advmacro Takeki Sunakawa RCK

Ramsey-Cass-Koopmans model

Takeki Sunakawa

Advanced Macroeconomics at Tohoku University

What is the Ramsey-Cass-Koopmans model?

Aka the neo-classical growth model, a model of long run economic growth. Developed by Frank Ramsey (1929) and later extended by David Cass (1969) and Tjalling Koopmans (1969).

The workhorse model of modern macroeconomics.

Why is the RCK model useful?

More reasonable explanation on the household’s decision on

saving/consumption à la Fisher. [The Solow model has an ad-hoc constant saving rate.]

The model is also a basis for real business cycle (RBC) models of short-run fluctuations, which will be later in the class.

Households

There are households and firms (no govt and foreign sector; but the government sector can be easily incorporated)

Household owns capital and firms, and decide how much to save and consume:

ct+ st= rtkt−1+ πt= yt. Decision on saving is now endogenous.

Be careful of the time subscripts.

Euler equation

As a result of the household’s optimization, we havethe Euler equation: M Ut= βM Ut+1(1 + rt+1− δ) .

The LHS is the benefit from one unit of consumption today. The RHS is the benefit from one unit of saving today and consuming the return on saving tomorrow.

β is discount factor. M Ut is the marginal utility from consumption in period t.

M RSt,t+1= M Ut/(βM Ut+1) is the marginal rate of substitution between consumption in period t and t + 1.

Marginal utility

Marginal utility is the benefit from one unit of consumption.

To link consumption to utility, we need a utility function of the household. For example,

Ut= log ct. Then we have

M Ut= 1/ct.

Firms

The representative firm can access to the Cobb-Douglas production function technology:

yt= Atk_t−1^α . As a result of profit maximization,

rt= αyt/kt−1, and

πt= yt− rtkt−1= (1 − α)yt.

Capital accumulation

Same as before;

kt− kt−1= st− δkt−1. Saving is equal to net investment,

st= kt− (1 − δ)kt−1. From the household’s budget constraint,

ct+ kt− (1 − δ)kt−1= yt.

Equilibrium

Now we have the following equilibrium conditions: 1 = β_c^ct

t+1^{(1 + rt+1− δ) ,}

rt= αyt/kt−1, yt= Atk^α_t−1, ct+ kt− (1 − δ)kt−1= yt.

There are four unknowns and four equations, so we can solve the model.

The two key equations

The equilibrium conditions are summarized as

ct+1

ct ^{= β 1 + αAt+1k}

α−1 t ^{− δ
,}

kt− kt−1= Atk^α_t−1− δkt−1− ct.

The steady state

In the steady state,

1 = β 1 + αAk^α−1− δ
, 0 = Ak^α− δk − c, hold.

The steady-state conditions can be solved for

k^∗=

 _αβA

1 − β(1 − δ)

_1−α¹ , c^∗= A(k^∗)^α− δk^∗. These equations are drawn on the (k, c) plane.

The Phase diagram

The transition dynamics

Let

gct≡ β 1 + αAtk_t−1^α−1− δ
− 1,

˜

gkt≡ Ak^α_t−1− δkt−1− ct.

If gct= ct/ct−1− 1 > 0, ct> ct−1holds; kt−1< k^∗ implies gc> 0.

If ˜gkt= kt− kt−1> 0, kt> kt−1 holds; ct> Ak^α_t−1− δkt−1implies ˜gk < 0. ct and ktare called jump variables, whereas kt−1 is state variable.

Saddle path to the steady state

As long as the state of the economy is on the saddle path, it converges to the steady state.

The state of the economy can diverge. There is a condition for the stability.

Two ways to solve the RCK model

A Robinson Crusoe (Social Planner) economy. A Competitive economy.

When the allocations and prices in these economies coincide each other? The second fundamental theorem of welfare economics.

A Robinson Crusoe economy

Consider an economy with only one individual.

The individual wants to maximize a lifetime utility of the form

∞

X

t=0

β^tlog ct

subject to

kt= (1 − δ)kt−1+ it, yt= Atk^α_t−1≥ ct+ it.

where ctis consumption, ktis capital, itis investment and ytis output. Parameters are given as β ∈ (0, 1) is discount factor, δ ∈ (0, 1] is depreciation rate, and α is capital share.

Lagrangean

We set up the Lagrangean as

L0≡

∞

X

t=0

β^tlog c_t− λt ct+ kt− (1 − δ)kt−1− Atk^α_t−1
.

λtis called the Lagrange multiplier, which measures the marginal utility of consumption.

Taking the derivatives of the Lagrangean and set them to zero

∂ct: λt= 1/ct,

∂kt: λt= βλt+1(1 + αAt+1k_t^α− δ) ,

∂λt: ct+ kt− (1 − δ)kt−1− Atk^α_t−1= 0. These are the necessary conditions for the equilibrium.

Transversality condition

The transversality condition is

t→∞lim ^β

t_{M U} tkt= 0,

where ktis the remaining resources and M Utconverts the value of ktto the unit in terms of utility.

The condition says that the planner must use all the resources and/or have no marginal benefit from consumption.

A competitive economy

In a competitive economy, there are consumers who provide labor to the market and firms who hire the labor at wage wtand rent capital at rate rt. All individuals are the same. We consider the “representative” agent’s problem.

Firms

The representative firm can access to the Cobb-Douglas production function technology:

yt= Atk_t−1^α . As a result of profit maximization,

rt= αyt/kt−1, and

πt= yt− rtkt−1= (1 − α)yt.

Households

An individual i ∈ [0, 1] maximizes:

∞

X

t=0

β^tu(cⁱ_t)

subject to

cⁱ_t+ kⁱ_t− (1 − δ)kⁱ_t−1= rtk_tⁱ+ π_tⁱ,

Lagrangean

We set up the Lagrangean as

Lⁱ₀≡

∞

X

t=0

β^tlog cⁱ_t− λt cⁱ_t+ k_tⁱ− (1 − δ)kⁱ_t−1− rtkⁱ_t−1− πⁱ_t
.

Taking the derivatives of the Lagrangean and set them to zero,

∂cⁱ_t: λt= 1/cⁱ_t,

∂k_tⁱ: λt= βλt+1(1 + rt+1− δ) ,

∂λt: cⁱ_t+ kⁱ_t− (1 − δ)kⁱ_t−1− rtkⁱ_t−1− πⁱ_t= 0.

Aggregation

The aggregation rules are ct=

✂ 1 0

cⁱ_tdi, kt=^✁₀¹kⁱ_tdi,

yt=

✂ 1 0

y_tⁱdi, πt=^✁₀¹πⁱ_tdi,

Note that πt+ rtkt−1= yt. Then we have 1 = β ^ct

ct+1

1 + αAt+1k^α−1_t − δ
, ct+ kt− (1 − δ)kt−1= Atk^α_t−1.

Second welfare theorem

The second fundamental theorem of welfare economics: If household preferences and firm production sets are convex, there is a complete set of markets with publicly known prices, and every agent acts as a price taker, then any Pareto optimal outcome can be achieved as a competitive equilibrium if appropriate lump-sum transfers of wealth are arranged.

Second welfare theorem, cont’d

The first fundamental theorem: Any Competitive Equilibrium allocation (C.E.) is necessarily Pareto Optimum allocation (P.O.).

The second fundamental theorem: Any P.O. can be achieved as a C.E. with a lump-sum transfer.

If the second welfare theorem holds, we only need to look at P.O. instead of C.E.

Counter-example: An economy with distortionary tax.

Applications

An economy with variable labor Incorporating trend

Notes on Newton’s method and deterministic simulation

An economy with variable labor

The social planner wants to maximize a lifetime utility

∞

X

t=0

β^tlog ct+ v(ht) subject to

kt= (1 − δ)kt−1+ it, yt= Atk^α_t−1h^1−α_t ≥ ct+ it.

where ctis consumption, ktis capital, itis investment, ytis output and htis hours worked. Parameters are given as β ∈ (0, 1) is discount factor, δ ∈ (0, 1] is depreciation rate, and α is capital share.

Labor disutility

v(ht) is called labor disutility: ht∈ (0, 1) and v(ht) is a concave function such that v(ht) → −∞ as ht→ 1.

There are two forms of labor disutility.

Divisible labor: v(ht) = B log(1 − ht): Everyone works for hthours. Indivisible labor: v(ht) = −Bht: Only a fraction of individuals works for h⁰ hours.

Indivisibility and lottery

Labor is indivisible: Individuals can either work full time, denoted by h0, or not at all.

If the consumption possibilities set is non-convex, competitive equilibria may not exist. We convexify the consumption possibilities set by requiring individuals to choose lotteries αt:

log ct+ Aαtlog(1 − h0)

Note log(1) = 0. Total hours worked (per capita) is ht= αth0.

Indivisibility and lottery, cont’d

It can be viewed as linear disutility from the point of view of “representative” household.

By substituting out αt, we have

Aαtlog(1 − h0) = A log(1 − h0)/h0ht,

= −Bht

where B = −A log(1 − h0)/h0> 0.

Lagrangean

We set up the Lagrangean as

L0≡

∞

X

t=0

β^tlog c_t+ v(ht) − λt ct+ kt− (1 − δ)kt−1− Atk^α_t−1h^1−α_t
.

λtis the Lagrange multiplier.

Taking the derivatives of the Lagrangean and set them to zero,

∂ct: λt= 1/ct,

∂ht: λt(1 − α)Atk_t−1^α h⁻_t^α= −v^′(ht),

∂kt: λt= βλt+1 1 + αAt+1k_t^α−1h^1−α_t − δ
,

∂λt: ct+ kt− (1 − δ)kt−1− Atk^α_t−1h^1−α_t = 0.

Incorporating trend

Let Zt≡ A

1 1−α

t ^{and γ ≡ Zt+1/Zt}. The production function becomes yt= k_t−1^α (Ztht)^1−α.

This is called the Harrod-neutral production function.

In the steady state, y and k exponentially grows at the rate γ.

Detrending

The equilibrium conditions are

yt= k^α_t−1(Ztht)^1−α, (1 − α)^yt

ht

= ^Bct 1 − ht

, 1 = β ^ct

ct+1



1 + α^yt+1 kt

− δ

 , ct+ kt− (1 − δ)kt−1− yt= 0.

yt, ctand kt−1 grows at γ, whereas the model variables have to be stationary.

Detrending, cont’d

Define ˜yt≡ yt/Zt, ˜ct≡ ct/Ztand ˜kt−1≡ kt−1/Zt,

˜

yt= ˜k_t−1^α h^1−α_t , (1 − α)^y˜t

ht

= ^B˜ct 1 − ht

, 1 = (β/γ) ^c˜t

˜ ct+1



1 + α^y˜t+1_˜ kt

− δ

 ,

˜

ct+ γ˜kt− (1 − δ)˜kt−1− ˜yt= 0.

The steady state

The steady state conditions are

1 = ˜β^1 + α˜y/˜k − δ^, 0 = ˜y − (γ − 1 + δ)˜k − ˜c, (1 − α)(˜y/h) = B˜c/(1 − h),

˜

y = ˜k^αh^1−α, where ˜β = β/γ.

Dynare solves for the steady state values numerically with an educated initial guess. In this model, the steady state values are also analytically obtained.

The steady state

Let h = 1/3, then we have

˜

y/˜k = α⁻¹( ˜β⁻¹− 1 + δ),

˜

c/˜k = ˜y/˜k − (γ − 1 + δ),

˜k =

 _αh1−α

β˜⁻¹− 1 + δ

_1−α¹ , B = (1 − α)(˜y/˜c)(1 − h)/h.

˜

y = (˜y/˜k)˜k and ˜c = (˜c/˜k)˜k are also obtained. Note that B is a normalization parameter.

Newton-Raphson method

We solve an equation f (x) = 0 for x.

First-order Taylor expansion: f (x) ≈ f (x0) + f^′(x0)(x − x0) = 0. The Newton-Raphson method updates xk

xk+1= xk− f^′(xk)⁻¹f (xk), until kxk+1− xkk ≤ ǫ.

Newton-Raphson method, cont’d

We solve a system of equations fi(x) = 0 for i = 1, ..., n for x= [x1, · · · , xn]^′.

n dimensional First-order Taylor expansion:

fi(x) ≈ fi(x0) + ∇fi(x0)(x − x0) = 0, for i = 1, ..., n where ∇fi(x0) = [f_i^′(x1), ..., f_i^′(xn)].

Newton-Raphson method, cont’d

Let f (x) = [f1(x), ..., fn(x)]^′ and

∇f (x) =











∇f1(x)

∇f2(x) ...

∇fn(x)











=















∂f1_(x)

∂x1

∂f1_(x)

∂x2 ^{· · ·}

∂f1_(x)

∂xn

∂f2_(x)

∂x1 ^{· · ·}

... ...

∂fn_(x)

∂x1 ^{· · ·}

∂fn_(x)

∂xn















is an (n × n) matrix called Jacobian. The Newton-Raphson method updates xk

x_k+1= xk− [∇f (xk)]⁻¹f(xk), until kxk+1− xkk ≤ ǫ.

Deterministic simulations

Dynare can be used for deterministic simulations with the assumption of perfect foresight.

The numerical problem consists of solving a nonlinear system of simultaneous equations in n endogenous variables in T periods.

To solve the system of nT equations, Dynare uses a Newton-type method, which is based on the Fair-Taylor (1983) algorithm.

My explanation is based on Hollinger (1996).

A nonlinear system

A nonlinear dynamic system is of a general form fi,t(yt−1, yt, yt+1) = 0, for i = 1, ..., n where yt= [y1,t, · · · , yn,t]^′.

For example, the basic RCK model is expressed as f1,t(yt−1, yt, yt+1) = β ^ct

ct+1

1 + αAt+1k^α−1_t − δ
− 1 = 0, f2,t(yt−1, yt, yt+1) = Atk_t−1^α + (1 − δ)kt−1− kt− ct= 0, where yt= [kt, ct]^′.

A nonlinear system, cont’d

The system is expressed compactly as a (n × 1) vector of equations:

f_t(zt) = ft(yt−1, yt, yt+1) =







f1,t(yt−1, yt, yt+1) ...

fn,t(yt−1, yt, yt+1)





^{= 0,}

where ft= [f1,t, · · · , fn,t]^′ and zt= [y^′_t−1, y^′_t, y^′_t+1]^′.

Solving the nonlinear system

We solve ft(yt−1, yt, yt+1) = 0 simultaneously for t = 1, ..., T .

yt−1= y0 is predetermined at t = 1. Also, yt+1= yT+1is predetermined at t = T (the boundary conditions).

Solving the nonlinear system, cont’d

We stack the system for T periods as

F (Z) =















 f_1(z1)

... f_t(zt)

... f_T(zT)

















= 0,

where F = [f₁^′, · · · , f_t^′, · · · , f_n^′]^′ and Z = [z^′₁, · · · , z^′_t, · · · , z^′_n]^′. This is a (nT × 1) vector of equations.

We solve F (Z) by the Newton-Raphson method for nT variables Z = [y₀^′, y^′₁, · · · , y^′_T, y_T^′₊₁],

Assignment #2 (due: October 28)

Let β = .96, δ = .1, α = .36. Consider the basic RCK model without labor.

1 _{Let k0}= 0.1. Compute the initial dynamics converging to the steady state.

2 Assume a distortionary tax on capital income after depreciation, c_{t+ kt−(1 − δ)kt}

−^{1≤ rtkt}−^{1− τ(rt− δ)kt}−^{1+ πt.}

Let τ = .2.

1 Solve for the equilibrium conditions in the competitive economy.

2 Solve for the steady state.

3 Compute the initial dynamics with k0^{= 0.1}.

4 Compare the results with the ones in the model without distortionary taxes, which you computed in (1).