An Axiomatic Approach to the Measurement of Comparative Female Disadvantage

Abstract

Female comparative disadvantage refers to the mismatch of the female with respect to achievements in different dimensions of human well-being in comparison with the corresponding achievements of the male. This paper axiomatically derives a general family of female comparative disadvantage indicators which has very important policy implications. The axioms employed are shown to be ‘independent’. An empirical illustration of the general index is provided using the UNDP data on mean years of schooling, life expectancy at birth and gross national income per capita in 2018. Results show that female comparative disadvantage is not necessarily related to standard measures of human development, such as the HDI, and is present even in countries reaching very high human development. The factor where policy intervention is needed the most is income.

Appendix

Proof of Theorem 1

In view of the strong focus axiom, for any $d \in N$, we can restrict attention on the censored profile $z^{*} = \left( {z_{1}^{ * } ,z_{2}^{ * } ,\ldots z_{d}^{ * } } \right)$. By applying the factor decomposability postulate to the profile $z^{ * }$, we get $F\left( z \right) = \sum\nolimits_{i = 1}^{d} {w_{i} } f\left( {z_{i}^{ * } } \right)$, where $f\left( {z_{i}^{ * } } \right) = F\left( {z_{i}^{ * } } \right)\,,$$1 \le i \le d$,$f:\Re_{ + }^{1} \to \Re_{ + }^{1} .$ Continuity of $F$ ensures that $f$ is continuous over its domain. In view of the normalization axiom, $f\left( 0 \right) = 0$ and $f\left( 1 \right) = 1$.

The proof of strict convexity of $f$ over $\left[ {0,1} \right]$ relies on the following theorem of Jensen (1906): Let $g:I \to \Re^{1}$ be a continuous function, $I$ being a non-degenerate interval in the set of real numbers $\Re^{1}$. Then $g$ is strictly convex if and only if it is midpoint strictly convex, that is, for arbitrary $p \ne q \in I$, $g\left( {\frac{p + q}{2}} \right) < \frac{g\left( p \right) + g\left( q \right)}{2}$. (See Niculescu & Persson, 2006, p.10.)

Now, for $t \in \left[ {0,1} \right]$, choose $c > 0$ such that $u\left( { = t - c} \right),v\left( { = u - c} \right) \in \left[ {0,\;1} \right]$. Aggravation sensitivity implies that

$$f\left( v \right) - f\left( u \right) > f\left( u \right) - f\left( t \right).$$

(12)

Note that $u = \frac{v + t}{2}$. Hence we can rewrite inequality (12) as

$$f\left( {\frac{t + v}{{2}}} \right) < \frac{f\left( t \right) + f\left( v \right)}{{2}}.$$

(13)

Since $t,v \in \left[ {{0,}\;{1}} \right]$ are arbitrary (as $c > 0$ is arbitrary), inequality (13) establishes mid-point strict convexity of $f$. This combined with continuity of $f$ over $\left[ {0,\;1} \right]$ demonstrates that $f$ fulfills the desired strict convexity property. This establishes the necessity part of the theorem. The sufficiency part is easy to verify. $\Delta$.

Proof of Proposition 1

Given $g\left( 0 \right) = 1$ and $g\left( 1 \right) = 0$, by strict convexity of $g$,$g\left( {\left( {1 - \theta } \right)} \right) = g\left( {\theta .0 + \left( {1 - \theta } \right).1} \right) < \theta g\left( 0 \right) + \left( {1 - \theta } \right)g\left( 1 \right) = \theta$, where $0 < \theta < 1$ is arbitrary. Hence for any $0 < u < 1$, $g\left( u \right) < \left( {1 - u} \right)$ which implies that $g\left( u \right)$ is decreasing at 0. Next, strict convexity of $g$ also implies that $g\left( u \right) > 0$ for all $0 < u < 1$. Otherwise, suppose for some $0 < t < 1\,,$ $g\left( t \right) = 0$. Then, given $g\left( 1 \right) = 0$, by strict convexity of $g$, $g\left( {\theta .t + \left( {1 - \theta } \right).1} \right) < \theta g\left( t \right) + \left( {1 - \theta } \right)g\left( 1 \right) = 0\,,$ where $0 < \theta < 1$. This inequality holds only if $g\left( {\theta .t + \left( {1 - \theta } \right).1} \right) < 0$, which is a contradiction to the assumption that $g$ is non-negative valued. Now, if $g$ is not strictly decreasing, then there exist at least three points $0 < t_{1} < t_{2} < t_{3} < 1$ such that $g\left( {t_{1} } \right) > g\left( {t_{2} } \right) > 0$ and $g\left( {t_{2} } \right) < g\left( {t_{3} } \right) > 0$. By construction and by convexity of $\left[ {0,1} \right]$, we can get $0 < \delta < 1$ such that $t_{3} = \delta .t_{2} + \left( {1 - \delta } \right).1$. Then by strict convexity of $g$, $g\left( {t_{2} } \right) < g\left( {t_{3} } \right) =$$g\left( {\delta .t_{2} + \left( {1 - \delta } \right).1} \right) < \delta g\left( {t_{2} } \right) + \left( {1 - \delta } \right)g\left( 1 \right) = \delta g\left( {t_{2} } \right)$, a contradiction. Since $g\left( 1 \right) = 0,$ it then follows that $g$ is strictly decreasing over $\left[ {0,1} \right]$. This completes the proof of the proposition.$\Delta$.

Proof of Theorem 2: Since in (2) the weights $w_{i} \ge 0$ obeying the restriction $\sum\nolimits_{i = 1}^{d} {w_{i} } = 1$, can be chosen arbitrarily for identifying a functional form that violates a particular postulate, in each part of the demonstration of the independence theorem, we can choose $w_{i} = \frac{1}{d}$ for all $i \in Q\,.$

(A) Consider the female disadvantage index given by

$$F_{1} \left( {z\,} \right) = \frac{1}{d}\left[ {\sum\limits_{{z_{i} < 1}} {\left( {1 - z_{i} } \right)}^{2} + \sum\limits_{{z_{i} \ge 1}} {\left( {\frac{{z_{i} - 1}}{{z_{i} + 1}}} \right)} } \right],$$

(14)

where $z \in \Re_{ + }^{d}$ is such that for at least one $i \in \left\{ {1,2,\ldots,d} \right\}$,$\,z_{i} > 1$ holds. Given such an FMA ratio profile $z$ with $\,z_{i} > 1$ choose $\,c > 0$ so that $\left( {\,z_{i} - c} \right) > 1$. Given this $z$, define $y \in \Re_{ + }^{d}$ as follows: $y_{i} = \left( {z_{i} - c} \right) > 1$ and $y_{k} = z_{k}$ for all $k \in \left\{ {1,2,\ldots,d} \right\}/\left\{ k \right\}$. It is easy to verify that $F_{1} \left( {y\,} \right) < F_{1} \left( {z\,} \right)$, which demonstrates violation of the strong focus axiom because of its dependence on FMA ratios that are greater than 1. However, $F_{1}$ fulfills the aggravation sensitivity, factor decomposability, normalization and continuity axioms.

(B) Because of linearity in $z_{i}^{ * }$ values, the Dijkstra–Hanmer index $F_{2} \left( {z\,} \right) = \frac{1}{d}\sum\nolimits_{i = 1}^{d} {\left( {1 - z_{i}^{ * } } \right)} ,$ where $z \in \Re_{ + }^{d}$, is not aggravation sensitive, although it is strongly focused, factor decomposable, normalized and continuous.

(C) For any $z \in \Re_{ + }^{d}$, let $\hat{z}^{*}$ be the censored FMA ratio profile associated with the non-increasingly ordered permutation of $z$, that is,$z_{1} \le z_{2} \le \cdots \le z_{d}$ so that $\left( {1 - \hat{z}_{1}^{ * } } \right) \ge \left( {1 - \hat{z}_{2}^{ * } } \right) \ge \cdots \ge \left( {1 - \hat{z}_{d}^{ * } } \right)$. Then for any $z \in \Re_{ + }^{d}$, the index $F_{3} \left( {z\,} \right) = \frac{1}{{d^{2} }}\sum\nolimits_{i = 1}^{d} {\left( {2i - 1} \right)\left( {1 - \hat{z}_{i}^{ * } } \right)}$ that employs Gini-type aggregation is a transgressor of the factor decomposability postulate since it is a rank ordered weighted average of $\left( {1 - \hat{z}_{i}^{ * } } \right)$ ratio values where the weights themselves are dependent of the entire distribution of ratios. However, $F_{3}$ is an abider of the all the remaining four axioms considered in the theorem statement.

(D) For any $z \in \Re_{ + }^{d}$, the sum $F_{4} \left( {z\,} \right) = \sum\nolimits_{i = 1}^{d} {\left( {1 - z_{i}^{ * } } \right)^{2} }$ of squared deviations of $z_{i}^{ * }$ from unity is a violator of part (ii) of the normalization postulate. (Note that since the normalization principle consists of two separate parts, violation of one part should serve our purpose.) Evidently, $F_{4}$ is aggravation sensitive, strongly focused, factor decomposable and continuous.

(E) Consider the following form of the disadvantaged index

$$F_{5} \left( {z\,} \right) = \left\{ {\begin{array}{*{20}l} {\frac{1}{d}\frac{1}{2}\sum\limits_{i = 1}^{n} {\left( {1 - z_{i}^{ * } } \right)^{2} } } \hfill & {if\,\,z \ne 01^{n} } \hfill \\ 1 \hfill & {if\,\,\,z = 01^{n} .} \hfill \\ \end{array} } \right.$$

(15)

where $z = 01^{n}$ is the n-coordinated vector of zeroes and $z \ne 01^{n}$ means that at least one coordinate of z is non-zero. Evidently, $F_{5}$ is discontinuous at $z = 01^{n}$. However, it is strongly focused, aggravation sensitive, factor decomposable and normalized. This completes the proof of the theorem.$\Delta$.