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Competitive Equilibria of Chang Model

Co-author: Sebastian Graves

Overview

This lecture describes how Chang [Cha98] analyzed competitive equilibria and a best competitive equilibrium called a Ramsey plan

He did this by

  • characterizing a competitive equilibrium recursively in a way also employed in the dynamic Stackelberg problems and Calvo model lectures to pose Stackelberg problems in linear economies, and then
  • appropriately adapting an argument of Abreu, Pearce, and Stachetti [APS90] to describe key features of the set of competitive equilibria

Roberto Chang [Cha98] chose a model of Calvo [Cal78] as a simple structure that conveys ideas that apply more broadly

A textbook version of Chang’s model appears in chapter 25 of [LS18]

This lecture and Credible Government Policies in Chang Model can be viewed as more sophisticated and complete treatments of the topics discussed in Ramsey plans, time inconsistency, sustainable plans

Both this lecture and Credible Government Policies in Chang Model make extensive use of an idea to which we apply the nickname dynamic programming squared

In dynamic programming squared problems there are typically two interrelated Bellman equations

  • A Bellman equation for a set of agents or followers with value or value function $ v_a $
  • A Bellman equation for a principal or Ramsey planner or Stackelberg leader with value or value function $ v_p $ in which $ v_a $ appears as an argument

We encountered problems with this structure in dynamic Stackelberg problems, optimal taxation with state-contingent debt, and other lectures

The setting

First, we introduce some notation

For a sequence of scalars $ \vec z \equiv \{z_t\}_{t=0}^\infty $, let $ \vec z^t = (z_0, \ldots , z_t) $, $ \vec z_t = (z_t, z_{t+1}, \ldots ) $

An infinitely lived representative agent and an infinitely lived government exist at dates $ t = 0, 1, \ldots $

The objects in play are

  • an initial quantity $ M_{-1} $ of nominal money holdings
  • a sequence of inverse money growth rates $ \vec h $ and an associated sequence of nominal money holdings $ \vec M $
  • a sequence of values of money $ \vec q $
  • a sequence of real money holdings $ \vec m $
  • a sequence of total tax collections $ \vec x $
  • a sequence of per capita rates of consumption $ \vec c $
  • a sequence of per capita incomes $ \vec y $

A benevolent government chooses sequences $ (\vec M, \vec h, \vec x) $ subject to a sequence of budget constraints and other constraints imposed by competitive equilibrium

Given tax collection and price of money sequences, a representative household chooses sequences $ (\vec c, \vec m) $ of consumption and real balances

In competitive equilibrium, the price of money sequence $ \vec q $ clears markets, thereby reconciling decisions of the government and the representative household

Chang adopts a version of a model that [Cal78] designed to exhibit time-inconsistency of a Ramsey policy in a simple and transparent setting

By influencing the representative household’s expectations, government actions at time $ t $ affect components of household utilities for periods $ s $ before $ t $

When setting a path for monetary expansion rates, the government takes into account how the household’s anticipations of the government’s future actions affect the household’s current decisions

The ultimate source of time inconsistency is that a time $ 0 $ Ramsey planner takes these effects into account in designing a plan of government actions for $ t \geq 0 $

Setting

The household’s problem

A representative household faces a nonnegative value of money sequence $ \vec q $ and sequences $ \vec y, \vec x $ of income and total tax collections, respectively

The household chooses nonnegative sequences $ \vec c, \vec M $ of consumption and nominal balances, respectively, to maximize

$$ \sum_{t=0}^\infty \beta^t \left[ u(c_t) + v(q_t M_t ) \right] \tag{1} $$

subject to

$$ q_t M_t \leq y_t + q_t M_{t-1} - c_t - x_t \tag{2} $$

and

$$ q_t M_t \leq \bar m \tag{3} $$

Here $ q_t $ is the reciprocal of the price level at $ t $, which we can also call the value of money

Chang [Cha98] assumes that

  • $ u: \mathbb{R}_+ \rightarrow \mathbb{R} $ is twice continuously differentiable, strictly concave, and strictly increasing;
  • $ v: \mathbb{R}_+ \rightarrow \mathbb{R} $ is twice continuously differentiable and strictly concave;
  • $ u'(c)_{c \rightarrow 0} = \lim_{m \rightarrow 0} v'(m) = +\infty $;
  • there is a finite level $ m= m^f $ such that $ v'(m^f) =0 $

The household carries real balances out of a period equal to $ m_t = q_t M_t $

Inequality (2) is the household’s time $ t $ budget constraint

It tells how real balances $ q_t M_t $ carried out of period $ t $ depend on income, consumption, taxes, and real balances $ q_t M_{t-1} $ carried into the period

Equation (3) imposes an exogenous upper bound $ \bar m $ on the household’s choice of real balances, where $ \bar m \geq m^f $

Government

The government chooses a sequence of inverse money growth rates with time $ t $ component $ h_t \equiv {M_{t-1}\over M_t} \in \Pi \equiv [ \underline \pi, \overline \pi] $, where $ 0 < \underline \pi < 1 < { 1 \over \beta } \leq \overline \pi $

The government faces a sequence of budget constraints with time $ t $ component

$$ -x_t = q_t (M_t - M_{t-1}) $$

which by using the definitions of $ m_t $ and $ h_t $ can also be expressed as

$$ -x_t = m_t (1-h_t) \tag{4} $$

The restrictions $ m_t \in [0, \bar m] $ and $ h_t \in \Pi $ evidently imply that $ x_t \in X \equiv [(\underline \pi -1)\bar m, (\overline \pi -1) \bar m] $

We define the set $ E \equiv [0,\bar m] \times \Pi \times X $, so that we require that $ (m, h, x) \in E $

To represent the idea that taxes are distorting, Chang makes the following assumption about outcomes for per capita output:

$$ y_t = f(x_t), \tag{5} $$

where $ f: \mathbb{R}\rightarrow \mathbb{R} $ satisfies $ f(x) > 0 $, is twice continuously differentiable, $ f''(x) < 0 $, and $ f(x) = f(-x) $ for all $ x \in \mathbb{R} $, so that subsidies and taxes are equally distorting

Calvo’s and Chang’s purpose is not to model the causes of tax distortions in any detail but simply to summarize the outcome of those distortions via the function $ f(x) $

A key part of the specification is that tax distortions are increasing in the absolute value of tax revenues

Ramsey plan: A Ramsey plan is a competitive equilibrium that maximizes (1)

Within-period timing of decisions is as follows:

  • first, the government chooses $ h_t $ and $ x_t $;
  • then given $ \vec q $ and its expectations about future values of $ x $ and $ y $’s, the household chooses $ M_t $ and therefore $ m_t $ because $ m_t = q_t M_t $;
  • then output $ y_t = f(x_t) $ is realized;
  • finally $ c_t = y_t $

This within-period timing confronts the government with choices framed by how the private sector wants to respond when the government takes time $ t $ actions that differ from what the private sector had expected

This consideration will be important in lecture credible government policies when we study credible government policies

The model is designed to focus on the intertemporal trade-offs between the welfare benefits of deflation and the welfare costs associated with the high tax collections required to retire money at a rate that delivers deflation

A benevolent time $ 0 $ government can promote utility generating increases in real balances only by imposing sufficiently large distorting tax collections

To promote the welfare increasing effects of high real balances, the government wants to induce gradual deflation

Household’s problem

Given $ M_{-1} $ and $ \{q_t\}_{t=0}^\infty $, the household’s problem is

$$ \begin{aligned} \mathcal{L} & = \max_{\vec c, \vec M} \min_{\vec \lambda, \vec \mu} \sum_{t=0}^\infty \beta^t \bigl\{ u(c_t) + v(M_t q_t) + \lambda_t [ y_t - c_t - x_t + q_t M_{t-1} - q_t M_t ] \\ & \quad \quad \quad + \mu_t [\bar m - q_t M_t] \bigr\} \end{aligned} $$

First-order conditions with respect to $ c_t $ and $ M_t $, respectively, are

$$ \begin{aligned} u'(c_t) & = \lambda_t \\ q_t [ u'(c_t) - v'(M_t q_t) ] & \leq \beta u'(c_{t+1}) q_{t+1} , \quad = \ {\rm if} \ M_t q_t < \bar m \end{aligned} $$

The last equation expresses Karush-Kuhn-Tucker complementary slackness conditions (see here)

These insist that the inequality is an equality at an interior solution for $ M_t $

Using $ h_t = {M_{t-1}\over M_t} $ and $ q_t = {m_t \over M_t} $ in these first-order conditions and rearranging implies

$$ m_t [u'(c_t) - v'(m_t) ] \leq \beta u'(f(x_{t+1})) m_{t+1} h_{t+1}, \quad = \text{ if } m_t < \bar m \tag{6} $$

Define the following key variable

$$ \theta_{t+1} \equiv u'(f(x_{t+1})) m_{t+1} h_{t+1} \tag{7} $$

This is real money balances at time $ t+1 $ measured in units of marginal utility, which Chang refers to as ‘the marginal utility of real balances’

From the standpoint of the household at time $ t $, equation (7) shows that $ \theta_{t+1} $ intermediates the influences of $ (\vec x_{t+1}, \vec m_{t+1}) $ on the household’s choice of real balances $ m_t $

By “intermediates” we mean that the future paths $ (\vec x_{t+1}, \vec m_{t+1}) $ influence $ m_t $ entirely through their effects on the scalar $ \theta_{t+1} $

The observation that the one dimensional promised marginal utility of real balances $ \theta_{t+1} $ functions in this way is an important step in constructing a class of competitive equilibria that have a recursive representation

A closely related observation pervaded the analysis of Stackelberg plans in lecture dynamic Stackelberg problems

Competitive equilibrium

Definition:

  • A government policy is a pair of sequences $ (\vec h,\vec x) $ where $ h_t \in \Pi \ \forall t \geq 0 $
  • A price system is a nonnegative value of money sequence $ \vec q $
  • An allocation is a triple of nonnegative sequences $ (\vec c, \vec m, \vec y) $

It is required that time $ t $ components $ (m_t, x_t, h_t) \in E $

Definition:

Given $ M_{-1} $, a government policy $ (\vec h, \vec x) $, price system $ \vec q $, and allocation $ (\vec c, \vec m, \vec y) $ are said to be a competitive equilibrium if

  • $ m_t = q_t M_t $ and $ y_t = f(x_t) $
  • The government budget constraint is satisfied
  • Given $ \vec q, \vec x, \vec y $, $ (\vec c, \vec m) $ solves the household’s problem

Inventory of objects in play

Chang constructs the following objects

  1. A set $ \Omega $ of initial marginal utilities of money $ \theta_0 $
    • Let $ \Omega $ denote the set of initial promised marginal utilities of money $ \theta_0 $ associated with competitive equilibria
    • Chang exploits the fact that a competitive equilibrium consists of a first period outcome $ (h_0, m_0, x_0) $ and a continuation competitive equilibrium with marginal utility of money $ \theta _1 \in \Omega $
  2. Competitive equilibria that have a recursive representation
    • A competitive equilibrium with a recursive representation consists of an initial $ \theta_0 $ and a four-tuple of functions $ (h, m, x, \Psi) $ mapping $ \theta $ into this period’s $ (h, m, x) $ and next period’s $ \theta $, respectively
    • A competitive equilibrium can be represented recursively by iterating on
      $$ \begin{split} h_t & = h(\theta_t) \\ m_t & = m(\theta_t) \\ x_t & = x(\theta_t) \\ \theta_{t+1} & = \Psi(\theta_t) \end{split} \tag{8} $$ starting from $ \theta_0 $
      The range and domain of $ \Psi(\cdot) $ are both $ \Omega $
  3. A recursive representation of a Ramsey plan
    • A recursive representation of a Ramsey plan is a recursive competitive equilibrium $ \theta_0, (h, m, x, \Psi) $ that, among all recursive competitive equilibria, maximizes $ \sum_{t=0}^\infty \beta^t \left[ u(c_t) + v(q_t M_t ) \right] $
    • The Ramsey planner chooses $ \theta_0, (h, m, x, \Psi) $ from among the set of recursive competitive equilibria at time $ 0 $
    • Iterations on the function $ \Psi $ determine subsequent $ \theta_t $’s that summarize the aspects of the continuation competitive equilibria that influence the household’s decisions
    • At time $ 0 $, the Ramsey planner commits to this implied sequence $ \{\theta_t\}_{t=0}^\infty $ and therefore to an associated sequence of continuation competitive equilibria
  4. A characterization of time-inconsistency of a Ramsey plan
    • Imagine that after a ‘revolution’ at time $ t \geq 1 $, a new Ramsey planner is given the opportunity to ignore history and solve a brand new Ramsey plan
    • This new planner would want to reset the $ \theta_t $ associated with the original Ramsey plan to $ \theta_0 $
    • The incentive to reinitialize $ \theta_t $ associated with this revolution experiment indicates the time-inconsistency of the Ramsey plan
    • By resetting $ \theta $ to $ \theta_0 $, the new planner avoids the costs at time $ t $ that the original Ramsey planner must pay to reap the beneficial effects that the original Ramsey plan for $ s \geq t $ had achieved via its influence on the household’s decisions for $ s = 0, \ldots, t-1 $

Analysis

A competitive equilibrium is a triple of sequences $ (\vec m, \vec x, \vec h) \in E^\infty $ that satisfies (2), (3), and (6)

Chang works with a set of competitive equilibria defined as follows

Definition: $ CE = \bigl\{ (\vec m, \vec x, \vec h) \in E^\infty $ such that (2), (3), and (6) are satisfied $ \bigr\} $

$ CE $ is not empty because there exists a competitive equilibrium with $ h_t =1 $ for all $ t \geq 1 $, namely, an equilibrium with a constant money supply and constant price level

Chang establishes that $ CE $ is also compact

Chang makes the following key observation that combines ideas of Abreu, Pearce, and Stacchetti [APS90] with insights of Kydland and Prescott [KP80a]

Proposition: The continuation of a competitive equilibrium is a competitive equilibrium

That is, $ (\vec m, \vec x, \vec h) \in CE $ implies that $ (\vec m_t, \vec x_t, \vec h_t) \in CE \ \forall \ t \geq 1 $

(Lecture dynamic Stackelberg problems also used a version of this insight)

We can now state that a Ramsey problem is to

$$ \max_{(\vec m, \vec x, \vec h) \in E^\infty} \sum_{t=0}^\infty \beta^t \left[ u(c_t) + v(m_t) \right] $$

subject to restrictions (2), (3), and (6)

Evidently, associated with any competitive equilibrium $ (m_0, x_0) $ is an implied value of $ \theta_0 = u'(f(x_0))(m_0 + x_0) $

To bring out a recursive structure inherent in the Ramsey problem, Chang defines the set

$$ \Omega = \left\{ \theta \in \mathbb{R} \ \text{ such that } \ \theta = u'(f(x_0)) (m_0 + x_0) \ \text{ for some } \ (\vec m, \vec x, \vec h) \in CE \right\} $$

Equation (6) inherits from the household’s Euler equation for money holdings the property that the value of $ m_0 $ consistent with the representative household’s choices depends on $ (\vec h_1, \vec m_1) $

This dependence is captured in the definition above by making $ \Omega $ be the set of first period values of $ \theta_0 $ satisfying $ \theta_0 = u'(f(x_0)) (m_0 + x_0) $ for first period component $ (m_0,h_0) $ of competitive equilibrium sequences $ (\vec m, \vec x, \vec h) $

Chang establishes that $ \Omega $ is a nonempty and compact subset of $ \mathbb{R}_+ $

Next Chang advances:

Definition: $ \Gamma(\theta) = \{ (\vec m, \vec x, \vec h) \in CE | \theta = u'(f(x_0))(m_0 + x_0) \} $

Thus, $ \Gamma(\theta) $ is the set of competitive equilibrium sequences $ (\vec m, \vec x, \vec h) $ whose first period components $ (m_0, h_0) $ deliver the prescribed value $ \theta $ for first period marginal utility

If we knew the sets $ \Omega, \Gamma(\theta) $, we could use the following two-step procedure to find at least the value of the Ramsey outcome to the representative household

  1. Find the indirect value function $ w(\theta) $ defined as
$$ w(\theta) = \max_{(\vec m, \vec x, \vec h) \in \Gamma(\theta)} \sum_{t=0}^\infty \beta^t \left[ u(f(x_t)) + v(m_t) \right] $$
  1. Compute the value of the Ramsey outcome by solving $ \max_{\theta \in \Omega} w(\theta) $

Thus, Chang states the following

Proposition:

$ w(\theta) $ satisfies the Bellman equation

$$ w(\theta) = \max_{x,m,h,\theta'} \bigl\{ u(f(x)) + v(m) + \beta w(\theta') \bigr\} \tag{9} $$

where maximization is subject to

$$ (m,x,h) \in E \ {\rm and} \ \theta' \in \Omega \tag{10} $$

and

$$ \theta = u'(f(x)) (m+x) \tag{11} $$

and

$$ -x = m(1-h) \tag{12} $$

and

$$ m \cdot [ u'(f(x)) - v'(m) ] \leq \beta \theta' , \quad = \ {\rm if} \ m < \bar m \tag{13} $$

Before we use this proposition to recover a recursive representation of the Ramsey plan, note that the proposition relies on knowing the set $ \Omega $

To find $ \Omega $, Chang uses the insights of Kydland and Prescott [KP80a] together with a method based on the Abreu, Pearce, and Stacchetti [APS90] iteration to convergence on an operator $ B $ that maps continuation values into values

We want an operator that maps a continuation $ \theta $ into a current $ \theta $

Chang lets $ Q $ be a nonempty, bounded subset of $ \mathbb{R} $

Elements of the set $ Q $ are taken to be candidate values for continuation marginal utilities

Chang defines an operator

$$ B(Q) = \theta \in \mathbb{R} \ \text{ such that there is } \ (m,x,h, \theta') \in E \times Q $$

such that (11), (12), and (13) hold

Thus, $ B(Q) $ is the set of first period $ \theta $’s attainable with $ (m,x,h) \in E $ and some $ \theta' \in Q $

Proposition:

  1. $ Q \subset B(Q) $ implies $ B(Q) \subset \Omega $ (‘self-generation’)
  2. $ \Omega = B(\Omega) $ (‘factorization’)

The proposition characterizes $ \Omega $ as the largest fixed point of $ B $

It is easy to establish that $ B(Q) $ is a monotone operator

This property allows Chang to compute $ \Omega $ as the limit of iterations on $ B $ provided that iterations begin from a sufficiently large initial set

Some useful notation

Let $ \vec h^t = (h_0, h_1, \ldots, h_t) $ denote a history of inverse money creation rates with time $ t $ component $ h_t \in \Pi $

A government strategy $ \sigma=\{\sigma_t\}_{t=0}^\infty $ is a $ \sigma_0 \in \Pi $ and for $ t \geq 1 $ a sequence of functions $ \sigma_t: \Pi^{t-1} \rightarrow \Pi $

Chang restricts the government’s choice of strategies to the following space:

$$ CE_\pi = \{ {\vec h} \in \Pi^\infty: \text{ there is some } \ (\vec m, \vec x) \ \text{ such that } \ (\vec m, \vec x, \vec h) \in CE \} $$

In words, $ CE_\pi $ is the set of money growth sequences consistent with the existence of competitive equilibria

Chang observes that $ CE_\pi $ is nonempty and compact

Definition: $ \sigma $ is said to be admissible if for all $ t \geq 1 $ and after any history $ \vec h^{t-1} $, the continuation $ \vec h_t $ implied by $ \sigma $ belongs to $ CE_\pi $

Admissibility of $ \sigma $ means that anticipated policy choices associated with $ \sigma $ are consistent with the existence of competitive equilibria after each possible subsequent history

After any history $ \vec h^{t-1} $, admissibility restricts the government’s choice in period $ t $ to the set

$$ CE_\pi^0 = \{ h \in \Pi: {\rm there \ is } \ \vec h \in CE_\pi \ {\rm with } \ h=h_0 \} . $$

In words, $ CE_\pi^0 $ is the set of all first period money growth rates $ h=h_0 $, each of which is consistent with the existence of a sequence of money growth rates $ \vec h $ starting from $ h_0 $ in the initial period and for which a competitive equilibrium exists

Remark: $ CE_\pi^0 = \{h \in \Pi: \text{ there is } \ (m,\theta') \in [0, \bar m] \times \Omega \ \text{ such that } \ m u'[ f((h-1)m) - v'(m)] \leq \beta \theta' \ \text{ with equality if } \ m < \bar m \} $

Definition: An allocation rule is a sequence of functions $ \vec \alpha = \{\alpha_t\}_{t=0}^\infty $ such that $ \alpha_t: \Pi^t \rightarrow [0, \bar m] \times X $

Thus, the time $ t $ component of $ \alpha_t(h^t) $ is a pair of functions $ (m_t(h^t), x_t(h^t)) $

Definition: Given an admissible government strategy $ \sigma $, an allocation rule $ \alpha $ is called competitive if given any history $ \vec h^{t-1} $ and $ h_t \in CE_\pi^0 $, the continuations of $ \sigma $ and $ \alpha $ after $ (\vec h^{t-1},h_t) $ induce a competitive equilibrium sequence

Another operator

At this point it is convenient to introduce another operator that can be used to compute a Ramsey plan

For computing a Ramsey plan, this operator is wasteful because it works with a state vector that is bigger than necessary

We introduce this operator because it helps to prepare the way for Chang’s operator called $ \tilde D(Z) $ that we shall describe in lecture credible government policies

It is also useful because a fixed point of the operator to be defined here provides a good guess for an initial set from which to initiate iterations on Chang’s set-to-set operator $ \tilde D(Z) $ to be described in lecture credible government policies

Let $ S $ be the set of all pairs $ (w, \theta) $ of competitive equilibrium values and associated initial marginal utilities

Let $ W $ be a bounded set of values in $ \mathbb{R} $

Let $ Z $ be a nonempty subset of $ W \times \Omega $

Think of using pairs $ (w', \theta') $ drawn from $ Z $ as candidate continuation value, $ \theta $ pairs

Define the operator

$$ D(Z) = \Bigl\{ (w,\theta): {\rm there \ is } \ h \in CE_\pi^0 $$
$$ \text{ and a four-tuple } \ (m(h), x(h), w'(h), \theta'(h)) \in [0,\bar m]\times X \times Z $$

such that

$$ w = u(f(x( h))) + v(m( h)) + \beta w'( h) \tag{14} $$

$$ \theta = u'(f(x( h))) ( m( h) + x( h)) \tag{15} $$

$$ x(h) = m(h) (h-1) \tag{16} $$

$$ m(h) (u'(f(x(h))) - v'(m(h))) \leq \beta \theta'(h) \tag{17} $$

$$ \quad \quad \ \text{ with equality if } m(h) < \bar m \Bigr\} $$

It is possible to establish

Proposition:

  1. If $ Z \subset D(Z) $, then $ D(Z) \subset S $ (‘self-generation’)
  2. $ S = D(S) $ (‘factorization’)

Proposition:

  1. Monotonicity of $ D $: $ Z \subset Z' $ implies $ D(Z) \subset D(Z') $
  2. $ Z $ compact implies that $ D(Z) $ is compact

It can be shown that $ S $ is compact and that therefore there exists a $ (w, \theta) $ pair within this set that attains the highest possible value $ w $

This $ (w, \theta) $ pair i associated with a Ramsey plan

Further, we can compute $ S $ by iterating to convergence on $ D $ provided that one begins with a sufficiently large initial set $ S_0 $

As a very useful by-product, the algorithm that finds the largest fixed point $ S = D(S) $ also produces the Ramsey plan, its value $ w $, and the associated competitive equilibrium

Calculating all $ (w,\theta) $ pairs in $ CE $

Above we have defined the $ D(Z) $ operator as:

$$ D(Z) = \{ (w,\theta): \exists h \in CE^0_\pi \text{ and } (m(h),x(h),w'(h),\theta'(h)) \in [0,\bar m] \times X \times Z $$

such that

$$ w = u(f(x(h))) + v(m(h)) + \beta w'(h) $$$$ \theta = u'(f(x(h)))(m(h) + x(h)) $$$$ x(h) = m(h)(h-1) $$$$ m(h)(u'(f(x(h))) - v'(m(h))) \leq \beta \theta'(h) \text{ (with equality if } m(h) < \bar m) \} $$

We noted that the set $ S $ can be found by iterating to convergence on $ D $, provided that we start with a sufficiently large initial set $ S_0 $

Our implementation builds on ideas in this notebook

To find $ S $ we use a numerical algorithm called the outer hyperplane approximation algorithm

It was invented by Judd, Yeltekin, Conklin [JYC03]

This algorithm constructs the smallest convex set that contains the fixed point of the $ D(S) $ operator

Given that we are finding the smallest convex set that contains $ S $, we can represent it on a computer as the intersection of a finite number of half-spaces

Let $ H $ be a set of subgradients, and $ C $ be a set of hyperplane levels

We approximate $ S $ by:

$$ \tilde S = \{(w,\theta)| H \cdot (w,\theta) \leq C \} $$

A key feature of this algorithm is that we discretize the action space, i.e., we create a grid of possible values for $ m $ and $ h $ (note that $ x $ is implied by $ m $ and $ h $). This discretization simplifies computation of $ \tilde S $ by allowing us to find it by solving a sequence of linear programs

The outer hyperplane approximation algorithm proceeds as follows:

  1. Initialize subgradients, $ H $, and hyperplane levels, $ C_0 $
  2. Given a set of subgradients, $ H $, and hyperplane levels, $ C_t $, for each subgradient $ h_i \in H $:
    • Solve a linear program (described below) for each action in the action space
    • Find the maximum and update the corresponding hyperplane level, $ C_{i,t+1} $
  3. If $ |C_{t+1}-C_t| > \epsilon $, return to 2

Step 1 simply creates a large initial set $ S_0 $

Given some set $ S_t $, Step 2 then constructs the set $ S_{t+1} = D(S_t) $. The linear program in Step 2 is designed to construct a set $ S_{t+1} $ that is as large as possible while satisfying the constraints of the $ D(S) $ operator

To do this, for each subgradient $ h_i $, and for each point in the action space $ (m_j,h_j) $, we solve the following problem:

$$ \max_{[w',\theta']} h_i \cdot (w,\theta) $$

subject to

$$ H \cdot (w',\theta') \leq C_t $$$$ w = u(f(x_j)) + v(m_j) + \beta w' $$$$ \theta = u'(f(x_j))(m_j + x_j) $$$$ x_j = m_j(h_j-1) $$$$ m_j(u'(f(x_j)) - v'(m_j)) \leq \beta \theta'\hspace{2mm} (= \text{if } m_j < \bar m) $$

This problem maximizes the hyperplane level for a given set of actions

The second part of Step 2 then finds the maximum possible hyperplane level across the action space

The algorithm constructs a sequence of progressively smaller sets $ S_{t+1} \subset S_t \subset S_{t-1} \cdots \subset S_0 $

Step 3 ends the algorithm when the difference between these sets is small enough

We have created a Python class that solves the model assuming the following functional forms:

$$ u(c) = log(c) $$$$ v(m) = \frac{1}{500}(m \bar m - 0.5m^2)^{0.5} $$$$ f(x) = 180 - (0.4x)^2 $$

The remaining parameters $ \{\beta, \bar m, \underline h, \bar h\} $ are then variables to be specified for an instance of the Chang class

Below we use the class to solve the model and plot the resulting equilibrium set, once with $ \beta = 0.3 $ and once with $ \beta = 0.8 $

(Here we have set the number of subgradients to 10 in order to speed up the code for now - we can increase accuracy by increasing the number of subgradients)

Note: this code requires the polytope package

The package can be installed in a terminal/command prompt with pip

In [1]:
!pip install polytope
Requirement already satisfied: polytope in /home/quantecon/anaconda3/lib/python3.7/site-packages (0.2.1)
Requirement already satisfied: numpy>=1.10.0 in /home/quantecon/anaconda3/lib/python3.7/site-packages (from polytope) (1.15.1)
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In [2]:
"""
Author: Sebastian Graves

Provides a class called ChangModel to solve different 
parameterizations of the Chang (1998) model.
"""

import numpy as np
import quantecon as qe
import time

from scipy.spatial import ConvexHull
from scipy.optimize import linprog, minimize, minimize_scalar
from scipy.interpolate import UnivariateSpline
import numpy.polynomial.chebyshev as cheb


class ChangModel:
    """
    Class to solve for the competitive and sustainable sets in the Chang (1998)
    model, for different parameterizations.
    """

    def __init__(self, β, mbar, h_min, h_max, n_h, n_m, N_g):
        # Record parameters
        self.β, self.mbar, self.h_min, self.h_max = β, mbar, h_min, h_max
        self.n_h, self.n_m, self.N_g = n_h, n_m, N_g

        # Create other parameters
        self.m_min = 1e-9
        self.m_max = self.mbar
        self.N_a = self.n_h*self.n_m

        # Utility and production functions
        uc = lambda c: np.log(c)
        uc_p = lambda c: 1/c
        v = lambda m: 1/500 * (mbar * m - 0.5 * m**2)**0.5
        v_p = lambda m: 0.5/500 * (mbar * m - 0.5 * m**2)**(-0.5) * (mbar - m)
        u = lambda h, m: uc(f(h, m)) + v(m)

        def f(h, m):
            x = m * (h - 1)
            f = 180 - (0.4 * x)**2
            return f

        def θ(h, m):
            x = m * (h - 1)
            θ = uc_p(f(h, m)) * (m + x)
            return θ

        # Create set of possible action combinations, A
        A1 = np.linspace(h_min, h_max, n_h).reshape(n_h, 1)
        A2 = np.linspace(self.m_min, self.m_max, n_m).reshape(n_m, 1)
        self.A = np.concatenate((np.kron(np.ones((n_m, 1)), A1),
                                 np.kron(A2, np.ones((n_h, 1)))), axis=1)

        # Pre-compute utility and output vectors
        self.euler_vec = -np.multiply(self.A[:, 1], uc_p(f(self.A[:, 0], self.A[:, 1])) - v_p(self.A[:, 1]))
        self.u_vec = u(self.A[:, 0], self.A[:, 1])
        self.Θ_vec = θ(self.A[:, 0], self.A[:, 1])
        self.f_vec = f(self.A[:, 0], self.A[:, 1])
        self.bell_vec = np.multiply(uc_p(f(self.A[:, 0],
                                   self.A[:, 1])),
                                   np.multiply(self.A[:, 1],
                                   (self.A[:, 0] - 1))) + np.multiply(self.A[:, 1],
                                   v_p(self.A[:, 1]))

        # Find extrema of (w, θ) space for initial guess of equilibrium sets
        p_vec = np.zeros(self.N_a)
        w_vec = np.zeros(self.N_a)
        for i in range(self.N_a):
            p_vec[i] = self.Θ_vec[i]
            w_vec[i] = self.u_vec[i]/(1 - β)

        w_space = np.array([min(w_vec[~np.isinf(w_vec)]),
                            max(w_vec[~np.isinf(w_vec)])])
        p_space = np.array([0, max(p_vec[~np.isinf(w_vec)])])
        self.p_space = p_space

        # Set up hyperplane levels and gradients for iterations
        def SG_H_V(N, w_space, p_space):
            """
            This function  initializes the subgradients, hyperplane levels,
            and extreme points of the value set by choosing an appropriate
            origin and radius. It is based on a similar function in QuantEcon's Games.jl
            """

            # First, create unit circle. Want points placed on [0, 2π]
            inc = 2 * np.pi / N
            degrees = np.arange(0, 2 * np.pi, inc)

            # Points on circle
            H = np.zeros((N, 2))
            for i in range(N):
                x = degrees[i]
                H[i, 0] = np.cos(x)
                H[i, 1] = np.sin(x)

            # Then calculate origin and radius
            o = np.array([np.mean(w_space), np.mean(p_space)])
            r1 = max((max(w_space) - o[0])**2, (o[0] - min(w_space))**2)
            r2 = max((max(p_space) - o[1])**2, (o[1] - min(p_space))**2)
            r = np.sqrt(r1 + r2)

            # Now calculate vertices
            Z = np.zeros((2, N))
            for i in range(N):
                Z[0, i] = o[0] + r*H.T[0, i]
                Z[1, i] = o[1] + r*H.T[1, i]

            # Corresponding hyperplane levels
            C = np.zeros(N)
            for i in range(N):
                C[i] = np.dot(Z[:, i], H[i, :])

            return C, H, Z

        C, self.H, Z = SG_H_V(N_g, w_space, p_space)
        C = C.reshape(N_g, 1)
        self.c0_c, self.c0_s, self.c1_c, self.c1_s = np.copy(C), np.copy(C), np.copy(C), np.copy(C)
        self.z0_s, self.z0_c, self.z1_s, self.z1_c = np.copy(Z), np.copy(Z), np.copy(Z), np.copy(Z)

        self.w_bnds_s, self.w_bnds_c = (w_space[0], w_space[1]), (w_space[0], w_space[1])
        self.p_bnds_s, self.p_bnds_c = (p_space[0], p_space[1]), (p_space[0], p_space[1])

        # Create dictionaries to save equilibrium set for each iteration
        self.c_dic_s, self.c_dic_c = {}, {}
        self.c_dic_s[0], self.c_dic_c[0] = self.c0_s, self.c0_c

    def solve_worst_spe(self):
        """
        Method to solve for BR(Z). See p.449 of Chang (1998)
        """

        p_vec = np.full(self.N_a, np.nan)
        c = [1, 0]

        # Pre-compute constraints
        aineq_mbar = np.vstack((self.H, np.array([0, -self.β])))
        bineq_mbar = np.vstack((self.c0_s, 0))

        aineq = self.H
        bineq = self.c0_s
        aeq = [[0, -self.β]]

        for j in range(self.N_a):
            # Only try if consumption is possible
            if self.f_vec[j] > 0:
                # If m = mbar, use inequality constraint
                if self.A[j, 1] == self.mbar:
                    bineq_mbar[-1] = self.euler_vec[j]
                    res = linprog(c, A_ub=aineq_mbar, b_ub=bineq_mbar, 
                                  bounds=(self.w_bnds_s, self.p_bnds_s))
                else:
                    beq = self.euler_vec[j]
                    res = linprog(c, A_ub=aineq, b_ub=bineq, A_eq=aeq, b_eq=beq,
                                  bounds=(self.w_bnds_s, self.p_bnds_s))
                if res.status == 0:
                    p_vec[j] = self.u_vec[j] + self.β * res.x[0]

        # Max over h and min over other variables (see Chang (1998) p.449)
        self.br_z = np.nanmax(np.nanmin(p_vec.reshape(self.n_m, self.n_h), 0))

    def solve_subgradient(self):
        """
        Method to solve for E(Z). See p.449 of Chang (1998)
        """

        # Pre-compute constraints
        aineq_C_mbar = np.vstack((self.H, np.array([0, -self.β])))
        bineq_C_mbar = np.vstack((self.c0_c, 0))

        aineq_C = self.H
        bineq_C = self.c0_c
        aeq_C = [[0, -self.β]]

        aineq_S_mbar = np.vstack((np.vstack((self.H, np.array([0, -self.β]))),
                                  np.array([-self.β, 0])))
        bineq_S_mbar = np.vstack((self.c0_s, np.zeros((2, 1))))

        aineq_S = np.vstack((self.H, np.array([-self.β, 0])))
        bineq_S = np.vstack((self.c0_s, 0))
        aeq_S = [[0, -self.β]]

        # Update maximal hyperplane level
        for i in range(self.N_g):
            c_a1a2_c, t_a1a2_c = np.full(self.N_a, -np.inf), np.zeros((self.N_a, 2))
            c_a1a2_s, t_a1a2_s = np.full(self.N_a, -np.inf), np.zeros((self.N_a, 2))

            c = [-self.H[i, 0], -self.H[i, 1]]

            for j in range(self.N_a):
                # Only try if consumption is possible
                if self.f_vec[j] > 0:
                
                    # COMPETITIVE EQUILIBRIA
                    # If m = mbar, use inequality constraint
                    if self.A[j, 1] == self.mbar:
                        bineq_C_mbar[-1] = self.euler_vec[j]
                        res = linprog(c, A_ub=aineq_C_mbar, b_ub=bineq_C_mbar,
                                      bounds=(self.w_bnds_c, self.p_bnds_c))
                    # If m < mbar, use equality constraint
                    else:
                        beq_C = self.euler_vec[j]
                        res = linprog(c, A_ub=aineq_C, b_ub=bineq_C, A_eq = aeq_C,
                                      b_eq = beq_C, bounds=(self.w_bnds_c, self.p_bnds_c))
                    if res.status == 0:
                        c_a1a2_c[j] = self.H[i, 0]*(self.u_vec[j] + self.β * res.x[0]) + self.H[i, 1] * self.Θ_vec[j]
                        t_a1a2_c[j] = res.x

                    # SUSTAINABLE EQUILIBRIA
                    # If m = mbar, use inequality constraint
                    if self.A[j, 1] == self.mbar:
                        bineq_S_mbar[-2] = self.euler_vec[j]
                        bineq_S_mbar[-1] = self.u_vec[j] - self.br_z
                        res = linprog(c, A_ub=aineq_S_mbar, b_ub=bineq_S_mbar, 
                                      bounds=(self.w_bnds_s, self.p_bnds_s))
                    # If m < mbar, use equality constraint
                    else:
                        bineq_S[-1] = self.u_vec[j] - self.br_z
                        beq_S = self.euler_vec[j]
                        res = linprog(c, A_ub=aineq_S, b_ub=bineq_S, A_eq = aeq_S,
                                      b_eq = beq_S, bounds=(self.w_bnds_s, self.p_bnds_s))
                    if res.status == 0:
                        c_a1a2_s[j] = self.H[i, 0] * (self.u_vec[j] + self.β*res.x[0]) + self.H[i, 1] * self.Θ_vec[j]
                        t_a1a2_s[j] = res.x

            idx_c = np.where(c_a1a2_c == max(c_a1a2_c))[0][0]
            self.z1_c[:, i] = np.array([self.u_vec[idx_c] + self.β * t_a1a2_c[idx_c, 0],
                                              self.Θ_vec[idx_c]])

            idx_s = np.where(c_a1a2_s == max(c_a1a2_s))[0][0]
            self.z1_s[:, i] = np.array([self.u_vec[idx_s] + self.β*t_a1a2_s[idx_s, 0],
                                        self.Θ_vec[idx_s]])

        for i in range(self.N_g):
            self.c1_c[i] = np.dot(self.z1_c[:, i], self.H[i, :])
            self.c1_s[i] = np.dot(self.z1_s[:, i], self.H[i, :])

    def solve_sustainable(self, tol=1e-5, max_iter=250):
        """
        Method to solve for the competitive and sustainable equilibrium sets.
        """

        t = time.time()
        diff = tol + 1
        iters = 0

        print('### --------------- ###')
        print('Solving Chang Model Using Outer Hyperplane Approximation')
        print('### --------------- ### \n')

        print('Maximum difference when updating hyperplane levels:')

        while diff > tol and iters < max_iter:
            iters = iters + 1
            self.solve_worst_spe()
            self.solve_subgradient()
            diff = max(np.maximum(abs(self.c0_c - self.c1_c),
                       abs(self.c0_s - self.c1_s)))
            print(diff)

            # Update hyperplane levels
            self.c0_c, self.c0_s = np.copy(self.c1_c), np.copy(self.c1_s)

            # Update bounds for w and θ
            wmin_c, wmax_c = np.min(self.z1_c, axis=1)[0], np.max(self.z1_c, axis=1)[0]
            pmin_c, pmax_c = np.min(self.z1_c, axis=1)[1], np.max(self.z1_c, axis=1)[1]

            wmin_s, wmax_s = np.min(self.z1_s, axis=1)[0], np.max(self.z1_s, axis=1)[0]
            pmin_S, pmax_S = np.min(self.z1_s, axis=1)[1], np.max(self.z1_s, axis=1)[1]

            self.w_bnds_s, self.w_bnds_c = (wmin_s, wmax_s), (wmin_c, wmax_c)
            self.p_bnds_s, self.p_bnds_c = (pmin_S, pmax_S), (pmin_c, pmax_c)

            # Save iteration
            self.c_dic_c[iters], self.c_dic_s[iters] = np.copy(self.c1_c), np.copy(self.c1_s)
            self.iters = iters

        elapsed = time.time() - t
        print('Convergence achieved after {} iterations and {} seconds'.format(iters, round(elapsed, 2)))

    def solve_bellman(self, θ_min, θ_max, order, disp=False, tol=1e-7, maxiters=100):
        """
        Continuous Method to solve the Bellman equation in section 25.3
        """
        mbar = self.mbar
        
        # Utility and production functions
        uc = lambda c: np.log(c)
        uc_p = lambda c: 1 / c
        v = lambda m: 1 / 500 * (mbar * m - 0.5 * m**2)**0.5
        v_p = lambda m: 0.5/500 * (mbar*m - 0.5 * m**2)**(-0.5) * (mbar - m)
        u = lambda h, m: uc(f(h, m)) + v(m)
 
        def f(h, m):
            x = m * (h - 1)
            f = 180 - (0.4 * x)**2
            return f
 
        def θ(h, m):
            x = m * (h - 1)
            θ = uc_p(f(h, m)) * (m + x)
            return θ
 
        # Bounds for Maximization
        lb1 = np.array([self.h_min, 0, θ_min])
        ub1 = np.array([self.h_max, self.mbar - 1e-5, θ_max])
        lb2 = np.array([self.h_min, θ_min])
        ub2 = np.array([self.h_max, θ_max])
 
        # Initialize Value Function coefficents
        # Calculate roots of Chebyshev polynomial
        k = np.linspace(order, 1, order)
        roots = np.cos((2 * k - 1) * np.pi / (2 * order))
        # Scale to approximation space
        s = θ_min + (roots - -1) / 2 * (θ_max - θ_min)
        # Create basis matrix
        Φ = cheb.chebvander(roots, order - 1)
        c = np.zeros(Φ.shape[0])
 
        # Function to minimize and constraints
        def p_fun(x):
            scale = -1 + 2 * (x[2] - θ_min)/(θ_max - θ_min)
            p_fun = - (u(x[0], x[1]) + self.β * np.dot(cheb.chebvander(scale, order - 1), c))
            return p_fun
 
        def p_fun2(x):
            scale = -1 + 2*(x[1] - θ_min)/(θ_max - θ_min)
            p_fun = - (u(x[0],mbar) + self.β * np.dot(cheb.chebvander(scale, order - 1), c))
            return p_fun
 
        cons1 = ({'type': 'eq',   'fun': lambda x: uc_p(f(x[0], x[1])) * x[1] * (x[0] - 1) + v_p(x[1]) * x[1] + self.β * x[2] - θ},
                 {'type': 'eq',   'fun': lambda x: uc_p(f(x[0], x[1])) * x[0] * x[1] - θ})
        cons2 = ({'type': 'ineq', 'fun': lambda x: uc_p(f(x[0], mbar)) * mbar * (x[0] - 1) + v_p(mbar) * mbar + self.β * x[1] - θ},
                 {'type': 'eq',   'fun': lambda x: uc_p(f(x[0], mbar)) * x[0] * mbar - θ})
 
        bnds1 = np.concatenate([lb1.reshape(3, 1), ub1.reshape(3, 1)], axis=1)
        bnds2 = np.concatenate([lb2.reshape(2, 1), ub2.reshape(2, 1)], axis=1)
 
        # Bellman Iterations
        diff = 1
        iters = 1
 
        while diff > tol:
        # 1. Maximization, given value function guess
            p_iter1 = np.zeros(order)
            for i in range(order):
                θ = s[i]
                res = minimize(p_fun,
                               lb1 + (ub1-lb1) / 2,
                               method='SLSQP',
                               bounds=bnds1,
                               constraints=cons1,
                               tol=1e-10)
                if res.success == True:
                    p_iter1[i] = -p_fun(res.x)
                res = minimize(p_fun2,
                               lb2 + (ub2-lb2) / 2,
                               method='SLSQP',
                               bounds=bnds2,
                               constraints=cons2,
                               tol=1e-10)
                if -p_fun2(res.x) > p_iter1[i] and res.success == True:
                    p_iter1[i] = -p_fun2(res.x)

            # 2. Bellman updating of Value Function coefficients
            c1 = np.linalg.solve(Φ, p_iter1)
            # 3. Compute distance and update
            diff = np.linalg.norm(c - c1)
            if bool(disp == True):
                print(diff)
            c = np.copy(c1)
            iters = iters + 1
            if iters > maxiters:
                print('Convergence failed after {} iterations'.format(maxiters))
                break

        self.θ_grid = s
        self.p_iter = p_iter1
        self.Φ = Φ
        self.c = c
        print('Convergence achieved after {} iterations'.format(iters))

        # Check residuals
        θ_grid_fine = np.linspace(θ_min, θ_max, 100)
        resid_grid = np.zeros(100)
        p_grid = np.zeros(100)
        θ_prime_grid = np.zeros(100)
        m_grid = np.zeros(100)
        h_grid = np.zeros(100)
        for i in range(100):
            θ = θ_grid_fine[i]
            res = minimize(p_fun,
                           lb1 + (ub1-lb1) / 2,
                           method='SLSQP',
                           bounds=bnds1,
                           constraints=cons1,
                           tol=1e-10)
            if res.success == True:
                p = -p_fun(res.x)
                p_grid[i] = p
                θ_prime_grid[i] = res.x[2]
                h_grid[i] = res.x[0]
                m_grid[i] = res.x[1]
            res = minimize(p_fun2,
                           lb2 + (ub2-lb2)/2,
                           method='SLSQP',
                           bounds=bnds2,
                           constraints=cons2,
                           tol=1e-10)
            if -p_fun2(res.x) > p and res.success == True:
                p = -p_fun2(res.x)
                p_grid[i] = p
                θ_prime_grid[i] = res.x[1]
                h_grid[i] = res.x[0]
                m_grid[i] = self.mbar
            scale = -1 + 2 * (θ - θ_min)/(θ_max - θ_min)
            resid_grid[i] = np.dot(cheb.chebvander(scale, order-1), c) - p

        self.resid_grid = resid_grid
        self.θ_grid_fine = θ_grid_fine
        self.θ_prime_grid = θ_prime_grid
        self.m_grid = m_grid
        self.h_grid = h_grid
        self.p_grid = p_grid
        self.x_grid = m_grid * (h_grid - 1)

        # Simulate
        θ_series = np.zeros(31)
        m_series = np.zeros(30)
        h_series = np.zeros(30)

        # Find initial θ
        def ValFun(x):
            scale = -1 + 2*(x - θ_min)/(θ_max - θ_min)
            p_fun = np.dot(cheb.chebvander(scale, order - 1), c)
            return -p_fun

        res = minimize(ValFun,
                      (θ_min + θ_max)/2,
                      bounds=[(θ_min, θ_max)])
        θ_series[0] = res.x

        # Simulate
        for i in range(30):
            θ = θ_series[i]
            res = minimize(p_fun,
                           lb1 + (ub1-lb1)/2,
                           method='SLSQP',
                           bounds=bnds1,
                           constraints=cons1,
                           tol=1e-10)
            if res.success == True:
                p = -p_fun(res.x)
                h_series[i] = res.x[0]
                m_series[i] = res.x[1]
                θ_series[i+1] = res.x[2]
            res2 = minimize(p_fun2,
                            lb2 + (ub2-lb2)/2,
                            method='SLSQP',
                            bounds=bnds2,
                            constraints=cons2,
                            tol=1e-10)
            if -p_fun2(res2.x) > p and res2.success == True:
                h_series[i] = res2.x[0]
                m_series[i] = self.mbar
                θ_series[i+1] = res2.x[1]

        self.θ_series = θ_series
        self.m_series = m_series
        self.h_series = h_series
        self.x_series = m_series * (h_series - 1)
In [3]:
ch1 = ChangModel(β=0.3, mbar=30, h_min=0.9, h_max=2, n_h=8, n_m=35, N_g=10)
ch1.solve_sustainable()
### --------------- ###
Solving Chang Model Using Outer Hyperplane Approximation
### --------------- ### 

Maximum difference when updating hyperplane levels:
[1.91679545]
[0.66781649]
[0.49234789]
[0.3241217]
[0.19022279]
[0.10862838]
[0.05817151]
[0.02620056]
[0.01836386]
[0.01415009]
[0.00297077]
[0.00089123]
[0.00026737]
[8.02108797e-05]
[2.40632639e-05]
[7.21897917e-06]
Convergence achieved after 16 iterations and 329.72 seconds
In [4]:
import polytope
import matplotlib.pyplot as plt
%matplotlib inline


def plot_competitive(ChangModel):
    """
    Method that only plots competitive equilibrium set
    """
    poly_C = polytope.Polytope(ChangModel.H, ChangModel.c1_c)
    ext_C = polytope.extreme(poly_C)

    fig, ax = plt.subplots(figsize=(7, 5))

    ax.set_xlabel('w', fontsize=16)
    ax.set_ylabel(r"$\theta$", fontsize=18)

    ax.fill(ext_C[:,0], ext_C[:,1], 'r', zorder=0)
    ChangModel.min_theta = min(ext_C[:, 1])
    ChangModel.max_theta = max(ext_C[:, 1])

    # Add point showing Ramsey Plan
    idx_Ramsey = np.where(ext_C[:, 0] == max(ext_C[:, 0]))[0][0]
    R = ext_C[idx_Ramsey, :]
    ax.scatter(R[0], R[1], 150, 'black', 'o', zorder=1)
    w_min = min(ext_C[:, 0])

    # Label Ramsey Plan slightly to the right of the point
    ax.annotate("R", xy=(R[0], R[1]), xytext=(R[0] + 0.03 * (R[0] - w_min),
                R[1]), fontsize=18)

    plt.tight_layout()
    plt.show()

plot_competitive(ch1)
`polytope` failed to import `cvxopt.glpk`.
will use `scipy.optimize.linprog`
In [5]:
ch2 = ChangModel(β=0.8, mbar=30, h_min=0.9, h_max=1/0.8, n_h=8, n_m=35, N_g=10)
ch2.solve_sustainable()
### --------------- ###
Solving Chang Model Using Outer Hyperplane Approximation
### --------------- ### 

Maximum difference when updating hyperplane levels:
[0.06369]
[0.02476]
[0.02153]
[0.01915]
[0.01795]
[0.01642]
[0.01507]
[0.01284]
[0.01106]
[0.00694]
[0.0085]
[0.00781]
[0.00433]
[0.00492]
[0.00303]
[0.00182]
[0.00638]
[0.00116]
[0.00093]
[0.00075]
[0.0006]
[0.00494]
[0.00038]
[0.00121]
[0.00024]
[0.0002]
[0.00016]
[0.00013]
[0.0001]
[0.00008]
[0.00006]
[0.00005]
[0.00004]
[0.00003]
[0.00003]
[0.00002]
[0.00002]
[0.00001]
[0.00001]
[0.00001]
Convergence achieved after 40 iterations and 804.19 seconds
In [6]:
plot_competitive(ch2)

Solving a continuation Ramsey planner’s Bellman equation

In this section we solve the Bellman equation confronting a continuation Ramsey planner

The construction of a Ramsey plan is decomposed into a two subproblems in Ramsey plans, time inconsistency, sustainable plans and dynamic Stackelberg problems

  • Subproblem 1 is faced by a sequence of continuation Ramsey planners at $ t \geq 1 $
  • Subproblem 2 is faced by a Ramsey planner at $ t = 0 $

The problem is:

$$ J(\theta) = \max_{m,x,h,\theta'} u(f(x)) + v(m) + \beta J(\theta') $$

subject to:

$$ \theta \leq u'(f(x))x + v'(m)m + \beta \theta' $$$$ \theta = u'(f(x))(m + x ) $$$$ x = m(h-1) $$$$ (m,x,h) \in E $$$$ \theta' \in \Omega $$

To solve this Bellman equation, we must know the set $ \Omega $

We have solved the Bellman equation for the two sets of parameter values for which we computed the equilibrium value sets above

Hence for these parameter configurations, we know the bounds of $ \Omega $

The two sets of parameters differ only in the level of $ \beta $

From the figures earlier in this lecture, we know that when $ \beta = 0.3 $, $ \Omega = [0.0088,0.0499] $, and when $ \beta = 0.8 $, $ \Omega = [0.0395,0.2193] $

In [7]:
ch1 = ChangModel(β=0.3, mbar=30, h_min=0.99, h_max=1/0.3, n_h=8, n_m=35, N_g=50)
ch2 = ChangModel(β=0.8, mbar=30, h_min=0.1, h_max=1/0.8, n_h=20, n_m=50, N_g=50)
/home/quantecon/anaconda3/lib/python3.7/site-packages/ipykernel_launcher.py:35: RuntimeWarning: invalid value encountered in log
In [8]:
ch1.solve_bellman(θ_min=0.01, θ_max=0.0499, order=30, tol=1e-6)
ch2.solve_bellman(θ_min=0.045, θ_max=0.15, order=30, tol=1e-6)
/home/quantecon/anaconda3/lib/python3.7/site-packages/ipykernel_launcher.py:290: RuntimeWarning: invalid value encountered in log
Convergence achieved after 15 iterations
Convergence achieved after 72 iterations

First, a quick check that our approximations of the value functions are good

We do this by calculating the residuals between iterates on the value function on a fine grid:

In [9]:
max(abs(ch1.resid_grid)), max(abs(ch2.resid_grid))
Out[9]:
(6.463130481471069e-06, 6.875604405820468e-07)

The value functions plotted below trace out the right edges of the sets of equilibrium values plotted above

In [10]:
fig, axes = plt.subplots(1, 2, figsize=(12, 4))

for ax, model in zip(axes, (ch1, ch2)):
    ax.plot(model.θ_grid, model.p_iter)
    ax.set(xlabel=r"$\theta$",
           ylabel=r"$J(\theta)$",
           title=rf"$\beta = {model.β}$")

plt.show()

The next figure plots the optimal policy functions; values of $ \theta',m,x,h $ for each value of the state $ \theta $:

In [11]:
for model in (ch1, ch2):

    fig, axes = plt.subplots(2, 2, figsize=(12, 6), sharex=True)
    fig.suptitle(rf"$\beta = {model.β}$", fontsize=16)

    plots = [model.θ_prime_grid, model.m_grid,
             model.h_grid, model.x_grid]
    labels = [r"$\theta'$", "$m$", "$h$", "$x$"]

    for ax, plot, label in zip(axes.flatten(), plots, labels):
        ax.plot(model.θ_grid_fine, plot)
        ax.set_xlabel(r"$\theta$", fontsize=14)
        ax.set_ylabel(label, fontsize=14)

    plt.show()

With the first set of parameter values, the value of $ \theta' $ chosen by the Ramsey planner quickly hits the upper limit of $ \Omega $

But with the second set of parameters it converges to a value in the interior of the set

Consequently, the choice of $ \bar \theta $ is clearly important with the first set of parameter values

One way of seeing this is plotting $ \theta'(\theta) $ for each set of parameters

With the first set of parameter values, this function does not intersect the 45-degree line until $ \bar \theta $, whereas in the second set of parameter values, it intersects in the interior

In [12]:
fig, axes = plt.subplots(1, 2, figsize=(12, 4))

for ax, model in zip(axes, (ch1, ch2)):
    ax.plot(model.θ_grid_fine, model.θ_prime_grid, label=r"$\theta'(\theta)$")
    ax.plot(model.θ_grid_fine, model.θ_grid_fine, label=r"$\theta$")
    ax.set(xlabel=r"$\theta$", title=rf"$\beta = {model.β}$")

axes[0].legend()
plt.show()

Subproblem 2 is equivalent to the planner choosing the initial value of $ \theta $ (i.e. the value which maximizes the value function)

From this starting point, we can then trace out the paths for $ \{\theta_t,m_t,h_t,x_t\}_{t=0}^\infty $ that support this equilibrium

These are shown below for both sets of parameters

In [13]:
for model in (ch1, ch2):

    fig, axes = plt.subplots(2, 2, figsize=(12, 6))
    fig.suptitle(rf"$\beta = {model.β}$")

    plots = [model.θ_series, model.m_series, model.h_series, model.x_series]
    labels = [r"$\theta$", "$m$", "$h$", "$x$"]

    for ax, plot, label in zip(axes.flatten(), plots, labels):
        ax.plot(plot)
        ax.set(xlabel='t', ylabel=label)

    plt.show()

Next Steps

In Credible Government Policies in Chang Model we shall find a subset of competitive equilibria that are sustainable in the sense that a sequence of government administrations that chooses sequentially, rather than once and for all at time $ 0 $ will choose to implement them

In the process of constructing them, we shall construct another, smaller set of competitive equilibria