Macroeconomics:Day 10

Kenji Sato

January 18, 2017

Ramsey–Cass–Koopmans Model

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.

Additional features

The Ramsey model incorporates the following considerations:

Assumptions for firms

Assumptions for households

Consumption decision

\[ U=\int_{0}^{\infty}e^{-\rho t}u(c(t))\frac{L(t)}{H}dt \]

Consumption decision (cont’d)

\[ \beta := \rho-n-(1-\theta)g>0 \]

under which \(U < \infty\).

Firm’s optimization

The profit maximization. Profit is \(F(K, AL) - \bar r K - wL\), where \(\bar r\) is gross rate of return from the capital owner’s viewpoint. \(r(t) + \delta\), \(r(t)\) is net rate of return and \(\delta\) the depreciation rate.

\[ \max_{K, L} F(K, AL) - (r + \delta) K - w L \]

The first order conditions are given as below:

\[ \begin{aligned} \frac{\partial F}{\partial K} &= r + \delta, \\ \frac{\partial F}{\partial L} &= w \end{aligned} \]

Firm’s optimization

We will use a different symbol for capital per unit of effective labor: \(\hat k = K/AL\). Small symbols are for per-labor variables; small symbols with hat for per-effective-labor variables.

Note that with \(f(\hat k) = F(\hat k, 1) = F(K, AL) / AL\), \(f'(\hat k) = \partial F / \partial K\). We thus have

\[ f'(\hat k) = r + \delta \]


\[ w = A\left[f(\hat{k})-\hat{k}f'(\hat{k})\right], \] namely,

\[ \hat w = \frac{w}{A} = f(\hat{k})-\hat{k}f'(\hat{k}) \]

Households’ budget constraint

The household divides income between consumption \(c(t)L(t)\) and saving \(S(t)\). Each household is subject to the budget constraint:

\[ \dot{S}(t) + c(t)\frac{L(t)}{H} \le r(t)S(t)+w(t)\frac{L(t)}{H} \]

Market clearing (Capital market)

The capital market clearing condition is

\[ H\cdot S(t)=K(t),\qquad t\ge0, \]

from which we get

\[ \begin{aligned} \dot{K}(t) &\le r(t)K(t) + w(t)L(t) - c(t)L(t)\\ &= F(K(t),A(t)L(t)) - \delta K(t) - c(t)L(t). \end{aligned} \]


\[ \frac{\dot{K}(t)}{A(t)L(t)}\le f(\hat{k}(t))-\delta\hat{k}(t)-\hat{c}(t) \]

where \(\hat{c}(t) = c(t) / A(t)\) is consumption per unit of effective labor.

Capital accumulation in the Ramsey model

The inequality

\[ \frac{\dot{K}(t)}{A(t)L(t)}\le f(\hat{k}(t))-\delta\hat{k}(t)-\hat{c}(t) \]

gets us

\[ \begin{aligned} \dot{\hat{k}}(t) &= \frac{\dot{K}(t)}{A(t)L(t)}-(g+n)\hat{k}(t)\\ &\le f(\hat{k}(t))-(\delta+g+n)\hat{k}(t)-\hat{c}(t) \end{aligned} \]

Utility maximization

Utility function

\[ \begin{aligned} u(c(t)) &= \frac{c(t)^{1-\theta}}{1-\theta} \\ &= \frac{[A(t)\hat{c}(t)]^{1-\theta}}{1-\theta} \\ &= \frac{[A(0)e^{gt}\hat{c}(t)]^{1-\theta}}{1-\theta} \\ &= A(0)^{1-\theta}e^{(1-\theta)gt}\frac{\hat{c}(t)^{1-\theta}}{1-\theta}. \end{aligned} \]

Utility function (cont’d)

Since \(L(t)=L(0)e^{nt}\),

\[ \begin{aligned} U &=\int_{0}^{\infty}e^{-\rho t}\frac{c(t)^{1-\theta}}{1-\theta}\frac{L(t)}{H}dt\\ &=\int_{0}^{\infty} e^{-\rho t}\left[A(0)^{1-\theta}e^{(1-\theta)gt} \frac{\hat{c}(t)^{1-\theta}}{1-\theta}\right]\frac{L(0)e^{nt}}{H}dt\\ &=A(0)^{1-\theta}\frac{L(0)}{H}\int_{0}^{\infty}e^{-[\rho-n-(1-\theta)g]t} \frac{\hat{c}(t)^{1-\theta}}{1-\theta}dt\\ &=:B\int_{0}^{\infty}e^{-\beta t}\frac{\hat{c}(t)^{1-\theta}}{1-\theta}dt \end{aligned} \]

where \(\beta=\rho-n-(1-\theta)g > 0\) by assumption.

Optimal growth

The Ramsey model boils down to the optimal growth problem:

\[ \begin{aligned} &\max \int_{0}^{\infty} Be^{-\beta t}\frac{\hat{c}(t)^{1-\theta}}{1-\theta}dt\\ &\text{subject to }\\ &\quad\dot{\hat{k}}(t)=f(\hat{k}(t))-(\delta+g+n)\hat{k}(t)-\hat{c}(t). \end{aligned} \]

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.

Dynamics of the model

For a given \(\hat k (0) = K(0)/A(0)L(0)\), there is only one initial \(\hat c (0) = C(0) / A(0) L(0)\) such that the optimal path from \((\hat k(0), \hat c(0))\) converges to the steady state \((\hat k^\star, \hat c^\star)\).

In the steady state (if the economy starts from it)

\[ \begin{aligned} C(t) &= A(t) L(t) \hat c^\star \\ K(t) &= A(t) L(t) \hat k^\star \end{aligned} \]

The growth rates:

Implications of the Solow model are maintained.

Treatment in Romer 4e

Romer’s textbook employs a Lagrangian method after introducing the lifetime budget constraint.

Households’ budget constraint (in present terms)

The flow budget constraint written in present terms:

\[ e^{-R(t)}\dot{S}(t) \le r(t)e^{-R(t)}S(t)+e^{-R(t)}w(t)\frac{L(t)}{H}-e^{-R(t)}c(t)\frac{L(t)}{H}, \]


\[ R(t)=\int_{0}^{t}r(s)ds,\qquad\dot{R}(t)=r(t) \]

Lifetime budget constraint

Integrate the flow budget constraint:

\[ e^{-R(t)}\dot{S}(t) \le r(t)e^{-R(t)}S(t)+e^{-R(t)}w(t) \frac{L(t)}{H}-e^{-R(t)}c(t)\frac{L(t)}{H}, \]

The left-hand side becomes:

\[ \begin{aligned} \int_{0}^{\infty}e^{-R(t)}\dot{S}(t)dt &= \left[e^{-R(t)}S(t)\right]_{0}^{\infty}-\int_{0}^{\infty}-r(t)e^{-R(t)}S(t)dt\\ &= \lim_{t\to\infty}e^{-R(t)}S(t)- S(0)+\int_{0}^{\infty}r(t)e^{-R(t)}S(t)dt. \end{aligned} \]

The right-hand side becomes:

\[ \int_{0}^{\infty}\left[r(t)e^{-R(t)}S(t)+e^{-R(t)}w(t) \frac{L(t)}{H}-e^{-R(t)}c(t)\frac{L(t)}{H}\right]dt \]

Life-time budget constraint (cont’d)

We get

\[ \lim_{t\to\infty}e^{-R(t)}S(t)-S(0) \le \int_{0}^{\infty}e^{-R(t)}w(t)\frac{L(t)}{H}dt - \int_{0}^{\infty}e^{-R(t)}c(t)\frac{L(t)}{H}dt \]

By assuming \(\lim_{t\to\infty}e^{-R(t)}S(t)\ge 0\) we get

\[ \int_{0}^{\infty}e^{-R(t)}c(t)\frac{L(t)}{H}dt \le S(0)+\int_{0}^{\infty}e^{-R(t)}w(t)\frac{L(t)}{H}dt \]

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.

No-Ponzi Game condition

The condition

\[ \lim_{t\to\infty}e^{-R(t)}S(t) \ge 0 \]

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.

Optimization problem

The consumer’s problem is thus

\[ \begin{aligned} &\max \int_{0}^{\infty} B e^{-\beta t}\frac{\hat{c}(t)^{1-\theta}}{1-\theta}dt\\ &\text{subject to }\\&\quad\int_{0}^{\infty}e^{-R(t)+(n+g)t}\hat{c}(t)dt \le\hat{k}(0)+\int_{0}^{\infty}e^{-R(t)+(n+g)t}\hat{w}(t)dt. \end{aligned} \]

Romer defines the Lagrangian function and bypasses (sacrifycing mathematical rigor) the technical stuff.