A.3. Properties of the Relative Gender-Equality Index E

In this subsection we examine the properties of the relative gender-equality index E, defined as X_ede/, for the isoelastic social valuation function V(X). We have

and

 
Hence

From now on we assume that p_f = p_m = ½, and we define z = (X_f/X_m) as the ratio of female to male achievement. Then

 
It is not difficult to check that, for any Î > 0, the index of relative equality E is maximized at z = 1, and its maximum value is E = 1. In general, we have 0 E(z)  1 for z  0.
By differentiating E(z) with respect to z, and simplifying, we obtain

 
Given Î > 0, it follows that E'(z) > 0 for z < 1, E'(z) = 0 for z = 1, and E'(z) < 0 for z > 1. By further differentiation with respect to z we can show that E"(1) < 0, i.e. z = 1 maximizes E(z).

Manipulating the second derivative of E(z), it can be seen that E"(z) = 0 when
 

Inspecting the left- and right-hand sides of this equation, it is evident that its solution must occur at a value of z > 1.²⁵ To the left of this point of inflection in E(z) the second derivative E"(z) will be negative, and to the right of this point the second derivative E"(z) will be positive. The shape of E(z) for different values of Î is shown in the figure below.

For  = 0, there is no concern for equality and E = 1 for all values of z.

For 0 < < 1, we have E  (1/2)^/(1-) both as z  0 and as z  .

For  = 1, we have X_ede as the geometric mean of X_f and X_m, and

0 as z  0 and as z  .

For  > 1, we can again check from equation (6) that E  0 both as z 0 and as z  .

For  = 2, in particular, we have X_ede as the harmonic mean of X_f and X_m, and

E = 4/[1 + z][1 + (1/z)]

0 as z  0 and as z  .

Irrespective of , the value of E(z) will obviously be the same as z  0 and as z ; the index of relative equality E is clearly symmetric (given p_f = p_m = ½) in z and (1/z). From equation (6) it is easy to verify that E(z) = E(1/z) for all z  0. For any Π 0, E as a function of z must therefore always satisfy the following properties:
 

(1) E(z) = E(1/z) 0 for all z  0.

(2) E(z) is maximized at z = 1, and E(1) = 1.

Thus within this framework we cannot impose arbitrary functional forms for E(z) for z 0, such as E(z) = z or E(z) = 1-, which violate properties (1) and (2).

 
25. For certain parameter values of , this equation can be solved immediately. Thus the solution is z = 2 for  = 2, and z = 2.1547 for both  = 1/2 and  = 1. (For these parameter values, the equation is a quadratic.)