Let the wellbeing increase in the
\(\ell\)-th dimension, everything else remaining the same, i.e.
\(GeMS_{i\ell }\) increases. let us calculate
\(\frac{\partial GeMS_i}{\partial GeMS_{i\ell }}\). We have :
$$\begin{aligned} GeMS_i= & {} 1-\left[ \frac{1}{k} \sum _j (1-GeMS_{j,i})^{g(\gamma _i)}\right] ^{\frac{1}{g(\gamma _i)}} \end{aligned}$$
(9)
$$\begin{aligned}= & {} 1- B(\gamma _i)^{\frac{1}{g(\gamma _i)}} \end{aligned}$$
(10)
with
\(g(\gamma _i) > 1\). Hence
$$\begin{aligned} \frac{\partial GeMS_i}{\partial GeMS_{i\ell }} = - \; \frac{\partial B(\gamma _i)^{\frac{1}{g(\gamma _i)}}}{\partial GeMS_{i\ell }} \end{aligned}$$
(11)
Let us derive
\(\frac{\partial B(\gamma _i)^{\frac{1}{g(\gamma _i)}}}{\partial GeMS_{i\ell }}\). We first write:
$$\begin{aligned} ln \, B(\gamma _i)^{\frac{1}{g(\gamma _i)}} = {\frac{1}{g(\gamma _i)}} ln \, B(\gamma _i) \end{aligned}$$
(12)
Deriving both sides, we get
$$\begin{aligned} \frac{1}{B(\gamma _i)^{\frac{1}{g(\gamma _i)}}} \frac{\partial B(\gamma _i)^{\frac{1}{g(\gamma _i)}}}{\partial GeMS_{i\ell }}= & {} \frac{1}{g(\gamma _i)} \frac{\partial ln B(\gamma _i)}{\partial GeMS_{i\ell }} \\&- \frac{1}{g(\gamma _i)^2} \frac{\partial g(\gamma _i)}{\partial GeMS_{i\ell }} ln \, B(\gamma _i) \end{aligned}$$
\(\Rightarrow\)$$\begin{aligned} \frac{\partial B(\gamma _i)^{\frac{1}{g(\gamma _i)}}}{\partial GeMS_{i\ell }}=\, & {} B(\gamma _i)^{\frac{1}{g(\gamma _i)}} \left[ \frac{1}{g(\gamma _i)} \frac{1}{B(\gamma _i)} \frac{\partial B(\gamma _i)}{\partial GeMS_{i\ell }} \right. \nonumber \\&\left. - \frac{1}{g(\gamma _i)^2} \frac{\partial g(\gamma _i)}{\partial GeMS_{i\ell }} ln \, B(\gamma _i) \right] \end{aligned}$$
(13)
Since
\(B(\gamma _i)^{\frac{1}{g(\gamma _i)}}\) is always positive, the sign of
\(\frac{\partial B(\gamma _i)^{\frac{1}{g(\gamma _i)}}}{\partial GeMS_{i\ell }}\) is that of the expression in the square brackets in (
13). Let us now examine the bracketed expression in greater depth. We have:
$$\begin{aligned} \frac{1}{g(\gamma _i)} \frac{1}{B(\gamma _i)} \frac{\partial B(\gamma _i)}{\partial GeMS_{i\ell }} - \frac{1}{g(\gamma _i)^2} \frac{\partial g(\gamma _i)}{\partial GeMS_{i\ell }} ln \, B(\gamma _i) \end{aligned}$$
(14)
Once again, noting that
\(g(\cdot )\) is always positive (
\(>1\)), we can ignore
\(\frac{1}{g}\) in both terms, and we are left with the following expression to sign:
$$\begin{aligned} \frac{1}{B(\gamma _i)} \frac{\partial B(\gamma _i)}{\partial GeMS_{i\ell }} - \frac{1}{g(\gamma _i)} \frac{\partial g(\gamma _i)}{\partial GeMS_{i\ell }} ln \, B(\gamma _i) \end{aligned}$$
(15)
We have
$$\begin{aligned} \frac{\partial B(\gamma _i)}{\partial GeMS_{i\ell }}= & {} \frac{1}{k} \sum _j \frac{\partial }{\partial GeMS_{i\ell }} \left[ (1-GeMS_{j,i})^{g(\gamma _i)}\right] \nonumber \\= & {} \frac{1}{k} \sum _j \left[ (1-GeMS_{j,i})^{g(\gamma _i)} ln(1-GeMS_{j,i}) \frac{\partial g(\gamma _i)}{\partial GeMS_{i\ell }}\right. \nonumber \\&\left. + \delta _{j\ell } \, g(\gamma _i) (1-GeMS_{i\ell })^{-1} \frac{\partial (1-GeMS_{i\ell })}{\partial GeMS_{i\ell }} \right] \end{aligned}$$
(16)
with
$$\begin{aligned} \delta _{j\ell } = \left\{ \begin{array}{lll} 1 \; \text {if} \; j=\ell \\ 0 \; \text {if} \; j\ne \ell \end{array} \right. \end{aligned}$$
Substituting (
16) into (
15), and noting that
\(\frac{\partial (1-GeMS_{i\ell })}{\partial GeMS_{i\ell }}\) is
\(-1\), we get
$$\begin{aligned}&\frac{1}{B} \left[ \frac{1}{k} \sum _j (1-GeMS_{j,i})^{g(\gamma _i)} ln(1-GeMS_{j,i}) \frac{\partial g(\gamma _i)}{\partial GeMS_{i\ell }} \right] \nonumber \\&- \frac{1}{B} \left[ \frac{1}{k} \sum _j \delta _{j\ell } \, g(\gamma _i) (1-GeMS_{i\ell })^{-1} \right] - \frac{1}{g(\gamma _i)} \frac{\partial g(\gamma _i)}{\partial GeMS_{i\ell }} ln \, B(\gamma _i) \end{aligned}$$
(17)
Let us first note that (
17) has three terms and that the second term is always negative
$$\begin{aligned} - \frac{1}{B} \left[ \frac{1}{k} \sum _j \delta _{j\ell } \, g(\gamma _i) (1-GeMS_{i\ell })^{-1} \right] = - \underbrace{\frac{1}{B}}_{>0} \underbrace{\frac{1}{k}}_{>0} \underbrace{g(\gamma _i)}_{>0} \underbrace{(1-GeMS_{i\ell })^{-1}}_{>0} < 0 \end{aligned}$$
Now, let us consider the first and the third terms together:
$$\begin{aligned} \frac{1}{B} \left[ \frac{1}{k} \sum _j (1-GeMS_{j,i})^{g(\gamma _i)} ln(1-GeMS_{j,i}) \frac{\partial g(\gamma _i)}{\partial GeMS_{i\ell }} \right] - \frac{1}{g(\gamma _i)} \frac{\partial g(\gamma _i)}{\partial GeMS_{i\ell }} ln \, B(\gamma _i) \end{aligned}$$
Let us take out
\(\frac{\partial g(\gamma _i)}{\partial GeMS_{i\ell }}\) as a common factor as it does not depend on the index of the sum, and omit it in the comparison. However, noting that it is
\(<0\), we have to remember that whatever sign we get at the end will have to be reversed when we add this factor again. This gives
$$\begin{aligned} \frac{1}{B} \left[ \frac{1}{k} \sum _j (1-GeMS_{j,i})^{g(\gamma _i)} ln(1-GeMS_{j,i}) \right] - \frac{1}{g(\gamma _i)} ln \, B(\gamma _i) \end{aligned}$$
Let us
denote
\(a_j \equiv (1-GeMS_j)^{g(\gamma _i)}\). Then
\(ln \,a_j = g(\gamma _i) ln \, (1-GeMS_j)\) implying that
\(ln \, (1-GeMS_j) = \frac{1}{g(\gamma _i)} \ ln \, a_j\). Thus the expression becomes
$$\begin{aligned} \frac{1}{B} \left[ \frac{1}{k} \sum _j a_j \frac{1}{g(\gamma _i)} \ ln \ a_j \right] - \frac{1}{g(\gamma _i)} \ ln \, B(\gamma _i) \end{aligned}$$
Since we are only determining the sign of the expression, let us multiply both terms by
Bk (which is positive) and noting that
\(B k = \sum _j a_j\) and
\(ln \, B = ln \, \frac{\sum _j a_j}{k}\), we get
$$\begin{aligned} \frac{1}{g(\gamma _i)} \left[ \sum _j a_j \ ln \, a_j \right] - \frac{1}{g(\gamma _i)} \left( \sum _j a_j \right) \left( ln \, \frac{\sum _j a_j}{k} \right) \end{aligned}$$
Ignoring the common factor
\(\frac{1}{g(\gamma _i)}\) which is positive, we get
$$\begin{aligned} \left[ \sum _j a_j \ ln \, a_j \right] - \left( \sum _j a_j \right) \left( ln \, \frac{\sum _j a_j}{k} \right) \end{aligned}$$
Using the log sum inequality i.e.
\(\sum _j a_j ln \, \frac{a_j}{b_j} \ge \left( \sum _j a_j\right) ln \, \frac{\sum _j a_j}{\sum _j b_j}\), and setting
\(a_j \equiv a_j\) defined above,
\(b_j \equiv 1\) such that
\(\sum _j b_j =k\), we get
$$\begin{aligned} \left[ \sum _j a_j ln \, a_j \right] - \left( \sum _j a_j \right) \left( ln \, \frac{\sum _j a_j}{k} \right) \ge 0 \end{aligned}$$
Now, going backwards, we can get
$$\begin{aligned} \frac{1}{B} \left[ \frac{1}{k} \sum _j (1-GeMS_{j,i})^{g(\gamma _i)} ln(1-GeMS_{j,i}) \right] - \frac{1}{g(\gamma _i)} \frac{\partial g(\gamma _i)}{\partial GeMS_{i\ell }} ln \, B(\gamma _i) \ge 0 \end{aligned}$$
and hence the first and third terms combined of (
17) is negative:
$$\begin{aligned} \frac{1}{B} \left[ \frac{1}{k} \sum _j (1-GeMS_{j,i})^{g(\gamma _i)} ln(1-GeMS_{j,i}) \frac{\partial g(\gamma _i)}{\partial GeMS_{i\ell }} \right] - \frac{1}{g(\gamma _i)} \frac{\partial g(\gamma _i)}{\partial GeMS_{i\ell }} ln \, B(\gamma _i) \le 0 \end{aligned}$$
Now adding the second term of (
17) which was already shown to be also negative, we get
$$\begin{aligned}&\frac{1}{B} \left[ \frac{1}{k} \sum _j (1-GeMS_{j,i})^{g(\gamma _i)} ln(1-GeMS_{j,i}) \frac{\partial g(\gamma _i)}{\partial GeMS_{i\ell }} \right] \\ -&\frac{1}{B} \left[ \frac{1}{k} \sum _j \delta _{j\ell } \, g(\gamma _i) (1-GeMS_{i\ell })^{-1} \right] - \frac{1}{g(\gamma _i)} \frac{\partial g(\gamma _i)}{\partial GeMS_{i\ell }} ln \, B(\gamma _i) < 0 \end{aligned}$$
Thus we have shown that (
17)
\(< 0\). In other words, (
15)
\(< 0\) as it is the same expression, and therefore (
14)
\(< 0\) as (
14) = (
15)
\(\frac{1}{g}\). This in turn gives (
13)
\(< 0\) as (
13) = (
14)
\(B(\gamma _i)^{\frac{1}{g(\gamma _i)}}\). Finally, we have
\(\frac{\partial GeMS_i}{\partial GeMS_{i\ell }} > 0\) as it is -(
13). Thus we see that
\(GeMS_i\) is monotonic increasing in
\(GeMS_{i\ell }\).
An important remark: Note that the above derivation assumes differentiability of
\(g(\cdot )\) with respect to each element of
\(\gamma _i\). But we know that there is one point at which
\(g(\cdot )\) is not differentiable i.e. at
\(\mu _{b,i} \equiv \iota ' \gamma _i = z_b\). So at this point, instead of calculating the change in
\(GeMs_i\) as a differential, we should calculate it as a discrete change. That is, go through
\(\Delta GeMS_i\). If all derivatives exist, then we have
$$\begin{aligned} \Delta GeMS_i = \frac{\partial GeMS_i}{\partial GeMS_{i\ell }} \Delta GeMS_{i \ell } \end{aligned}$$
Now, for all terms in
\(\frac{\partial GeMS_i}{\partial GeMS_{i\ell }}\) that do not contain the derivative
\(\frac{\partial g(\gamma _i)}{\partial GeMS_{i\ell }}\), the derivation remains the same with the additional multiplication by
\(\Delta GeMS_{i \ell }\). For the terms containing
\(\frac{\partial g(\gamma _i)}{\partial GeMS_{i\ell }}\), one should replace,
\(\frac{\partial g(\gamma _i)}{\partial GeMS_{i\ell }} \Delta GeMS_{i \ell }\) by
\(\Delta g(z_b)\). At
\(z_b\), when
\(GeMS_{i\ell }\) becomes
\(GeMS_{i\ell } + \Delta GeMS_{i \ell }\),
\(z_b\) becomes
\(z_b + \Delta GeMS_{i \ell }\).