Ranking distributions of an ordinal variable

Abstract

We establish an equivalence between three criteria for comparing distributions of an ordinal variable taking finitely many values. The first criterion is the possibility of going from one distribution to the other by a finite sequence of increments and/or Hammond transfers. The latter transfers are like the Pigou–Dalton ones, but without the requirement that the amount transferred be fixed. The second criterion is the unanimity of all comparisons of the distributions performed by a class of additively separable social evaluation functions. The third criterion is a new statistical test based on a weighted recursion of the cumulative distribution. We also identify an exact test for the possibility of going from one distribution to another by a finite sequence of Hammond transfers only. An illustration of the usefulness of our approach for evaluating distributions of self-reported happiness level is also provided.

Appendix: Proofs

1.1 Proposition 1

Let s and $s^{\prime }$ be two societies such that $F(i;s)\le F(i;s^{\prime })$ holds for all $i\in \{1,\dots ,k-1\}$. It follows that $F(1;s)\le F(1;s^{\prime })$ and:

$$\begin{aligned} \sum _{h=1}^{i-1}\left( 2^{i-h-1}\right) F(h;s)+F(i;s)\le \,\sum _{h=1}^{i-1}\left( 2^{i-h-1}\right) F(h;s^{\prime })+F(i;s^{\prime }),\forall i\in \{2,\dots ,k\} \end{aligned}$$

so that, thanks to Expressions (10) and (11), $H(i;s)\le H(i;s^{\prime })$ holds for all $i\in \{1,\dots ,k-1\}$. To establish the ${\bar{H}}$-dominance of s by $s^{\prime }$, it suffices to notice that the requirement $F(i;s)\le F(i;s^{\prime })$ for all $i\in \{1,\dots ,k-1\}$ can alternatively be written as ${\bar{F}}(i;s)\ge {\bar{F}} (i,s^{\prime })$ for all $i\in \{1,\dots ,k-1\}$. That implies ${\bar{F}} (k-1;s)\ge {\bar{F}}(k-1;s^{\prime })$ and:

$$\begin{aligned}&\sum _{h=i+1}^{k-1}\left( 2^{h-i-1}\right) {\bar{F}}(h;s)+{\bar{F}}(i;s)\\&\quad \ge \sum _{h=i+1}^{k-1}\left( 2^{h-i-1}\right) {\bar{F}}(h;s^{\prime })+{\bar{F}} (i;s^{\prime }),\ \forall i\in \{1,\dots ,k-2\} \end{aligned}$$

so that, thanks to Expressions (15) and (16), ${\bar{H}} (i;s^{\prime })\le H(i;s)$ for all $i\in \{1,\dots ,k-1\}$.

1.2 Propositions 2 and 3

For Proposition 2, let s be a society obtained from $s^{\prime }$ by an increment. By Definition 1, there exists some $j\in \{1,\dots ,k-1\}$ such that we have $n_{j}^{s}=n_{j}^{s^{\prime }}-1$ and $n_{j+1}^{s}=n_{j+1}^{s^{\prime }}+1$ and that, for all $h\in \{1,\dots ,k\}$ such that $h\ne j,j+1$, we have $n_{h}^{s}=n_{h}^{s^{ \prime }}$. It follows that:

$$\begin{aligned} \sum _{h=1}^k n^s_h\,\alpha _h = \sum _{h=1}^k n^{s^\prime }_h\,\alpha _h + \left( \alpha _{j+1} - \alpha _{j} \right) . \end{aligned}$$

Thus, inequality $\sum _{h=1}^k n^s_h\,\alpha _h \ge \sum _{h=1}^k n^{s^\prime }_h\,\alpha _h$ holds if and only if $\left( \alpha _{j+1} - \alpha _{j} \right) \ge 0$. As this inequality must hold for any $j\in \{1,\dots ,k-1\}$ , this completes the proof of Proposition 2. The argument for Proposition 3 is similar (with Definition 1 replaced by Definition 2).

1.3 Lemma 1

Observe first that:

$$\begin{aligned} \sum _{h=1}^{k}n_{h}^{s}\,\alpha _{h}=\left\{ \begin{array}{ll} &{} n_{1}^{s}\,\alpha _{1} \\ + &{} n_{2}^{s}\,\alpha _{2} \\ + &{} \cdots \\ + &{} n_{k}^{s}\,\alpha _{k}, \end{array} \right. \end{aligned}$$

(27)

or equivalently:

$$\begin{aligned} \sum _{h=1}^{k}n_{h}^{s}\,\alpha _{h}=\left\{ \begin{array}{ll} &{} n_{1}^{s}\,\alpha _{1} \\ + &{} n_{2}^{s}\,\alpha _{1}+n_{2}^{s}\,\left[ \alpha _{2}-\alpha _{1}\right] \\ + &{} n_{3}^{s}\,\alpha _{1}+n_{3}^{s}\,\left[ \alpha _{2}-\alpha _{1}\right] +n_{3}^{s}\,\left[ \alpha _{3}-\alpha _{2}\right] \\ + &{} \cdots \\ + &{} n_{k}^{s}\,\alpha _{1}+n_{k}^{s}\,\left[ \alpha _{2}-\alpha _{1}\right] +n_{k}^{s}\,\left[ \alpha _{3}-\alpha _{2}\right] +\cdots n_{k}^{s}\,\left[ \alpha _{k}-\alpha _{k-1}\right] , \end{array} \right. \end{aligned}$$

(28)

hence:

$$\begin{aligned} \sum _{h=1}^{k}n_{h}^{s}\,\alpha _{h}=\left\{ \begin{array}{ll} &{} n\,\alpha _{1} \\ + &{} (n-n_{1}^{s})\left[ \alpha _{2}-\alpha _{1}\right] \\ + &{} \left[ n-(n_{1}^{s}+n_{2}^{s})\right] \left[ \alpha _{3}-\alpha _{2} \right] \\ + &{} \cdots \\ + &{} \left[ n-\sum _{h=1}^{k-1}n_{h}^{s}\right] \left[ \alpha _{k}-\alpha _{k-1}\right] , \end{array} \right. \end{aligned}$$

(29)

from which one obtains:

$$\begin{aligned} \frac{1}{n}\sum _{h=1}^{k}n_{h}^{s}\,\alpha _{h}= & {} [\alpha _{1}+(\alpha _{k}-\alpha _{1})]-\sum \limits _{h=1}^{k-1}F(h;s)[\alpha _{h+1}-\alpha _{h}] \end{aligned}$$

(30)

$$\begin{aligned}= & {} \alpha _{k}-\sum \limits _{h=1}^{k-1}F(h;s)[\alpha _{h+1}-\alpha _{h}], \end{aligned}$$

(31)

as required by Eq. (19). Now, by reconsidering Eq. (29) and recalling that ${{\bar{F}}}(i;s)=1-F(i;s)=\left( n-\sum _{h=1}^{i}n_{h}^{s}\right) /n$ for every $i\in {\mathcal {C}}$, one immediately obtains Eq. (20). We must now establish Eq. (21). For this sake, one can notice that, for any $t\in \{2,\dots ,k-1\}$, one has:

$$\begin{aligned} \sum _{h=1}^{k}n_{h}^{s}\,\alpha _{h}=\sum _{h=1}^{t}n_{h}^{s}\,\alpha _{h}+\sum _{h=t+1}^{k}n_{h}^{s}\,\alpha _{h}. \end{aligned}$$

(32)

If one successively decomposes the two terms on the right hand of (32), one obtains for the first one:

$$\begin{aligned} \sum _{h=1}^{t}n_{h}^{s}\,\alpha _{h}=\left\{ \begin{array}{ll} &{} n_{1}^{s}\,\alpha _{1} \\ + &{} n_{2}^{s}\,\alpha _{1}+n_{2}^{s}\,\left[ \alpha _{2}-\alpha _{1}\right] \\ + &{} n_{3}^{s}\,\alpha _{1}+n_{3}^{s}\,\left[ \alpha _{2}-\alpha _{1}\right] +n_{3}^{s}\,\left[ \alpha _{3}-\alpha _{2}\right] \\ + &{} \cdots \\ + &{} n_{t}^{s}\,\alpha _{1}+n_{t}^{s}\left[ \alpha _{2}-\alpha _{1}\right] +n_{i}^{s}\left[ \alpha _{3}-\alpha _{2}\right] +\cdots n_{t}^{s}\left[ \alpha _{t}-\alpha _{t-1}\right] , \end{array} \right. \end{aligned}$$

One has therefore:

$$\begin{aligned} \sum _{h=1}^{t}n_{h}^{s}\,\alpha _{h}=\left\{ \begin{array}{ll} &{} \left( \sum \nolimits _{h=1}^{t}n_{h}^{s}\right) \alpha _{1} \\ + &{} \left[ \sum \nolimits _{h=1}^{t}n_{h}^{s}-n_{1}^{s}\right] \left[ \alpha _{2}-\alpha _{1}\right] \\ + &{} \left[ \sum \nolimits _{h=1}^{t}n_{h}^{s}-(n_{1}^{s}+n_{2}^{s})\right] \left[ \alpha _{3}-\alpha _{2}\right] \\ + &{} \cdots \\ + &{} \left[ \sum \nolimits _{h=1}^{t}n_{h}^{s}-\sum \nolimits _{h=1}^{t-1}n_{h}^{s}\right] \left[ \alpha _{t}-\alpha _{t-1}\right] , \end{array} \right. \end{aligned}$$

or equivalently:

$$\begin{aligned} \frac{1}{n}\sum _{h=1}^{t}n_{h}^{s}\,\alpha _{h}=\left( \frac{1}{n} \sum _{h=1}^{t}n_{h}^{s}\right) \alpha _{t}-\sum _{h=1}^{t-1}F(h;s)\left[ \alpha _{h+1}-\alpha _{h}\right] . \end{aligned}$$

(33)

For the second term of (32), the successive decomposition yields:

$$\begin{aligned} \sum _{h=t+1}^{k}n_{h}^{s}\,\alpha _{h}=\left\{ \begin{array}{ll} &{} n_{t+1}^{s}\,\alpha _{t+1} \\ + &{} n_{t+2}^{s}\,\alpha _{t+1}+n_{t+2}^{s}\,\left[ \alpha _{t+2}-\alpha _{t+1}\right] \\ + &{} n_{t+3}^{s}\,\alpha _{t+1}+n_{t+3}^{s}\,\left[ \alpha _{t+2}-\alpha _{t+1}\right] +n_{t+3}^{s}\,\left[ \alpha _{t+3}-\alpha _{t+2}\right] \\ + &{} \cdots \\ + &{} n_{k}^{s}\,\alpha _{t+1}+n_{k}^{s}\,\left[ \alpha _{t+2}-\alpha _{t+1} \right] +n_{k}^{s}\,\left[ \alpha _{t+3}-\alpha _{t+2}\right] +\cdots n_{k}^{s}\left[ \alpha _{k}-\alpha _{k-1}\right] , \end{array} \right. \end{aligned}$$

This can be written as:

$$\begin{aligned} \sum _{h=t+1}^{k}n_{h}^{s}\,\alpha _{h}=\left\{ \begin{array}{ll} &{} \left( \sum \nolimits _{h=t+1}^{k}n_{h}^{s}\right) \alpha _{t+1} \\ + &{} \left( \sum \nolimits _{h=t+2}^{k}n_{h}^{s}\right) \left[ \alpha _{t+2}-\alpha _{t+1}\right] \\ + &{} \left( \sum \nolimits _{h=t+3}^{k}n_{h}^{s}\right) \left[ \alpha _{t+3}-\alpha _{t+2}\right] \\ + &{} \cdots \\ + &{} n_{k}^{s}\left[ \alpha _{k}-\alpha _{k-1}\right] , \end{array} \right. \end{aligned}$$

or equivalently:

$$\begin{aligned} \frac{1}{n}\sum _{h=t+1}^{k}n_{h}^{s}\,\alpha _{h}=\left( \frac{1}{n} \sum _{h=t+1}^{k}n_{h}^{s}\right) \alpha _{t+1}+\sum _{h=t+1}^{k-1}{{\bar{F}}} (h;s)\left[ \alpha _{h+1}-\alpha _{h}\right] . \end{aligned}$$

(34)

By summing Eqs. (33) and (34), one concludes that:

$$\begin{aligned} \frac{1}{n}\sum _{h={1}}^{k}n_{h}^{s}\,\alpha _{h}= & {} \left( \frac{1}{n} \sum _{h=1}^{t}n_{h}^{s}\right) \alpha _{t}+\left( \frac{1}{n} \sum _{h=t+1}^{k}n_{h}^{s}\right) \alpha _{t+1} \nonumber \\&-\sum _{h=1}^{t-1}F(h;s)\left[ \alpha _{h+1}-\alpha _{h}\right] +\sum _{h=t+1}^{k-1}{{\bar{F}}}(h;s)\left[ \alpha _{h+1}-\alpha _{h}\right] . \qquad \quad \end{aligned}$$

(35)

This equality can be further simplified, by observing that:

$$\begin{aligned} \left( \frac{1}{n}\sum _{h=1}^{t}n_{h}^{s}\right) \alpha _{t}+\left( \frac{1}{ n}\sum _{h=t+1}^{k}n_{h}^{s}\right) \alpha _{t+1}= & {} \frac{1}{n}\left( n-\sum _{h=t+1}^{k}n_{h}^{s}\right) \alpha _{t}+\left( \frac{1}{n} \sum _{h=t+1}^{k}n_{h}^{s}\right) \alpha _{t+1} \nonumber \\= & {} \alpha _{t}+{{\bar{F}}}(t;s)[\alpha _{t+1}-\alpha _{t}]. \end{aligned}$$

(36)

Equation (21) is then obtained from the reintroduction of (36) into (35).

1.4 Theorem 1

The equivalence between Statements (a) and (c) of this theorem is well known in the literature. We therefore only prove the equivalence between Statements (b) and (c). Using Eq. (19) of Lemma 1, one has:

$$\begin{aligned} \frac{1}{n}\left[ \sum _{h=1}^{k}n^{s}_{h}\,\alpha _{h}-\sum _{h=1}^{k}n_{h}^{s^\prime }\,\alpha _{h}\right] =\sum _{h=1}^{k-1} \left[ F(h;s^{\prime })-F(h;s)\right] \left[ \alpha _{h+1}-\alpha _{h}\right] . \quad \end{aligned}$$

(37)

Hence, if $F(h;s^{\prime })-F(h;s)\ge 0$ for every $h\in {\mathcal {C}}$ and $(\alpha _{1},\dots ,\alpha _{k})\in {\mathcal {A}}_{F}$, then $\sum _{h=1}^{k}n^{s}_{h}\,\alpha _{h}\ge \sum _{h=1}^{k}n^{s^\prime }_{h}\,\alpha _{h}$. To establish the converse implication, define, for every $i\in \{1,\dots ,k-1\}$, the list of k numbers $\alpha ^{i}=(\alpha _{1}^{i},\dots ,\alpha _{k}^{i})$ to be such that $\alpha _{h}^{i}=0$ for $h=1,\dots ,i$ and $\alpha _{h}^{i}=1$ for $h \in \{i+1,\dots ,k\}$. We note that $\alpha ^{i}\in {\mathcal {A}}_{F}$ for any $i\in \{1,\dots ,k-1\}$. If Inequality (6) holds for all lists of numbers $(\alpha _{1},\dots ,\alpha _{k})\in {\mathcal {A}}_{F}$, then one must have $\sum _{h=1}^{k}n^{s}_{h}\,\alpha _{h}^{i} \ge \sum _{h=1}^{k}n^{s^\prime }_{h}\,\alpha _{h}^{i}$ for any $i \in \{1,\dots ,k-1\}$ or, equivalently:

$$\begin{aligned} \left( n-\sum _{h=1}^{i}n^{s}_{h} \right) = \sum _{h=i+1}^{k}n^{s}_{h} \ge \sum _{h=i+1}^{k}n^{s^\prime }_{h} = \left( n-\sum _{h=1}^{i}n^{s^\prime }_{h} \right) , \end{aligned}$$

(38)

which, in turn, is equivalent to $\sum _{h=1}^{i}n^{s}_{h}\le \sum _{h=1}^{i}n^{s^\prime }_{h}$, as required.

1.5 Proposition 4

Suppose that society s has been obtained from society $s^{\prime }$ by means of a Hammond transfer as per Definition 3. This means that there are categories $1\le g<i\le j<l\le k$ for which one has:

$$\begin{aligned} \sum _{h=1}^{k}n^{s}_{h}\,\alpha _{h}=\sum _{h=1}^{k}\alpha _{h}\,n_{h}^{s^\prime }-\alpha _{g}+\alpha _{i}+\alpha _{j}-\alpha _{l}. \end{aligned}$$

(39)

Hence, if Inequality (6) holds for s and $s^{\prime }$, one have $\sum _{h=1}^{k}n_{h}^{s}\,\alpha _{h}-\sum _{h=1}^{k}n_{h}^{s^\prime }\,\alpha _{h}=(\alpha _{i}-\alpha _{g})-(\alpha _{l}-\alpha _{j})\ge 0$ for all categories $1\le g<i\le j<l\le k$, which is precisely the definition of the set ${\mathcal {H}}$.

1.6 Proposition 5

Assume that the list of numbers $(\alpha _{1},\dots ,\alpha _{k})$ belongs to ${{\mathcal {H}}}$ and, therefore, satisfies $\alpha _{i}-\alpha _{g}\ge \alpha _{l}-\alpha _{j}$ for all $1\le g<i\le j<l\le k$. This implies in particular that $\alpha _{i+1}-\alpha _{i}\ge \alpha _{k}-\alpha _{i+1}$ for any $i\in \{1,2,\dots ,k-2\}$. Let $t=\min \{i=1,\dots ,k\,|\,\alpha _{i+1}-\alpha _{i}\le 0\}$ (using the convention that $\alpha _{k+1}=\alpha _{k}$). Such a t clearly exists under this convention, because $k\in$ $\{i=1,\dots ,k\,|\,\alpha _{i+1}-\alpha _{i}\le 0\}$. If $t=k$, then the fact that $\alpha _{i+1}-\alpha _{i}\ge \alpha _{k}-\alpha _{i+1}$ holds for any $i\in \{1,2,\dots ,k-2\}$ implies that $\alpha _{i+1}-\alpha _{i}\ge \alpha _{t}-\alpha _{i+1}$ for all $i\in \{1,2,\dots ,t-1\}$ and (trivially) that $\alpha _{i^{\prime }+1}-\alpha _{i^{\prime }}\le \alpha _{i^{\prime }}-\alpha _{t}$ holds for all $i^{\prime }\in \{t,\dots ,k-1\}=\varnothing$. Notice that if $t=k$, then one has $\alpha _{i+1}-\alpha _{i}\ge \alpha _{k}-\alpha _{i+1}$ $>0$ for any $i\in \{1,2,\dots ,k-2\}$. (The alphas are increasing with respect to the categories.) If $t=1$, then the set $\{i=1,2,\dots ,t-1\}$ is empty so that one must simply verify that $\alpha _{i^{\prime }+1}-\alpha _{i^{\prime }}\le \alpha _{i^{\prime }}-\alpha _{1}$, for $i^{\prime }\in \{1,\dots ,k-1\}$. But this results immediately from the definition of t (if $i^{\prime }=1)$ or from applying the requirement that $\alpha _{i}-\alpha _{g}\ge \alpha _{l}-\alpha _{j}$ for all $1\le g<i\le j<l\le k$ to the particular case where $g=1$, $i=j=i^{\prime }$ $>1$ and $l=i^{\prime }+1$ (otherwise). Notice that if $t=1$, then one has by definition that $0\ge \alpha _{2}-\alpha _{1}\ge \alpha _{j}-\alpha _{j-1}$ for every $j\in \{3,\dots ,k\}$ so that the alphas are decreasing with the categories. Assume now that $t\in \{2,\dots ,k-1\}$. We must check first that $\alpha _{i+1}-\alpha _{i}\ge \alpha _{t}-\alpha _{i+1}$ for all $i=1,2,\dots ,t-1$ . The case where $i=t-1$ is proved by observing that, by definition of t, one has $\alpha _{t}-\alpha _{t-1}>0=\alpha _{t}-\alpha _{t}$. The case where $i<t-1$ (if any) is proved by applying the statement $\alpha _{i}-\alpha _{g}\ge \alpha _{l}-\alpha _{j}$ for all $1\le g<i\le j<l\le k$ to the particular case where $g=i\in \{1,\dots ,t-2\}$ $i=j=i+1$ and $l=t$. To check that the inequality $\alpha _{i^{\prime }+1}-\alpha _{i^{\prime }}\le \alpha _{i^{\prime }}-\alpha _{t}$ holds for all $i^{\prime }\in \{t^{\prime },\dots ,k-1\}$, simply observe that, for $i^{\prime }=t$, the inequality is obtained from the very definition of t and, for $i^{\prime }>t$, it results from applying the fact that $\alpha _{i}-\alpha _{g}\ge \alpha _{l}-\alpha _{j}$ for all $1\le g<i\le j<l\le k$ to the particular case where $g=t$, $i=j=i^{\prime }$ and $l=i^{\prime }+1.$

Conversely, consider any list $(\alpha _{1},\dots ,\alpha _{k})$ for which there exists a $t\in \{1,\dots ,k\}$ such that:

$$\begin{aligned} \alpha _{i+1}-\alpha _{i}\ge \alpha _{t}-\alpha _{i+1} \end{aligned}$$

(40)

holds for all $i\in \{1,2,\dots ,t-1\}$ (if any) and:

$$\begin{aligned} \alpha _{i^{\prime }+1}-\alpha _{i^{\prime }}\le \alpha _{i^{\prime }}-\alpha _{t} \end{aligned}$$

(41)

holds for all $i^{\prime }\in \{t,\dots ,k-1\}$ (if any). Notice that applying Inequality (40) to $i=t-1$ implies that $\alpha _{t}-\alpha _{t-1}\ge \alpha _{t}-\alpha _{t}=0$. Combining this recursively with Inequality (40) implies in turn that $\alpha _{2}-\alpha _{1}\ge \alpha _{3}-\alpha _{2}\ge \cdots \ge \alpha _{t}-\alpha _{t-1}\ge 0$ so that the list of numbers $(\alpha _{1},\dots ,\alpha _{k})$ is increasing from 1 up to t. Similarly, applying Inequality (41) to $i^{\prime }=t$ implies that $\alpha _{t+1}-\alpha _{t}\le \alpha _{t}-\alpha _{t}=0$. Combining this recursively with Inequality (41) satisfied for all $i^{\prime }\in \{t,\dots ,k-1\}$ (if any) leads to the conclusion that $\alpha _{k}-\alpha _{k-1}\le \alpha _{k-1}-\alpha _{k-2}\le \cdots \le \alpha _{t+1}-\alpha _{t}\le 0$ so that the list of numbers $(\alpha _{1},\dots ,\alpha _{k})$ is decreasing from t up to k. Consider then any four integers g, i, j and l satisfying $1\le g<i\le j<l\le k$. Five cases need to be distinguished:

(i) $g\ge t\ge 1$, then one has:

$$\begin{aligned} \alpha _{l}-\alpha _{j}= & {} (\alpha _{l}-\alpha _{l-1})+(\alpha _{l-1}-\alpha _{l-2})+\cdots +(\alpha _{j+1}-\alpha _{j}) \\\le & {} \alpha _{j+1}-\alpha _{j}\text { (because the }\alpha _{h}\text { are decreasing above }t\text {)} \\\le & {} \alpha _{j}-\alpha _{t}\text { (by Inequality }(41)) \\= & {} \alpha _{j}-\alpha _{i}+\alpha _{i}-\alpha _{g}+\alpha _{g}-\alpha _{t} \text { (for any integer }g,\text { }i,\text { }j\text {)} \\\le & {} \alpha _{i}-\alpha _{g}\text { (because the }\alpha _{h}\text { are decreasing above }t\text {)}. \end{aligned}$$

(ii) $g<t\le i\le j<l\le k$. Then one has:

$$\begin{aligned} \alpha _{l}-\alpha _{j}= & {} (\alpha _{l}-\alpha _{l-1})+(\alpha _{l-1}-\alpha _{l-2})+\cdots +(\alpha _{j+1}-\alpha _{j}) \\\le & {} \alpha _{j+1}-\alpha _{j}\text { (because the }\alpha _{h}\text { are decreasing above }t\text {)} \\\le & {} \alpha _{j}-\alpha _{t}\text { (by Inequality }(41)) \\= & {} \alpha _{j}-\alpha _{i}+\alpha _{i}-\alpha _{g}+\alpha _{g}-\alpha _{t} \text {(for any integer }g,\text { }i,\text { }j\text {)} \\\le & {} \alpha _{i}-\alpha _{g}\text { (because }\alpha _{j}-\alpha _{i}\le 0 \text { and }\alpha _{g}-\alpha _{t}\le 0\text {)}. \end{aligned}$$

(iii) $g<i<t\le j<l\le k$. Then one has:

$$\begin{aligned} \alpha _{l}-\alpha _{j}= & {} (\alpha _{l}-\alpha _{l-1})+(\alpha _{l-1}-\alpha _{l-2})+\cdots +(\alpha _{j+1}-\alpha _{j}) \\\le & {} \alpha _{j+1}-\alpha _{j}\text { (because the }\alpha _{h}\text { are decreasing above }t\text {)} \\\le & {} \alpha _{j}-\alpha _{t}\text { (by Inequality }(41)) \\\le & {} 0\text { (because the }\alpha _{h}\text { are decreasing above }t\text {) } \\\le & {} \alpha _{i}-\alpha _{g}\text { (because the }\alpha _{h}\text { are increasing below }t\text {)}. \end{aligned}$$

(iv) $g<i\le j<t\le l\le k$. Then one has:

$$\begin{aligned} \alpha _{i}-\alpha _{g}= & {} (\alpha _{i}-\alpha _{i-1})+(\alpha _{i-1}-\alpha _{i-2})+\cdots +(\alpha _{g+2}-\alpha _{g+1})+(\alpha _{g+1}-\alpha _{g}) \\\ge & {} \alpha _{g+1}-\alpha _{g}\text { (because the }\alpha _{h}\text { are increasing below }t\text {)} \\\ge & {} \alpha _{t}-\alpha _{g+1}\text { (by Inequality }(40)) \\= & {} \alpha _{t}-\alpha _{l}+\alpha _{l}-\alpha _{j}+\alpha _{j}-\alpha _{g+1} \text { (for any }g+1\le j<l\le k\text {)} \\\ge & {} \alpha _{l}-\alpha _{j}\text { (because }\alpha _{t}-\alpha _{l}\ge 0 \text { and }\alpha _{j}-\alpha _{g+1}\ge 0\text {)}. \end{aligned}$$

(v) $l<t\le k$ In this case, one has:

$$\begin{aligned} \alpha _{i}-\alpha _{g}= & {} (\alpha _{i}-\alpha _{i-1})+(\alpha _{i-1}-\alpha _{i-2})+\cdots +(\alpha _{g+2}-\alpha _{g+1})+(\alpha _{g+1}-\alpha _{g}) \\\ge & {} \alpha _{g+1}-\alpha _{g}\text { (because the }\alpha _{h}\text { are increasing below }t\text {)} \\\ge & {} \alpha _{t}-\alpha _{g+1}\text { (by Inequality }(40)) \\= & {} \alpha _{t}-\alpha _{l}+\alpha _{l}-\alpha _{j}+\alpha _{j}-\alpha _{g+1} \text { (for any }g+1\le j<l\le k\text {)} \\\ge & {} \alpha _{l}-\alpha _{j}\text { (because the }\alpha _{h}\text { are increasing below }t\text {)}. \end{aligned}$$

Hence, any list of k numbers $(\alpha _{1},\dots ,\alpha _{k})$ or which there exists a $t\in \{1,\dots ,k\}$ such that Inequalities (41) and (40) belongs to ${\mathcal {H}}$.

1.7 Theorem 3

1.7.1 Statement (a) implies Statement (b)

Suppose s has been obtained from $s^{\prime }$ by means of an increment. It then follows from Proposition 2 that Inequality (6) holds for all ordered lists of k real numbers $(\alpha _{1},\dots ,\alpha _{k})$ in the set ${\mathcal {A}}_{F}$. This inequality holds therefore in particular for all such lists that belong to ${\mathcal {A}}_{H}\subset {\mathcal {A}}_{F}$. If, on the other hand, s has been obtained from $s^{\prime }$ by means of a Hammond transfer, we know from Proposition 4 that Inequality (6) holds for all ordered lists of k real numbers $(\alpha _{1},\dots ,\alpha _{k})$ in the set ${\mathcal {H}}$ and, therefore, for all ordered list of k real numbers in the set ${\mathcal {A}}_{H}\subset {\mathcal {H}}$. The implication then follows from any finite repetition of these two elementary implications.

1.7.2 Statement (b) implies Statement (c)

Assume that Inequality (6) holds for all $(\alpha _{1},\dots ,\alpha _{k})\in {\mathcal {A}}_{{\mathcal {H}}}$. For any $i\in \{1,\dots ,k\}$, define the ordered list of k numbers $(\alpha _{1}^{i},\dots ,\alpha _{k}^{i})$ by:

$$\begin{aligned} \alpha _{h}^{i}= & {} -(2^{i-h}),\text { for }h\in \{1,\dots ,i\} \\ \alpha _{h}^{i}= & {} 0,\text { for }h\in \{i+1,\dots ,k\}. \end{aligned}$$

Let us first show that the ordered list $(\alpha _{1}^{i},\dots ,\alpha _{k}^{i})$ of real numbers thus defined belongs to ${\mathcal {A}}_{H}$ for every $i\in {\mathcal {C}}$. Thanks to Proposition 6, this amounts to show that these real numbers satisfy:

$$\begin{aligned} \alpha _{h+1}^{i}-\alpha _{h}^{i}\ge \alpha _{k}^{i}-\alpha _{h+1}^{i} \end{aligned}$$

(42)

for every $h\in \{1,\dots ,k-1\}$. If $h\ge i+1$, then one has:

$$\begin{aligned} \alpha _{h+1}^{i}-\alpha _{h}^{i}=0-0=\alpha _{k}^{i}-\alpha _{h+1}^{i} \end{aligned}$$

so that Inequality (42) holds for that case. If $h=i$, then

$$\begin{aligned} \alpha _{i+1}^{i}-\alpha _{i}^{i}=0+2^{0}>0-0=\alpha _{k}^{i}-\alpha _{i+1}^{i} \end{aligned}$$

so that (42) holds also for that case. If finally $h<i$, then one has:

$$\begin{aligned} \alpha _{h+1}^{i}-\alpha _{h}^{i}= & {} -2^{i-h-1}+2^{i-h} \\= & {} 2^{i-h-1} \\= & {} 0-(-2^{i-h-1}) \\= & {} \alpha _{k}^{i}-\alpha _{h+1}^{i} \end{aligned}$$

so that (42) holds for this case as well. Since the ordered list $(\alpha _{1}^{i},\dots ,\alpha _{k}^{i})$ of real numbers belongs to ${\mathcal {A}}_{H}$ for every $i\in {\mathcal {C}}$, Inequality (6) must hold for any such ordered list of numbers. Hence, for every $i\in {\mathcal {C}}$, Inequality $\sum _{h=1}^{k}n_{h}^{s}\,\alpha _{h}^{i}\ge \sum _{h=1}^{k}n_{h}^{s^{\prime }}\alpha _{h}^{i}$ is equivalent to:

$$\begin{aligned} \sum _{h=1}^{i}(2^{i-h})n_{h}^{s}\le \sum _{h=1}^{i}(2^{i-h})n_{h}^{s^{\prime }} \end{aligned}$$

which is nothing else than the condition for H-dominance, as expressed by Eq. (9).

1.7.3 Statement (c) implies Statement (a)

Assume that $H(i;s)\le H(i;s^{\prime })$ for all $i \in \{1,\dots ,k-1\}$. We know from Proposition 1 that $F(i;s)\le F(i;s^{\prime })$ for all $i \in \{1,\dots ,k-1\}$ implies that $H(i;s)\le H(i;s^{\prime })$ for all $i \in \{1,2,\dots ,k-1\}$. If it is the case that, for all $i \in \{1,\dots ,k-1\}$, one has both $H(i;s)\le H(i;s^{\prime })$ and $F(i;s)\le F(i;s^{\prime })$, we conclude from Theorem 1 that s can be obtained from $s^{\prime }$ by means of a finite sequence of increments and the proof is done. In the following, we therefore assume that $H(i;s)\le H(i;s^{\prime })$ holds for all $i \in \{1,2,\dots ,k-1\}$ but that there exists some $g\in \{1,\dots ,k-1\}$ for which one has $F(g;s)-F(g;s^{\prime })>0$.

Step 1 Define then the index h by:

$$\begin{aligned} h=\min \left\{ g\mid F(g;s)-F(g;s^{\prime })>0\right\} . \end{aligned}$$

(43)

Given that index h, one can also define the index l by:

$$\begin{aligned} l=\min \left\{ g>h\mid F(j;s)-F(j;s^{\prime })\le 0,\forall j\in \{g,g+1,...,k\}\right\} . \end{aligned}$$

(44)

Such a l exists because $F(k;s)-F(k;s^{\prime })=0$. Notice that, by definition of l, one has:

$$\begin{aligned} F(l-1;s)-F(l-1;s^{\prime })>0\quad \text {and}\quad F(l;s)-F(l;s^{\prime })\le 0, \end{aligned}$$

(45)

Hence, one has [using the definition of F provided by (7)], that $n_{l}^{s}<n_{l}^{s^{\prime }}$. We now establish the existence of a unique $i\in \{1,\dots ,h-1\}$ such that:

$$\begin{aligned} F(i;s)-F(i;s^{\prime })<0\quad \text {and}\quad F(g;s)-F(g,s^{\prime })=0,\forall g<i. \end{aligned}$$

(46)

Indeed, by assumption, we have $H(g;s)\le H(g;s^{\prime })$ for all $g=1,2,\dots ,k-1$. It follows that, thanks to Expression (10), Inequality $H(1;s)\le H(1;s^{\prime })$ is equivalent to:

$$\begin{aligned} F(1;s)<F(1;s^{\prime }), \end{aligned}$$

(47)

or:

$$\begin{aligned} F(1;s)=F(1;s^{\prime }). \end{aligned}$$

(48)

If Case (47) holds, then the existence of some $i\in \{1,\dots ,h-1\}$ for which Expression (46) holds is established (in that case, $i=1$). Now assume that Case (48) holds. Since $H(2;s)\le H(2;s^{\prime })$ also holds, we must have $H(2;s)<H(2;s^{\prime })$ or $H(2;s)=H(2;s^{\prime })$ which is, respectively, equivalent—thanks to Expression (11)—to:

$$\begin{aligned} 2F(1;s)+F(2;s)<2F(1;s^{\prime })+F(2;s^{\prime }), \end{aligned}$$

(49)

or:

$$\begin{aligned} 2F(1;s)+F(2;s)=2F(1;s^{\prime })+F(2;s^{\prime }). \end{aligned}$$

(50)

Again, if we are in Case (49), we can conclude [since $F(1;s)=F(1;s^{\prime })$] that $F(2;s)<F(2;s^{\prime })$, which establishes the existence of some $i\in \{1,\dots ,h-1\}$ for which Expression (46) holds (in that case, $i=2$). If we are in Case (50), we iterate in the same fashion using the definition of H provided by (11). We notice that the index i for which (46) holds must be strictly smaller than h, as defined in (43), because assuming otherwise will contradict the fact that $H(g;s)\le H(g;s^{\prime })$ holds for all $g\in \{1,\dots ,k-1\}$. We finally note that, because of the definition of F provided by (7), the definition of the index i just provided entails that:

$$\begin{aligned} n_{i}^{s}<n_{i}^{s^{\prime }},\quad \text {and}\quad n_{g}^{s}=n_{g}^{s^{\prime }},\ \forall g\in \{1,\dots ,i-1\}. \end{aligned}$$

(51)

Step 2 Given the indices h and l as defined in (43) and (44), respectively, we now proceed by defining a new society—$s^{1}$ say—obtained from $s^{\prime }$ by means of a Hammond transfer and such that $H(g;s)\le H(g;s^{1})\le H(g;s^{\prime })$, for every $g\in \{1,\dots ,k-1\}$, with a strict inequality between $s^{1}$ and $s^{\prime }$ for at least one g. For this sake, we define the numbers $\delta _{1}$, $\delta _{2}$ and $\delta$ by:

$$\begin{aligned} \delta _{1}=n[F(i;s^{\prime })-F(i;s)],\quad \delta _{2}=n[F(l-1;s)-F(l-1;s^{\prime })]\quad \text {and}\quad \delta =\min (\delta _{1},\delta _{2}). \nonumber \\ \end{aligned}$$

(52)

We note that, by the very definition of the index i, one has $\delta _{1}=n_{i}^{s^{\prime }}-n_{i}^{s}>0$. We notice also that, thanks to (45) and the definition of the index l, one has $0<\delta _{2}\le n_{l}^{s^{\prime }}-n_{l}^{s}$. Define then the society $s^{1}$ by:

$$\begin{aligned} n_{g}^{s^{1}}=n_{g}^{s^{\prime }},\ \forall \ g\ne i,i+1,l\,; \end{aligned}$$

$$\begin{aligned} n_{i}^{s^{1}}=n_{i}^{s^{\prime }}-\delta \ ;\ n_{i+1}^{s^{1}}=n_{i+1}^{s^{\prime }}+2\delta \ ;\ n_{l}^{s^{1}}=n_{l}^{s^{\prime }}-\delta \,; \end{aligned}$$

(53a)

It is clear that $s^{1}$ has been obtained from $s^{\prime }$ by $\delta$ Hammond transfers as per Definition 3 where the indices g,  i,  j and l of this definition are, here, i, $i+1$, $i+1$ and l, respectively. After simple manipulation and the definition of $s^{1}$ in (53), we observe that:

$$\begin{aligned} F(g;s^{1})-F(g;s^{\prime })=\left\{ \begin{array}{ll} 0 &{}\quad \text {for}\ g\in \{1,\dots ,i-1\}, \\ -\delta /n &{}\quad \text {for}\ g=i, \\ +\delta /n &{}\quad \text {for}\ g\in \{i+1,\dots ,l-1\}, \\ 0 &{}\quad \text {for}\ g\in \{l,\dots ,k\}. \end{array} \right. \end{aligned}$$

Again, after simple manipulation, one deduces that:

$$\begin{aligned} H(g;s^{1})-H(g;s^{\prime })=\left\{ \begin{array}{ll} 0 &{}\quad \text {for}\ g\in \{1,\dots ,i-1\}, \\ -\delta /n &{}\quad \text {for}\ g=i, \\ 0 &{}\quad \text {for}\ g\in \{i+1,\dots ,l-1\}, \\ -\left( 2^{g-l}\right) \delta /n &{}\quad \text {for}\ g\in \{l,\dots ,k\}. \end{array} \right. \end{aligned}$$

(54)

Recalling that $\delta >0$ thanks to Expression (52), Expression (54) thus confirms that $H(g;s^{1})\le H(g;s^{\prime })$, for every $g\in \{1,\dots ,k-1\}$, with strict inequalities for $g\in {i}\cup \{l,\dots ,k\}$. Let us now verify that $H(g;s)\le H(g;s^{1})$, for all $g\in \{1,\dots ,k-1\}$. Because $[H(\cdot ,s)-H(\cdot ,s^{1})]=[H(\cdot ,s)-H(\cdot ,s^{\prime })]-[H(\cdot ,s^{1})-H(\cdot ,s^{\prime })]$, we first notice that Expression (54) entails that:

$$\begin{aligned} H(g,s)-H(g,s^{1})=\left\{ \begin{array}{ll} H(g,s)-H(g,s^{\prime }) &{}\quad \text {for}\ g\in \{1,\dots ,i-1\}, \\ H(g,s)-H(g,s^{\prime })+\delta /n &{}\quad \text {for}\ g=i, \\ H(g,s)-H(g,s^{\prime }) &{}\quad \text {for}\ g\in \{i+1,\dots ,l-1\}, \\ H(g,s)-H(g,s^{\prime })+\left( 2^{g-l}\right) \delta /n &{}\quad \text {for}\ g\in \{l,\dots ,k\}. \end{array} \right. \nonumber \\ \end{aligned}$$

(55)

The fact that $H(g;s)-H(g,s^{\prime })\le 0$ holds for all $g\in \{1,\dots ,k-1\}$ entails that $H(g;s)-H(g;s^{1})\le 0$ for all $g\in \{1,2,\dots ,i-1\}\cup \{i+1,\dots ,l-1\}$. Consider now the case $g=i$. Using (12), we know that:

$$\begin{aligned} H(i;s)-H(i,s^{\prime })=2\,[H(i-1;s)-H(i-1;s^{\prime })]+(n_{i}^{s}-n_{i}^{s^{\prime }})/n. \end{aligned}$$

(56)

By definition of i, one has $F(g;s)-F(g;s^{\prime })=0$ for all $g<i$, so that the first term in the right-hand side of Eq. (56) is 0. Recalling then from (52) that $\delta _{1}=n_{i}^{s^{\prime }}-n_{i}^{s}>0$ and that $\delta =\min (\delta _{1},\delta _{2})$, it follows that:

$$\begin{aligned} n_{i}^{s}-n_{i}^{s^{\prime }}+\delta \le 0. \end{aligned}$$

By combining Eqs. (55) and (56), we conclude that:

$$\begin{aligned} H(i,s)-H(i,s^{1})=H(i,s)-H(i,s^{\prime })+\delta /n=(n_{i}^{s}-n_{i}^{s^{\prime }}+\delta )/n\le 0. \qquad \end{aligned}$$

(57)

Consider finally the case where $g\in \{l,\dots ,k-1\}$. By using Eq. (12), and recalling that $\delta _{2}=n[F(l-1;s)-F(l-1;s^{\prime })]$, one has:

$$\begin{aligned} H(l;s)-H(l;s^{\prime })=2[H(l-1;s)-H(l-1;s^{\prime })]+[F(l;s)-F(l;s^{\prime })]-\delta _{2}\,/n. \nonumber \\ \end{aligned}$$

(58)

Combining (58) with the last line of (55), and remembering that $\delta \le \delta _{2}$, one obtains:

$$\begin{aligned} H(l,s)-H(l,s^{1})= & {} 2[(H(l-1;s)-H(l-1;s^{\prime })]+[F(l;s)-F(l;s^{\prime })]\nonumber \\&+(\delta -\delta _{2})/n\le 0. \end{aligned}$$

(59)

Finally, using successive applications of Eq. (12), one obtains, for any $g\in \{l+1,\dots ,k-1\}$:

$$\begin{aligned} H(g;s)-H(g;s^{\prime })= & {} \big (2^{g-l+1}\big )[H(l-1;s)-H(l-1;s^{\prime })] \\&+\sum \limits _{j=l}^{g-1}\big (2^{g-j-1}\big )[F(j;s)-F(j;s^{\prime })] \\&+F(g;s)-F(g;s^{\prime })-\big (2^{g-l}\big )\delta _{2}/n \\\le & {} 0\, \end{aligned}$$

since $F(j;s)\le F(j;s^{\prime })$ for all $j\in \{l,l+1,...,k\}$. Combined with the last line of (55) and the fact that $\delta \le \delta _{2}$, this completes the proof that $H(g;s)-H(g;s^{1})\le 0$ for all $g\in \{1,\dots ,k-1\}$. Hence, we have found a society $s^{1}$ obtained from society $s^{\prime }$ by means of a non-trivial Hammond transfers that is H-dominated by s.

Step 3 We now show that, in moving from $s^{\prime }$ to s, one has ‘brought to naught’ at least one of the differences $\left| F(h;s)-F(h;s^{\prime })\right|$ that distinguishes s from $s^{\prime }$. That is to say, we establish the existence of some $g \in \{1,\dots ,k-1\}$ for which one has $\left| F(g;s)-F(g;s^{1})\right| =0$ and $\left| F(g;s)-F(g;s^{\prime })\right| >0$. This is easily seen from the fact that, in the construction of $s^{1}$, one has either:

$$\begin{aligned} \delta =\delta _{1}=n_{i}^{s^{\prime }}-n_{i}^{s}. \end{aligned}$$

(60)

or:

$$\begin{aligned} \delta =\delta _{2}=n[F(l-1;s)-F(l-1;s^{\prime })]. \end{aligned}$$

(61)

If we are in Case (60), one has $F(i;s)-F(i;s^{1})=0$ and $F(i;s)-F(i;s^{\prime })<0$, by definition of the index i. If we are in Case (61), then $F(l-1;s)-F(l-1;s^{1}) = F(l-1,s)-F(l-1,s^{\prime })+\delta _{2}/n = 0$, and $F(l-1,s)-F(l-1,s^{\prime })>0$, by definition of the index l.

Now, if $s=s^{1}$, then the proof is complete. If s is distinct from $s^{1}$ but s first-order dominates $s^{1}$, then we conclude that society s can be obtained from society $s^{\prime }$ by means of a finite sequence of one Hammond transfer and a collection of increments (using Theorem 1). If s is distinct from $s^{1}$ and s does not first-order dominate $s^{1}$, then we can find three categories i, h and l just as in the preceding steps and construct a new distribution—say $s^{2}$—that can be obtained from distribution $s^{1}$ by means of an (integer number of) Hammond transfers and that is H-dominated by s and so on. More generally, after a finite number—t say—of iterations, we will find a distribution $s^{t}$ obtained from $s^{\prime }$ by means of t Hammond transfers such that s H-dominates $s^{t}$. In that case, we will have either $s=s^{t}$ or s first-order dominates $s^{t}$. Since there are finitely many differences of the kind $\left| F(g;s)-F(g;s^{\prime })\right|$ to bring to naught, the number t must be finite. This completes the proof.

1.8 Lemma 2

Step 1 Using (19) in Lemma 1, we have, for any society s:

$$\begin{aligned} \frac{1}{n}\sum _{h=1}^{k}n_{h}^{s}\,\alpha _{h}=\alpha _{k}-\sum _{h=1}^{k-1}F(h;s)\left[ \alpha _{h+1}-\alpha _{h}\right] . \end{aligned}$$

(62)

In order to simplify the notation we let, in this proof, $F_h = F(h; s)$ and $\theta _h = (\alpha _{h+1} - \alpha _{h})$ so that:

$$\begin{aligned} \frac{1}{n}\sum _{i=1}^{k}n_{h}^{s}\,\alpha _{h}=\alpha _k - \sum _{h=1}^{k-1} F_h\,\theta _h. \end{aligned}$$

(63)

Letting $\vartheta _{h}=\theta _{h}-\sum _{j=h+1}^{k-1}\theta _{j}$ for all $h \in \{1,\dots ,k-2\}$ and $\vartheta _{k-1}=\theta _{k-1}$, we rewrite each term of the sum $\sum _{h=1}^{k-1} F_h\,\theta _h$ in (63) as follows:

${\underline{\hbox {For }h=1}}$:

$$\begin{aligned} F_1\,\theta _1= & {} F_1 \left[ \theta _1 - \sum _{j=2}^{k-1} \theta _j \right] + F_1\,\theta _2 + F_1\,\theta _3 + \cdots + F_1\,\theta _{k-1} \nonumber \\= & {} F_1 \left[ \theta _1 - \sum _{j=2}^{k-1} \theta _j \right] + F_1 \left[ \theta _2 - \sum _{j=3}^{k-1} \theta _j \right] + 2\,F_1\,\theta _3 + \cdots + 2\,F_1\,\theta _{k-1} \nonumber \\= & {} F_1 \left[ \theta _1 - \sum _{j=2}^{k-1} \theta _j \right] + F_1 \left[ \theta _2 - \sum _{j=3}^{k-1} \theta _j \right] + 2\,F_1 \left[ \theta _3 - \sum _{j=4}^{k-1} \theta _j \right] + \cdots + 4\,F_1\,\theta _{k-1} \nonumber \\= & {} \cdots \nonumber \\= & {} F_1\,\vartheta _1 + F_1\,\vartheta _2 + 2\,F_1\,\vartheta _3 + \left( 2^{2}\right) \,F_1\,\vartheta _4 + \cdots + \left( 2^{k-3}\right) F_1\,\vartheta _{k-1}. \end{aligned}$$

(64)

${\underline{\hbox {For }h=2}}$:

$$\begin{aligned} F_2\,\theta _2= & {} F_2 \left[ \theta _2 - \sum _{j=3}^{k-1} \theta _j \right] + F_2\,\theta _3 + F_2\,\theta _4 + \cdots + F_2\,\theta _{k-1} \nonumber \\= & {} F_2 \left[ \theta _2 - \sum _{j=3}^{k-1} \theta _j \right] + F_2 \left[ \theta _3 - \sum _{j=4}^{k-1} \theta _j \right] + 2\,F_2\,\theta _4 + \cdots + 2\,F_2\,\theta _{k-1} \nonumber \\= & {} F_2 \left[ \theta _2 - \sum _{j=3}^{k-1} \theta _j \right] + F_2 \left[ \theta _3 - \sum _{j=4}^{k-1} \theta _j \right] + 2\,F_2 \left[ \theta _4 - \sum _{j=5}^{k-1} \theta _j \right] + \cdots + 4\,F_2\,\theta _{k-1} \nonumber \\= & {} \cdots \nonumber \\= & {} F_2\,\vartheta _2 + F_2\,\vartheta _3 + 2\,F_2\,\vartheta _4 + \left( 2^{2}\right) \,F_2\,\vartheta _5 + \cdots + \left( 2^{k-4}\right) F_2\,\vartheta _{k-1}. \end{aligned}$$

(65)

More generally, we have $F_{k-1}\,\theta _{k-1} = F_{k-1}\,\vartheta _{k-1}$ and:

$$\begin{aligned} F_h\,\theta _h = F_h\,\vartheta _h + F_h \sum _{j=i+1}^{k-1} \left( 2^{j-i-1}\right) \vartheta _j,\quad \forall i \in \{1,\dots ,k-2\}. \end{aligned}$$

(66)

Hence, one can write:

(67)

Remembering that $\vartheta _{k-1}=\theta _{k-1}$ and $\vartheta _{h}=\theta _{h}-\sum _{j=h+1}^{k-1}\theta _{j}$ for all $h \in \{1,\dots ,k-2\}$, one can use Eq. (11) and sum vertically the decomposition (67) to obtain:

$$\begin{aligned} \sum _{h=1}^{k-1} F_h\,\theta _h = \sum _{h=1}^{k-2} H(h; s) \left[ \theta _h - \sum _{j=h+1}^{k-1} \theta _j \right] + H(k-1; s)\,\theta _{k-1}. \end{aligned}$$

(68)

Since $\frac{1}{n} \sum _{h=1}^{k} n^s_{h}\,\alpha _{h} = \alpha _k - \sum _{h=1}^{k-1} F_h\,\theta _h$, one obtains finally:

$$\begin{aligned} \frac{1}{n} \sum _{h=1}^{k} n^s_{h}\,\alpha _{h} = \alpha _{k} - \sum _{h=1}^{k-2} H(h; s) \left[ \theta _h - \sum _{j=h+1}^{k-1} \theta _j \right] - H(k-1; s) \theta _{k-1}. \end{aligned}$$

(69)

Step 2. In a symmetric fashion, one obtains from Eq. (20) in Lemma 1:

$$\begin{aligned} \frac{1}{n} \sum _{h=1}^{k} n^s_{h}\,\alpha _{h} = \alpha _1 + \sum _{h=1}^{k-1} {\bar{F}}(h; s) \left[ \alpha _{h+1} - \alpha _{h} \right] . \end{aligned}$$

(70)

Letting ${\bar{F}}_h = {\bar{F}}(h; s)$ and $\theta _h = (\alpha _{h+1} - \alpha _{h})$, we have:

$$\begin{aligned} \frac{1}{n} \sum _{h=1}^{k} n^s_{h}\,\alpha _{h} = \alpha _1 + \sum _{h=1}^{k-1} {\bar{F}}_h\,\theta _h. \end{aligned}$$

(71)

Now, letting $\beta _{1} = \theta _{1}$ and $\beta _h = (\theta _h - \sum _{j=1}^{h-1} \theta _j)$ for all $h \in \{2,\dots ,k-1\}$, we propose to rewrite $\sum _{h=1}^{k-1} {\bar{F}}_h\,\theta _h$ in (71) as follows:

${\underline{\hbox {For }h=k-1}}$:

$$\begin{aligned} {\bar{F}}_{k-1}\,\theta _{k-1}= & {} {\bar{F}}_{k-1} \left[ \theta _{k-1} - \sum _{j=1}^{k-2} \theta _j \right] + {\bar{F}}_{k-1}\,\theta _{k-2} + {\bar{F}} _{k-1}\,\theta _{k-3} + \cdots + {\bar{F}}_{k-1}\,\theta _{1} \nonumber \\= & {} {\bar{F}}_{k-1} \left[ \theta _{k-1} - \sum _{j=1}^{k-2} \theta _j \right] + {\bar{F}}_{k-1} \left[ \theta _{k-2} - \sum _{j=1}^{k-3} \theta _j \right] \nonumber \\&+ 2\, {\bar{F}}_{k-1}\,\theta _{k-3} + \cdots + 2\,{\bar{F}}_{k-1}\,\theta _{1} \nonumber \\= & {} {\bar{F}}_{k-1} \left[ \theta _{k-1} - \sum _{j=1}^{k-2} \theta _j \right] + {\bar{F}}_{k-1} \left[ \theta _{k-2} - \sum _{j=1}^{k-3} \theta _j \right] \nonumber \\&+ 2\, {\bar{F}}_{k-1} \left[ \theta _{k-3} - \sum _{j=1}^{k-4} \theta _j \right] + \cdots + 4\,{\bar{F}}_{k-1}\,\theta _{1} \nonumber \\= & {} \cdots \nonumber \\= & {} {\bar{F}}_{k-1}\,\beta _{k-1} + {\bar{F}}_{k-1}\,\beta _{k-2} + 2\,{\bar{F}} _{k-1}\,\beta _{k-3} \nonumber \\&+ \left( 2^{2}\right) \,{\bar{F}}_{k-1}\,\beta _{k-4} + \cdots + \left( 2^{k-3}\right) {\bar{F}}_{k-1}\,\beta _{1}. \end{aligned}$$

(72)

${\underline{\hbox {For }h=k-2}}$:

$$\begin{aligned} {\bar{F}}_{k-2}\,\theta _{k-2}= & {} {\bar{F}}_{k-2} \left[ \theta _{k-2} - \sum _{j=1}^{k-3} \theta _j \right] + {\bar{F}}_{k-2}\,\theta _{k-3} + {\bar{F}} _{k-2}\,\theta _{k-4} + \cdots + {\bar{F}}_{k-2}\,\theta _1 \nonumber \\= & {} {\bar{F}}_{k-2} \left[ \theta _{k-2} - \sum _{j=1}^{k-3} \theta _j \right] + {\bar{F}}_{k-2} \left[ \theta _{k-3} - \sum _{j=1}^{k-4} \theta _j \right] \nonumber \\&+ 2\, {\bar{F}}_{k-2}\,\theta _{k-4} + \cdots + 2\,{\bar{F}}_{k-2}\,\theta _1 \nonumber \\= & {} {\bar{F}}_{k-2} \left[ \theta _{k-2} - \sum _{j=1}^{k-3} \theta _j \right] + {\bar{F}}_{k-2} \left[ \theta _{k-3} - \sum _{j=1}^{k-4} \theta _j \right] \nonumber \\&+ 2\, {\bar{F}}_{k-2} \left[ \theta _{k-4} - \sum _{j=1}^{k-5} \theta _j \right] + \cdots + 4\,{\bar{F}}_{k-2}\,\theta _1 \nonumber \\= & {} \cdots \nonumber \\= & {} {\bar{F}}_{k-2}\,\beta _{k-2} + {\bar{F}}_{k-2}\,\beta _{k-3} + 2\,{\bar{F}} _{k-2}\,\beta _{k-4} \nonumber \\&+ \left( 2^{2}\right) \,{\bar{F}}_{k-2}\,\beta _{k-5} + \cdots + \left( 2^{k-4}\right) {\bar{F}}_{k-2}\,\beta _1. \end{aligned}$$

(73)

More generally, one has ${\bar{F}}_{1}\,\theta _{1} = {\bar{F}}_{1}\,\beta _{1}$ and:

$$\begin{aligned} {\bar{F}}_h\,\theta _h = {\bar{F}}_h\,\beta _h + {\bar{F}}_h \sum _{j=1}^{h-1} \left( 2^{h-j-1}\right) \beta _j,\quad \forall h \in \{2,\dots ,k-1\}. \end{aligned}$$

(74)

Hence, one can conclude that:

(75)

Using (16), and summing vertically the previous equation, one obtains:

$$\begin{aligned} \sum _{h=1}^{k-1} {\bar{F}}_h\,\theta _h = {\bar{H}}(1; s)\,\theta _{1} + \sum _{h=2}^{k-1} {\bar{H}}(h; s) \left[ \theta _h - \sum _{j=1}^{h-1} \theta _j \right] . \end{aligned}$$

(76)

Since $\frac{1}{n} \sum _{h=1}^{k} n^s_{h}\,\alpha _{h} = \alpha _1 + \sum _{h=1}^{k-1} {\bar{F}}_h\,\theta _h$, we finally obtain:

$$\begin{aligned} \frac{1}{n} \sum _{h=1}^{k} n^s_{h}\,\alpha _{h} = \alpha _1 + {\bar{H}}(1; s)\,\theta _{1} + \sum _{h=2}^{k-1} {\bar{H}}(h; s)\left[ \theta _h - \sum _{j=1}^{h-1} \theta _j \right] . \end{aligned}$$

(77)

Step 3. From Eq. (21) in Lemma 1, one has, for any $t\in \{2,\dots ,k-1\}$:

$$\begin{aligned} \frac{1}{n}\sum _{h=1}^{k}n_{h}^{s}\,\alpha _{h}=\alpha _{t}-\sum _{h=1}^{t-1}F_{h}\theta _{h}+\sum _{h=t}^{k-1}{\bar{F}}_{h}\theta _{h}. \end{aligned}$$

(78)

By using Eq. (68) and replacing category k by category t in this equation, we obtain:

$$\begin{aligned} \sum _{h=1}^{t-1}F_{h}\theta _{h}=\sum _{h=1}^{t-2}H(h;s)\left[ \theta _{h}-\sum _{j=h+1}^{t-1}\theta _{j}\right] +H(t-1;s)\,\theta _{t-1}. \end{aligned}$$

(79)

Symmetrically, replacing category 1 by category t in Eq. (76) enables one to write::

$$\begin{aligned} \sum _{h=t}^{k-1}{\bar{F}}_{h}\,\theta _{h}={\bar{H}}(t;s)\,\theta _{t}+\sum _{h=t+1}^{k-1}{\bar{H}}(h;s)\left[ \theta _{h}-\sum _{j=t}^{h-1}\theta _{j}\right] . \end{aligned}$$

(80)

Combining Eqs. (78), (79) and (80), one gets finally the desired result.

1.9 Theorem 5

1.9.1 Statement (a) implies Statement (b)

That results immediately from the definition of the set ${\mathcal {H}}$ (using Proposition 4).

1.9.2 Statement (b) implies Statement (a)

The proof builds on the dual theory for convex cones, recently investigated by Muller and Scarsini (2012), but adapted to our fully discrete framework. An more detailed description of the proof—stated here for completeness—can be found in Magdalou (2018). (See in particular the discussion on p. 17.) First, we define by ${\mathcal {E}}$ the set of all lists $\alpha =(\alpha _{1},\dots ,\alpha _{k})\in {\mathbb {R}}^{k}$ which assign numerical evaluations $\alpha _{h}$ to each category $h\in {\mathcal {C}}$, and such that there exist $i,j\in {\mathcal {C}}$ with $\alpha _{i}\ne \alpha _{j}$ . We then denote by ${\mathcal {M}}$ the set of all lists $m=(m_{1},\dots ,m_{k})\in {\mathbb {R}}^{k}$ such that $\sum _{h=1}^{k}m_{h}=0$. For all $m\in {\mathcal {M}}$ and all $\alpha \in {\mathcal {E}}$, we define the bilinear mapping $b(\cdot ,\cdot )$ by $b(m,\alpha )=-\sum _{h=1}^{k}m_{h}\alpha _{h}$. The pair $({\mathcal {M}},{\mathcal {E}})$ is thus, under the bilinear mapping $b(\cdot ,\cdot )$, a dual pair, what we denote $({\mathcal {M}},{\mathcal {E}};b)$. The dual pair $({\mathcal {M}}, {\mathcal {E}};b)$ is moreover strict and is the sense that for each $0\ne m\in {\mathcal {M}}$, there is an $\alpha \in {\mathcal {E}}$ with $b(m,\alpha )\ne 0$ and, for each $0\ne \alpha \in {\mathcal {E}}$, there is a $m\in {\mathcal {M}}$ with $b(m,\alpha )\ne 0$.

We then recall that any two societies s and $s^{\prime }$ are characterized by the lists $n^{s}=(n_{1}^{s},\dots ,n_{k}^{s})$ and $n^{s^{\prime }}=(n_{1}^{s^{\prime }},\dots ,n_{k}^{s^{\prime }})$, respectively, which indicate the number of agents in each category in ${\mathcal {C}}$, and such that $\sum _{h=1}^{k}n_{h}^{s}= \sum _{h=1}^{k}n_{h}^{s^{\prime }}=n$. We define by ${\mathcal {T}}\subset {\mathcal {M}}$ the set of all lists $(n^{s}-n^{s^{\prime }})$ such that s can be obtained from $s^{\prime }$ by means of only one Hammond transfer.

Consider now two particular societies s and $s^{\prime }$ and assume that Statement (b) is true, so that inequality $\sum _{h=1}^{k}n_{h}^{s}\alpha _{h}\ge \sum _{h=1}^{k}n_{h}^{s^\prime }\alpha _{h}$, or equivalently $b(n^{s} - n^{s^{\prime }},\alpha ) \le 0$, holds for all lists $\alpha \in {\mathcal {H}}$. We have to establish that s can be obtained from $s^{\prime }$ by means of a finite sequence of Hammond transfers. Consider first the set of all constant lists $\alpha \in {\mathcal {H}}$, such that $\alpha _{i+1} = \alpha _{i}$ for all $i \in \{1,\dots ,k-1\}$. The only information provided by this subclass of ${\mathcal {H}}$ is that $\sum _{h=1}^{k}n_{h}^{s} = \sum _{h=1}^{k}n_{h}^{s^\prime }$, which is actually known by assumption. Hence, we restrict attention to the set $\tilde{\mathcal { H}}$ of all lists $\alpha \in {\mathcal {H}}$, such that there exist $i, j \in {\mathcal {C}}$ with $\alpha _{i} \ne \alpha _{j}$.

The polar cone of $\tilde{{\mathcal {H}}} \subset {\mathcal {E}}$ under the duality $({\mathcal {M}},{\mathcal {E}};b)$, which is denoted by $\tilde{ {\mathcal {H}}}^{\circ }$, is defined by:

$$\begin{aligned} \tilde{{\mathcal {H}}}^{\circ } = \left\{ m \in {\mathcal {M}}\, |\, b(m,\alpha ) \le 0,\ \forall \alpha \in \tilde{{\mathcal {H}}} \right\} . \end{aligned}$$

(81)

By definition, $(n^{s} - n^{s^{\prime }}) \in {\mathcal {M}}$. Because we have assumed that $b(n^{s} - n^{s^{\prime }},\alpha ) \le 0$ holds for all $\alpha \in {\mathcal {H}}$, this also holds for all $\alpha \in \tilde{{\mathcal {H}}}$ (as $\tilde{{\mathcal {H}}} \subset {\mathcal {H}}$). Hence, we also have $(n^{s} - n^{s^{\prime }}) \in \tilde{{\mathcal {H}}}^{\circ }$. Now, the polar cone of the set ${\mathcal {T}} \subset {\mathcal {M}}$ under the duality $({\mathcal {M}}, {\mathcal {E}};b)$, which is denoted by ${\mathcal {T}}^{\circ }$, is defined by:

$$\begin{aligned} {\mathcal {T}}^{\circ } = \left\{ \alpha \in {\mathcal {E}}\, |\, b(m,\alpha ) \le 0,\ \forall m \in {\mathcal {T}} \right\} . \end{aligned}$$

(82)

Thanks to Proposition 4, one immediately deduces that $\tilde{{\mathcal {H}}} = {\mathcal {T}}^{\circ }$. Because $(n^{s} - n^{s^{\prime }}) \in \tilde{{\mathcal {H}}}^{\circ }$ and $\tilde{{\mathcal {H}}} = {\mathcal {T}}^{\circ }$, one concludes that $(n^{s} - n^{s^{\prime }}) \in {\mathcal {T}}^{\circ \circ }$, where ${\mathcal {T}}^{\circ \circ }$ is the bipolar cone of ${\mathcal {T}}$.

We then let ${\mathcal {D}}_{o}({\mathcal {T}})=\text {co}\{\lambda m\,|\,\lambda \in {\mathbb {R}}_{+},m\in {\mathcal {T}}\}$, where co indicates the convex hull of the set. As ${\mathcal {T}}$ is a discrete and finite set, ${\mathcal {D}}_{o}({\mathcal {T}})$ is closed. By applying the bipolar theorem, one deduces that ${\mathcal {T}}^{\circ \circ }={\mathcal {D}}_{o}( {\mathcal {T}})$. One also remarks that, by definition, ${\mathcal {T}}\subset {\mathbb {Z}}^{k}$ where ${\mathbb {Z}}$ is the set of integers. Thus, $(n^{s}-n^{s^{\prime }})\in {\mathcal {D}}_{o}({\mathcal {T}})\cap {\mathbb {Z}}^{k}$. By applying the notion of minimal Hilbert basis [see Magdalou (2018)], it can be shown that, actually, there exists a positive and finite integer t such that $(n^{s}-n^{s^{\prime }})=\sum _{h=1}^{t}\lambda _{h}m_{h}$, where $m_{h}\in {\mathcal {T}}$ and $\lambda _{h}\in {\mathbb {Z}}_{+}$—with ${\mathbb {Z}}_{+}$ the nonnegative orthant of ${\mathbb {Z}}$—for all $h\in \{1,\dots ,t\}$. That concludes the proof.

1.9.3 Statement (b) implies Statement (c)

Assume that the inequality $\sum _{h=1}^{k}n_{h}^{s}\alpha _{h}\ge \sum _{h=1}^{k}n_{h}^{s^\prime }\alpha _{h}$ holds for all lists $(\alpha _{1},\dots ,\alpha _{k})\in {\mathcal {H}}$. This implies in particular that the inequality holds for all $(\alpha _{1},\dots ,\alpha _{k})\in {\mathcal {A}}_{H}$ (as ${\mathcal {A}}_{H} \subset {\mathcal {H}}$). It then follows from Theorem  3 that society s H-dominates society $s^{\prime }$. Similarly, the fact that the inequality $\sum _{h=1}^{k}n_{h}^{s}\alpha _{h}\ge \sum _{h=1}^{k}n_{h}^{s^\prime }\alpha _{h}$ holds for all lists $(\alpha _{1},\dots ,\alpha _{k})\in {\mathcal {H}}$ implies in particular that it holds for all $(\alpha _{1},\dots ,\alpha _{k})\in {\mathcal {A}}_{{\bar{H}}}$ (as ${\mathcal {A}}_{{\bar{H}}} \subset {\mathcal {H}}$). Hence, thanks to Theorem 4, society s ${\overline{H}}$-dominates society $s^{\prime }$ . We conclude that society s both H-dominates and ${\bar{H}}$-dominates society $s^{\prime }$.

1.9.4 Statement (c) implies Statement (b)

Assume that $H(h,s)\le H(h,s^{\prime })$ and ${\bar{H}}(h,s)\le {\bar{H}}(h,s^{\prime })$ for every category $h \in \{1,\dots ,k-1\}$. Thanks to Proposition 5, we have to show that $\sum _{h=1}^{k}n_{h}^{s} \alpha _{h}\ge \sum _{h=1}^{k}n_{h}^{s^\prime }\alpha _{h}$ must hold as well for $(\alpha _{1},\dots ,\alpha _{k})\in {\mathbb {R}}^{k}$ for which there exists an integer $t\in \{1,\dots ,k\}$ such that $(\alpha _{i+1}-\alpha _{i})\ge (\alpha _{t}-\alpha _{i+1})$, for all $i \in \{1,\dots ,t-1\}$ (if any) and $(\alpha _{i^{\prime }+1}-\alpha _{i^{\prime }})\le (\alpha _{i^{\prime }}-\alpha _{t})$, for all $i^{\prime }\in \{t,\dots ,k-1\}$ (again if this set is non-empty).

Consider that such a t exists. If $t=k$, we know from Proposition  6 that $(\alpha _{1},\dots ,\alpha _{k}) \in {\mathcal {A}}_{H}$. Because by assumption $H(h,s)\le H(h,s^{\prime })$ for every category $h \in \{1,\dots ,k-1\}$ we know, thanks to Theorem 3, that Inequality $\sum _{h=1}^{k}n_{h}^{s}\alpha _{h}\ge \sum _{h=1}^{k}n_{h}^{s^\prime }\alpha _{h}$ holds. Symmetrically, if $t=1$, it follows from Proposition 7 that $(\alpha _{1},\dots ,\alpha _{k}) \in {\mathcal {A}}_{{\bar{H}}}$. Because by assumption ${\bar{H}}(h,s)\le {\bar{H}}(h,s^{\prime })$ for every category $h \in \{1,\dots ,k-1\}$ we know, thanks to Theorem 4, that Inequality $\sum _{h=1}^{k}n_{h}^{s} \alpha _{h}\ge \sum _{h=1}^{k}n_{h}^{s^\prime }\alpha _{h}$ holds again. Finally, consider that $t \in \{2,\dots ,k-1\}$. By definition of t, we have:

$$\begin{aligned} \theta _{t-1} \ge 0,&\quad and \quad \left[ \theta _h - \sum _{j=h+1}^{t-1} \theta _j \right] \ge 0, \forall h \in \{ 1, \dots , t-2\}\,; \end{aligned}$$

(83)

$$\begin{aligned} \theta _t \le 0,&\quad and \quad \left[ \theta _h - \sum _{j=t}^{h-1} \theta _j \right] \le 0, \forall h \in \{ t+1, \dots , k-1 \}. \end{aligned}$$

(84)

Letting $\theta _{h}=(\alpha _{h+1}-\alpha _{h})$ for every $h \in \{1,\dots ,k-1\}$ we know, thanks to the decomposition result of Lemma 2, that:

$$\begin{aligned} \frac{1}{n} \left[ \sum _{h=1}^k n^s_h\,\alpha _h - \sum _{h=1}^k n_{h}^{s^{\prime }}\,\alpha _h \right]= & {} \sum _{h=1}^{t-2} \left[ H(h; s^{\prime }) - H(h; s) \right] \left[ \theta _h - \sum _{j=h+1}^{t-1} \theta _j \right] \nonumber \\+ & {} \left[ H(t-1; s^{\prime }) - H(t-1; s) \right] \theta _{t-1} \nonumber \\+ & {} \left[ {\bar{H}}(t; s) - {\bar{H}}(t; s^{\prime }) \right] \theta _{t} \nonumber \\+ & {} \sum _{h=t+1}^{k-1} \left[ {\bar{H}}(h; s) - {\bar{H}}(h; s^{\prime }) \right] \left[ \theta _h - \sum _{j=t}^{h-1} \theta _j \right] . \nonumber \\ \end{aligned}$$

(85)

Since by assumption $H(h,s)\le H(h,s^{\prime })$ and ${\bar{H}}(h,s)\le {\bar{H}}(h,s^{\prime })$ for every category $h \in \{1,\dots ,k-1\}$, combining this information with Eqs. (83) and (84) leads to the required conclusion that Expression (85) is nonnegative.

1.10 Proposition 8

Assume that s and $s^{\prime }$ are two distinct societies for which $H(h,s)\le H(h,s^{\prime })$ holds for all categories $h\in \{1,\dots ,k-1\}$. It follows from the recursive definition of the H-curve provided by Eqs. (10) and (12) that the smallest $i\in \{1,2,\dots ,k\}$ for which $n_{i}^{s}\ne n_{i}^{s^{\prime }}$ is such that $n_{i}^{s}<n_{i}^{s^{\prime }}$. But this implies that $s\succeq _{L}s^{\prime }$.

1.11 Proposition 9

Step 1. As a preliminary of the proof, we first notice that, for any society s defined on the grid ${\mathcal {C}}(t)$, one has, for any $t\in \{0,1,\dots \}$:

$$\begin{aligned} n^{s} \left( \frac{2\,i+1}{2^{t+1}} \right) =0, \end{aligned}$$

(86)

and:

$$\begin{aligned} H^{t+1}\left( \frac{2\,i+1}{2^{t+1}};s\right) =2\,H^{t+1}\left( \frac{i}{2^{t}} ;s\right) . \end{aligned}$$

(87)

Indeed, we notice that:

$$\begin{aligned} {\mathcal {C}}(t) = \left\{ \frac{1}{2^t},\frac{2}{2^t},\dots , \frac{(2^t)k}{2^t } \right\} ,\quad \text {and}\quad {\mathcal {C}}(t+1) = \left\{ \frac{1}{ 2^{t+1}},\frac{2}{2^{t+1}},\dots ,\frac{(2^{t+1})k}{2^{t+1}} \right\} ,\nonumber \\ \end{aligned}$$

(88)

and that:

$$\begin{aligned} \frac{i}{2^{t}}=\frac{2\,i}{2^{t+1}},\quad \forall \in \{1,2,\dots ,(2^{t})k\} . \end{aligned}$$

(89)

Equation (86) then follows from the fact that $n^{s}(x)=$ 0 for all $x \notin {\mathcal {C}}(t)$, while Eq. (87) is an immediate consequence of Eqs. (86) and (89) and the fact that, thanks to Expression (12), one has:

$$\begin{aligned} H^{t+1}\left( \frac{2\,i+1}{2^{t+1}};s \right) =2\,H^{t+1}\left( \frac{2\,i}{ 2^{t+1}};s \right) +n^{s}\left( \frac{2\,i+1}{2^{t+1}} \right) /n,\quad \forall i \in \{0,1,\dots ,(2^{t})k-1\}. \end{aligned}$$

Step 2. We now observe that, for any society s:

$$\begin{aligned} H^{t+1}\left( \frac{i}{2^{t}};s \right) =\sum _{h=1}^{i-1}\left( 2^{2(i-h)-1}\right) {H^{t}\left( \frac{h}{2^{t}};s \right) }+{H^{t}\left( \frac{ i}{2^{t}};s \right) },\quad \forall i \in \{0,1,\dots ,(2^{t})k\}. \nonumber \\ \end{aligned}$$

(90)

Indeed, from Eq. (26) applied to the grid ${\mathcal {C}}(t+1)$, we know that:

$$\begin{aligned} H^{t+1}\left( x;s\right) =\frac{1}{n}\sum _{h=1}^{j}\left( 2^{j-h}\right) n^{s}\left( \frac{h}{{2^{t+1}}}\right) , \end{aligned}$$

for any $x\in {\mathcal {C}}(t+1)$, and $j=x2^{t+1}$ . Applying this to $x = i / 2^{t}$ for any $i \in \{1,\dots ,(2^{t})k\}$ yields:

$$\begin{aligned} H^{t+1}\left( \frac{i}{2^{t}};s \right) =\frac{1}{n}\sum _{h=1}^{2i}\left( 2^{2i-h}\right) n^{s}\left( \frac{h}{{2^{t+1}}}\right) . \end{aligned}$$

(91)

Expression (90) can then be obtained from (91) and the following observations (made only for $i=1,2,3$, but easily extendable to any other i). For $i=1,2,3$, and recalling that $n^{s}\left( \frac{h}{{2^{t+1}}}\right) = 0$ as soon as h is uneven, Expression (91) writes indeed as:

$$\begin{aligned} H^{t+1}\left( \frac{1}{2^{t}};s\right)= & {} \frac{1}{n}\left[ 2n^{s}\left( \frac{1 }{{2^{t+1}}}\right) +n^{s}\left( \frac{2}{{2^{t+1}}}\right) \right] \nonumber \\= & {} \frac{1}{n}n^{s}\left( \frac{2}{{2^{t+1}}}\right) \nonumber \\= & {} \frac{1}{n}n^{s}\left( \frac{1}{{2^{t}}}\right) , \end{aligned}$$

(92)

$$\begin{aligned} H^{t+1}\left( \frac{2}{2^{t}};s\right)= & {} \frac{1}{n}\left[ 8n^{s}\left( \frac{1 }{{2^{t+1}}}\right) +4n^{s}\left( \frac{2}{{2^{t+1}}}\right) +2n^{s}\left( \frac{ 3}{{2^{t+1}}}\right) +n^{s}\left( \frac{4}{{2^{t+1}}}\right) \right] \nonumber \\= & {} \frac{1}{n}\left[ 4n^{s}\left( \frac{2}{{2^{t+1}}}\right) +n^{s}\left( \frac{4 }{{2^{t+1}}}\right) \right] \nonumber \\= & {} \frac{1}{n}\left[ 4n^{s}\left( \frac{1}{{2^{t}}}\right) +n^{s}\left( \frac{2}{ {2^{t}}}\right) \right] , \end{aligned}$$

(93)

$$\begin{aligned} H^{t+1}\left( \frac{3}{2^{t}};s\right)= & {} \frac{1}{n}\left[ 32n^{s}\left( \frac{ 1}{{2^{t+1}}}\right) +16n^{s}\left( \frac{2}{{2^{t+1}}}\right) +8n^{s}\left( \frac{3}{{2^{t+1}}}\right) \right. \nonumber \\&+ \left. 4n^{s}\left( \frac{4}{{2^{t+1}}}\right) +2n^{s}\left( \frac{5}{{ 2^{t+1}}}\right) +n^{s}\left( \frac{6}{{2^{t+1}}}\right) \right] \nonumber \\= & {} \frac{1}{n}\left[ 16n^{s}\left( \frac{2}{{2^{t+1}}}\right) +4n^{s}\left( \frac{4}{{2^{t+1}}}\right) +n^{s}\left( \frac{6}{{2^{t+1}}}\right) \right] \nonumber \\= & {} \frac{1}{n}\left[ 16n^{s}\left( \frac{1}{{2^{t}}}\right) +4n^{s}\left( \frac{2 }{{2^{t}}}\right) +n^{s}\left( \frac{3}{{2^{t}}}\right) \right] . \end{aligned}$$

(94)

Now, applying Eq. (26) to the grid ${\mathcal {C}}(t)$, one has:

$$\begin{aligned} H^{t}\left( \frac{1}{2^{t}};s\right)= & {} \frac{1}{n}n^{s}\left( \frac{1}{{2^{t}} }\right) \end{aligned}$$

(95)

$$\begin{aligned} H^{t}\left( \frac{2}{2^{t}};s\right)= & {} \frac{1}{n}\left[ 2n^{s}\left( \frac{1}{ {2^{t}}}\right) +n^{s}\left( \frac{2}{{2^{t}}}\right) \right] \end{aligned}$$

(96)

$$\begin{aligned} H^{t}\left( \frac{3}{2^{t}};s\right)= & {} \frac{1}{n}\left[ 4n^{s}\left( \frac{1}{ {2^{t}}}\right) +2n^{s}\left( \frac{2}{{2^{t}}}\right) +n^{s}\left( \frac{3}{{ 2^{t}}}\right) \right] , \end{aligned}$$

(97)

so that Expression (90) for $i=1,2,3$ results from combining (92)–(94) with (95)–(97).

Step 3. In order to prove the result, consider two societies s and $s^{\prime }$ and assume that society s $H^{t}$ dominates society $s^{\prime }$ so that:

$$\begin{aligned} H^{t}\left( \frac{i}{2^{t}};s\right) \le H^{t}\left( \frac{i}{2^{t}} ;s^{\prime }\right) \, \end{aligned}$$

holds for all $i\in \{1,\dots ,(2^{t})k\}$. Taking any such i, one has in particular:

$$\begin{aligned} H^{t}\left( \frac{h}{2^{t}};s\right) \le H^{t}\left( \frac{h}{2^{t}} ;s^{\prime }\right) \, \end{aligned}$$

for any $h\in \{1,\dots ,i\}$. Hence, using (90):

$$\begin{aligned} H^{t+1}\left( \frac{i}{2^{t}};s\right) \le H^{t+1}\left( \frac{i}{2^{t}} ;s^{\prime }\right) \end{aligned}$$

for all $i\in \{1,\dots ,(2^{t})k\}$ which implies, thanks to (87), that society s $H^{t+1}$-dominates society $s^{\prime }$.

1.12 Theorem 6

The proof that Statement (a) of the theorem implies Statement (b) is established in Proposition 8 (by using $t=0$). In order to prove the converse implication, consider two arbitrary societies s and $s^{\prime }$ such that $s\succeq _{L}s^{\prime }$. Because of Proposition 9, we only have to show that there exists a nonnegative integer t for which s $H^{t}$-dominates $s^{\prime }$ holds or, equivalently thanks to Theorem 3, that s can be obtained from $s^{\prime }$ by means of a finite sequence of increments and/or Hammond transfers on the grid ${\mathcal {C}}(t)$. Since $s\succeq _{L}s^{\prime }$, there is by Definition 4 an index $i\in \{1,2,\dots ,k\}$ such that $n_{h}^{s}=n^{s}(h)=n^{s^{\prime }}(h)=n_{h}^{s^\prime }$ for all $h \in \{1,2,\dots ,i-1\}$ and $n_{i}^{s}=n^{s}(i)<n^{s^{\prime }}(i)=n_{i}^{s^\prime }$. Given this index i, consider a society $s^{\prime \prime }$ such that:

$$\begin{aligned} n_{h}^{s^{\prime \prime }}= & {} n_{h}^{s},\ \forall h \in \{1,\dots ,i\}\,; \\ n_{i+1}^{s^{\prime \prime }}= & {} \sum _{h=i+1}^{k}n_{h}^{s}\,;\\ n_{h}^{s^{\prime \prime }}= & {} 0,\ \forall h\in \{i+2,\dots ,k\}. \end{aligned}$$

Notice that $\sum _{h=1}^{k}n_{h}^{s^{\prime \prime }}=n$ and that $F(i;s)\le F(i;s^{\prime \prime })\,$ for all $i\in \{1,\dots ,k\}$ so that, by Theorem (1), s can be obtained from $s^{\prime \prime }$ by means of a finite sequence of increments. We also observe that:

$$\begin{aligned}&n_{h}^{s^\prime }=n_{h}^{s^{\prime \prime }},\ \forall h\in \{1,\dots ,i-1\}\,; \end{aligned}$$

(98)

$$\begin{aligned}&n_{i}^{s^\prime }-n_{i}^{s^{\prime \prime }}>0\,;\quad n_{i+1}^{s^\prime }-n_{i+1}^{s^{\prime \prime }}<0\,; \end{aligned}$$

(99)

$$\begin{aligned}&n_{h}^{s^\prime }\ge n_{h}^{s^{\prime \prime }}=0,\ \forall h\in \{i+2,\dots ,k\}. \end{aligned}$$

(100)

Define, for any $h\in {\mathcal {C}}$, the number $\delta _{h}$ by:

$$\begin{aligned} \delta _{h}=n_{h}^{s^\prime }-n_{h}^{s^{\prime \prime }}. \end{aligned}$$

It is clear that $\delta _{h}$ so defined is an integer (which may be positive or negative). Since $\sum _{h=i}^{k}n_{h}^{s^\prime }=\sum _{h=i}^{k}n_{h}^{s^{\prime \prime }}$, one can write:

$$\begin{aligned} \delta _{i}+\delta _{i+1}=-\sum _{h=i+2}^{k}\delta _{h}. \end{aligned}$$

(101)

Since, by (100), $\delta _{h}\ge 0$ for all $h \in \{i+2,\dots ,k\}$, one observes that $\delta _{i}+\delta _{i+1}\le 0$. We consider two cases.

Case 1: $\delta _{i}+\delta _{i+1}=0$. In that case, we conclude from (101) that $\sum _{h=i+2}^{k}\delta _{h}=0$ and, thanks to (100), that $n_{h}^{s^\prime }=n_{h}^{s^{\prime \prime }}$ for all $h\in \{i+2,\dots ,k\}$ . Hence, one has $n_{h}^{s^\prime }=n_{h}^{s^{\prime \prime }}$ for all $h=\{1,\dots ,i-1\}\cap \{i+2,\dots ,k\}$ and $\delta _{i}=n_{i}^{s^\prime }-n_{i}^{s^{\prime \prime }}=n_{i+1}^{s^{\prime \prime }}-n_{i+1}^{s^{\prime }}>0$. Hence, $s^{\prime \prime }$ can be obtained from $s^{\prime }$ by means of $\delta _{i}$ increments from i to $i+1$. We conclude that s first-order dominates $s^{\prime \prime }$, but also $s^{\prime \prime }$ first-order dominates $s^{\prime }$, which implies that s $H^{t}$-dominates $s^{\prime }$ for all $t\in \{0,1,\dots \}$.

Case 2: $\delta _{i}+\delta _{i+1}<0$. In that case, we deduce from (101) that there is an $h\in \{i+2,\dots ,k\}$ such that $\delta _{h}>0$ or, equivalently, that $n_{h}^{s^{\prime }}>n_{h}^{s^{\prime \prime }}=0$. From (98)–(100), one immediately observes that $s^{\prime \prime }$ can be obtained from $s^{\prime }$ by means of $\delta _{i}$ increments from category i to category $i+1$, and $(-\delta _{i+1})$ decrements ($\delta _{i+1}$ is a negative integer), from each category $h>i+1$ for which $n_{h}^{s^{\prime }}>0$ to category $i+1$. However, more decrements than increments are required ($(-\delta _{i+1})>\delta _{i}$, so that increments and decrements cannot be matched one by one to produce Hammond transfers—and only Hammond transfers—in order to obtain, on the initial grid ${\mathcal {C}}$, $s^{\prime \prime }$ from $s^{\prime }$. Yet, we can match the increments with the decrements if an appropriate refinement of the grid between i and $i+1\$ can be performed. First, staying on the initial grid ${\mathcal {C}}$, and starting from $s^{\prime }$, we can combine $(\delta _{i}-1)$ increments (from i to $i+1$) to the same number of decrements starting from one or several categories h above $i+1$ and bringing the agents from these categories to $i+1$ . This generates immediately $(\delta _{i}-1)$ Hammond transfers. In order to complete the move from $s^{\prime }$ to s by means of Hammond transfers, we need to match the last $[\delta _{i}-(\delta _{i}-1)]=1$ increment from i to $i+1$ with the remaining $[(-\delta _{i+1})-(\delta _{i}-1)]>1$ decrements that are required from each category $h>i+1$ where the number of agents remains strictly positive to the category $i+1$. Whatever is the number $[(-\delta _{i+1})-(\delta _{i}-1)]>1$, it is clearly possible to refine the grid ${\mathcal {C}}$ in such a way as to obtain at least $[(-\delta _{i+1})-(\delta _{i}-1)]$ adjacent categories between i and $i+1$. Once this refinement is obtained, one can then proceed in decomposing the last increment from i to $i+1$ into $[(-\delta _{i+1})-(\delta _{i}-1)]$ “small” increments between adjacent intermediate categories, each of which being matched with a decrement from each category $h>i+1$ for which there is a strictly positive number of agents. Hence, it is possible to achieve $s^{\prime \prime }$ from s by using Hammond transfers only (provided that a suitable refinement of the grid be performed). Hence, there exists a nonnegative integer t such that $s^{\prime \prime }$ can be obtained from $s^{\prime }$ by means of exactly $(-\delta _{i+1})$ Hammond transfers on the grid ${\mathcal {C}}(t)$ (recalling that a transformation on the grid ${\mathcal {C}}$ is also a transformation on the grid ${\mathcal {C}}(t)$). We then conclude that society s first-order dominates society $s^{\prime \prime }$ which in turn $H^{t}$-dominates society $s^{\prime }$ and this completes the proof.