Social Choice. Jan-Michael van Linthoudt

Transcription

1 Social Choice Jan-Michael van Linthoudt Summer term 2017 Version: March 15, 2018

2 CONTENTS Remarks 1 0 Introduction 2 1 The Case of 2 Alternatives Examples for social choice rules Properties of social choice rules Universal Domain (UD) Weak Pareto (WP) Dictatorial (D) Non-Dictatorial (ND) Anonymity (AN) Neutrality (N) Positive Responsiveness (PR) More than 2 Alternatives Extension of simple majority voting to the case of more than 2 alternatives Condorcet Winner Method Staging Procedure Copeland Rule Plurality voting The Arrow Impossible Theorem Social Welfare Function I

3 CONTENTS Possible Properties of SWF Social Decision Function Liberalism (L) A possibility Result II

4 REMARKS First of all some clarifications on notation: preference refers to strict preference, whereas weak preference is always called as such for shorter notation a b c will be used for a b b c to make the statements easier to read dom(f) stands for the domain of a map/function f x y for two vectors x, y R n means that x i y i i = 1,..., n and x i > y i for some i Feedback In case you have comments or feedback please send them to wiwi-tutorial@outlook.com. This includes errors which are definitely still hiding in this document or comments on better wording/explanation/... Any comment helps to improve these lecture notes and is highly appreciated. 1

5 CHAPTER 0 INTRODUCTION In standard microeconomic theory the decision making unit is always an individual person or firm. Social Choice however is about how groups of such decision makers choose among a set of alternatives. Examples would be how a family chooses from elements of a family budget set, a directory board of a shareholding company from a set of financial policies or society from various allocations (e.g. level of inequality). 2

6 CHAPTER 1 THE CASE OF 2 ALTERNATIVES Let X = {x, y} be the set of alternatives (x y). Let I = {1, 2,..., n} be the set of individuals, so n is the number of individuals making up the group (n 2). We describe preferences of individual i I over X by a real number α i {1, 1, 0}, with the following interpretation: α i = 1 means i prefers x over y α i = 1 α i = 0 means i prefers y over x means i is indifferent between x and y Definition (Preference Profile). A profile of preferences is a list α = (α 1, α 2,..., α n ) {1, 1, 0} I. Thus {1, 1, 0} I is the set of all preference profiles, denoted by P. Definition (Social Choice Rule). A social choice rule is a map/function f with dom(f) P and it takes values in {1, 1, 0} with the following interpretation: f(α) = 1 then x is socially preferred to y (x wins ) f(α) = 1 then y is socially preferred to x (y wins ) f(α) = 0 then x and y are socially indifferent ( tie between x and y) Notation: given α P : n + (α) = #{i I : α i = 1} n (α) = #{i I : α i = 1} n 0 (α) = #{i I : α i = 0} 1.1 Examples for social choice rules (1) simple majority voting: 1 if n + (α) > n (α) f(α) = 1 if n + (α) < n (α) 0 if n + (α) = n (α) which is equivalent to ( ) f(α) = sgn α i i I (2) weighted majority voting: let β = (β 1, β 2,..., β n ) be a vector of weights with β i 0 i I ( ) f(α) = sgn β i α i = sgn(β α) i I 3

7 Chapter 1. The Case of 2 Alternatives If β 0 one could normalize (i.e. i I β i = 1) by taking (3) absolute majority voting: 1 if n + (α) > n 2 f(α) = 1 if n (α) > n 2 0 otherwise β i i I βi instead of β i. (4) constant rules: e.g. traditional code f + (α) = 1 α P f (α) = 1 α P f 0 (α) = 0 α P Remark Simple majority voting is a special case of weighted majority voting, each individual getting the same positive weight Simple and absolute majority voting differ. Example: n = 4, α = (1, 1, 1, 0). Then for simple majority voting: f(α) = 1 whereas for absolute majority voting: f(α) = Properties of social choice rules Universal Domain (UD) Definition. The rule applies to all α P, i.e. domf = P. Remark All examples from above satisfy UD. An example which does not fulfill UD would be: which is not defined at α = (0,..., 0). 1 if n + (α) > n 2 f(α) = 1 0 if n (α) > n 2 if n + (α) = n (α) = n 2 (1.1) Weak Pareto (WP) Definition. Unanimity of strict preferences is respected by the rule, i.e. f ( (1,..., 1) ) = 1 as well as f ( ( 1,..., 1) ) = 1 Remark WP is satisfied by simple and absolute majority voting, also by weighted majority voting if β 0, but not satisfied by constant rules Dictatorial (D) Definition. There exists an individual i d I, called dictator, for which f(α) = 1 in case α id = 1 and f(α) = 1 in case α id = 1 4

8 Chapter 1. The Case of 2 Alternatives Examples for a dictatorial social choice rules with dictator i d are: just what the dictator chooses f(α) = α id α P only what the dictator chooses in case of strict preference for the dictator, if the dictator is indifferent then simple majority voting is applied: 1 if α id = 1 f(α) = 1 if α id = 1 sgn( i I α i) if α id = 0 Remark If a social choice rule is dictatorial, then it is also WP. Proof. For f ( (1,..., 1) ) it is that f(α id ) = 1, thus f ( (1,..., 1) ) = 1 (as the dictator is also prefers x over y) For f ( ( 1,..., 1) ) it is that f(α id ) = 1, thus f ( ( 1,..., 1) ) = 1 (as the dictator is also prefers y over x) (1.2) Non-Dictatorial (ND) Definition. There is no individual which is a dictator. Alternatively: For all individuals i there exists a preference profile α P such that f(α) α i Anonymity (AN) Definition. The names of the individuals play no role for determining the social preference (the role does not discriminate between individuals, i.e. it treats individuals symmetrically) formally: if α, α dom(f) such that n + (α) = n + (α ) and n (α) = n (α ), then f(α) = f(α ) Remark Simple and absolute majority voting satisfy AN. Also the 3 constant rules satisfy AN. However weighted majority voting and dictatorial rules do not satisfy AN. Proof. Simple majority voting: Suppose n + (α) = n + (α ) and n (α) = n (α ), then: f(α) = 1 n + (α) > n (α) n + (α ) > n (α ) f(α ) = 1 f(α) = 1 n + (α) < n (α) n + (α ) < n (α ) f(α ) = 1 f(α) = 0 n + (α) = n (α) n + (α ) = n (α ) f(α ) = 0 Absolute majority voting: Suppose n + (α) = n + (α ) and n (α) = n (α ), then: f(α) = 1 n + (α) > n 2 n +(α ) > n 2 f(α ) = 1 f(α) = 1 n (α) > n 2 n (α ) > n 2 f(α ) = 1 f(α) = 0 n 2 max{n +(α), n (α)} n 2 max{n +(α ), n (α )} f(α ) = 0 Constant rules: As the rules always yield the same outcome, no permutation will give a different choice. Weighted majority voting: Example: α = (1, 1, 1), α = ( 1, 1, 1) and β = (1, 0.1, 0.1), then: 3 i=1 β iα i = 1 thus f(α) = 1 3 i=1 β iα i = 0.8 thus f(α ) = 1 Dictatorial rule: (Is a special case of weighted majority voting, weight 1), Example: i d = 1 α = (1, 1), α = ( 1, 1), then: f(α) = 1 but f(α ) = 1 5

9 Chapter 1. The Case of 2 Alternatives Neutrality (N) Definition. The rule treats alternatives symmetrically (there is no bias in favor of one of the alternatives) formally: if α, α dom(f), then f( α) = f(α) Remark If f is neutral and f(o) is defined, then f(o) = 0. Proof. o = o f(o) = f( o) N = f(o); as f(o) = f(o) it must be f(o) = 0. Examples: 1 if n + (α) > 3 4 f(α) = n 0 otherwise is not neutral, as f(1,..., 1) = 1, but f( 1,..., 1) = 0 f(1,..., 1) Remark Simple majority voting and absolute majority voting is neutral; f + and f are not neutral, but f 0 is neutral; dictatorial social choice rules can, but need not be neutral. Proof. Simple majority voting: n + (α) = n ( α) and n (α) = n + ( α), so f(α) = 1 n + (α) > n (α) n + ( α) < n ( α) f( α) = 1 f(α) = 1 n + (α) < n (α) n + ( α) > n ( α) f( α) = 1 f(α) = 0 n + (α) = n (α) n + ( α) = n ( α) f( α) = 0 absolute majority voting: n + (α) = n ( α) and n (α) = n + ( α), so f(α) = 1 n + (α) > n 2 n ( α) > n 2 f( α ) = 1 f(α) = 1 n (α) > n 2 n +( α) > n 2 f( α ) = 1 f(α) = 0 n 2 max{n +(α), n (α)} n 2 max{n +( α), n ( α)} f(α ) = 0 constant rules: f + (α) = 1 but also f + ( α) = 1, so f + (α) = 1 1 = f + ( α) f (α) = 1 but also f ( α) = 1, so f (α) = 1 1 = f ( α) f 0 (α) = 1 and also f 0 ( α) = 0, so f 0 (α) = 0 = f 0 ( α) dictatorial: neutral example: f(α) = α id non-neural example: 1 if α id {1, 0} f(α) = 1 if α id = 1 is not neutral as according to Remark it must be that f(o) = 0, however here f(o) = Positive Responsiveness (PR) Definition. Whenever α, α dom(f) such that α α but α α and f(α) 0, then f(α ) = 1 Remark f is PR, also f +, but f 0 is not PR. Simple majority voting satisfies PR, absolute majority voting does not. Dictatorial rules can but need not be PR. 6

10 Chapter 1. The Case of 2 Alternatives Proof. Constant rules: f : as f(α) 0 is always wrong (ex falso quodlibet) f + : premise and conclusion are always fulfilled f 0 : the premise is always true, however the conclusion can never be fulfilled Simple majority voting: suppose α α, α α and f(α) 0 implying n i=1 α i = 0 and as α α also n i=1 α i > 0 and thus f(α ) = 1 Absolute majority voting: Example: α = (1, 0, 0, 1) f(α) = 0, but α = (1, 1, 0, 1) α f(α ) = 0 Dictatorial rule: not PR: f(α) = α id. Example: Let i d = 1, then f((0,..., 0)) = 0, but f((0, 1,..., 1)) = 0 PR: Take Rule (1.2); suppose i d = 1, α α, α α, f(α) 0, which implies α 1 = 1 (α 1 = 0 n i=1 α i 0); as α α: α 1 = 1 (α 1 = 0 n i=1 α i > 0) so f(α ) = 1 Remark If (1,..., 1), ( 1,..., 1) dom(f) then PR + N implies WP Proof. By contradiction: Suppose that f(1,..., 1) 0, then N implies f( 1,..., 1) 0. As (1,..., 1) ( 1,..., 1) and unequal PR implies f(1,..., 1) = 1 Thus f(1,..., 1) = 1 N f( 1,..., 1) = 1 Remark Without N, PR does not imply WP, even together with UD and AN Proof. Example: 1 if α {1,..., 1} f(α) = 0 otherwise which fulfills UD, AN and PR; but f(1,..., 1) = 0 1. N fails at every α P Remark If α dom(f) implies α dom(f) then PR and N imply negative responsiveness in the sense: if α, α dom(f) with α α, α α and f(α) 0 then f(α ) = 1. Proof. α α and f( α) 0 implies by PR f( α) = 1. By N: f( ( α )) = f(α ) 1. Theorem 1.1 (May s Theorem). If there are 2 alternatives, then a collective choice rule f satisfies UD, AN, N and PR if and only if f is simple majority voting. Proof. : already seen : let f be UD, AN, N and PR let α P : n + (α) = n (α). Then n + ( α) = n (α) and n ( α) = n + (α), or put differently n + ( α) = n ( α) as well as n + (α) = n + ( α), so by UD and AN: f( α) = f(α) and by N: f( α) = f(α). Together this implies f(α) = f(α) so it must be that f(α) = 0. let α P : n + (α) > n (α). Let b = n + (α) n (α), so b > 0; let B I : #B = b and α i = 1 i B; letting α P be given by α i 0 if i B = if i I \ B α i 7

11 Chapter 1. The Case of 2 Alternatives i I, then n + (α ) = n (α ) (making excess indifferent), so from before it follows that f(α ) = 0. Now as α α, α α by UD and PR it follows that f(α) = 1. let α P : n + (α) < n (α). Then n + ( α) > n ( α), so from before we know f( α) = 1. By UD and N it follows that f(α) = 1. Remark The 4 conditions in May s Theorem are independent, i.e. given any 3 of them the remaining 1 does not hold. Proof. By counterexamples: absolute majority voting: AN, UD, N but PR fails f + : UD, AN, PR but N fails Rule (1.2): UD, N, PR but AN fails Rule (1.1): AN, N, PR but UD fails The following table summarizes findings from above: SC rule UD WP AN N PR simple majority voting weighted majority voting if β 0 skipped skipped absolute majority voting constant rules only f 0 only f + and f dictatorial rules possible possible 8

12 CHAPTER 2 MORE THAN 2 ALTERNATIVES X is the set of alternative, now we allow for #X 2 I = {1,..., n}, n = #I is as before the number of individuals in the population Each individual has a reflexive, complete and transitive preference relation i on X with the interpretation x i y means i prefers x to y or is indifferent to y. We also say x is as least as good as y to individual i or i weakly prefers x to y. First some revision of some microeconomic basics needed in the following: a (binary) relation on X is a subset of X X (i.e. (x, y) : x, y X, thus it is a set of ordered pairs of elements of X) a subset of all ordered pairs reflexive: x x x X complete: x, y X with x y : x y y x transitive: x y, y z x z x, y, z X Example: on R has all 3 properties; > on R is complete and transitive but not reflexive Remark If x 1,..., x n X with n 3 then if is transitive, it follows by induction that if x 1 x 2, x 2 x 3,..., x n 1 x n then x 1 x n. Proof. Induction base: It is true for n = 3. Induction hypothesis: Suppose it is true for an n 3. Induction step: Need to show it is also true for n + 1. Let x 1,..., x n+1 X : x 1 x 2, x 2 x 3,..., x n x n+1. By induction hypothesis: x 1 x n, so together with x n x n+1 and transitivity of : x 1 x n+1. Notation: i denotes the indifference relation derived from i, i.e. x i y x i y y i x and say x and y are treated as indifferent/equally good by individual i i denotes the strict preference relation derived from i, i.e. x i y x i y (x i y) and say i (strictly) prefers x to y Note: x i y x i y (x i y) x i y x i y x i y 9

13 Chapter 2. More than 2 Alternatives Theorem 2.1. Suppose is transitive. Then for and as derived from as above it holds: (a) is transitive (c) x y y z x z (b) is transitive (d) x y y z x z Proof. (a) assume x y, y z, so for transitivity one needs to show x z x z z x. As x y x y and y z y z, transitivity of implies that x z so the first part is shown. Still need to show z x, this can be done indirectly: assume z x: then y z and the assumption z x trans y x (x y y x) therefore it must be that z x (b) assume x y, y z, so for transitivity one needs to show x z x z z x. As x y x y y x and y z y z z y transitivity of implies x z z x (c) assume x y, y z, one needs to show x z x z z x. As x y x y and y z y z, transitivity of implies that x z so the first part is shown. The second one is again proven by contradiction: assume z x: then y z and the trans assumption z x y x (x y y x) therefore it must be that z x (d) assume x y, y z, one needs to show x z x z z x. As x y x y and y z y z, transitivity of implies that x z so the first part is shown. The second one is again proven by contradiction: assume z x: then the assumption z x and x y trans z y (y z z y) therefore it must be that z x Definition (greatest element). Let be a preference relation on X. x X is called a greatest element (in X) for if x y y X Such elements are also called an optimal/best choice. Remark Let X be a non-empty and finite set of alternatives. If there is a reflexive, complete and transitive preference relation on X, then there exists a greatest element in X for. Proof. By induction. Induction base: For n = 1 the statement is true by reflexivity of. For n = 2 it is true by completeness of (and reflexivity). Induction hypothesis: Suppose for an n 2 it holds that every set with n elements has a greatest element for. Induction step: Need to show it is also true for n + 1. Let X be a set with n + 1 elements; pick any x X and let Y = X \ {x}, so #Y = n. By induction hypothesis y Y that is 10

14 Chapter 2. More than 2 Alternatives a greatest element in Y. Now there are 2 cases: (1) y x: then y x x X (2) y x: then by completeness of it follows that x y. Let y Y : y y, so as y is a greatest element in Y : y y. As is transitive: x y, reflexivity gives x x, so x y y X. So in both cases there exists a greatest element in X. Notation: If X = {a, b, c,...} then i :..., b c a,... means that individual i prefers b to c and c to a, so by transitivity also b to a. i :..., [b, c],... means that individual i is indifferent between b and c. 2.1 Extension of simple majority voting to the case of more than 2 alternatives Definition (Condorcet Winner). An alternative x X is a Condorcet Winner if it wins or ties against any other alternative y X in simple majority voting, i.e. if #{i I : x i y} #{i I : y i x} y X \ {x}. Remark A Condorcet Winner need not be unique. Proof. Example: n = 2, X = {a, b, c} 1 : a b c 2 : b a c Here a and b are Condorcet Winner, c is not. It could be that at a certain time only a subset V X is available for choice. V is then called an agenda. Definition (Property α). A choice rule is said to satisfy property α, if V V X, x V and x is in the choice set given V, then x is in the choice set given V. Property α is a property for contraction consistency. If an social choice rule does not fulfill property α it can be hard to interpret the final choice as the best social choice in X. E.g. if x i V thus also x i V and let x i be in the choice set given V, but not in the choice set given V (even though it is still available for choice) Condorcet Winner Method Definition (Condorcet Winner Method). The Condorcet Winner Method chooses the Condorcet Winners out of X. The Condorcet Winner Method is nice in terms of properties discussed earlier 1 : It fulfills Anonymity, is symmetric among alternatives, Neutral in case of two alternatives, Weak Pareto 1 See Section (1.2) 11

15 Chapter 2. More than 2 Alternatives Efficient in the sense that if x, y X and all individuals prefer x to y then y cannot be a Condorcet Winner. In fact if x is strictly preferred by any individual to any y x, then x is the unique Condorcet Winner. However there is a problem: Remark The Condorcet Winner Method does not always work (not for any constellation of preferences over X), i.e. it does not satisfy UD. Proof. Example: A cake has to be shared among 3 individuals, each of them preferring a larger to a smaller share. Then X = {(α 1, α 2, α 3 ) R 3 : 3 i=1 α i = 1}. For any i = 1, 2, 3, x = (α 1, α 2, α 3 ) X and y = (β 1, β 2, β 3 ) X it holds that: x i y α i β i and x i y α i > β i. For any x X there is an i I : α i > 0, w.l.o.g let α 1 > 0. Let y X be given by β 1 = 0, β 2 = α 2 + α1 2, β 3 = α 3 + α1 2. Then β 2 > α 2 and β 3 > α 3, so y 2 x and y 3 x, i.e. y wins against x in simple majority voting. This argument holds for any alternative x X, so no x X is a Condorcet Winner. Definition (Condorcet Paradox). The Condorcet Paradox says that even though all individuals have transitive preferences, collective preferences can be cyclic. It is also known as the paradox of voting. Example: n = 3, X = {a, b, c} 1 : a b c 2 : b c a 3 : c a b Here a looses against c, c against b but b looses against a. Thus there is no Condorcet Winner. Remark Voting cycles can also occur when qualified majority is used instead of simple majority voting. Proof. Example: Suppose x wins against y x if the number of individuals preferring x to y is larger than twice the number of individuals with the opposite preference and we say there is a tie if none of x, y gets such a qualified majority. Let n = 4 and X = {a, b, c, d} 1 : a b c d 2 : b c d a 3 : c d a b 4 : d a b c Now a wins against b, b against c, c against d and d against a. So there is again no Condorcet Winner. Remark Voting cycles can also occur when minority voting is used. Proof. Example: An alternative wins if it gets at least 40% of the vote. Let n = 5 and X = {a, b} 1 : a b 2 : a b 3 : a b 4 : b a 5 : b a Now a wins against b and b wins (also) against b. Again no Condorcet Winner. 12

16 Chapter 2. More than 2 Alternatives The voting paradox may or may not appear. What is the probability that a Condorcet Winner exists? We assume in the following that all preferences are strict, i.e. no indifferences between any 2 alternatives. Example: Assume there are 3 alternatives. Then there are 6 possible sets of preference relations. Assume additionally that there are 3 individuals, then there are 6 3 = 216 possible preference profiles. One may calculate that there are exactly 12 possibilities of individual preferences for which there is no Condorcet Winner. So if all profiles of preferences are equally likely, the probability that there is no Condorcet Winner is = For other numbers of alternatives and individuals, the corresponding probabilities that there is a Condorcet Winner under the assumption that all preferences are strict see the following table: 2 Number of individuals Number of alternatives Remark For n = 4 and #X = 3 there is always a Condorcet Winner if preferences are strict. Proof. (Exercise) Let X = {x, y, z}. As there are more individuals than alternatives, one alternative is best for at least two people. W.l.o.g let x be the best choice for at least individual 1 and 2. Then x ties or wins against any alternative in simple majority voting as at least 2 people, which is at least half of the people vote for x. So x is a Condorcet Winner Staging Procedure The staging procedure (also called staging majority voting ) assumes that #X = N (so X finite) and that one can order alternatives just by assigning numbers to them as X = {x 1, x 2,..., x N } which is also called a managing order. One then proceeds in stages as follows: Stage 1: simple majority voting between x 1 and x 2 Stage 2: simple majority voting between x 3 and the alternative chosen in stage 1. Stage N 1: simple majority voting between x N and the alternative chosen in stage N 2 There is of course a flaw with this procedure, namely that there could be ties. This means one needs a tie breaking rule. In the following we use the rule: In case of a tie the alternative with the lower level is chosen. For notational purpose: a b is short for a defeats b ; a =. b is short for a ties against b However, there are a lot of problems with the staging procedure: 2 Kelly, Jerry S. (1988): Social Choice - An Introduction; page 21 13

17 Chapter 2. More than 2 Alternatives (1) outcome may depend on managing order Example: n = 3, X = {a, b, c} 1 : a b c 2 : b c a 3 : c a b Order 1 x 1 = a, x 2 = b, x 3 = c: Stage 1: a b, Stage 2: c a, thus c is finally chosen Order 2 x 1 = c, x 2 = a, x 3 = b: Stage 1: c a, Stage 2: b c, thus b is finally chosen (2) not Condorcet consistent, i.e. there could be a Condorcet Winner that is not finally chosen Example: n = 3, X = {x 1, x 2, x 3, x 4, x 5 } 1 : x 1 x 2 x 4 x 3 x 5 2 : x 3 x 4 [x 1 x 2 ] x 5 3 : x 5 x 2 x 4 x 3 x 1 Here x 2 is the unique Condorcet Winner, but: Stage 1: x 1. = x2, so x 1 is chosen, Stage 2: x 3 x 1, Stage 3: x 4 x 3, Stage 4: x 4 x 5, thus x 4 is finally chosen (3) Weak Pareto property fails, i.e. an alternative could be chosen even though there is another one which all individuals prefer Example: n = 3, X = {x 1, x 2, x 3, x 4 } 1 : x 2 x 1 x 4 x 3 2 : x 3 x 2 x 1 x 4 3 : x 1 x 4 x 3 x 2 Stage 1: x 2 x 1, Stage 2: x 3 x 2, Stage 3: x 4 x 3, thus x 4 is finally chosen, even though all individuals strictly prefer x 1 to x 4 (4) not strategy proof, i.e. there are possibly incentives for individuals to manipulate the final outcome by reporting wrong preferences Example: n = 4, X = {x 1, x 2, x 3, x 4 } 1 : x 1 x 2 x 4 x 3 2 : x 2 x 3 x 4 x 1 3 : x 3 x 4 x 1 x 2 4 : x 4 x 3 x 2 x 1 Stage 1: x 1. = x2, so x 1 is chosen, Stage 2: x 3 x 1, Stage 3: x 3. = x4, so x 3 is finally chosen. For individual 1 this is the worst possible outcome. Suppose individual 1 votes strategically by 1. : x 2 x 1 x 4 x 3, then Stage 1: x 2 x 1, Stage 2: x 2 = x3, so x 2 is chosen,. Stage 3: x 2 = x4, so x 2 is finally chosen, which suits individual 1 more (5) not contraction consistent, i.e. it does not satisfy property α Example: same as in (4), know that x 3 is finally chosen. Let V = {x 2, x 3, x 4 }, then: Stage 1: x 2. = x3, so x 2 is chosen, Stage 2: x2. = x 4, so x 2 is finally chosen, event though x 3 V, i.e. is still available for choice Copeland Rule Let X be finite, for each alternative x X let n(x) be the number of those alternatives against which x wins or ties in simple majority voting. Then we choose the set which consists of those alternatives for which n(x) is maximal in the set {n(y) : y X}. 14

18 Chapter 2. More than 2 Alternatives Example: n = 3, X = {x 1, x 2, x 3, x 4 } 1 : x 1 x 2 x 3 x 4 2 : x 2 x 3 x 4 x 1 3 : x 3 x 4 x 1 x 2 4 : x 4 x 3 x 2 x 1 Then: n(x 1 ) = 1, n(x 2 ) = 3, n(x 3 ) = 3, n(x 4 ) = 2, so x 2 and x 3 make up the choice set. Properties: (1) coincides with simple majority voting in case of 2 alternatives (2) Condorcet consistent let x be a Condorcet Winner, i.e. it wins or ties against every y X \ {x}, so it gets the maximal possible number of points, namely n(x) = #X 1 (3) Weak Pareto suppose x i y i I, let z X such that z x z y. If y ties or wins against z in simple majority voting, then so does x, because preferences of individuals are transitive, thus n(x) n(y). But as x beats y and not vice versa, n(x) > n(y), so n(y) won t be maximal and thus y not chosen. (4) property α fails Example: n = 3, X = {a, b, c} 1 : a b c 2 : b c a 3 : c a b so n(a) = n(b) = n(c) = 1, thus the choice set is the entire set X. Let V = {a, b}, then the choice set only contains a, even though b is still available and belongs to the choice set given X. (5) problems with strategy proofness Suppose that if the choice set consists of more than one element the element with the lowest level is finally chosen. In the introductory example the choice set was originally {x 2, x 3 }, so with the rule just stated, x 2 would be chosen. However if 4 misrepresents its preferences as 4 : x 4 x 3 x 1 x 2 one gets n(x 1 ) = 1, n(x 2 ) = 2, n(x 3 ) = 3, n(x 4 ) = 2, so now x 3 is chosen. For 4 x 3 is better than x 2 which was previously chosen Plurality voting Let X be finite. Then for any V X let k V (x) for x X be the number of individuals for which x is a greatest element in V for their respective preference (i.e. the number of individuals for which x is in the topmost indifference class). Those elements in V are chosen for which k V (x) is maximal. Example: Let X = {a, b, c, d}, n = 21 3 individuals: a b c d 5 individuals: a c b d 7 individuals: b d c a 15

19 Chapter 2. More than 2 Alternatives 6 individuals: c b d a for V = X a is chosen, as k V (a) = 8, k V (b) = 7, k V (c) = 6, k V (d) = 0 for V = {a, b, d} b is chosen, as k V (a) = 8, k V (b) = 13, k V (d) = 0 Properties: (1) coincides with simple majority voting in case of 2 alternatives (2) property α fails see example above (3) WP holds if x, y V X and x i y i I, then y is definitely not chosen for in this case k V (y) = 0 (4) not Condorcet consistent in example above, c is the unique Condorcet Winner (in fact a looses against any other alternative in simple majority voting) 16

20 CHAPTER 3 THE ARROW IMPOSSIBLE THEOREM As before X is a set of alternatives, #X 2, I the set of individuals with n 2. Each individual i has a reflexive, complete, transitive preference relation i on X. R stands for the set of all reflexive, complete, transitive preference relations on X, thus i R. R I is the set of all logically possible profiles (constellations) of individual preference realtions on X, i.e. of lists ( 1, 2,..., n ) of preference relations of R in X. For abbreviation we often write u for such lists, thus u R I, i.e. u = ( 1, 2,..., n ) with i R i = 1,..., n 3.1 Social Welfare Function Definition (Social Welfare Function). A social welfare function (SWF) defined on a set A R I is a rule f assigning to each u = ( 1,..., n ) A an element f(u) R, i.e. f(u) is a reflexive, complete, transitive preference relation on X with the interpretation of social preference relation. For short: a SWF is a function f with domf R I, taking values in R. Then x f(u) y means x is socially at least as good as y or x is weakly socially preferred to y. f P (u) stands for the strict relation induced by f(u), i.e. x f P (u) y x f(u) y (y f(u) x); we read this as x is socially better than y or x is socially (strictly) preferred to y f I (u) stands for the strict relation induced by f(u), i.e. x f I (u) y x f(u) y y f(u) x; we read this as x and y are socially equally good or x and y are socially indifferent Possible Properties of SWF Universal Domain (UD) domf = R I Examples: (1) Let f be given by f(u) = 1, then domf = R I and individual 1 is a dictator. (2) Method Of Majority Decision (MMD): for u = ( 1,..., n ) R I let f(u) be defined by x f(u) y #{i I : x i y} #{i I : y i x} #{i I : x i y} #{i I : 17

21 Chapter 3. The Arrow Impossible Theorem y i x} Problem with MMD: Let X = {a, b, c}, n = 3 and u be given by 1 : a b c 2 : b c a 1 : c a b so a f(u) b and b f(u) c, but (a f(u) c), thus f(u) is not transitive, so f is not a SWF. If we want to have a SWF which is MMD we have to take out of R I some profiles, i.e. MMD does not satisfy (UD). Weak Pareto (WP) for any u = ( 1,..., n ) domf, if for some x, y X with x i y i I then x f P (u) y. Example: for any u = ( 1,..., n ) domf, let f(u) = 1, can take domf = R I Example: let R, let f : R I R be given by f(u) = (similar to constant rule). Let x, y X with x y such that x f(u) y. Define u = ( 1,..., n ) as follows: for any i I: y i y, y i z z X \ {y} and u i v u, v X \ {y}, then y i z z X \ {y} and u i v u, v X \ {y}, so in particular y i x i I Anonymity (AN) (names of individuals do not matter for social preference), let u = ( 1,..., n ) domf and u = ( 1,..., n) domf, if u is just a rearrangement of elements of u among the individuals formally: if there is a bijection π : I I such that i= π(i) i I then f(u) = f(u ). Non-Dictatorial (ND) There is no i I such that for any u = ( 1,..., n ) domf and any x, y X, if x i y then (automatically) x f P (u) y Remark (UD) + (A) (ND) Proof. Let x, y X, x y, i, j I, i j, u = ( 1,..., n ) such that x i y and y j x. Let u = ( 1,..., n) be given by i= j, j= i and k= k k I \ {i, j} (so i and j exchange their preferences while everyone else stays the same). Then by (UD) we have u, u domf. By (AN): f(u) = f(u ), so neither i nor j is a dictator. Remark MMD implies (ND) Proof. Take 2 individuals i, j I, i j; x, y X with x y and x i x, x i z z X \ {x}, u i v u, v X \ {x}, which implies x i y. On the other hand: y j y, y j z z X \ {y}, u j v u, v X \ {y}, which implies y j x. For h I with h i, h j take complete indifference on X. If f is MMD, then for this profile u of preferences one gets x f I (u) y, x f P (u) z z X \ {x, y}, y f P (u) z z X \ {x, y}, as well as u f I (u) v u, v X \ {x, y}. This implies that f(u) R, i.e., u domf. But x f I (u) y implies that neither i nor j is a dictator. 18

22 Chapter 3. The Arrow Impossible Theorem Binary Independence of irrelevant alternatives (BI) For any two x, y X, and any u, u domf the following holds: If for all individuals i I x i y x i y and y i x y i x, then x f(u) y x f(u ) y and y f(u) x y f(u ) x. I.e. the social ranking between two alternatives x, y X does only depend on the individual rankings with respect to exactly these two alternatives and not the ranking of any third alternative. Justifications: (1) Normative: If the social issue is just x against y, then presence or absence of third alternatives should not matter, only the issue at hand matters (2) Practical: To determine the social choice (or social preference) relative to some subset of alternatives, no information is needed on how individuals rank alternatives in this set against those outside. (3) Without (BI) incentive compatibility may fail, i.e. there is an incentive to manipulate the voting procedure. Neutrality (N) Alternatives are treated symmetrically. For any u, u domf and any x, y, a, b X if for all i I x i y a i b, y i x b i a, then x f(u) y a f(u ) b and y f(u) x b f(u ) a. Remark (N) (BI) Proof. Simply setting a = x and b = y in the definition of (N) gives: for any u, u domf and any x, y X if for all i I x i y x i y, y i x y i x, then x f(u) y x f(u ) y and y f(u) x y f(u ) x; which is just the definition of (BI). Remark If u domf is such that x i y i I, for some x, y X, then: (N) x f I (u) y Proof. Simply setting u = u and a = y, b = x in the definition of (N) gives: for any u domf and any x, y X if for all i I x i y y i x, y i x x i y, then x f(u) y y f(u) x and y f(u) x x f(u) y; where the last two expressions are simply x f I (u) y. Remark For any u, u domf and x, y X (N) as defined above reduces to Neutrality as defined in the case of two alternatives (#X = 2). Proof. own words set a = y, b = x. Positive Responsiveness (PR) For any x, y X and u, u domf: if for all i I x i y x i y and x i y x i y and for some individuals (x i y x i y) or (y i x x i y), then: x f(u) y x f P (u ) y So nobody changed his preferences from x to y, and at least one person changed preferences in favor of x. Remark For any u, u and x, y X (PR) as defined above reduces to Positive Responsiveness as defined in the case of two alternatives (#X = 2). Proof. own words 19

23 Chapter 3. The Arrow Impossible Theorem Remark (UD) + (PR) + (N) (WP) Proof. Let u = ( 1,..., n ) R I and x, y X : x i y i I. To show: x f P (u) y. Let u = ( 1,..., n) R I such that for all i I i is complete indifference. By (UD) u, u domf. By (N) and Remark it follows: x f I (u ) y. Now applying (PR), as for all i I x i y x i y ( ex falso qoudlibet ) and x i y x i y (as we assumed x i y and for some individuals (in fact for all) (x i y x i y), we get that x f I (u ) y implies x f P (u) y. Remark MMD satisfies almost everything we had so far: (ND), (AN), (WP), (N), (BI), (PR) Proof. own words Theorem 3.1 (May s Theorem). If there are more than 2 alternatives (#X 3) there is no SWF which satisfies (UD), (N), (PR) and (AN). Proof. By contradiction. Suppose there exists such a SWF f. (a) From before we know that (N) (BI). (b) For any 2 alternatives and any individual, any ranking of individual i of x against y is induced by some i R. Proof. If, say, x and y are treated indifferent by i take complete indifference on X for i. If x is better than y for i, define i by setting x i x, x i z u i v u, v X \ {x} z X \ {x} and (c) If we combine (b) and (UD) this implies for any two x, y X and any ranking of x against y, there is a social ranking of x against y. In this sense (UD) implies Universal Domain as defined earlier for #X = 2. (d) As already noted, for any u, u domf and any x, y X (N) implies Neutrality as defined earlier for #X = 2. (e) As already noted, for any u, u domf and any x, y X (PR) implies Positive Responsiveness as defined earlier for #X = 2. (f) For any u, u domf and any x, y X (A) + (UD) + (BI) implies Anonymity as defined earlier for #X = 2. Proof. Let u = ( 1,..., n ) and u = ( 1,..., n) be in domf, such that for some bijection π : I I the ranking of x against y for i in the new profile u is the ranking of x against y for π(i) in u. Let u = ( 1,..., n) be given by i = π(i). (UD) means u, u, u domf. (A): f(u) = f(u ), in particular we get that the social ranking of x against y is the same in u and u. (BI): The social ranking of x against y is the same for u and u. As the social ranking for x against y is the same for u and u as well as for u and u, it is the save in u and u. (g) Now we can apply May s Theorem for 2 alternatives: (a) and (c) to (f) imply that for any x, y X the social ranking of x against y is determined according to Simple Majority Voting, that is f must be what we called MMD. 20

24 Chapter 3. The Arrow Impossible Theorem (h) Let now x, y, z be distinct elements of X (these exist as we assumed #X 3). Let u = ( 1,..., n ) R I be such that x 1 y y 1 z, z 2 x x 2 y and x i y y i z i 1, 2. A f is MMD then y f(u) z and z f(u) x. However (y f(u) x) and this means f is not transitive, so f is not a SWF. If we want a SWF which satisfies (UD) we have to replace (AN), (N), (PR) by weaker conditions, say (AN) by (ND), (N) by (BI) and (PR) by (WP). Examples: (1) MMD satisfies (AN), (BI), (PR) but not (UD) (2) Let f(u) = R u R I, i.e. we have a constant rule. Then this satisfies (UD), (ND), (BI) but not (WP) (3) Let f(u) = 1 u R I, then this satisfies (UD), (BI), (WP), but not (ND) (4) (Global) Borda Rule: Assume X is finite, say X = {x 1,..., x k } with k 2. For each individual i I and each i, let b i (x, i ) = #{y X : y i x} and for each u = ( 1,..., n ) let b(x, u) = n i=1 b i(x, i ). The number b(x, u) is also called Borda Score or Borda Points at u R I. Then for each u = ( 1,..., n ) define f(u) by x f(u) y b(x, u) b(y, u). Remark The Borda Rule is a SWF that satisfies (UD), (ND), (WP), but not (BI), thus not strategy proof. Furthermore it is not Condorcet consistent. Proof. (UD): For each possible profile u R I f(u) is reflexive, complete and transitive just because on R has these properties. Thus the Borda Rule satisfies (UD). (ND): Let u = ( 1,..., n ) R I be given such that for i j: i : x 1 [x 2... x k ] j : x 2 [x 1 x 3... x k ] h : [x 1 x 2... x k ] h I \ {i, j} So b(x 1, u) = b(x 2, u) = 1, thus x 1 f I (u) x 2, so neither i nor j is a dictator. (WP): Let i I and x i y. Because i is transitive: if z i x, then z i y, so b i (x, i ) b i (y, i ). As x i y and x i x we have in fact b i (x, i ) < b i (y, i ). So for any profile u = ( 1,..., n ) R I if x i y i I, then b(x, u) < b(y, u), so x f P (u) y. (BI): Suppose X = {w, x, y, z}, n = 3 u: u : 1 : z x y w 1 : x y z w 2 : z x y w 2 : x y z w 3 : y z w x 3 : y x z w So b(w, u) = 8, b(x, u) = 5, So b(w, u ) = 9, b(x, u ) = 1, b(y, u) = 4, b(z, u) = 1 b(y, u ) = 2, b(z, u ) = 6 Thus: y f P (u) x Thus: x f P (u ) y u : 1 : x y z w 2 : x y z w 3 : y z w x So b(w, u ) = 8, b(x, u ) = 3, b(y, u ) = 2, b(z, u ) = 5 Thus: y f P (u ) x From u to u all individuals keep the same ranking of x against y, however the social ranking changes, thus (BI) is violated. 21

25 Chapter 3. The Arrow Impossible Theorem The transition from u to u shows that the Border Rule is not strategy proof, as (BI) fails. u shows that the Borda Rule is not Condorcet consistent (x being the Condorcet Winner). Definition ((Weak) Decisiveness). Let f be a SWF, x, y X and G I, then G is weakly decisive for x against y, if x f P (u) y whenever u domf is such that x i y i G and y j x j I \ G; denote this by x D G y G is decisive for x against y, if x f P (u) y whenever u domf is such that x i y i G; denote this by x D G y As the name suggests any G that is decisive is also weakly decisive. Remark Suppose f satisfies (WP), then x D I y for any x, y X. Let i I, if x D {i} y for any x, y X, then i is a dictator. Theorem 3.2 (Field Expansion Lemma). Suppose #X 3 and #I 2; let f be a SWF which satisfies (UD), (WP) and (BI), let G I, G Ø. Then: x D G y for some x, y X, x y a D G b for all a, b X Proof. (UD) implies domf = R I, this fact is used below without explicit reference Step 1: Let x, y X, x y and suppose x D G y. Then x D G z z X. Proof. Suppose z x, z y (otherwise nothing is to be shown). Let u = ( 1,..., n ) R I be such that x i y y i z i G and y j z z j x j I \ G. x D G y implies x f P (u) y; (WP) implies y f P (u) z; f(u) transitive implies x f P (u) z; (BI) implies x D G z Step 2: Let x, y X, x y and suppose x D G y. Then z D G y z X. Proof. Suppose z x, z y (otherwise nothing is to be shown). Let u = ( 1,..., n ) R I be such that z i x x i y i G and y j z z j x j I \ G. x D G y implies x f P (u) y; (WP) implies z f P (u) x; f(u) transitive implies z f P (u) y; (BI) implies z D G y Step 3: Let x, y, z X be pairwise distinct (i.e. x y, x z, y z) and suppose x D G y. Then: (i) w D G z w X and (ii) z D G w w X. Proof. (i): Step 1 implies x D G z. As x z, Step 2 gives w D G z w X (ii): Step 2 implies z D G y. As y z, Step 1 gives z D G w w X Step 4: Let x, y X, x y and suppose x D G y. Then a D G b for any a, b X. Proof. Suppose a b (otherwise nothing to is to be shown). Case 1, a = z: Then Step 3 (ii) with w = b gives a D G b. Case 2, b = z: Then Step 3 (i) with w = a gives a D G b. Case 3, a z, b z: Then Step 3 (ii) with w = b gives z D G b. Now Step 2 applied to z, b, a gives a D G b. Step 5: Let x, y X, x y and suppose x D G y. Then a D G b for any a, b X. Proof. Let u = ( 1,..., n) R I be such that a i b i G. To show a f P (u ) b. As #X 3, there exists a c X : c a, c b. Let u = ( 1,..., n ) R I be such that a i c c i b i G, c j a a j b j I \G and such that a j b a j b j I \G and b j a b j a j I \ G. To construct j from j for j I \ G, Step 4 gives a D G c implying a f P (u) c; (WP) implies c f P (u) b; f(u) transitive implies a f P (u) b; (BI) implies a f P (u ) b 22

26 Chapter 3. The Arrow Impossible Theorem Theorem 3.3 (Group Contraction Lemma). Suppose #X 3 and #I 2; let f be a SWF which satisfies (UD), (WP) and (BI), let G I, #G 2 and suppose x D G y for any x, y X. Then there is a G G, G Ø such that x D G y for any x, y X. Proof. As #G 2 we can separate G into 2 non-empty subgroups G 1 and G 2 (i.e. G 1 Ø, G 2 Ø, G 1 G 2 = Ø and G 1 G 2 = G). As #X 3 there are distinct a, b, c X. There is a u = ( 1,..., n ) R I such that a i b b i c i G 1 ; c i a a i b i G 2 and b j c c j a i I \ (G 1 G 2 ). x D G y for any x, y X implies a f P (u) b; f(u) is complete implies a f P (u) c or c f(u) a. Case 1, a f P (u) c: (BI) implies a D G1 c. By the Field Expansion Lemma this yields a D G1 b for any a, b X Case 2, c f(u) a: f(u) transitive implies c f P (u) b, so c D G2 b. By the Field Expansion Lemma: a D G2 b for any a, b X So either G 1 or G 2 is decisive for a against b for any a, b X. Theorem 3.4 (Arrow Impossibility Theorem). Suppose #X 3 and #I 2, then there is no SWF which fulfills all of (UD), (WP), (ND) and (BI). Proof. By contradiction: Let f be as SWF which satisfies (UD), (WP), (ND) and (BI). Then x D I y for any x, y X. As I is finite repeated application of the Group Contraction Lemma gives a G with #G = 1 such that x D G y for any x, y X. 3.2 Social Decision Function Let P R be the subset of strict preferences (i.e. no indifference between distinct alternatives. Definition (Social Decision Function). A social decision function (SDF) is a function f defined on some subset P I with values in the set of alternatives X. So given u domf, f(u) X can be interpreted as the chosen alternative, or the socially best alternative. A SDF is manipulatable if at a profile u = ( 1,..., n ) domf there is an individual i I and a i P such that ( 1,..., n) domf and such that f( 1,..., i,..., n ) i f( 1,..., i,..., n ). A SDF is dictatorial if there is an individual i I such that f( 1,..., n ) i x x X and all ( 1,..., n ) domf. Theorem 3.5 (Gibbard-Setterthwaile Theorem). Suppose #X 3 and #I 2, then a SDF f : P I X ( universal domain ) which is not manipulatable at any u P I (i.e. strategy proof ) and which is surjective, must be dictatorial. Proof. omitted Remark f is surjective holds if whenever x X is such that x is a best element for i of any i I, then f( 1,..., n ) = x. This is just Weak Pareto combined with Universal Domain. 23

27 Chapter 3. The Arrow Impossible Theorem Example: 2 individuals A and B, 1 exemplar of a book (Lady Chatterly s Love) and there is just time for one individual to read it. Thus there are 3 alternatives: x A... A reads the book, x B... B reads the book, x N... nobody reads the book. Preferences are as follows: A ( prude ): x N A x A A x B B ( lewd ): x A B x B B x N By (WP) we have x A f P (u) x B ; everybody has the right not to read the book: x N f P (u) x A ; everybody has the right to read the book if nobody else reads it: x B f P (u) x N. Thus x A f P (u) x B, x B f P (u) x N and x N f P (u) x A, so there is no socially best element Liberalism (L) Let f be a SWF. Then for any i I there are x, y X, x y such that i is decisive between x and y in both directions, i.e. for any u = ( 1,..., i,..., n ) domf it holds that x i y x f P (u)y and y i x y f P (u)x. Theorem 3.6 (The Liberalism Paradox). (also Sen s Paradox) There is no SWF that satisfies (UD), (WP) and (L). Proof. omitted 3.3 A possibility Result Definition (Linear Order). A linear order on X is a binary relation which is reflexive, transitive and such that for any x, y X with x y we have x y or y x but not both. Write x > y if x y and x y. Then we have for any x, y X: x = y or x > y or y > x. Definition (Single Peaked Preference Relation). Let be a linear order on X. Then a preference relation on X is called single peaked with respect to if there is a x X such that x z > y x y and y > z x x y; i.e. is strictly decreasing on {y : y x} and strictly increasing on {y : x y}. This x is called the peak for w.r.t. w x y z (a) single peaked w x y z (b) single peaked w x y z (c) not single peaked Figure 3.1: Various examples for single peaked preferences Remark The peak is unique. Proof. omitted Definition (Median Voter). An individual h is called median voter at the profile u = ( 1,..., n ) if #{i I : x i x h } n 2 and #{i I : x i x h } n 2 for i and h respectively. 24

28 Chapter 3. The Arrow Impossible Theorem Theorem 3.7 (Mean Voter Theorem). Let be a linear order on X and suppose u = ( 1,..., n ) is such that each i is single peaked w.r.t.. Then the peak of a median voter is a Condorcet Winner. Proof. omitted 25