How to Classify Countries Based on Their Level of Development

There are two kinds of people in the world: those who divide the world into two kinds of people, and those who don’t.

—Robert Benchley.

Abstract

The paper analyzes how the United Nations Development Programme, the World Bank, the International Monetary Fund and the World Trade Organization classify countries based on their level of development. These systems are found lacking in clarity with regard to their underlying rationale. The paper argues that a country classification system based on a transparent, data-driven methodology is preferable to one based on judgment or ad hoc rules. Such an alternative methodology is developed and used to construct classification systems using a variety of proxies for development attainment. The methodology provides a way to construct a linear approximation of a Lorenz curve such that the difference between the linear approximation and the actual Lorenz curve is minimized. The linear segments represent different categories of countries (e.g., low development and high development countries). The methodology has wider applicability; it can be used whenever there is a need to construct a classification system of relatively few categories from a large heterogeneous sample.

Appendices

Appendix 1

The sufficient second order conditions are that

$$ \frac{{\partial^{2} E_{3} }}{{\partial x_{0}^{2} }} = \frac{{x_{1} L^{\prime \prime} (x_{0} )}}{2} > 0 $$

$$ \frac{{\partial^{2} E_{3} }}{{\partial x_{1}^{2} }} = \frac{{(1 - x_{0} )L^{\prime \prime} (x_{1} )}}{2} > 0.$$

These conditions are met because $L(x)$ is strictly convex. An additional necessary second order condition is that the determinant of the Hessian matrix is non-negative:

$$ \left| {\begin{array}{*{20}c} {\frac{{\partial^{2} E_{3} }}{{\partial x_{0}^{2} }}} & {\frac{{\partial^{2} E_{3} }}{{\partial x_{1} \partial x_{0} }}} \\ {\frac{{\partial^{2} E_{3} }}{{\partial x_{0} \partial x_{1} }}} & {\frac{{\partial^{2} E_{3} }}{{\partial x_{1}^{2} }}} \\ \end{array} } \right| = \left| {\begin{array}{*{20}c} {\frac{{x_{1} L^{\prime \prime} (x_{0} )}}{2}} & {\frac{{L^{\prime \prime} (x_{0} ) - L^{\prime \prime} (x_{1} )}}{2}} \\ {\frac{{L^{\prime} (x_{0} ) - L^{\prime} (x_{1} )}}{2}} & {\frac{{(1 - x_{0} )L^{\prime \prime} (x_{1} )}}{2}} \\ \end{array} } \right| \ge^{?} 0 $$

This is equivalent to evaluating the sign of the following expression:

$$ x_{1} (1 - x_{0} )L^{\prime \prime} (x_{0} )L^{\prime \prime} (x_{1} ) - (L^{\prime} (x_{0} ) - L^\prime (x_{1} ))^{2} $$

When the first order conditions are satisfied, the second derivatives are given by:

$$ L^{\prime \prime} (x_{0} ) = \frac{{L\prime (x_{1} )x_{1} - L(x_{1} )}}{{x_{1}^{2} }} = \frac{{\frac{{1 - L(x_{0} )}}{{1 - x_{0} }}x_{1} - L(x_{1} )}}{{x_{1}^{2} }} = \frac{{1 - L(x_{0} )}}{{(1 - x_{0} )x_{1} }} - \frac{{L(x_{1} )}}{{x_{1}^{2} }} $$

$$ L^{\prime \prime} (x_{1} ) = \frac{{ - L\prime (x_{0} )(1 - x_{0} ) + 1 - L(x_{0} )}}{{(1 - x_{0} )^{2} }} = \frac{{ - \frac{{L(x_{1} )}}{{x_{1} }}(1 - x_{0} ) + 1 - L(x_{0} )}}{{(1 - x_{0} )^{2} }} = - \frac{{L(x_{1} )}}{{(1 - x_{0} )x_{1} }} + \frac{{1 - L(x_{0} )}}{{(1 - x_{0} )^{2} }} $$

Therefore, the first part of the expression can be rewritten as follows:

$$ \begin{aligned} &(1 - x_{0} )x_{1} L^{\prime \prime} (x_{0} )L^{\prime \prime} (x_{1} ) \\ & \quad = (1 - x_{0} )x_{1} \left\{ {\frac{{1 - L(x_{0} )}}{{(1 - x_{0} )x_{1} }} - \frac{{L(x_{1} )}}{{x_{1}^{2} }}} \right\}\left\{ { - \frac{{L(x_{1} )}}{{(1 - x_{0} )x_{1} }} + \frac{{1 - L(x_{0} )}}{{(1 - x_{0} )^{2} }}} \right\} \\ & \quad = \left\{ {1 - L(x_{0} ) - \frac{{L(x_{1} )(1 - x_{0} )}}{{x_{1} }}} \right\}\left\{ { - \frac{{L(x_{1} )}}{{(1 - x_{0} )x_{1} }} + \frac{{1 - L(x_{0} )}}{{(1 - x_{0} )^{2} }}} \right\} \\ & \quad = - \frac{{L(x_{1} )}}{{(1 - x_{0} )x_{1} }} + \frac{{1 - L(x_{0} )}}{{(1 - x_{0} )^{2} }} + \frac{{L(x_{0} )L(x_{1} )}}{{(1 - x_{0} )x_{1} }} - \frac{{L(x_{0} )[1 - L(x_{0} )]}}{{(1 - x_{0} )^{2} }} + \frac{{L^{2} (x_{1} )}}{{x_{1}^{2} }} - \frac{{L(x_{1} )[1 - L(x_{0} )]}}{{(1 - x_{0} )x_{1} }} \\ & \quad = \frac{{ - L(x_{1} ) + L(x_{0} )L(x_{1} ) - L(x_{1} )[1 - L(x_{0} )]}}{{(1 - x_{0} )x_{1} }} + \frac{{1 - L(x_{0} ) - L(x_{0} )[1 - L(x_{0} )]}}{{(1 - x_{0} )^{2} }} + \frac{{L^{2} (x_{1} )}}{{x_{1}^{2} }} \\ &\quad = \frac{{ - L(x_{1} )[1 - L(x_{0} )] - L(x_{1} )[1 - L(x_{0} )]}}{{(1 - x_{0} )x_{1} }} + \frac{{1 - L(x_{0} ) - L(x_{0} ) + L^{2} (x_{0} )}}{{(1 - x_{0} )^{2} }} + \frac{{L^{2} (x_{1} )}}{{x_{1}^{2} }} \\ & \quad = - 2\frac{{L(x_{1} )}}{{x_{1} }}\frac{{1 - L(x_{0} )}}{{1 - x_{0} }} + \frac{{(1 - L(x_{0} ))^{2} }}{{(1 - x_{0} )^{2} }} + \frac{{L^{2} (x_{1} )}}{{x_{1}^{2} }} \\ & \quad = \left[ {\frac{{L(x_{1} )}}{{x_{1} }} - \frac{{1 - L(x_{0} )}}{{1 - x_{0} }}} \right]^{2} \\ \end{aligned} $$

As the first part of the expression is equal to the second part, the determinant of the Hessian matrix is zero, and all necessary second order conditions are met. The corner solution is when $x_{0} = 0 \wedge x_{1} = 1$. That point, $E_{3} (0,1) = E_{1}$, maximizes the error. The interior point where the first order conditions are met is therefore a minimum.

Appendix 2

In a taxonomy with N categories, the error term associated with that taxonomy consists of the sum of N triangles and N−1 rectangles less the area under the Lorenz curve. The areas of the triangles are as follows:

$$ T_{0} = \frac{1}{2}x_{0} L(x_{0} ) $$

$$ T_{1} = \frac{1}{2}(x_{1} - x_{0} )(L(x_{1} ) - L(x_{0} )) = \frac{1}{2}x_{1} L(x_{1} ) - \frac{1}{2}x_{1} L(x_{0} ) - \frac{1}{2}x_{0} L(x_{1} ) + \frac{1}{2}x_{0} L(x_{0} ) $$

$$ T_{2} = \frac{1}{2}(x_{2} - x_{1} )(L(x_{2} ) - L(x_{1} )) = \frac{1}{2}x_{2} L(x_{2} ) - \frac{1}{2}x_{2} L(x_{1} ) - \frac{1}{2}x_{1} L(x_{2} ) + \frac{1}{2}x_{1} L(x_{1} ) $$

$$ T_{n - 2} = \frac{1}{2}(x_{n - 2} - x_{n - 3} )(L(x_{n - 2} ) - L(x_{n - 3} )) = \frac{1}{2}x_{n - 2} L(x_{n - 2} ) - \frac{1}{2}x_{n - 2} L(x_{n - 3} ) - \frac{1}{2}x_{n - 3} L(x_{n - 2} ) + \frac{1}{2}x_{n - 3} L(x_{n - 3} ) $$

$$ T_{n - 1} = \frac{1}{2}(1 - x_{n - 2} )(1 - L(x_{n - 2} )) = \frac{1}{2} - \frac{1}{2}L(x_{n - 2} ) - \frac{1}{2}x_{n - 2} + \frac{1}{2}x_{n - 2} L(x_{n - 2} ) $$

Therefore, the sum of the areas of the triangles is

$$ \sum\limits_{i = 0}^{n - 1} {T_{i} } = \sum\limits_{i = 0}^{n - 2} {x_{i} L(x_{i} ) - \frac{1}{2}\sum\limits_{i = 0}^{n - 3} {x_{i + 1} L(x_{i} ) - \frac{1}{2}\sum\limits_{i = 0}^{n - 3} {x_{i} L(x_{i + 1} )} } } - \frac{1}{2}L(x_{n - 2} ) - \frac{1}{2}x_{n - 2} + \frac{1}{2} $$

The areas of the $N - 1$ rectangles are as follows:

$$ R_{0} = (x_{1} - x_{0} )L(x_{0} ) = x_{1} L(x_{0} ) - x_{0} L(x_{0} ) $$

$$ R_{1} = (x_{2} - x_{1} )L(x_{1} ) = x_{2} L(x_{1} ) - x_{1} L(x_{1} ) $$

$$ R_{2} = (x_{3} - x_{2} )L(x_{2} ) = x_{3} L(x_{2} ) - x_{2} L(x_{2} ) $$

$$ R_{n - 3} = (x_{n - 2} - x_{n - 3} )L(x_{n - 3} ) = x_{n - 2} L(x_{n - 3} ) - x_{n - 2} L(x_{n - 2} ) $$

$$ R_{n - 2} = (1 - x_{n - 2} )L(x_{n - 2} ) = L(x_{n - 2} ) - x_{n - 2} L(x_{n - 2} ) $$

Therefore, the sum of the areas of the rectangles is

$$ \sum\limits_{i = 0}^{n - 2} {R_{i} } = \sum\limits_{i = 0}^{n - 3} {x_{i + 1} L(x_{i} ) + L(x_{n - 2} ) - \sum\limits_{i = 0}^{n - 2} {x_{i} L(x_{i} )} } $$

The general form for the error term for n categories is therefore given as:

$$ \begin{aligned} E_{n} & = \sum\limits_{i = 0}^{n - 1} {T_{i} } + \sum\limits_{i = 0}^{n - 2} {R_{i} } - \int\limits_{0}^{1} {L(x)} \\ & = \sum\limits_{i = 0}^{n - 2} {x_{i} L(x_{i} ) - \frac{1}{2}\sum\limits_{i = 0}^{n - 3} {x_{i + 1} L(x_{i} ) - \frac{1}{2}\sum\limits_{i = 0}^{n - 3} {x_{i} L(x_{i + 1} )} } } - \frac{1}{2}L(x_{n - 2} ) - \frac{1}{2}x_{n - 2} + \frac{1}{2} \\ & \quad + \sum\limits_{i = 0}^{n - 3} {x_{i + 1} L(x_{i} ) + L(x_{n - 2} ) - \sum\limits_{i = 0}^{n - 2} {x_{i} L(x_{i} )} } - \int\limits_{0}^{1} {L(x)} \\ & = \frac{1}{2}\sum\limits_{i = 0}^{n - 3} {x_{i + 1} L(x_{i} ) - \frac{1}{2}\sum\limits_{i = 0}^{n - 3} {x_{i} L(x_{i + 1} )} } + \frac{1}{2}L(x_{n - 2} ) - \frac{1}{2}x_{n - 2} + \frac{1}{2} - \int\limits_{0}^{1} {L(x)} \\ \end{aligned} $$

By defining $x_{n - 1} \equiv L(x_{n - 1} ) \equiv 1,$ and recalling the definition of $E_{1} ,$ the following more compact formulation obtains:

$$ E_{n} = E_{1} + \frac{1}{2}\sum\limits_{i = 0}^{n - 2} {x_{i + 1} L(x_{i} ) - \frac{1}{2}\sum\limits_{i = 0}^{n - 2} {x_{i} L(x_{i + 1} )} } $$

The first order conditions are as follows:

$$ \left( \begin{array} {l} \frac{{\partial E_{n} }}{{\partial x_{0} }} \hfill \\ \frac{{\partial E_{n} }}{{\partial x_{1} }} \hfill \\ \frac{{\partial E_{n} }}{{\partial x_{2} }} \hfill \\ \vdots \hfill \\ \frac{{\partial E_{n} }}{{\partial x_{n - 2} }} \hfill \\ \end{array} \right) = \frac{1}{2}\left( \begin{array} {l} x_{1} L^\prime (x_{0} ) - L(x_{1} ) \hfill \\ L(x_{0} ) + x_{2} L^\prime (x_{1} ) - x_{0} L^\prime (x_{1} ) - L(x_{2} ) \hfill \\ L(x_{1} ) + x_{3} L^\prime (x_{2} ) - x_{1} L^\prime (x_{2} ) - L(x_{3} ) \hfill \\ \vdots \hfill \\ L(x_{n - 3} ) + L^\prime (x_{n - 2} ) - x_{n - 3} L^\prime (x_{n - 2} ) - 1 \hfill \\ \end{array} \right) = 0 \Leftrightarrow \left( \begin{array} {l} L^\prime (x_{0} ) \hfill \\ L^\prime (x_{1} ) \hfill \\ L^\prime (x_{2} ) \hfill \\ \vdots \hfill \\ L^\prime (x_{n - 2} ) \hfill \\ \end{array} \right) = \left( \begin{array} {l} \frac{{L(x_{1} )}}{{x_{1} }} \hfill \\ \frac{{L(x_{2} ) - L(x_{0} )}}{{x_{2} - x_{0} }} \hfill \\ \frac{{L(x_{3} ) - L(x_{1} )}}{{x_{3} - x_{1} }} \hfill \\ \vdots \hfill \\ \frac{{1 - L(x_{n - 3} )}}{{1 - x_{n - 3} }} \hfill \\ \end{array} \right) $$