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1 EC9A4 Social Choice and Voting Lecture 3 EC9A4 Social Choice and Voting Lecture 3 Prof. Francesco Squintani

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Presentation on theme: "1 EC9A4 Social Choice and Voting Lecture 3 EC9A4 Social Choice and Voting Lecture 3 Prof. Francesco Squintani"— Presentation transcript:

1 1 EC9A4 Social Choice and Voting Lecture 3 EC9A4 Social Choice and Voting Lecture 3 Prof. Francesco Squintani

2 2 Summary from previous lectures We have defined the general set up of the social choice problem. We have shown that majority voting is particularly valuable to choose between two alternatives. We have proved Arrow’s theorem: the only transitive complete social rule satisfying weak Pareto, IIA and unrestricted domain is a dictatorial rule.

3 3 We have extended Arrow’ theorem to social choice functions. We have introduced the possibility of interpersonal comparisons of utility. We have described different concept of social welfare: the utilitarian Arrowian representation and the maximin Rawlsian representation.

4 4 Preview of the lecture We will introduce single-peaked utilities. We will prove Black’s theorem: Majority voting is socially fair when utilities are single-peaked. We will prove Median Voter Convergence in the Downsian model of elections. We will introduce the probabilistic voting model.

5 5 Single Peaked Preferences A binary relation > on the set X is a linear order on X if it reflexive (i.e. x > x for all x), transitive and total (for any x,y in X, either x > y or y > x, not both). The preference R(i) is single peaked wrt the linear order > if there is x such that if x>z>y, then z P(i) y; and if y’>z’>x, then z’ P(i) y’ The alternative x is the peak of satisfaction relative to the linear order >

6 6 x u y z x z’ y’ The single peak of this preference profile is x.

7 7 x u This preference profile does not have a single peak. yxzz’y’

8 8 The complete and reflexive R is acyclic if for every subset X’ of X, there is a maximal element {x in X’ : x R y for all y in X’} is nonempty. Every transitive relation R is acyclic. The Condorcet paradox violates acyclicity. There are rules that violate transitivity but not acyclicity.

9 9 An oligarchy is a set agents S, subset of N, such that x P y if and only if for every i in S, x P(i) y. Whenever #S>1, this is not transitive. Example: `z P(1) x,’ `P(1) y’ and `y P(2) z P(2) x.’ We obtain that `x R y’, `y R z,’ and `z P x’. The oligarchic rule is acyclic because there is always a set of allocations for which there is no unanimous better allocation in S.

10 10 Consider the majoritarian social rule: x R y if n(x,y) = #{i : x R(i) y} > N/2. This rule induces a complete relation, but we have seen that the relation may be intransitive. In the domain of single-peaked preference, we will show that it is acyclic. Denote by x*(i) the peak for agent i.

11 11 Agent n is the median agent for the single peaked profile of preferences (R(1), …, R(N)) if #{i : x*(i)>x*(n)}> N/2; #{i : x*(n)>x*(i)}>N/2. Black’s Theorem: If > is a linear order and R(i) is single peaked wrt > for all i, then the peak x*(n) cannot be defeated by majority by any alternative. The peak x*(n) is called the Condorcet winner, and it makes majority voting rule acyclic.

12 12 Proof. Take any y < x*(n). Consider any agent i with peak x*(i) > x*(n). Because X is linearly ordered by >, x*(i)>x*(n)> y. Because R(i) is single peaked with respect to >, we obtain that x*(n) R(i) y for all such agents i. By definition of median voter, #{i : x*(i) > x*(n)}> N/2; hence, x*(n) cannot be rejected by majority vote. The case for y > x*(n) is analogous.

13 13 x u x*(1) x*(2) x*(3) x*(4) x*(5) An example with n odd.

14 14 x u x*(1) x*(2) x*(3) x*(4) An example with n even. All the allocations between x*(2) and x*(3) cannot be rejected by majority voting.

15 15 The key feature of the above example is the unidimensionality of the set of allocations. 0 The space of alternative is the unit square X = [0,1] 2 There are 3 agents, with preferences: u 1 (x) = - 2x 1 - x 2 u 2 (x) = x 1 + 2x 2 u 3 (x) = x 1 – x 2 x1x1 x2x2 u1u1 u2u2 u3u3

16 16 If x 1 =0, then agents 1 and 3 prefer y = (1/2, x 2 ) to x. 0 x1x1 x2x2 u2u2 u3u3

17 17 0 If x 2 =1, then agents 1 and 3 prefer y = (x 1, 1/2) to x. x1x1 x2x2 u1u1 u3u3

18 18 0 x1x1 x2x2 u1u1 u2u2 If x 1 >0 and x 2 <1, then agents 1 and 2 prefer y = (x 1 - a, x 2 + a) to x.

19 19 Voting Models Voting models are non-cooperative games that model elections. They can study one-shot elections, or repeated elections. There may be 2 or more candidates. Candidates’ strategic decisions may include whether and when to run in the election, with which promised policy platform, campaign spending, and so on.

20 20 Downsian Elections There are 2 candidates, i=1,2. Candidates care only about winning the election. Candidates i=1,2 simultaneously commit to policies x i if elected. Policies are real numbers. There is a continuum of voters, with diverse ideologies k, with cumulative distribution F. The utility of a voter with ideology k if policy x is implemented is u(x,k)= L(|x-k|), with L’<0. After candidates choose platforms, each citizen votes, and candidate with the most votes wins. If x 1 = x 2, then the election is tied.

21 21 For any ideology distribution F, let the median policy m be such that 1/2 of voters' ideologies y > m & 1/2 of ideologies y < m : F (m) = 1/2. Median Voter Theorem: The unique Nash Equilibrium of the Downsian Election model is such that candidates i choose such that candidates i choose x i = m, and tie the election.

22 22 Proof. Fix any x 1 = x 2. Because L’<0, each voter with ideology k votes for the candidate i that minimizes |x i -k|. Hence, if x i < x j, candidate i’s vote share is F(½ ( x 1 + x 2 ) ), and candidate j’s is 1- F(½ ( x 1 + x 2 ) ). For either candidate i, if x i = m, then candidate j’s best response is BR j = {x j : |x j - m|< |x i - m|}. Candidate j wins the election. Hence, there cannot be any Nash equilibrium where either candidate i plays x i = m.

23 23 Suppose both candidates play x 1 = x 2 = m, then all voters are indifferent between x 1 and x 2, and the election is tied. If either candidate i deviates and plays x i = m, then she loses the election. So, there is unique Nash equilibrium: x 1 = x 2 = m.

24 24 Downsian Elections with Ideological Candidates Downsian Elections with Ideological Candidates Suppose that candidate i’s ideology is k Suppose that candidate i’s ideology is k i, with k. k 1 < m < k 2. The utility of candidate i if policy x is kk implemented is u(x, k i )= L(|x- k i |), with L’<0. Theorem: The unique Nash Equilibrium is such that candidates i choose such that candidates i choose x i = m, and tie.

25 25 Proof. Proof. Again, for any x 1 = x 2, if x i < x j, candidate i’s vote share is F(½ ( x 1 + x 2 ) ), and candidate j’s is 1- F(½ ( x 1 + x 2 ) ). Suppose that x 1 < m, then candidate 2 wins and implement x 2 by choosing x 2 in (x 1, 2m - x 1 ). Hence, if x 1 < 2m- k 2, then BR 2 (x 1 )={k 2 }, and if 2m- k 2 < x 1 < m, then BR 2 (x 1 ) is empty. If m < x 1 < k 2, then BR 2 (x 1 )=[x 1,infinity)., then If x 1 > k 2, then BR 2 (x 1 )={k 2 }.

26 26 The best response of candidate 1 is symmetric. Hence, we conclude that there cannot be any Nash Equilibrium with x 1 = m or x 2 = m. Suppose that candidate i chooses x i = m. Then, regardless of the choice x j, the implemented policy is m. Hence, BR j (x i )=(-infinity, infinity). We conclude that the unique Nash Equilibrium is such that x 1 = x 2 = m, and the election is tied.

27 27 Probabilistic Voting Suppose that candidates do not know the voter’s preferences. The median policy m is randomly distributed, with cdf G. Let M be the median of G. Theorem. If candidates care only about winning the election, then the unique Nash Equilibrium is candidates i choose such that both candidates i choose x i = M, and tie. Theorem. If candidates are ideological, with k k 1 < M < k 2, then there is no Nash Equilibrium candidates i choose such that both candidates i choose x i = M.

28 28 Proofs. For any x 1 = x 2, candidate i wins the election if and only if |x i - m|<|x j - m|. If x i < x j, then candidate i wins with probability Pr(½ ( x 1 + x 2 ) > m) = G (½ ( x 1 + x 2 ) ), and Pr(j wins) = 1 - G (½ ( x 1 + x 2 ) ). The probability that i wins increases in x i on (-infinity, x j ), decreases in x i on (x j, infinity), and is discontinuous at x i = x j unless x i = x j =m. Hence, if candidates only care about winning, there is a unique NE: x 1 = x 2 = M.

29 29 If instead candidates are ideological, there cannot be any equilibrium such that x 1 = x 2 = M. Suppose in fact that x 2 = M. If candidate 1 plays x 1 = M, then the implemented policy is M with probability 1. If 1 plays any x 1 in (k 1, M), she increases her expected utility, because x 1 is implemented with positive probability, and she prefers x 1 to M. Also, note that a Nash Equilibrium exists in this game, by “standard” general existence results.

30 30 Conclusion We have introduced single-peaked utilities. We have proved Black’s theorem: Majority voting is socially fair when utilities are single-peaked. We have proved Median Voter Convergence in the Downsian model of elections. We have shown that Median Convergence persists in the presence of either ideological parties or uncertainty on voters’preferences.

31 31 Preview of the next lecture We will introduce ideology in the probabilistic voting model, modelling `responsible parties’. We will derive the equilibrium in the model with responsible parties uncertain of the voters’ preferences. We will prove that responsible parties yield higher welfare to the electorate than opportunistic parties.


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