The greatest unhappiness of the least number

Abstract

We propose an alternative articulation of the Benthamite greatest-happiness-of-the-greatest-number principle. With ordinally measurable and interpersonally non-comparable utilities, the rule chooses those feasible alternatives that maximize the number of individuals who end up with their greatest element. This rule is tantamount to the plurality rule. Furthermore, in the spirit of Rawls’s maximin principle, we propose the greatest-unhappiness-of-the-least-number principle. In analogy to the greatest-happiness principle, the least-unhappiness principle is formally equivalent to the anti-plurality rule. Our main result is a characterization of the least-unhappiness principle.

Appendix

In this appendix, we show that the axioms of our theorem are independent. For each of the six properties, we define a collective choice rule that violates this property and satisfies the remaining five.

Anonymity Define, for all $N \in \mathcal{N}$, for all $R_N \in \mathcal{R}_N$ and for all $S \in \mathcal{X}$,

$$\begin{aligned} f(R_N,S) = \left\{ \begin{array}{ll} f^U(R_1,S) &{}\quad \text{ if } |N| = \{1,2\} \text{ and } |S| = 2, \\ f^U(R_N,S) &{}\quad \text{ otherwise. } \end{array} \right. \end{aligned}$$

To see that this collective choice correspondence violates anonymity, consider the following example. Let $S=\{x,y\}$ with $x \ne y$ and let $N = \{1,2\}$. The preferences $R_1$ and $R_2$ restricted to S are $x R_1 y$ and $y R_2 x$. Let $\pi \in \Pi$ be such that $\pi (2)=3$, $\pi (3)=2$ and $\pi (i)=i$ for all $i \in \mathbb {N} {\setminus } \{2,3\}$. Thus, we have $N_\pi = \{1,3\}$, $R_{\pi (1)} = R_1$ and $R_{\pi (2)} = R_3$. We obtain $f(R_N,S) = f^U(R_1,S) = \{x\}$ and $f(R_{N_\pi },S) = f^U((R_1,R_3),S) = \{x,y\} \ne \{x\}$, a contradiction to anonymity.

Neutrality is satisfied because the labels attached to the alternatives are irrelevant.

To show that reinforcement is satisfied, suppose that $N,N' \in \mathcal{N}$, $R_N \in \mathcal{R}_N$, $R_{N'} \in \mathcal{R}_{N'}$ and $S \in \mathcal{X}$ are such that $N \cap N' = \emptyset$ and $f(R_N,S) \cap f(R_{N'},S) \ne \emptyset$.

If N and $N'$ both differ from $\{1,2\}$ or if $|S| \ne 2$, the conclusion follows immediately because $f^U$ satisfies reinforcement.

In all remaining cases, one of the sets N and $N'$ is equal to $\{1,2\}$, the other set has an empty intersection with $\{1,2\}$, and $|S|=2$. Without loss of generality, suppose that $N=\{1,2\}$, $N \cap N' = \emptyset$ and $S=\{x,y\}$. Furthermore, without loss of generality, suppose that $x R_1 y$ so that $f(R_N,\{x,y\}) = f^U(R_1,\{x,y\}) = \{x\}$. It follows that

$$\begin{aligned} \ell (R_N,\{x,y\},x) \le \ell (R_N,\{x,y\},y) \end{aligned}$$

(10)

because agent 1 considers x to be better than y and there are only two agents in N. Also, we must have $f(R_N,\{x,y\}) \cap f(R_{N'},\{x,y\}) = \{x\}$ because this intersection is non-empty. This, in turn, implies that $x \in f(R_{N'},\{x,y\})$ and thus

$$\begin{aligned} \ell (R_{N'},\{x,y\},x) \le \ell (R_{N'},\{x,y\},y) \end{aligned}$$

(11)

because $f(R_{N'},\{x,y\}) = f^U(R_{N'},\{x,y\})$ by definition of f. Combining (10) and (11) and using the assumption that the intersection of N and $N'$ is empty, it follows that

$$\begin{aligned} \ell \left( (R_N,R_{N'}),\{x,y\},x \right) \le \ell \left( (R_N,R_{N'}),\{x,y\},y \right) \end{aligned}$$

and, because $f \left( (R_N,R_{N'}),\{x,y\} \right) = f^U \left( (R_N,R_{N'}),\{x,y\} \right)$ by definition of f, we obtain

$$\begin{aligned} f\left( (R_N,R_{N'}),\{x,y\} \right)= & {} f^U \left( (R_N,R_{N'}),\{x,y\} \right) \\= & {} \{x\} = f(R_N,\{x,y\}) \cap f(R_{N'},\{x,y\}) \end{aligned}$$

as was to be proven.

Individual-equality independence is satisfied because all relevant comparisons according to f involve the respective comparisons according to $f^U$. Finally, single-agent monotonicity and single-agent expansion are satisfied because f coincides with $f^U$ for single-agent societies.

Neutrality Let $x^* \in X$ and define, for all $N \in \mathcal{N}$, for all $R_N \in \mathcal{R}_N$ and for all $S \in \mathcal{X}$,

$$\begin{aligned} f(R_N,S) = \left\{ \begin{array}{ll} \{x\} &{}\quad \text{ if } S = \{x^*,x\} \text{ for } \text{ some } x \in X {\setminus } \{x^*\} \text{ and } \\ &{}\quad |N| \ge 2 \text{ and } \ell (R_N,\{x^*,x\},x^*) = \ell (R_N,\{x^*,x\},x) > 0, \\ \\ f^U(R_N,S) &{}\quad \text{ otherwise. } \end{array} \right. \end{aligned}$$

To see that this correspondence does not satisfy neutrality, let $N = \{1,2\}$, $S = \{x^*,x\}$ for some $x \in X {\setminus } \{x^*\}$, $x^* R_1 x$ and $x R_2 x^*$. Furthermore, let $y \in X {\setminus } \{x^*,x\}$ and consider the permutation $\sigma \in \Sigma$ such that $\sigma (x^*)=y$, $\sigma (y)=x^*$ and $\sigma (z)=z$ for all $z \in X {\setminus } \{x^*,x\}$. By definition, we have

$$\begin{aligned} y = \sigma (x^*) \, \sigma _1(R_1) \, \sigma (x) = x \; \; \text{ and } \; \; x = \sigma (x) \, \sigma _2(R_2) \, \sigma (y) = x^*. \end{aligned}$$

It follows that $f(R_N,\{x^*,x\}) = \{x\}$ and $f(R_N,\{x,y\}) = f^U(R_N,\{x,y\}) = \{x,y\}$, in contradiction to neutrality.

Anonymity is satisfied because the identities of the individuals are irrelevant in the definition of this collective choice correspondence.

To prove that f satisfies reinforcement, note first that we only have to consider cases in which $S = \{x^*,x\}$ for some $x \in X {\setminus } \{x^*\}$; in all other cases, only the second branch of the above definition applies and the conclusion of reinforcement follows because the axiom is satisfied by $f^U$. Moreover, we can restrict attention to cases in which the first branch of the definition applies for N or for $N'$. So suppose that $S = \{x^*,x\}$ for some $x \in X {\setminus } \{x^*\}$ and, moreover, that $N,N' \in \mathcal{N}$, $R_N \in \mathcal{R}_N$ and $R_{N'} \in \mathcal{R}_{N'}$ are such that $N \cap N' = \emptyset$ and $f(R_N,S) \cap f(R_{N'},S) \ne \emptyset$. Up to exchanging the roles of N and $N'$, we have the following two cases left to examine.

Case 1.

In this case, it follows that

$$\begin{aligned} f(R_N,\{x^*,x\}) = f(R_{N'},\{x^*,x\}) = f(R_N,\{x^*,x\}) \cap f(R_{N'},\{x^*,x\}) = \{x\} \end{aligned}$$

by definition of f. Because $\ell (R_N,\{x^*,x\},x^*) = \ell (R_N,\{x^*,x\},x) > 0$ and $\ell (R_{N'},\{x^*,x\},x^*) = \ell (R_{N'},\{x^*,x\},x) > 0$, it follows that

(recall that $N \cap N' = \emptyset$). Thus, because we also have $|N \cup N'| \ge 2$, the first branch of the definition of f applies and we obtain

$$\begin{aligned} f \left( (R_N, R_{N'}),\{x^*,x\},x^* \right) = \{x\} = f(R_N,\{x^*,x\}) \cap f(R_{N'},\{x^*,x\}). \end{aligned}$$

Case 2.

$$\begin{aligned}&|N| \ge 2 \; \text{ and } \; \ell (R_N,\{x^*,x\},x^*) = \ell (R_N,\{x^*,x\},x) > 0 \; \; \; \text{ and } \\&|N'| = 1 \; \text{ or } \; \ell (R_{N'},\{x^*,x\},x^*) \ne \ell (R_{N'},\{x^*,x\},x) \; \text{ or } \; \ell (R_{N'},\{x^*,x\},x^*) = 0. \end{aligned}$$

We now obtain

$$\begin{aligned} f(R_N,\{x^*,x\}) = \{x\} = f(R_N,\{x^*,x\}) \cap f(R_{N'},\{x^*,x\}) = \{x\} \end{aligned}$$

by definition of f and because the above intersection is assumed to be non-empty. Thus, $x \in f(R_{N'},\{x^*,x\})$ and it follows that $\ell (R_{N'},\{x^*,x\},x) \le \ell (R_{N'},\{x^*,x\},x^*)$. Hence,

If $\ell (R_{N'},\{x^*,x\},x) = \ell (R_{N'},\{x^*,x\},x^*)$, it follows that

$$\begin{aligned} f \left( (R_N, R_{N'}),\{x^*,x\} \right) = \{x\} = f(R_N,\{x^*,x\}) \cap f(R_{N'},\{x^*,x\}) \end{aligned}$$

irrespective of whether the first or the second branch applies to $(R_N,R_{N'})$.

If $\ell (R_{N'},\{x^*,x\},x) < \ell (R_{N'},\{x^*,x\},x^*)$, it follows that the first branch of the definition of f applies and, as in the previous case, we obtain

$$\begin{aligned} f \left( (R_N, R_{N'}),\{x^*,x\} \right) = \{x\} = f(R_N,\{x^*,x\}) \cap f(R_{N'},\{x^*,x\}). \end{aligned}$$

Individual-equality independence is satisfied because the axiom only imposes restrictions when there are at least three alternatives and, thus, f coincides with $f^U$ in all relevant cases. Single-agent monotonicity and single-agent expansion are satisfied because f coincides with $f^U$ whenever only a single agent is present.

Reinforcement Define, for all $N \in \mathcal{N}$, for all $R_N \in \mathcal{R}_N$ and for all $S \in \mathcal{X}$,

$$\begin{aligned} f(R_N,S) = \left\{ \begin{array}{ll} f^U(R_N,S) &{}\quad \text{ if } |N|=1 \text{ or } |S|=1, \\ \cup _{i \in N} f^U(R_i,S) &{}\quad \text{ otherwise. } \end{array} \right. \end{aligned}$$

This correspondence does not satisfy reinforcement. Consider the following example. Let $S = \{x,y,z\}$, $N = \{1\}$, $N' = \{2\}$, $x R_1 y R_1 z$ and $x R_2 z R_2 y$. By definition of f, we obtain

$$\begin{aligned} f(R_N,S) = f(R_1,S) = f^U(R_1,S) = \{x,y\} \end{aligned}$$

and

$$\begin{aligned} f(R_{N'},S) = f(R_2,S) = f^U(R_{N'},S) = \{x,z\}. \end{aligned}$$

Thus, $N \cap N' = \emptyset$ and $f(R_N,S) \cap f(R_{N'},S) = \{x\} \ne \emptyset$. By definition of f, we obtain

$$\begin{aligned} f \left( (R_N,R_{N'}),S \right)= & {} f \left( (R_1,R_2),S \right) = f(R_1,S) \cup f(R_2,S) \\= & {} \{x,y,z\} \ne \{x\} = f(R_N,S) \cap f(R_{N'},S), \end{aligned}$$

contradicting reinforcement.

That anonymity and neutrality are satisfied follows because neither the identities of the individuals nor the labels of the alternatives are relevant in the definition of f. Individual-equality independence is satisfied because the order of non-worst alternatives in the profiles is irrelevant. Single-agent monotonicity and single-agent expansion are satisfied because f coincides with $f^U$ in the single-agent case.

Individual-equality independence For our next example, some further definitions are required. For all $N \in \mathcal{N}$, for all $R_N \in \mathcal{R}_N$ and for all $S \in \mathcal{X}$, let

$$\begin{aligned} L^0(R_N,S) = S \end{aligned}$$

and define the following |S| sets recursively. For all $j \in \{1,\ldots ,|S|\}$, let

$$\begin{aligned} L^j(R_N,S)= & {} \{ x \in L^{j-1}(R_N,S) \mid \ell (R_N,L^{j-1}(R_N,S),x) \le \ell (R_N,L^{j-1}(R_N,S),y) \\&\quad \text{ for } \text{ all } y \in L^{j-1}(R_N,S)\}. \end{aligned}$$

That is, $L^1(R_N,S)$ contains the alternatives that minimize the number of agents i who end up with their worst alternative in S according to $R_i$. Furthermore, if $|S| > 1$, $L^j(R_N,S)$ identifies the set of alternatives that minimize the number of agents i who end up with their j-worst alternatives in $L^{j-1}(R_N,S)$ according to $R_i$ for all $j \in \{2,\ldots ,|S|\}$. Note that, in this iteration, alternatives that do not appear as minimizers at some stage are eliminated when moving to the following stage. Analogously, for all $N \in \mathcal{N}$, for all $R_N \in \mathcal{R}_N$, for all $S \in \mathcal{X}$, for all $z \in S$ and for all $j \in \{1,\ldots ,|S|\}$, define

$$\begin{aligned} \ell ^j(R_N,S,z) = |\{ i \in N \mid y R_i z \text{ for } \text{ all } y \in L^{j-1}(R_N,S) \}|. \end{aligned}$$

For future reference, note that, for all $N,N' \in \mathcal{N}$, for all $R_N \in \mathcal{R}_N$, for all $R_{N'} \in \mathcal{R}_{N'}$, for all $S \in \mathcal{X}$ such that $N \cap N' = \emptyset$ and $f(R_N,S) \cap f(R_{N'},S) \ne \emptyset$, and for all $j \in \{1,\ldots ,|S|\}$,

$$\begin{aligned} \ell ^j \left( (R_N,R_{N'}),S,z \right) = \ell ^j(R_N,S,z) + \ell ^j(R_{N'},S,z) \end{aligned}$$

(12)

for all $z \in S$ because $N \cap N' = \emptyset$.

To define the collective choice correspondence f relevant here, we proceed as follows.

• (a) If there exists $j \in \{1,\ldots ,|S|\}$ such that $|L^j(R_N,S)|=1$, let

  $$\begin{aligned} f(R_N,S)=L^{j^*}(R_N,S), \text{ where } j^* = \min \{j \in \{1,\ldots ,|S|\} \mid |L^j(R_N,S)|=1\}. \end{aligned}$$

• (b) If $|L^j(R_N,S)|>1$ for all $j \in \{1,\ldots ,|S|\}$, let

  $$\begin{aligned} f(R_N,S)=L^{|S|}(R_N,S). \end{aligned}$$

This correspondence f violates individual-equality independence. Suppose $S = \{x,y,z\}$, $N = \{1,2,3\}$,

$$\begin{aligned}&x R_1 y R_1 z,&\\&z R_2 x R_2 y,&\\&y R_3 z R_3 x,&\end{aligned}$$

and $y R'_1 x R'_1 z$. By definition, $f(R'_1,S) = f(R_1,S) = \{x,y\}$. Furthermore, we have

$$\begin{aligned} L^1 \left( (R_1,R_2,R_3),S \right) = L^2 \left( (R_1,R_2,R_3),S \right) = L^3 \left( (R_1,R_2,R_3),S \right) = \{x,y,z\} \end{aligned}$$

and, thus,

$$\begin{aligned} f \left( (R_1,R_2,R_3),S \right) = L^3 \left( (R_1,R_2,R_3),S \right) = \{x,y,z\} \end{aligned}$$

by definition of f. Furthermore,

$$\begin{aligned} L^1 \left( (R'_1,R_2,R_3),S \right) = \{x,y,z\} \; \text{ and } \; L^2 \left( (R'_1,R_2,R_3),S \right) = \{y\}. \end{aligned}$$

Thus, $|L^2 \left( (R'_1,R_2,R_3),S \right) | = 1$ and it follows that

$$\begin{aligned} f \left( (R'_1,R_2,R_3),S \right) = \{y\} \ne \{x,y,z\} = f \left( (R_1,R_2,R_3),S \right) , \end{aligned}$$

contradicting individual-equality independence.

That anonymity and neutrality are satisfied follows because neither the identities of the individuals nor the labels of the alternatives are relevant in the definition of f.

To prove that reinforcement is satisfied, suppose that $N,N' \in \mathcal{N}$, $R_N \in \mathcal{R}_N$, $R_{N'} \in \mathcal{R}_{N'}$ and $S \in \mathcal{X}$ are such that $N \cap N' = \emptyset$ and $f(R_N,S) \cap f(R_{N'},S) \ne \emptyset$. Up to exchanging the roles of N and $N'$, the following three cases exhaust all possibilities.

Case 1. There exist $j,j' \in \{1,\ldots ,|S|\}$ such that $|L^j(R_N,S)| = 1$ and $|L^{j'}(R_{N'},S)| = 1$. By definition of f, it follows that $f(R_N,S)$ and $f(R_{N'},S)$ are singletons and, because the intersection of these two sets is non-empty, each of the two sets must have some $x \in S$ as its only member. Let

$$\begin{aligned} j^* = \min \{j \in \{1,\ldots ,|S|\} \mid |L^j(R_N,S)| = 1\} \end{aligned}$$

and

$$\begin{aligned} j'^* = \min \{j' \in \{1,\ldots ,|S|\} \mid |L^{j'}(R_{N'},S)| = 1\}. \end{aligned}$$

Without loss of generality, suppose that $j^* \le j'^*$. It follows that

$$\begin{aligned} f(R_N,S)= & {} L^{j^*}(R_{N},S) = L^{j'^*}(R_{N'},S) = f(R_{N'},S)\\= & {} f(R_N,S) \cap f(R_{N'},S) = \{x\}. \end{aligned}$$

By definition of f, we have

$$\begin{aligned} \ell ^j(R_N,S,x)\le & {} \ell ^j(R_N,S,y) \; \; \;\,\hbox {for all } y \in L^{j-1}(R_N,S),\nonumber \\ \ell ^j(R_{N'},S,x)\le & {} \ell ^j(R_{N'},S,y) \;\;\;\hbox {for all } y \in L^{j-1}(R_{N'},S) \end{aligned}$$

(13)

for all $j \in \{1,\ldots ,j^*\} {\setminus } \{j^*\}$, and

$$\begin{aligned} \ell ^{j^*}(R_N,S,x)< & {} \ell ^{j^*}(R_N,S,y) \; \; \;\,\hbox { for all } y \in L^{j^*-1}(R_N,S) {\setminus } \{x\}, \\ \ell ^{j^*}(R_{N'},S,x)\le & {} \ell ^{j^*}(R_{N'},S,y) \; \; \;\hbox { for all } y \in L^{j^*-1}(R_{N'},S). \end{aligned}$$

Thus, using (12), we obtain

$$\begin{aligned} \ell ^j \left( (R_N,R_{N'}),S,x \right)= & {} \ell ^j(R_N,S,x) + \ell ^j(R_{N'},S,x), \\\le & {} \ell ^j(R_N,S,y) + \ell ^j(R_{N'},S,y) \\= & {} \ell ^j \left( (R_N,R_{N'}),S,y \right) \end{aligned}$$

for all $j \in \{1,\ldots ,j^*\} {\setminus } \{j^*\}$ and for all $y \in \left[ L^{j-1}(R_N,S) \cup L^{j-1}(R_{N'},S) \right] {\setminus } \{x\}$ and, moreover,

$$\begin{aligned} \ell ^{j^*} \left( (R_N,R_{N'}),S,x \right)= & {} \ell ^{j^*}(R_N,S,x) + \ell ^{j^*}(R_{N'},S,x), \nonumber \\< & {} \ell ^{j^*}(R_N,S,y) + \ell ^{j^*}(R_{N'},S,y)\nonumber \\= & {} \ell ^{j^*} \left( (R_N,R_{N'}),S,y \right) \end{aligned}$$

(14)

for all $y \in \left[ L^{j^*-1}(R_N,S) \cup L^{j^*}(R_{N'},S) \right] {\setminus } \{x\}$. This implies that $x \in f \left( (R_N,R_{N'}),S \right)$ by definition. Furthermore, because of the strict inequality in (14), no other element of S can be chosen; see also the argument employed in the ‘if’ part of the proof of Theorem 2. Thus,

$$\begin{aligned} f \left( (R_N,R_{N'}),S \right) = \{x\} = f(R_N,S) \cap f(R_{N'},S) \end{aligned}$$

as required.

Case 2. There exists $j \in \{1,\ldots ,|S|\}$ such that $|L^j(R_N,S)| = 1$ and $|L^{j'}(R_{N'},S)| > 1$ for all $j' \in \{1,\ldots ,|S|\}$. By definition of f, it follows that there exists $x \in S$ such that $f(R_N,S) = L^{j^*}(R_N,S) = x$, where

$$\begin{aligned} j^* = \min \{j \in \{1,\ldots ,|S|\} \mid |L^j(R_N,S)| = 1\}. \end{aligned}$$

Moreover, $f(R_{N'},S) = L^{|S|}(R_{N'},S)$ and, because $f(R_N,S) \cap f(R_{N'},S) \ne \emptyset$, we have $x \in f(R_{N'},S)$. Starting with (13), the remainder of the proof is identical to that of case 1.

Case 3. $|L^j(R_N,S)| > 1$ for all $j \in \{1,\ldots ,|S|\}$ and $|L^{j'}(R_{N'},S)| > 1$ for all $j' \in \{1,\ldots ,|S|\}$. By definition of f, $f(R_N,S) = L^{|S|}(R_N,S)$ and $f(R_{N'},S) = L^{|S|}(R_{N'},S)$. By assumption,

$$\begin{aligned} f(R_N,S) \cap f(R_{N'},S) \ne \emptyset . \end{aligned}$$

(15)

Now a slightly modified version of the argument employed in case 1 can be used; the only change required is that if the intersection in (15) contains more than one element, the strict-inequality argument applies to all members of the intersection versus all alternatives that are not in this intersection. See again the proof of Theorem 2 for details.

Single-agent monotonicity and single-agent expansion are satisfied because f coincides with $f^U$ in the single-agent case.

Single-agent monotonicity Let, for all $N \in \mathcal{N}$, for all $R_N \in \mathcal{R}_N$ and for all $S \in \mathcal{X}$,

$$\begin{aligned} f(R_N,S) = S. \end{aligned}$$

This collective choice correspondence does not satisfy single-agent monotonicity. Let $x,y \in X$ be distinct alternatives such that $x R_1 y$. By definition, we have

$$\begin{aligned} f(R_1,\{x,y\}) = \{x,y\} \ne \{x\}. \end{aligned}$$

That all other axioms are satisfied is immediate.

Single-agent expansion The correspondence $f = f^H$ does not satisfy single-agent expansion. Let $x,y,z \in X$ be distinct, let $S = \{y,z\}$ and suppose that $x R_1 y$ and $x R_1 z$. By definition, we have $f(R_1,S) = \{y\}$ and

$$\begin{aligned} f(R_1,S \cup \{x\}) = \{x\} \ne \{x,y\} = f(R_1,S) \cup \{x\}. \end{aligned}$$

That f satisfies anonymity, neutrality and reinforcement follows from Ching’s (1996) theorem (stated as Theorem 1 in the present paper).

To see that individual-equality independence is satisfied, apply the argument employed in the ‘if’-part of the proof of Theorem 2, replacing $\ell$ with g and $\le$ with $\ge$.

That single-agent monotonicity is satisfied is straightforward to verify.