An axiomatization of the human development index

Abstract

In 2010 the UNDP unveiled a new methodology for the calculation of the Human Development Index (HDI). In this paper I investigate the normative and practical properties of this change vis a vis the original formulation of the HDI in 1990. The main conceptual innovation of the new index can be summarized as follows: the new HDI penalizes both low and uneven achievements across all dimensions of human development, whereas the old formulation is not sensitive to such uneven development. In practice, however, both methodologies agree considerably in terms of how they rank countries, but when they differ, the new methodology produces results more consistent with what the HDI is intended to measure: human development and capabilities, as conceptualized by Sen (Commodities and capabilities. Elsevier, Oxford 1985).

Appendix

Proof of Theorem 1

$\left( \leftarrow \right) \hbox {Let}HDI\left( {h,e,y} \right) =I\left( {C_h \left( h \right) ,C_e \left( e \right) ,C_y \left( y \right) } \right)$ with $I\left( {C_h ,C_e ,C_y } \right) =C_h ^{1/3}\cdot C_e ^{1/3}\cdot C_y ^{1/3}$ and $C_h \left( h \right) =\frac{h-h^{o}}{h^{*}-h^{o}}, C_e \left( e \right) =\frac{e-e^{o}}{e^{*}-e^{o}}, C_y \left( y \right) =\frac{\log y-\log y^{o}}{\log y^{*}-\log y^{o}}$ .

Functions $C_h \left( h \right) , C_e \left( e \right)$ and $C_y \left( y \right)$ are all strictly increasing, and so is $I\left( {C_h ,C_e ,C_y } \right)$ in all its arguments. Therefore, Monotonicity holds. Now let (h,e,y), (h’,e’,y’) be in $\Omega$ with $h,h^{{\prime }}>h^{o};e,e^{{\prime }}>e^{o}and\,y,y^{{\prime }}>y^{o}$ and assume $C(h,e,y)\ge C(h,e^{{\prime }},y^{{\prime }})$. If we multiply each side by $\left[ {C_h \left( h^{{\prime }} \right) /C_h \left( h \right) } \right] ^{\frac{1}{3}}$ we get $C(h^{\prime },e,y)\ge C(h^{\prime },e^{{\prime }},y^{{\prime }})$. Now assume $C\left( {h,e,y} \right) \ge C\left( {h^{{\prime }},e,y^{{\prime }}} \right) .$ If we multiply each side by $\left[ {C_e \left( {e^{\prime }} \right) /C_e \left( e \right) } \right] ^{\frac{1}{3}}$ we get $C\left( {h,e^{{\prime }},y} \right) \ge C\left( {h^{{\prime }},e^{{\prime }},y^{{\prime }}} \right) .$ Now assume $C(h,e,y)\ge C(h^{\prime },e^{\prime },y)$. If we multiply each side by $\left[ {C_y \left( {y^{\prime }} \right) /C_y \left( y \right) } \right] ^{\frac{1}{3}}$ we get $C(h,e,y^{\prime })\ge C(h^{\prime },e^{\prime },y^{\prime })$. Therefore Independence holds. Since $C_h \left( {h^{o}} \right) =C_e \left( {e^{o}} \right) =C_y \left( {y^{o}} \right) =0$ and $I\left( {C_h ,C_e ,C_y } \right)$ is multiplicative, clearly Subsistence holds. Scale holds since $I\left( {c,c,c} \right) =c$. Now fix $(h,e,y)$  and consider feasible values for $\Delta h, \Delta e$ and $d_y$. Then $\Delta C_h \left( {h,\Delta h} \right) =\frac{\Delta h}{h^{*}-h^{o}}, \Delta C_e \left( {e,\Delta e} \right) =\frac{\Delta e}{e^{*}-e^{o}}$ and $\Delta C_y \left( {y,y \cdot d_y } \right) =\frac{\log \left( {1+d_y } \right) }{\log y^{*}-\log y^{o}}$. Therefore Partial Capabilities Growth holds. Aggregation Symmetry is straightforward to verify.

$\left( \rightarrow \right)$ Since $C$ satisfies Monotonicity, Subsistence and Independence it is a consequence of Theorem 1 in Herrero et al. (2010a) that

$$\begin{aligned} C\left( {h,e,y} \right) =f\left( h \right) \cdot g\left( e \right) \cdot m\left( y \right) \!, \end{aligned}$$

where $f\left( h \right) :H\rightarrow \left[ {0,1} \right] ,g\left( e \right) :E\rightarrow \left[ {0,1} \right] \mathrm{and}\,m\left( y \right) :Y\rightarrow \left[ {0,1} \right]$ are increasing functions such that $f\left( {h^{o}} \right) =g\left( {e^{o}} \right) =m\left( {y^{o}} \right) =0$ and $f\left( {h^{*}} \right) =g\left( {e^{*}} \right) =m\left( {y^{*}} \right) =1.$ By Scale and Aggregation Symmetry, $I\left( {C_h ,C_e ,C_y } \right) =C_h ^{1/3}\cdot C_e ^{1/3}\cdot C_y ^{1/3}$ with $C_h \left( h \right) =f\left( h \right) ^{3}, C_e \left( e \right) =g\left( h \right) ^{3}, C_y \left( y \right) =m\left( y \right) ^{3}$. Partial Capabilities Growth implies that (i) $C_h \left( h \right)$ is of the form $C_h \left( h \right) =\alpha _h +\beta _h h$ for some values for $\alpha _h$ and $\beta _h ,$ and $C_h \left( {h^{o}} \right) =0, C_h \left( {h^{*}} \right) =1$ implies that $C_h \left( h \right) =\frac{h-h^{o}}{h^{*}-h^{o}}$. (ii) $C_e \left( e \right)$ is of the form $C_e \left( e \right) =\alpha _e +\beta _e e$ and $C_e \left( {e^{o}} \right) =0, C_e \left( {e^{*}} \right) =1$ implies that $C_e \left( e \right) =\frac{e-e^{o}}{e^{*}-e^{o}}$. (iii) $C_y \left( y \right) =\frac{\log y-\log y^{o}}{\log y^{*}-\log y^{o}}$. To see this last step let $s\left( y \right) =C_y \left( y \right)$ and notice that Partial Capabilities Growth implies that

$$\begin{aligned} s\left( {y\left( {1+d_y } \right) } \right) =s\left( y \right) +s\left( {\left( {1+d_y } \right) y^{o}} \right) \!. \end{aligned}$$

Now let $a=d_y \cdot y$. Then $s\left( {\left( {1+\frac{a}{y}} \right) y} \right) =s\left( y \right) +s\left( {\left( {1+\frac{a}{y}} \right) y^{o}} \right) .$ The steps below are based on Lady (2005):

Step 1: Show $s^{\prime }\left( y \right) =\frac{K}{y}$ for some constant $K$. To see that this is so notice that

$$\begin{aligned} s^{{\prime }}\left( y \right)&= \mathop {\lim }\limits _{a\rightarrow 0} \frac{s\left( {y+a} \right) -s\left( y \right) }{a}=\mathop {\lim }\limits _{a\rightarrow 0} \frac{s\left( {y\left( {1+\frac{a}{y}} \right) } \right) -s\left( y \right) }{a}\\&= \mathop {\lim }\limits _{a\rightarrow 0} \frac{s\left( y \right) +s\left( {\left( {1+\frac{a}{y}} \right) y^{o}} \right) -s\left( y \right) }{a}\\&= \mathop {\lim }\limits _{a\rightarrow 0} \frac{s\left( {\left( {1+\frac{a}{y}} \right) y^{o}} \right) -s\left( {y^{o}} \right) }{a}=\mathop {\lim }\limits _{a\rightarrow 0} \frac{s\left( {y^{o}+\frac{a}{y}y^{o}} \right) -s\left( {y^{o}} \right) }{a}\\&= \mathop {\lim }\limits _{a\rightarrow 0} \frac{s\left( {y^{o}+\frac{y^{o}}{y}a} \right) -s\left( {y^{o}} \right) }{\frac{y}{y^{o}}\frac{y^{o}}{y}a}\\&= \frac{y^{o}}{\hbox {y}}\mathop {\lim }\limits _{a\rightarrow 0} \frac{s\left( {y^{o}+\frac{y^{o}}{y}a} \right) -s\left( {y^{o}} \right) }{\frac{y^{o}}{y}a}=\frac{y^{o}}{\hbox {y}}\mathop {\lim }\limits _{\frac{y^{o}}{y}a\rightarrow 0} \frac{s\left( {y^{o}+\frac{y^{o}}{y}a} \right) -s\left( {y^{o}} \right) }{\frac{y^{o}}{y}a}\\&= \frac{y^{o}}{\hbox {y}}s^{{\prime }}\left( {y^{o}} \right) \end{aligned}$$

with the desired constant $K$ given by $y^{o}s^{{\prime }}\left( {y^{o}} \right)$. Notice that K $\ne$ 0 because otherwise $s^{{\prime }}\left( y \right) =0$  and $s$ would be a constant function, which cannot be since $s\left( {y^{o}} \right) =0$ and $s\left( {y^{*}} \right) =1$. Indeed, the same argument shows that $K>0$ and that therefore $s$ is a strictly increasing function.

Step 2: Show that $s\left( {y^{r} \cdot y^{o}} \right) =rs\left( {y \cdot y^{o}} \right)$. To see this notice that, by the Chain Rule,

$\frac{d}{dy}s\left( {y^{r} \cdot y^{o}} \right) =\frac{K}{y^{r} \cdot y^{o}}y^{o}ry^{r-1}=r\frac{K}{y}=rs^{{\prime }}\left( y \right) ,$ and $s^{\prime }\left( {y \cdot y^{o}} \right) =\frac{K}{y \cdot y^{o}}y^{o}=s^{{\prime }}\left( y \right)$.

The implication is that, since $rs\left( {y \cdot y^{o}} \right)$ and $s\left( {y^{r} \cdot y^{o}} \right)$  have the same derivative, they differ by a constant, that is, $rs\left( {y \cdot y^{o}} \right) =s\left( {y^{r} \cdot y^{o}} \right) +Q$, for some number $Q$. Letting $y=1$ shows $Q =0$, since $s\left( {y^{o}} \right) =0$.

Step 3: Show that $\ell =s(y)$ if and only if $\left( {\frac{y^{*}}{y^{o}}} \right) ^{\ell }=\frac{y}{y^{o}}$, that is, $\ell$ is the logarithm of $\frac{y}{y^{o}}$ with respect to the base $\frac{y^{*}}{y^{o}}.$

To see this notice that, by the conclusion from Step 2 above,$\left( {\frac{y^{*}}{y^{o}}} \right) ^{\ell }=\frac{y}{y^{o}}$ implies $s\left( y \right) =s\left( {\left( {\frac{y^{*}}{y^{o}}} \right) ^{\ell }y^{o}} \right) =\ell \cdot s\left( {\frac{y^{*}}{y^{o}} \cdot y^{o}} \right) =\ell \cdot s\left( {y^{*}} \right) =\ell .$

On the other hand, $\ell =s(y)$ implies, as shown above, that $s\left( y \right) =s\left( {\left( {\frac{y^{*}}{y^{o}}} \right) ^{\ell }y^{o}} \right)$ and since $s$ is strictly increasing it follows that $y=\left( {\frac{y^{*}}{y^{o}}} \right) ^{\ell }y^{o}$.

I have thus shown that $C_y \left( y \right) =s\left( y \right) =\log _{\frac{y^{*}}{y^{o}}} \frac{y}{y^{o}}=\frac{\log y-\log y^{o}}{\log y^{*}-\log y^{o}}$ and the proof is complete.

To separate the properties consider the following indices:

• (1) The $HDI_a$. Satisfies Monotonicity, Independence, Scale, Aggregation Symmetry and Partial Capabilities Growth but not Subsistence.

• (2) $C\left( {h,e,y} \right) =0.$ Satisfies Subsistence, Independence, Scale, Aggregation Symmetry and Partial Capabilities Growth but not Monotonicity.

• (3) $C\left( {h,e,y} \right) =\frac{h-h^{o}}{h^{*}-h^{o}}\cdot \frac{e-e^{o}}{e^{*}-e^{o}}\cdot \frac{\log y-\log y^{o}}{\log y^{*}-\log y^{o}}\quad .$ Satisfies Subsistence, Independence, Monotonicity, Partial Capabilities Growth and Aggregation Symmetry but not Scale.

• (4) $C\left( {h,e,y} \right) =\frac{\log h-\log h^{o}}{\log h^{*}-\log h^{o}}^{1/3}\cdot \frac{\log e-\log e^{o}}{\log e^{*}-\log e^{o}}^{1/3}\cdot \frac{y-y^{o}}{y^{*}-y^{o}}^{1/3}.$ Satisfies Subsistence, Independence, Monotonicity, Aggregation Symmetry, Scale, but not Partial Capabilities Growth.

• (5) $C\left( {h,e,y} \right) =min\left\{ {\frac{h-h^{o}}{h^{*}-h^{o}},\frac{e-e^{o}}{e^{*}-e^{o}},\frac{\log y-\log y^{o}}{\log y^{*}-\log y^{o}}} \right\}$ Satisfies Subsistence, Partial Capabilities Growth, Monotonicity, Aggregation Symmetry, Scale, but not Independence.

• (6) $C\left( {h,e,y} \right) =\left( {\frac{h-h^{o}}{h^{*}-h^{o}}} \right) ^{a}\cdot \left( {\frac{e-e^{o}}{e^{*}-e^{o}}} \right) ^{b}\cdot \left( {\frac{\log y-\log y^{o}}{\log y^{*}-\log y^{o}}} \right) ^{c}$ with $a+b+c=1$ and $a\ne b\ne c.$ Satisfies Subsistence, Independence, Monotonicity, Partial Capabilities Growth, Scale but not Aggregation Symmetry.

$\square$

Proof of Theorem 2

$\left( \leftarrow \right) \hbox {Let}\,\, HDI_a \left( {h,e,y} \right) =I_a \left( {C_h \left( h \right) ,C_e \left( e \right) ,C_y \left( y \right) } \right)$ with $I_a \left( {C_h ,C_e ,C_y } \right) =\frac{C_h +C_e +C_y }{3}$ and $C_h \left( h \right) =\frac{h-h^{o}}{h^{*}-h^{o}}, C_e \left( e \right) =\frac{e-e^{o}}{e^{*}-e^{o}}, C_y \left( y \right) =\frac{\log y-\log y^{o}}{\log y^{*}-\log y^{o}}$ .

Functions $C_h \left( h \right) , C_e \left( e \right)$ and $C_y \left( y \right)$ are all strictly increasing, and so is $I_a \left( {C_h ,C_e ,C_y } \right)$ in all its arguments. Therefore, Monotonicity holds. Now let (h,e,y), (h’,e’,y’) be in $\Omega$ with $h,h^{{\prime }}>h^{o};e,e^{{\prime }}>e^{o}\mathrm{and}\,y,y^{{\prime }}>y^{o}$ and assume $C(h,e,y)\ge C(h,e^{{\prime }},y^{{\prime }})$. If we add $\frac{1}{3}\left( {C_h \left( {h^{\prime }} \right) -C_h \left( h \right) } \right)$ to each side we get $C(h^{\prime },e,y)\ge C(h^{\prime },e^{{\prime }},y^{{\prime }})$ . Scale holds since $I_a \left( {c,c,c} \right) =c$. Now fix $(h,e,y)$and consider feasible values for $\Delta h, \Delta e$ and $d_y$. Then $\Delta C_h \left( {h,\Delta h,e,y} \right) =\frac{1}{3}\frac{\Delta h}{h^{*}-h^{o}}, \Delta C_e \left( {h,e,\Delta e,y} \right) =\frac{1}{3}\frac{\Delta e}{e^{*}-e^{o}}$ and $\Delta C_y \left( {h,e,y,y \cdot d_y } \right) =\frac{1}{3}\frac{\log \left( {1+d_y } \right) }{\log y^{*}-\log y^{o}}$. Thus Partial Capabilities Growth holds. Now fix ($C_h ,C_e ,C_y ) \quad$ and consider feasible values for $\Delta C_h \quad \Delta C_e$ and $\Delta C_y$. Then $\Delta I_h \left( {C_h ,\Delta C_h ,C_e ,C_y } \right) =\frac{1}{3}\Delta C_h , \quad \Delta I_e \left( {C_h ,C_e ,\Delta C_e ,C_y } \right) =\frac{1}{3}\Delta C_e$ and $\Delta I_y \left( {C_h ,C_e ,C_y ,\Delta C_y } \right) =\frac{1}{3}\Delta C_y .$ Therefore, Capabilities Growth Independence holds. Aggregation Symmetry is straightforward to verify.

$\left( \rightarrow \right)$ Since $C$ satisfies Monotonicity and Capabilities Growth Independence, for any pair $\left( {C_e ,C_y } \right) \in \left[ {0,1} \right] ^{2}$ there exists $\beta _h \in \mathbb{R }_{++}$ and $\alpha _h ,\gamma _h \in \mathbb{R }$ such that: $I\left( {C_h ,C_e ,C_y } \right) =\alpha _h +\beta _h \cdot C_h +\gamma _h \cdot I\left( {1,C_e ,C_y } \right)$. Now, again by Monotonicity and Capabilities Growth Independence, for any $C_y \in \left[ {0,1} \right]$ and given $C_h =1$ there exists $\beta _e \in \mathbb{R }_{++}$ and $\alpha _e ,\gamma _e \in \mathbb{R }$ such that $I\left( {1,C_e ,C_y } \right) =\alpha _e +\beta _e \cdot C_e +\gamma _e \cdot I\left( {1,1,C_y } \right)$. Monotonicity and Capabilities Growth Independence, again, imply that given $C_h =C_h =1$ there exists $\beta _y \in \mathbb{R }_{++}$ and $\alpha _y ,\gamma _y \in \mathbb{R }$ such that $I\left( {1,1,C_y } \right) =\alpha _y +\beta _y \cdot C_y +\gamma _y \cdot I\left( {1,1,1} \right)$. Nesting these implications we get

$$\begin{aligned} I\left( {C_h ,C_e ,C_y } \right)&= \left( {\alpha _h +\gamma _h \cdot \alpha _e +\gamma _h \cdot \gamma _e \cdot \left( {\alpha _h +\gamma _y } \right) } \right) \nonumber \\&+\left( {\beta _h \cdot C_h +\gamma _h \cdot \beta _e \cdot C_e +\gamma _h \cdot \gamma _e \cdot \beta _y \cdot C_y } \right) \!. \end{aligned}$$

Letting $C_h =C_e =C_y =0$ we learn that $\left( {\alpha _h +\gamma _h \cdot \alpha _e +\gamma _h \cdot \gamma _e \cdot \left( {\alpha _h +\gamma _y } \right) } \right) =0.$ By Aggregation Symmetry, $\beta _h =\gamma _h \cdot \beta _e =\gamma _h \cdot \gamma _e \cdot \beta _y$and Scale implies that $\beta _h +\gamma _h \cdot \beta _e +\gamma _h \cdot \gamma _e \cdot \beta _y =1.$ The consequence is that $I\left( {C_h ,C_e ,C_y } \right) =\frac{C_h +C_e +C_y }{3}$. From this point on, the proof that the partial capabilities indices $C_h \left( h \right) , C_e \left( e \right)$ and $C_y \left( y \right)$ are of the desired form follows the same steps as in the proof of Theorem 1 above. I omit the details here.

To separate the properties consider the following indices:

• (1) The new $HDI$. Satisfies Monotonicity, Aggregation Symmetry, Scale and Partial Capabilities Growth but not Capabilities Growth Independence.

• (2) $C\left( {h,e,y} \right) =0.$ Satisfies Capabilities Growth Independence, Aggregation Symmetry, Scale and Partial Capabilities Growth but not Monotonicity.

• (3) $C\left( {h,e,y} \right) =\frac{h-h^{o}}{h^{*}-h^{o}}+\frac{e-e^{o}}{e^{*}-e^{o}}+\frac{\log y-\log y^{o}}{\log y^{*}-\log y^{o}}\quad$. Satisfies Capabilities Growth Independence, Monotonicity, Partial Capabilities Growth and Aggregation Symmetry but not Scale.

• (4) $C\left( {h,e,y} \right) =\frac{1}{3}\frac{\log h-\log h^{o}}{\log h^{*}-\log h^{o}}+\frac{1}{3}\frac{\log e-\log e^{o}}{\log e^{*}-\log e^{o}}+\frac{1}{3}\frac{y-y^{o}}{y^{*}-y^{o}}$. Satisfies Capabilities Growth Independence, Monotonicity, Scale and Aggregation Symmetry but not Partial Capabilities Growth.

• (5) $C\left( {h,e,y} \right) =a\frac{h-h^{o}}{h^{*}-h^{o}}+b\frac{e-e^{o}}{e^{*}-e^{o}}+c\frac{\log y-\log y^{o}}{\log y^{*}-\log y^{o}}$ with $a+b+c=1$ and $a\ne b\ne c.$ Satisfies Capabilities Growth Independence, Monotonicity, Partial Capabilities Growth and Scale but not Aggregation Symmetry.

$\square$