

, yielding interesting rate

Per-period budget constraint

and in per capita terms:

w(t)a(t) r(t)

(t) = r(t)A(t) + w(t)L(t) − c(t)L(t)A

(t) = (r(t) − n)a(t) + w(t) − c(t)a

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Time varying coe�cientsTime varying coe�cientsMath refresh 3: Linear non-autonomous differential equations

Steady state (may not exist):

General solution

Boundary condition yields solution

(t) = m(t)x(t) + b(t)x

x(t) = −b(t)/m(t)

x(t) = d + b(s) exp − m(v)dv ds exp m(s)ds[ ∫0

t

( ∫0

s

) ] (∫0

t

)

d = x(0) = x 0

x(t) = x + b(s) exp − m(v)dv ds exp m(s)ds[ 0 ∫0

t

( ∫0

s

) ] (∫0

t

)

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Lifetime budget constraintLifetime budget constraintIn our case

This yields the solution

m(t) = r(t) − n

b(t) = w(t) − c(t)

a(t) = a + w(s) − c(s) exp − (r(v) − n)dv ds[ 0 ∫0

t

[ ] ( ∫0

s

) ]

× exp (r(s) − n)ds(∫0

t

)

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Limiting conditionsLimiting conditionsNo-Ponzi condition:

Take limit when of -period lifetime budget constraint to obtain

a(t) exp − (r(s) − n)ds =t→∞lim ( ∫

0

t

) 0

T → ∞ T

a =0 c(t) − w(t) exp − (r(s) − n)ds dt∫0

∞

[ ] ( ∫0

t

)

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Optimization problemOptimization problemHow to solve the optimization problem?

use per-period budget constraintconstruct Hamiltonian, the continuous-time analog to Lagrangianapply maximum principle from optimal control theory

Agent's problem

a(t),c(t) [ ]t=0∞

max

subj. to

exp − ρ − n t u(c(t))dt∫0

∞

( ( ) )

(t) = (r(t) − n)a(t) + w(t) − c(t)a

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Optimal controlOptimal controlConsumer lives between periods 0 and 1

Lagrangian

a(t),c(t) [ ]t=01

max

subj. to

exp −ρt u(c(t))dt∫0

1

( )

(t) = ra(t) + w − c(t)a

a(t) ≥ 0, a > 00

L = exp −ρt u(c(t))dt∫0

1

( )

+ λ(t) ra(t) + w − c(t) − (t) dt∫0

1

[ a ]

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Finite solutionFinite solutionIntegration by parts

Optimality conditions

Alternatively

λ(t) (t)dt = λ(1)a(1) − λ(0)a(0) − (t)a(t)dt∫0

1

a ∫0

1

λ

exp −ρt u (c(t)) = λ(t)( ) ′

(t) = −rλ(t)λ

u (c(t)) = μ(t)′

ρμ(t) − (t) = rμ(t)μ

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In�nite horizonIn�nite horizonMore general infinite-horizon optimal control problem:

subject to

and

Notice payoff depends on time only through exponential discounting

W (x(t), y(t)) ≡ exp(−ρt)f(x(t), y(t))dtx(t),y(t)max ∫

0

∞

(t) = g(t,x(t), y(t))x

x(0) = x and b(t)x(t) ≥ x 0t→∞lim 1

f

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Hamiltonian functionHamiltonian functionHamiltonian

Current-value Hamiltonian

Assumptions:

and are weakly monotone in and and continuouslydifferentiableoptimal state variable satisfies:

which is needed for the transversality condition to hold

H(t,x, y,λ) = exp(−ρt)f(x, y) + λg(t,x, y)

(t,x, y,μ) =H f(x, y) + μg(t,x, y)

f g x y

(t)x

(t) = x ∈ R or (t)/ (t) = χ ∈ Rt→∞lim x ∗

t→∞lim x x

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Maximum PrincipleMaximum PrincipleTheorem: Maximum Principle, discounted infinite-horizon problems

Let be the current-value Hamiltonian of the typicaldiscounted infinite-horizon optimal control problem. Then the optimalcontrol pair satisfies the following necessary conditions:

and the transversality condition

t,x, y,μH ( )

( (t), (t))x y

t, (t), (t),μ(t) = 0 ∀ t ∈ RHy ( x y ) +

ρμ(t) − (t) = t, (t), (t),μ(t) ∀ t ∈ Rμ Hx ( x y ) +

(t) = t, (t), (t),μ(t) ∀ t ∈ R x Hμ ( x y ) +

x(0) = x and b(t)x(t) ≥ x 0t→∞lim 1

exp(−ρt)μ(t) (t) = 0.t→∞lim x

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InitutionInitutionThe "co-state" variable tracks the value of an additional increment of the

state variable

Above satisfies co-state evolution equation (integration by parts)

Notice we need an envelope condition for this to make sense

μ

x

μ(t) = exp(−ρs) (t +∫0

∞

Hx s)ds

(t)μ = exp(−ρs) (t + s)dx∫0

∞

Hx

= exp(−ρs) (t + s) − ρ exp(−ρs) (t + s)ds[ Hx ]0

∞∫

0

∞

Hx

= − (t) + ρμ(t)Hx

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Su�cient conditionsSu�cient conditionsLoosely speaking, this is not a issue for concave problems

Practical strategy:

use the necessary conditions in Maximum Theorem to construct acandidate solutionverify concavity for sufficiency

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OptimizationOptimizationCurrent-value Hamiltonian for Ramsey model:

Candidate solution:

is concave in since it is the sum of a concave function of

and a linear function of

(a, c,μ) =H u(c) + μ (r − n)a + w − c[ ]

0 = (a, c,μ) =Hc u (c) −′ μ

(ρ − n)μ − =μ (a, c,μ) =Ha μ(r − n)

=a (r − n)a + w − c

exp(−(ρ −t→∞lim n)t)μ(t)a(t) = 0

(a, c,μ)H (a, c) c

(a, c)

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Characterize solutionCharacterize solutionSecond optimality condition

First optimality condition

taking the time derivative and dividing

Thus we have eliminated and have a differential equation in

=μ

μ−(r − ρ)

u (c) =′ μ

=u (c)′

u (c)c′′

c

c =

μ

μ−(r − ρ)

μ c

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Utility elasticityUtility elasticityEuler equation:

where

Special case (CRRA):

=c

c (r −

ε (c)u

1ρ)

ε (c) ≡u −

u (c)′

u (c)c′′

u(c) = ⇒1 − θ

c − 11−θ

ε (c) =u θ ∀ c

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Consumption pathConsumption pathImplied consumption function:

Let be the average interest rate between dates and

Once again for the case of CRRA

c(t) = c(0) exp ds(∫0

t

ε (c(s))u

r(s) − ρ)

(t)r 0 t

(t) = r(s)dsrt

1∫

0

t

c(t) = c(0) exp t((θ

(t) − ρr) )

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Initial conditionInitial conditionWhere is coming from? Recall the lifetime budget constraint

Plugging in for the case of CRRA utility

c(0)

a =0 c(t) − w(t) exp(−( (t) −∫0

∞

[ ] r n)t)dt

c(t)

c(0) =

exp − + − n t dt∫0∞ ( (

θ

(θ−1) (t)r

θρ ) )

a + w(t) exp − (t) − n t dt0 ∫0∞ ( (r ) )

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Limiting conditionsLimiting conditionsTransversality condition ensures no-Ponzi schemes

Solution to equation

substituting into the transversality condition

μ

μ(t) = μ(0) exp − r(s) − ρ ds( ∫0

t

( ) )

= u (c(0)) exp − r(s) − ρ ds′ ( ∫0

t

( ) )

a(t) exp − (r(s) − n)ds = 0t→∞lim ( ∫

0

t

)

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EquilibriumEquilibriumOptimization yields pair which implies a competitive

equilibrium pair

Since , transversality condition is also equivalent to

Firms are standard:

(a(t), c(t))(k(t), c(t))

a(t) = k(t)

k(t) exp − (r(s) − n)ds = 0t→∞lim ( ∫

0

t

)

r(t) = R(t) − δ = f (k(t)) − δ′

w(t) = f(k(t)) − f (k(t))k(t)′

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Steady stateSteady stateSteady-state equilibrium is an equilibrium path in which capital-labor ratioand consumption are constant, thus:

Steady state Ramsey continuous time:

Compare with Solow model:

population growth has no impact on

when , and do not depend on

affects the transitional dynamics

= 0 and = 0c k

f k = ρ + δ′ ( ∗)

c = f k − (n + δ)k∗ ( ∗) ∗

k∗

g = 0 k∗ c∗ u ⋅( )u ⋅( )

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Cobb-DouglasCobb-DouglasHere we have . Therefore

So we can see that the investment rate is

f(k) = kα

k∗

c∗

= (ρ + δ

α)

1−α1

= (k ) − (n + δ)k∗ α ∗

= (k ) 1 − α∗ α [ (ρ + δ

n + δ)]

s = α <(ρ + δ

n + δ) α

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DynamicsDynamicsTwo differential equations:

Initial condition on capital:

Boundary condition at infinity (transversality)

Goal is to set so that boundary at infinity holds

this is surprisingly tough!

k

c

c

= f(k) − (n + δ)k − c

= f (k) − δ − ρε (c)u

1[ ′ ]

k(0) = k >0 0

k(t) exp − f (k(s)) − δ − n ds = 0t→∞lim ( ∫

0

t

( ′ ) )

c(0)

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Phase diagramPhase diagram

c

k

= 0 ⇒ f (k) = δ + ρ′

= 0 ⇒ c = f(k) − (n + δ)k

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UniquenessUniquenessWhy is the stable arm unique?

if started below stable arm, capital would eventually reach the

maximum level (zero consumption) violatestransversality conditionif started above stable arm, eventually we would get violates feasibility

c(0)>k k gold →

c(0) k < 0 →

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StabilityStabilityTheorem: Saddle path stability for linear problems

Consider the linear differential equation over

with initial value . Here is an matrix and is a length

vector. Let be the steady state, so that .

If of the eigenvalues of have negative real parts, then there exists

an -dimensional subspace such that for any , there is

a unique solution with .

x ∈ Rn

(t) =x Ax(t) + b

x(0) = x 0 A n × n b n

x∗ Ax +∗ b = 0

m ≤ n A

m M ⊂ Rn x(0) ∈ M

x(t) → x∗

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Non-linear systemsNon-linear systemsFor the non-linear system

we can define the Jacobian of and consider the matrix .

An analogous result follows for the linearized system

(t) =x G(x(t))

J(x) G A = J(x )∗

(t) =x J(x )(x(t) −∗ x )∗

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LinearizationLinearizationLinearizing system of differential equations around

Jacobian is given by:

Roots of associated characteristic polynomial confirm saddle-path

k , c( ∗ ∗)

≃dt

d(k − k )∗

≃dt

d(c − c )∗

f k − n − δ k − k − c − c( ′ ( ∗) ) ( ∗) ( ∗)

k − k + c − cε cu ( ∗)

c f k∗ ′′ ( ∗)( ∗)

ε cu ( ∗)f k − δ − ρ′ ( ∗)

( ∗)

J(k , c ) =

∗ ∗ [ρ − n

ε cu( ∗ )c f k∗ ′′( ∗ )

−10

]

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TechnologyTechnologyLet and use production function

As in Solow model define

where

/A =A g

Y = F (K,AL)

y ≡ ≡AL

Yf(k)

k ≡

AL

K

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ExistenceExistenceIs utility function consistent with existence of BGP?

Check Euler equation:

if , then requires

Proposition: Balanced growth in Ramsey model

Balanced growth in the neoclassical model requires that asymptotically (as ) all technological change is purely labor augmenting and the

elasticity of intertemporal substitution, , tends to a constant .

=c

c (r −

ε (c)u

1ρ)

r(t) → r∗ (t)/c(t) →c g c ε (c(t)) →u ε u

t → ∞ε (c(t))u ε u

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ExampleExampleUsing CRRA preferences with

But as with , also grows along the BGP. Bounded utility?

c(t) ≡ C(t)/L(t)

exp −(ρ − n)t dt∫0

∞

( )1 − θ

c(t)1−θ

y(t) c(t)

(t) ≡ ≡ c~A(t)L(t)

C(t)A(t)c(t)

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ExampleExampleLifetime utility

Assumption

exp −(ρ − n)t dt∫0

∞

( )1 − θ

c(t)1−θ

= exp −(ρ − n − g(1 − θ))t dt∫0

∞

( )1 − θ

(t)c~ 1−θ

→ ρ − n > g(1 − θ)

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DynamicsDynamicsLaw of motion for capital in Ramsey model with technology growth:

Euler equation:

=k f(k) − −c~ (n + g + δ)k

=c~c~

−c

cg = (r −

θ

1ρ − θg)

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Steady stateSteady stateTransversality condition:

Steady state:

so now plays a role in determining and

k(t) exp − f (k(s)) − g − δ − n ds = 0t→∞lim ( ∫

0

t

[ ′ ] )

f k = ρ + δ + θg′ ( ∗)

u(⋅) k∗ y (t) =∗ A(t)f(k )∗

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ExtensionsExtensions

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Role of policyRole of policyIf , , and are same across countries, then the only theory of

differences in is differences in levels of

That is OK, but not very interesting because is a residual

A possibility is to introduce differences in across countries viadifferences in policy

ρ δ θ g

y (t)∗ A(t)

A(t)

k∗

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TaxationTaxationIntroduce linear capital gains tax where is taxed at the rate and theproceeds are redistributed lump sum back to consumers

New Euler equation:

so in steady state

ra τ

r = (1 − τ)[f (k) − δ]′

= (1 − τ)[f (k) − δ] − ρ − θg

c~c~

θ

1( ′ )

f k = δ +

′ ( ∗)1 − τ

ρ + θg

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E�ectsE�ectsSuppose economy is in steady state and tax declines to

What are the comparative dynamics?

new steady-state equilibrium that is saddle-path stable

since ,

equilibrium growth rate will still be

Suppose change in tax is unanticipated and occurs at some date

at time curve shifts to the right

now previous steady state is above new stable armconsumption must drop immediately and then slowly increase alongnew stable arm

k ,( ∗ c~∗) τ <′ τ

k ,( ∗∗ c~∗∗)τ <′ τ k >∗∗ k∗

g

T

T d (t)/dt =c~ 0 k > k( ∗∗ ∗)c~∗

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ramsey model chapter 3 - doug hanleydoughanley.com/grad_macro/lectures/chapter_3.pdf · 2/16/2020...

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