Macroeconomics:Day 7

Kenji Sato

January 6, 2017

Departure from Solow

The following two are the workhorse models of modern macroeconomic analyses.

Very important to acquire working knowledge of these models. Without it, difficult to appreciate any recent research papers in the field of macroeconomics.

Today, we will study the Ramsey model.


Suppose that you have just heard that your tax you had already paid (say, 10,000 Yen) will be reimbursed next month.

What would you do with the windfall gains?

  1. Nothing. Wait until you have the cash.
  2. Go shopping or go for a dring this weekend, expecting the refund.

Why Solow not enough?

In the Solow model, the saving rate is exogenous. Take a tax hike for example. In the Solow model, consumers reduce consumption at the time of enforcment of the new tax system.

But our governments would inform the change of tax well before the enforcement so that we can prepare for the up-coming hard days.

Similarly, it is also possible that consumers consume more after knowing that they will have to pay less to the government.

Why Solow not enough? (Cont’d)

So, the Solow model is missing an important aspect of our economic activity: Expectation formation.

Although dealing with the problem of expectations is beyond the scope of this course, the Ramsey model will serve as the skelton of more elaborate models.

Optimal Growth Model

We firstly study the so-called one-sector optimal growth model, which is mathematically equivalent to the Ramsey model.

There is only one person (representative agent) in the economy who owns everything and wants to determine his/her capital accumulation path along which his/her utility from consumption is maximized.

This problem is sometimes called as the social planner’s problem.

Discrete-time or Continuous-time

As in the study of the Solow model, we will go back and forth between continuous-time and discrete-time models. This is because in general there are the following trade-offs between them:

Discrete-time or Continuous-time (cont’d)

Since (in my opinion) continuous-time models are approximation. Economic activities happen in discrete time. It is desirable to understand

Let’s first define feasibility and optimality in continuous time.


Recall the capital accumulation in the Solow model,

\[ \dot k(t) = \color{red}{ s f(k(t))} - \delta k(t). \]

Since \(s\) in not exogenously given in the optimal growth model, the correct formulation is

\[ \dot k(t) = \color{red}{f(k(t)) - c(t)} - \delta k(t). \]

Since consuming negative amount is impossible, it must hold that

\[ \dot k(t) \leq f(k(t)) - \delta k(t). \]

Feasibility (cont’d)

If the capital accumulation path \(t \mapsto k(t)\) satisfies

\[ \dot k(t) \leq f(k(t)) - \delta k(t) \quad \text{and} \quad k(t) \ge 0 \]

\(k(\cdot)\) and \(c(\cdot)\) are said to be feasible.

Notice that along a feasible path, consumption stream is always nonnegative

\[ c(t) \ge 0\qquad \text{for all } t \ge 0 \]

Instantaneous utility function

Suppose that at any moment in time, the representative consumer’s consumption \(c(t)\) gives him utility of amount \(u(c(t))\).

This function, \(u: \mathbb{R} \to \mathbb{R}: c \mapsto u(c)\), is called the instantaneous utility function.

Instantaneous utility function (cont’d)

Assumption: \(u\) is twice differentiable, satisfying

We will probably need more but these are the most fundamental assumptions.

Total utility

The representative consumer’s total utility of a consumption stream \(t \mapsto c(t)\) is given by

\[ \int_0^\infty e^{-\rho t} u(c(t)) dt, \]

where \(\rho > 0\) is called the discount rate.

Remember the definition of the integral?

Total utility (cont’d)

The improper Riemann integral: \[ \int_0^\infty e^{-\rho t} u(c(t)) dt = \lim_{T \to \infty} \int_0^T e^{-\rho t} u(c(t)) dt. \]

The Riemann integral: \[ \int_0^T e^{-\rho t} u(c(t)) dt = \lim_{N \to \infty} \sum_{n = 0}^{N - 1} e^{-\rho \tau_n} u(c(\tau_n))(t_{n+1} - t_n), \] where \(0 = t_0 < t_1 < \cdots < t_{N-1} < t_N = T\) is a partition of interval \([0,T]\).

As \(N \to \infty\), \(\max (t_{n+1} - t_n) \to 0\). This limit must not depend on a particular choice of \(\tau_n \in (t_n, t_{n+1})\).

Discrete-time analog

Let \(\tau_n = t_n\) and \(t_{n+1} - t_n = 1\) for all \(n\). Then, \(e^{-\rho \tau_n} = e^{-\rho n}\) and \[ \sum_{n = 0}^{N-1} (t_{n+1} - t_n) e^{-\rho \tau_n} u(c(\tau_n)), \] translates to \[ \sum_{n = 0}^{N-1} (e^{-\rho})^n u(c_n) \] with \(c_n = c(\tau_n)\). As you will see later in the course, the total utility in the discrete-time model will be defined by \[ \sum_{n = 0}^{\infty} \beta^n u(c_n) \]

Discount rate

From the expression \[ \sum_{n = 0}^{N-1} (e^{-\rho})^n u(c_n) \] you can see what \(\rho\) means. Roughly speaking, \[ u(c_n) = (t_{n+1} - t_n) u(c_n) \] is the total utility from consumption during year \(n\).

If \(c_n = c_m\) for \(n < m\), \(c_n\) gives greater effect on total utility (from the perspective of today) than \(c_m\).

Beer is tastier today than tommow, from today’s perspective.

Preference over different consumption streams

The representative agent compares two consumption streams \(c(\cdot)\) and \(\tilde c(\cdot)\) by the above utility;

i.e. he prefers \(c(\cdot)\) to \(\tilde c(\cdot)\) if

\[ \int_0^\infty e^{-\rho t} u(c(t)) dt \ge \int_0^\infty e^{-\rho t} u(\tilde c(t)) dt \]


A pair of a capital accumulation path \(k(\cdot)\) and a consumption stream \(c(\cdot)\) is said to be optimal if they are feasible and the total utility \[ \int_0^\infty e^{-\rho t} u(c(t)) dt \] is maximized over all feasible consumption streams.

Optimal Growth

We have a formulation of the optimal growth problem as follows:

\[ \begin{aligned} & \max \int_0^\infty e^{-\rho t} u(c(t)) dt \\ &\text{subject to } \\ &\qquad c(t) = f(k(t)) - \delta k(t) - \dot k(t) \ge 0, \\ &\qquad k(t) \ge 0, \\ &\qquad k(0): \text{given}. \end{aligned} \]

Trade-off between consumption and investment

Reshape the resource constraint into \[ c(t) + \dot k(t) = f(k(t)) - \delta k(t). \]

It is possible to increase the amount of consumption, \(c(t)\), at any point in time by reducing investment, \(\dot k(t)\), at the same moment. Since \(\dot k(t) < 0\) is allowable (disinvestment), \(c(t)\) can be arbitrarily large.

Trade-off between consumption and investment (cont’d)

If \(\dot k(t)\) is too small, however, the amount of capital will be reduced. This results in the reduced consumption in the future because the representative agent cannot continue \(\dot k(t) < - \epsilon < 0\) forever because \(k(t) \ge 0\) must hold.

The optimal growth model (and the Ramsey model) is a problem of intertemporal resource allocation.

You need to sacrifice future consumption to consume today.

Solving the optimal growth problem

We study the solution method next week. Today, we make clear what we mean by solution and draw a sketch of our journey.

The optimal growth model is a problem to find a function \(c(\cdot)\) or equivalently \(\dot k(\cdot)\) (Why equivalent?) that maximizes the total utility.

For the representative agent to make a viable decision, the optimal \(c(t)\) must not depend on the future information; i.e., \(c(t)\) is a function of \(\{(c(\tau) \mid \tau < t\}\) and \(\{ k(\tau) \mid \tau \le t \}\)

Solving the optimal growth problem (cont’d)

We included \(k(t)\) in the information set. This is because at time \(t\), \(k(t)\) should be known in a similar way like \(k(0)\) is known at \(t = 0\).

(At time \(t\), we know the beginning-of-period capital \(k(t)\) of period \([t, t+\Delta t)\). In contrast, we must determine the amount of consumption \(c(t)\Delta t\) that will happen in the same period.)


Jargons you should learn: - Variables like \(k\) are called predetermined or state variables. - Variables like \(c\) are called non-predetermined or control variables.

What solution means

To solve a model is to write down non-predetermined variables by

FYI: Modern macroeconomic models may include expectation terms to the list (and eliminate them by restricting shock components to VAR; See the beautiful paper of Blanchard and Khan 1980 in Econometrica and their successors such as Klein 2000 in Journal of Economic Dynamics and Control).

Differential equations governing dynamics

The first step to determine the dynamics of a model is to obtain the differential equation that the state and control variables obey: i.e., to obtain something like

\[ \begin{aligned} \dot k (t) &= ... \\ \dot c (t) &= ... \end{aligned} \]

In our particular case, the …’s only contain the current (\(t\)) value of the predetermined and nonpredetermined variables and fundamentals.

Differential equations governing dynamics (cont’d)

In fact, you already know the dynamics of \(k\):

\[ \begin{aligned} \dot k (t) &= f(k(t)) - \delta k(t) - c(t) \\ \dot c (t) &= ... \end{aligned} \]

The dynamics of \(k(\cdot)\) is determined by the capital accumulation, which is intuitive.

Our effort will be put to derivation of the dynamics of \(c(t)\).

Euler Equation

As you will see later, the full dynamics of \(c(\cdot)\) and \(k(\cdot)\) is governed by

\[ \begin{aligned} \dot k (t) &= f(k(t)) - \delta k(t) - c(t) \\ \dot c (t) &= \frac{u'(c(t))[f'(k(t)) - \delta - \rho]}{- u''(c(t))} = \frac{u'(c(t))(r(t) - \rho)}{- u''(c(t))}, \end{aligned} \]

where \(r(t) = f'(k(t)) - \delta\) is net rate of return on capital.

When the rental rate (instantaneous reward of owning capital) is greater than the discount rate (instantaneous cost of not consuming), the representative agent decide to spend this profits to increase consumption somewhat.

Problem solved?

At this stage, the problem is not solved yet. Because we have two dimensional dynamical system with only one initial condition \(k(0)\).

We need one more!

As it will turn out, the dynamics is fully determined by the terminal condition called transversality condition, which states that the representative agent will use up all his capital in the infinite future.

If he died with capital left unconsumed, he would fail to maximize utility.

Transversality condition

It is known that, under a mild set of conditions (See Kamihigashi 2001, Econometrica), transversality condition (not die before the present value of capital is zero)

\[ \lim_{t\to\infty}-e^{-\rho t}u'(c(t))k(t)=0 \]

is necessary for optimality; i.e., among infinitely many paths that satisfy the above mentioned system of differential equations, those that satisfy the transversality condition are optimal.

In our case, we can specify only one solution.