Partial lecture notes THE PROBABILITY OF VIOLATING ARROW S CONDITIONS

Partial lecture notes THE PROBABILITY OF VIOLATING ARROW S CONDITIONS 1

B. Preference Aggregation Rules 3. Anti-Plurality a. Assign zero points to a voter's last preference and one point to all other preferences. Sum the points across voters and rank alternatives in descending order of the sums. Ex: 4 voters 2 voters 4 voters A A C C B A B C B

B. Preference Aggregation Rules 3. Anti-Plurality a. Assign zero points to a voter's last preference and one point to all other preferences. Sum the points across voters and rank alternatives in descending order of the sums. Ex: 4 voters 2 voters 4 voters A (1) A (1) C (1) C (1) B (1) A (1) B (0) C (0) B (0) A = 7 B = 2 C = 8 Rank: C > A > B b. Note: anti-plurality can violate P and IIA, but never T or ND.

B. Preference Aggregation Rules 4. Hare (alternative vote) a. Begin with a plurality vote. Eliminate the alternative with the fewest plurality points (eliminating tieing alternatives if there is a tie for fewest points). Repeat the process on the remaining alternatives until all alternatives are eliminated. Alternatives are ranked in reverse order of elimination (i.e., the alternative(s) eliminated last is preferred most in the social ranking). Ex: 4 voters 2 voters 4 voters A A C C B A B C B Plurality Vote: A = 6 B = 0 C = 4 Therefore eliminate B.

B. Preference Aggregation Rules 4. Hare (alternative vote) a. Begin with a plurality vote. Eliminate the alternative with the fewest plurality points (eliminating tieing alternatives if there is a tie for fewest points). Repeat the process on the remaining alternatives until all alternatives are eliminated. Alternatives are ranked in reverse order of elimination (i.e., the alternative(s) eliminated last is preferred most in the social ranking). Ex: 4 voters 2 voters 4 voters A A C C C A Plurality Vote: A = 6 C = 4 Therefore eliminate C. Note: Hare can violate P and IIA, but never T or ND. Rank: A > C > B.

B. Preference Aggregation Rules 5. Nanson skip

B. Preference Aggregation Rules 6. Pairwise Majority (a particular type) a. Consider all possible two-alternative comparisons. For any pair of alternatives {a; b}, a is socially preferred to b if more voters prefer a to b. Alternatives a and b are socially indifferent if they tie. Rankings are then derived from the pairwise comparisons. Ex: 4 voters 2 voters 4 voters A A C C B A B C B A vs B -> A wins A vs C -> A wins B vs C -> C wins Rank: A > B > C. Note: Pairwise Majority can violate T, but never P, ND, or IIA.

B. Preference Aggregation Rules 7. Copeland a. Conduct a vote using pairwise majority. For each pair assign 1 point to the alternative that wins the pairwise comparison and.5 points if they tie. Rank alternatives in descending order of total points. Ex: 4 voters 2 voters 4 voters A A C C B A B C B A vs B -> A wins. 1 point A. A vs C -> A wins. 2 points A. B vs C -> C wins. 1 point B. Rank: A > B = C. Note: Copeland can violate IIA, but never P, T, or ND.

C. Assumptions Notation: A is the number of alternatives (e.g., candidates). N is the number of voters. ρ is a preference profile. is the set of all possible preference profiles with strict orderings among N individuals. We assume individuals only have strict preferences. Note: individuals cannot be indifferent, but society can. p 6 is the proportion of individuals with preference I>R>D.

D. Theorems 1. For A 3 and N > 1, plurality, anti-plurality, and Hare will violate IIA for all a. This means that for all populations, with any number of candidates and at least three alternatives, plurality, antiplurality, and Hare will always violate IIA. b. Hence, these rules will always violate violate Arrow s conditions. Ex: 4 voters 2 voters 4 voters A A C C B A B C B Plurality: A>C>B. With one exception we can always get plurality to violate IIA by moving the middle ranked alternative to the top, C in this case, in every preference order.

D. Theorems 1. For A 3 and N > 1, plurality, anti-plurality, and Hare will violate IIA for all a. This means that for all populations, with any number of candidates and at least three alternatives, plurality, antiplurality, and Hare will always violate IIA. b. Hence, these rules will always violate violate Arrow s conditions. Ex: 4 voters 2 voters 4 voters A A C C B A C C B C B Plurality: A>C>B. With one exception we can always get plurality to violate IIA by moving the middle ranked alternative to the top, C in this case, in every preference order.

D. Theorems 1. For A 3 and N > 1, plurality, anti-plurality, and Hare will violate IIA for all a. This means that for all populations, with any number of candidates and at least three alternatives, plurality, antiplurality, and Hare will always violate IIA. b. Hence, these rules will always violate violate Arrow s conditions. Ex: 4 voters 2 voters 4 voters C C C A A A B B B Plurality: A>C>B. With one exception we can always get plurality to violate IIA by moving the middle ranked alternative to the top, C in this case, in every preference order.

D. Theorems 1. For A 3 and N > 1, plurality, anti-plurality, and Hare will violate IIA for all a. This means that for all populations, with any number of candidates and at least three alternatives, plurality, antiplurality, and Hare will always violate IIA. b. Hence, these rules will always violate violate Arrow s conditions. Ex: 4 voters 2 voters 4 voters C C C A A A B B B Which gives us C > A=B, violating IIA because the relationship between A and B switched. Plurality: A>C>B. With one exception we can always get plurality to violate IIA by moving the middle ranked alternative to the top, C in this case, in every preference order.

D. Theorems 1. For A 3 and N > 1, plurality, anti-plurality, and Hare will violate IIA for all a. This means that for all populations, with any number of candidates and at least three alternatives, plurality, antiplurality, and Hare will always violate IIA. b. Hence, these rules will always violate violate Arrow s conditions. The one exception is when plurality produces a tie using the initial profile (i.e., A=B=C). In that case, we can get a violation of IIA by moving one person s second preference to the top. Suppose that is someone who prefers A> i B > I C. By moving B to the top, plurality produces B>A=C, changing the relationship between B and C.

D. Theorems 1. For A 1 and N 1, there exists a for which any PAR that satisfies Pareto does not violate IIA. a. Since Borda and Copeland adhere to Pareto, there always exists at least one preference profile where they don t violate IIA. b. Combined, theorems 1 & 2 suggest that Borda and Copeland will be better at IIA then plurality, anti-plurality, and Hare. c. Since, Borda and Copeland don t violate any of Arrow s remaining conditions, this means that they will always be more likely to adhere to Arrow s conditions than plurality, antiplurality, and Hare!

1. To determine whether Borda and Copeland are more likely to violate IIA than pairwise majority rule violates transitivity, and to show the magnitude of some of these differences, we are going to show some simulations. 2. A simulation asks a computer to repeatedly draw preference profiles and determine whether there is a violation of one of the criteria in each draw. It then reports the frequency of violations across draws. a. Like repeatedly flipping an uneven coin to figure out the probability that it lands on heads. 53% heads.

3. We first assume a distribution of voters and candidates. Our results will only hold true for the distributions specified. We then randomly draw voters from the distributions and determine the ranking for that draw as well as whether the voting rule adheres to each of the criteria on that draw. 4. We then repeat the process 1 million times and report the frequency that each voting rule adheres to or violates each criterion. We are going to illustrate the simulation using two PARs (plurality and Borda) and a single draw of individual preferences.

Case: A=3, N=5, IC preferences. A=3 implies there are A! = 6 linear preferences.

Case: A=3, N=5, IC preferences. A=3 implies there are A! = 6 linear preferences. Impartial Culture Condition: Each individual preference order is equally likely.

Case: A=3, N=5, IC preferences. A=3 implies there are A! = 6 linear preferences. 1. Randomly assign the 5 individuals to one of the preferences. 2. Vote: Impartial Culture Condition: Each individual preference order is equally likely. Plurality: D > R > I. Borda: D: 9+2+1 = 12 R: 6+3+3 = 12 I: 3+1+2 = 6 D = R > I.

3. See if we violate Pareto in this draw. a. Plurality: D > R > I.

3. See if we violate Pareto in this draw. a. Plurality: D > R > I. a. Everyone prefers R > I.

3. See if we violate Pareto in this draw. a. Plurality: D > R > I. a. Everyone prefers R > I. And so does society using plurality. b. Borda: D = R > I.

3. See if we violate Pareto in this draw. a. Plurality: D > R > I. a. Everyone prefers R > I. And so does society using plurality. b. Borda: D = R > I. a. Everyone prefers R > I. And so does society using Borda. Neither PAR violated Pareto

4. See if we violate Transitivity in this draw. a. Plurality: D > R > I. a. Is D > I by plurality? b. Borda: D = R > I. D: 9+2+1 = 12 R: 6+3+3 = 12 I: 3+1+2 = 6 Yes does not violate transitivity. D R and R I implies D I which is true, so it does not violate transitivity.

5. See if we violate IIA in this draw. New Profile a. Plurality: D > R > I. a. Move R to the top of everyone s preference order, but keep their individual preferences between I and D the same. b. New profile: R > D = I.

5. See if we violate IIA in this draw. New Profile a. Plurality: D > R > I. a. Move R to the top of everyone s preference order, but keep their individual preferences between I and D the same. b. New profile: R > D = I, which violates IIA.

5. See if we violate IIA in this draw. New Profile a. Borda: D = R > I. a. Move R to the top of everyone s preference order, but keep their individual preferences between I and D the same. b. New profile: R > D > I. D: 6+2+1 = 12 R: 9+3+3 = 15 I: 3+1+2 = 6

5. See if we violate IIA in this draw. New Profile a. Borda: D = R > I. a. Move R to the top of everyone s preference order, but keep their individual preferences between I and D the same. b. New profile: R > D > I. D: 6+2+1 = 12 R: 9+3+3 = 15 I: 3+1+2 = 6 Since D > I in both cases, we do not have a violation of IIA.

5. See if we violate IIA in this draw. a. But we don t know whether Borda violates IIA for the preference profile drawn, because we haven t considered all possible changes in the irrelevant pairs. b. The program proceeds by checking a large number of changes in the irrelevant pairs details in the readings.

6. See if we violate ND in this draw. a. None of our PARs will violate ND so we can skip this step.

1. To get the A = 3, N = 5 results for the distribution of individuals assumed, we repeat the process 1 million times and report the frequencies. 2. The frequencies approximate the true probabilities for that number of candidates, that number of voters, and the distributions.