Kenji Sato
January 18, 2017
Today, we are going to study the Ramsey–Cass–Koopmans model (Ramsey model, hereafter), which is the simplest general equilibrium macroeconomic model.
As it will turn out, the structure of the dynamics is identical to the optimal growth model.
The Ramsey model incorporates the following considerations:
U=∫∞0e−ρtu(c(t))L(t)Hdt
β:=ρ−n−(1−θ)g>0
under which U<∞.
The profit maximization. Profit is F(K,AL)−ˉrK−wL, where ˉr is gross rate of return from the capital owner’s viewpoint. r(t)+δ, r(t) is net rate of return and δ the depreciation rate.
maxK,LF(K,AL)−(r+δ)K−wL
The first order conditions are given as below:
∂F∂K=r+δ,∂F∂L=w
We will use a different symbol for capital per unit of effective labor: ˆk=K/AL. Small symbols are for per-labor variables; small symbols with hat for per-effective-labor variables.
Note that with f(ˆk)=F(ˆk,1)=F(K,AL)/AL, f′(ˆk)=∂F/∂K. We thus have
f′(ˆk)=r+δ
and
w=A[f(ˆk)−ˆkf′(ˆk)], namely,
ˆw=wA=f(ˆk)−ˆkf′(ˆk)
The household divides income between consumption c(t)L(t) and saving S(t). Each household is subject to the budget constraint:
˙S(t)+c(t)L(t)H≤r(t)S(t)+w(t)L(t)H
The capital market clearing condition is
H⋅S(t)=K(t),t≥0,
from which we get
˙K(t)≤r(t)K(t)+w(t)L(t)−c(t)L(t)=F(K(t),A(t)L(t))−δK(t)−c(t)L(t).
or
˙K(t)A(t)L(t)≤f(ˆk(t))−δˆk(t)−ˆc(t)
where ˆc(t)=c(t)/A(t) is consumption per unit of effective labor.
The inequality
˙K(t)A(t)L(t)≤f(ˆk(t))−δˆk(t)−ˆc(t)
gets us
˙ˆk(t)=˙K(t)A(t)L(t)−(g+n)ˆk(t)≤f(ˆk(t))−(δ+g+n)ˆk(t)−ˆc(t)
Utility function
u(c(t))=c(t)1−θ1−θ=[A(t)ˆc(t)]1−θ1−θ=[A(0)egtˆc(t)]1−θ1−θ=A(0)1−θe(1−θ)gtˆc(t)1−θ1−θ.
Since L(t)=L(0)ent,
U=∫∞0e−ρtc(t)1−θ1−θL(t)Hdt=∫∞0e−ρt[A(0)1−θe(1−θ)gtˆc(t)1−θ1−θ]L(0)entHdt=A(0)1−θL(0)H∫∞0e−[ρ−n−(1−θ)g]tˆc(t)1−θ1−θdt=:B∫∞0e−βtˆc(t)1−θ1−θdt
where β=ρ−n−(1−θ)g>0 by assumption.
The Ramsey model boils down to the optimal growth problem:
max∫∞0Be−βtˆc(t)1−θ1−θdtsubject to ˙ˆk(t)=f(ˆk(t))−(δ+g+n)ˆk(t)−ˆc(t).
Note that B before the total utility doesn’t change the optimality conditions and so you can safely remove it. Check this fact by solving the problem without removing B.
Excercise 1: Derive the differential equations that govern the dynamics of the model.
For a given ˆk(0)=K(0)/A(0)L(0), there is only one initial ˆc(0)=C(0)/A(0)L(0) such that the optimal path from (ˆk(0),ˆc(0)) converges to the steady state (ˆk⋆,ˆc⋆).
In the steady state (if the economy starts from it)
C(t)=A(t)L(t)ˆc⋆K(t)=A(t)L(t)ˆk⋆
The growth rates:
Implications of the Solow model are maintained.
Romer’s textbook employs a Lagrangian method after introducing the lifetime budget constraint.
The flow budget constraint written in present terms:
e−R(t)˙S(t)≤r(t)e−R(t)S(t)+e−R(t)w(t)L(t)H−e−R(t)c(t)L(t)H,
where
R(t)=∫t0r(s)ds,˙R(t)=r(t)
Integrate the flow budget constraint:
e−R(t)˙S(t)≤r(t)e−R(t)S(t)+e−R(t)w(t)L(t)H−e−R(t)c(t)L(t)H,
The left-hand side becomes:
∫∞0e−R(t)˙S(t)dt=[e−R(t)S(t)]∞0−∫∞0−r(t)e−R(t)S(t)dt=limt→∞e−R(t)S(t)−S(0)+∫∞0r(t)e−R(t)S(t)dt.
The right-hand side becomes:
∫∞0[r(t)e−R(t)S(t)+e−R(t)w(t)L(t)H−e−R(t)c(t)L(t)H]dt
We get
limt→∞e−R(t)S(t)−S(0)≤∫∞0e−R(t)w(t)L(t)Hdt−∫∞0e−R(t)c(t)L(t)Hdt
By assuming limt→∞e−R(t)S(t)≥0 we get
∫∞0e−R(t)c(t)L(t)Hdt≤S(0)+∫∞0e−R(t)w(t)L(t)Hdt
This inequality states that the present value of the consumption of a household must not exceed initial saving plus the present value of their lifetime income.
The condition
limt→∞e−R(t)S(t)≥0
is called the No-Ponzi Game condition. This condition is violated if the household always borrows money to repay their borrowings; the interest accrue and present value of borrowing in the infinite future becomes nonzero.
Such a financial scheme is excluded.
NB: NPG condition is different from the transversality condition (a few professional authors are confused!). NPG is an assumption for the model, while transversality is for the model’s optimal path.
The consumer’s problem is thus
max∫∞0Be−βtˆc(t)1−θ1−θdtsubject to ∫∞0e−R(t)+(n+g)tˆc(t)dt≤ˆk(0)+∫∞0e−R(t)+(n+g)tˆw(t)dt.
Romer defines the Lagrangian function and bypasses (sacrifycing mathematical rigor) the technical stuff.
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