Collective choice in Aristotle

“At its most general and fundamental level Aristotle’s analysis of the polis is a highly abstract exercise in rational choice theory.” (Schofield 2005, p. 318).

Abstract

In his Politics VI 3, 1318a–b, Aristotle discusses constitutional procedures for achieving justice in a society where its classes have different views on it. He analyzes the case of a society consisting in two groups, the poor and the rich, each holding a specific understanding of justice (democratic or oligarchic). In this paper we give, first, a non-formal summary of this section of Politics. Then we approach it in the framework of social choice theory and argue that a social rule for selecting between alternatives may be extracted from it. As Aristotle argued, this rule is consistent with the views on justice and equality of the supporters of both democracy and oligarchy. Finally, we study its properties, as well as some extensions of it when multiple classes are allowed or more than two alternatives are present.

Appendices

Appendix 1

The appendix includes the proofs of the Lemmas 1, 2 and 4 and the Theorems 1 and 2.

Lemma 1

The social choice function w is representative.

Proof

Starting with the group G = {j₁, j₂,… j_m} we shall construct another group G* with the property that w(p_G) = μ(p_G*). To do this, replace in G each member j_i of it with a group G_i defined as follows. Suppose that W(j_i) = a_i. Then G_i = $\{ \{ j_{i} \} ,\{ \{ j_{i} \} \} , \ldots \underbrace {{\{ \ldots \{ }}_{\substack{ a_{i} \\ times } }j_{i} \underbrace {{\} \ldots \} }}_{\substack{ a_{i} \\ times } }\}$. So, j_i is replaced by a set consisting in a number of a_i sets, and for each of them μ gives the same value as R(j_i). Now put G* = $\cup_{i = 1}^{m} G_{i}$.We can immediately check that w(p_G) = μ(p_G*). □

Lemma 2

Ar is representative.

Proof

The proof consists in defining a (higher-order) society H with the property that for each profile p we have: Ar(p_G) = μ(p_H) (here we assume that H is such that all the individuals occurring in it are members of G). To do this, let

$$H = \left\{ {\left\{ {X,Y} \right\},\left\{ {\left\{ {X,Y} \right\}} \right\},G*} \right\}$$

where, as we saw above, G* is a group with the property that w(p_G) = μ(p_G*). The group H includes two occurrences of the society {X, Y}. To differentiate them, the latter occurrence was constructed as the singleton consisting in this very society.

It can be proved that for each profile p we have:

$$Ar\left( {p_{G} } \right) =\upmu\left( {p_{H} } \right)$$

We shall consider three cases:

• Case 1: Ar(p_G) = 1. By definition, we have three subcases:

  • Subcase 1a: μ(p_X) = 1 and μ(p_Y) ≥ 0. We have: μ(p_{{{X, Y}, {{X, Y}}, G*}}) = μ(μ(μ(p_X), μ(p_Y)), μ(μ(μ(p_X), μ(μ(p_Y))), μ(p_G*)) = μ(μ(μ(1), μ(1/0)), μ(μ(μ(1), μ(μ(1/0))), μ(p_G*)) = μ(μ(1, 1/0), μ(μ(1), μ(1/0)), μ(p_G*)) = μ(μ(1, 1/0), μ(1, 1/0), μ(p_G*)) = μ(1, 1, μ(p_G*)) = 1.

  • Subcase 1b: μ(p_X) ≥ 0 and μ(p_Y) = 1: it is similar to subcase 1a.

  • Subcase 1c: μ(p_X) = − μ(p_Y) and w(p_G) = 1, i.e. μ(p_G*) = 1. Then we have both μ(μ(p_X), μ(p_Y)) = 0 and μ(μ(μ(p_X), μ(μ(p_Y))), therefore μ(p_{{{X, Y}, {{X, Y}}, G*}}) = μ(0, 0, μ(p_G*)) = μ(p_G*). But μ(p_G*) = 1 and so μ(p_{{{X, Y}, {{X, Y}}, G*}}) = μ(0, 0, 1) = 1.

• Case 2: Ar(p_G) = − 1. This is analogous to case 1.

• Case 3. Ar(p_G) = 0. Just like in case 1c, we get μ(p_{{{X, Y}, {{X, Y}}, G*}}) = μ(0, 0, 0) = 0. □

Lemma 4

$Ar_{\upmu}^{[3]}$ is representative.

The proof consists simply in constructing the society H³ = {{X, M, Y}, {{X, M, Y}}, G*}. Clearly, we have $Ar_{\upmu}^{[3]}$(p_G) = μ($p_{{H^{3} }}$).

Theorem 1

Ar satisfiesMon,NandMR.

Proof

I shall sketch the proofs for N and Mon, and then focus on the proof that Ar satisfies MR.²³ For N, we need to show that μ($p_{H}^{'}$) = − μ(p_H). Suppose that all the members of the group G have reversed choices, i.e. at profile $p_{G}^{'}$ it holds that R′(j) = –R(j) for all j ∈ G. Observe that μ satisfies N (May 1952) for all first-order societies like X and Y. So μ($p_{X}^{'}$) = − μ(p_X) and μ($p_{Y}^{'}$) = − μ(p_Y). Thus μ($p_{{\{ X,Y\} }}^{'}$) = μ(μ($p_{X}^{'}$), μ($p_{Y}^{'}$)) = μ(− μ(p_X), − μ(p_Y)) = − μ(μ(p_X), μ(p_Y)) = − μ(p_{{X, Y}}). Now μ(p_{X}) = μ(μ(p_X)). So μ($p_{{\{ \{ X,Y\} \} }}^{'}$) = μ(μ($p_{{\{ X,Y\} }}^{'}$)) = μ(− μ(p_{{X, Y}})) = − μ(μ(p_{{X, Y}})). This result can also be used to show that μ($p_{G*}^{'}$) = − μ(p_G*). Finally, μ($p_{H}^{'}$) = μ(μ($p_{{\{ X,Y\} }}^{'}$),μ($p_{{\{ \{ X,Y\} \} }}^{'}$),μ($p_{G*}^{'}$)) = μ(− μ(p_{{X, Y}}), − μ(μ(p_{{X, Y}})), − μ(p_G*)) = − μ(p_H). The proof for Mon also uses the fact that μ satisfies it for first-order societies; it will be omitted here.

Moving to MR, notice first that if Ar(p_G) = 1 and R′(j) > R(j) for some individual j in G, then by the monotonicity of Ar we get Ar($p_{G}^{'}$) = 1. So let Ar(p_G) = 0 and R′(j) > R(j) for some individual j in G. By definition, Ar(p_G) = 0 when the following conditions hold: μ(p_X) = − μ(p_Y) and sgn ($\sum\limits_{i = 1}^{m} {W(j_{i} )R(j_{i} )}$) = 0. We have three cases:

• Case 1: μ(p_X) = − μ(p_Y) = 0. Then, since µ satisfies MR for all first-order groups, we get µ($p_{X}^{'}$) = 1 if j ∊ X or µ($p_{Y}^{'}$) = 1 if j ∊ Y. In each case µ($p_{{\{ X,Y\} }}^{'}$) = 1 and so µ($p_{H}^{'}$) = 1.

• Case 2: μ(p_X) = − μ(p_Y) and μ(p_X) = 1 and μ(p_Y) = − 1. Then, if j ∊ X, by monotonicity we have µ($p_{X}^{'}$) = 1 and also µ($p_{Y}^{'}$) = − 1. So µ($p_{X}^{'}$) = − µ($p_{Y}^{'}$). But then in the second step we get $\sum\nolimits_{i = 1}^{m} {W^{\prime}(j_{i} )R^{\prime}(j_{i} )}$ > 0 and so sgn ($\sum\nolimits_{i = 1}^{m} {W^{\prime}(j_{i} )R^{\prime}(j_{i} )}$) = 1. Therefore µ($p_{G*}^{'}$) = 1 which entails that µ($p_{H}^{'}$) = 1. If j ∊ Y, we have two subcases.

  • Subcase 2a: μ($p_{Y}^{'}$) = 0. Then µ($p_{{\{ X,Y\} }}^{'}$) = 1 and so µ($p_{H}^{'}$) = 1.

  • Subcase 2b: μ($p_{Y}^{'}$) = − 1. Then µ($p_{{\{ X,Y\} }}^{'}$) = 0. But as above we can show that µ($p_{G*}^{'}$) = 1 and thus µ($p_{H}^{'}$) = 1.

• Case 3: μ(p_X) = − μ(p_Y) and μ(p_X) = − 1 and μ(p_Y) = 1. It is similar to Case 2. □

Theorem 2

Let two profiles p_G = (G, W, n, R) and $p_{G}^{'}$ = (G, W′, n, R) differ only in that there is some individual j ∊ G such that W′(j) > W(j). If Ar(p_G) = 0 and j is not pivotal for Y, then Ar($p_{G}^{'}$) = R(j).

Proof

Two cases can be distinguished.

• Case 1: j ∊ X, i.e. j is rich. Then j ∊ X at $p_{G}^{'}$: if the individual j was rich at p_G, then clearly he is also rich at $p_{G}^{'}$. Since the individual choice function are the same in the two profiles we have both μ(p_X) = μ($p_{X}^{'}$) and μ(p_Y) = μ($p_{Y}^{'}$). Since f(p_G) = 0, we have μ(p_X) = − μ(p_Y) and consequently we get μ($p_{X}^{'}$) = − μ($p_{Y}^{'}$). So μ($p_{{\{ X,Y\} }}^{'}$) = 0. Moreover, f(p_G) = 0 entails that $\sum\nolimits_{i = 1}^{m} {W(j_{i} )R(j_{i} )}$ = 0. But, since W′(j) > W(j) we get $\sum\nolimits_{i = 1}^{m} {W^{\prime}(j_{i} )R(j_{i} )}$ = $\sum\nolimits_{i = 1}^{m} {W(j_{i} )R(j_{i} )}$ + (W′(j) − W(j))R(j) and so μ($p_{G*}^{'}$) = sgn($\sum\nolimits_{i = 1}^{m} {W^{\prime}(j_{i} )R(j_{i} )}$) = sgn((W′(j) − W(j))R(j)) = R(j). So μ($p_{H}^{'}$) = R(j).

• Case 2. j ∊ Y, i.e. j is poor. We have two subcases.

  • Subcase 2a: j ∊ Y at $p_{G}^{'}$. Although his property increases, j remains poor. The proof is just like in case 1.

  • Subcase 2b: j ∊ X at $p_{G}^{'}$. This means that the property of j increases enough to move j from the class of the poor to the class of the rich. Let R(j) = 1 (the same argument works for R(j) = − 1, therefore it will be omitted). Suppose that μ(p_Y) = − 1. Then clearly μ($p_{{Y - \{ j\} }}^{'}$) = − 1 and μ($p_{{X \cup \{ j\} }}^{'}$) = 1. So μ($p_{{\{ X,Y\} }}^{'}$) = 0. But, as argued in Case 1, μ($p_{G*}^{'}$) = R(j) = 1 and so μ($p_{H}^{'}$) = 1. Secondly, let μ(p_Y) = 0 and therefore μ(p_X) = 0. Then of course μ($p_{{Y - \{ j\} }}^{'}$) = − 1 and μ($p_{{X \cup \{ j\} }}^{'}$) = 1, which again entails that μ($p_{{\{ X,Y\} }}^{'}$) = 0 and, as above, μ($p_{G*}^{'}$) = R(j) = 1; consequently, μ($p_{H}^{'}$) = 1. Third, suppose that μ(p_Y) = 1 and therefore μ(p_X) = − 1. Since j is not pivotal for Y, we have μ(p_{Y – {j}}) = 1. But given that μ(p_X) = − 1 and by supposition R(j) = 1, it follows that μ(p_{X ∪ {j}}) ≤ 0.Then μ($p_{{\{ X,Y\} }}^{'}$) ≥ 0. If μ($p_{{\{ X,Y\} }}^{'}$) = 1, then immediately μ($p_{G}^{'}$) = 1; if μ($p_{{\{ X,Y\} }}^{'}$) = 0, then apply again the above argument to get μ($p_{G*}^{'}$) = 1 and thus μ($p_{H}^{'}$) = 1 = R(j). Finally, note that if R(j) = 0 then i’s vote does not influence the social decision and so we have f($p_{G}^{'}$) = f(p_G) = 0 = R(j). □

Appendix 2

When more than two alternatives are available, the rule Ar may of course produce undesirable results. Social circles are the most obvious example. But it also induces other paradoxical results. One of them is a reversal effect. We shall present below one example in this sense. Clearly, it is a consequence of the fact that Ar does not satisfy the independence property.

Suppose there are three alternatives x, y and z. Again, let the group G consist in 30 members, of whom 20 are poor and 10 are rich. Let 2 poor citizens rank the alternatives in the order zyx, and the remaining 18 in the order yzx. Then all 20 poor citizens prefer y to x and z to x, while 18 prefer y to z. By the majority rule, the social order is yzx. The preferences of the rich are a bit more diversified: 1 has the order xyz, 5 have xzy, and 4 have zxy. Then all 10 citizens prefer x to y, 4 prefer x to z and 1 prefers y to z. By the majority rule, the social order is xzy.

The two classes have opposing choices: the alternative preferred by the rich is y, while the alternative preferred by the poor is x. The alternative z is irrelevant. We need to move to the second step to appeal to the weighted function in order to reach a social decision. Again, let n = 1, and suppose that W(j) = 3 for all j ∊ X and W(j) = 1 for all j ∊ Y. This means that in the second step each poor citizen has 1 vote, and each rich citizen has 3 votes. Given the preferences we already established, x gets 30 votes when compared with y, while z gets 38 votes when compared with x and 29 votes when compared with y. Since the total number of votes is 50, it follows that the social order we get in this second step is zxy.

But this is counterintuitive, because in the first step neither the poor no the rich preferred alternative z to the other two alternatives. The rule allows for a reversal in social preference in the move from the first to the second step.