Conference 2Intermediate Microeconomics - Fall 2018September 11, 2018Problem 1: Variations on French presidential elections2017 French presidential elections’ first round yielded the following results for the first fivecandidates:Emmanuel Macron (EM): 24.01%Marine Le Pen (MP): 21.30%Francois Fillon (FF): 20.01%Jean-Luc Melenchon (JM): 18.58%Benoit Hamon (BH): 6.36%Which is analogous to ranking EM MP FF JM BH.Imagine there were only 14 citizens that voted in France in 2017. These 14 citizens had thefollowing preferences:5 people 3 people 2 people 4 peopleEM F F JM MPF F MP BH JMBH EM EM EMMP BH F F F FJM JM MP BH1. Check that if a rule similar to the real rule is applied, that is if each person is to give itsvote for its favorite candidate, and candidates are to be ranked according to their numberof votes, then the historical ranking is obtained.EM gets 5 votes, MP gets 4, FF gets 3, JM gets 2 and BH gets 0. The final ranking is EM MP FF JM BH.2. What ranking does the Borda rule yield?Here are the Borda scores for each of the candidates:EM: 5 × 1 + 3 × 3 + 2 × 3 + 4 × 3 = 32FF: 5 × 2 + 3 × 1 + 2 × 4 + 4 × 4 = 37BH: 5 × 3 + 3 × 4 + 2 × 2 + 4 × 5 = 511MP: 5 × 4 + 3 × 2 + 2 × 5 + 4 × 1 = 40JM: 5 × 5 + 3 × 5 + 2 × 1 + 4 × 2 = 50Hence the ranking is EM FF MP JM BH3. What ranking does the Hare system yield? (several rounds of plurality voting are organized;at the end of each round, the candidate with the fewest votes is eliminated, and the remainingcandidates move on to the following round. The winner is the only remaining candidate atthe end of this process. Candidates are ranked according to the round they were eliminatedin.)The rounds in the Hare system go as follows:EM gets 5 votes, FF gets 3, JM gets 2, MP gets 4 and BH gets 0. BH is eliminatedEM gets 5 votes, FF gets 3, JM gets 2 and MP gets 4. JM is eliminatedEM gets 7 votes, FF gets 3, and MP gets 4. FF is eliminatedEM and MP get 7 votes, they tie for the first place.Hence the ranking is EM ∼ MP FF JM BH4. what ranking does the Coombs system yield? (several rounds of ’reversed” plurality votingare organized: citizens vote to exclude their least prefered candidate. At the end of eachround, the candidate with the most votes is eliminated, and the remaining candidates moveon to the following round. The winner is the only remaining candidate at the end of thisprocess. Candidates are ranked according to the round they were eliminated in.The rounds in the Coombs system go as follows:EM gets 0 votes, FF gets 0, JM gets 8, MP gets 2 and BH gets 4. JM is eliminatedEM gets 0 votes, FF gets 0, MP gets 7 and BH gets 7. BH and MP are eliminatedEM gets 3 votes and FF gets 11. FF is eliminatedHence the ranking is EM FF BH ∼ MP JM5. what ranking does the Nanson system yield? (After a Borda count, eliminate all candidateswith Borda scores above average. The remaining candidates move on to the following round.The winner is the only remaining candidate at the end of this process. Candidates are rankedaccording to the round they were eliminated in.)The Borda scores for each of the candidates are the following:EM: 5 × 1 + 3 × 3 + 2 × 3 + 4 × 3 = 32FF: 5 × 2 + 3 × 1 + 2 × 4 + 4 × 4 = 37BH: 5 × 3 + 3 × 4 + 2 × 2 + 4 × 5 = 51MP: 5 × 4 + 3 × 2 + 2 × 5 + 4 × 1 = 40JM: 5 × 5 + 3 × 5 + 2 × 1 + 4 × 2 = 50The average score is 42. Therefore BH and JM are eliminated.Recomputing Borda scores for EM, FF and MP only yields:EM: 5 × 1 + 3 × 3 + 2 × 1 + 4 × 2 = 252FF: 5 × 2 + 3 × 1 + 2 × 2 + 4 × 3 = 29MP: 5 × 3 + 3 × 2 + 2 × 3 + 4 × 1 = 31The average score is ∼ 28. Therefore MP and FF are eliminated, only EM remains.Problem 2: The Median Voter Theorem and its proof.Remember the social choice problem’s set up:Let I be the set of votersLet A be the set of alternatives voters choose fromEach voter has a rational (i.e. complete and transitive) preference ordering <ion A. Weassume this ordering is strictP = (<i)i∈Iis the preference profileThe collectivity adopts a majority voting system to determine the social choiceA version of the median voter theorem states as follows:Black (1948): Suppose #I is odd. If there exists a complete and transitive strict order <Aon A such that P is single peaked with respect to (A, <) then the favorite policy of the medianvoter πmis a Condorcet winner, that is:πmMVπ ∀π ∈ A (1)1. P is single peaked if all voters’ preferences are single peaked. Voter i has single peakedpreferences if there exists some policy a that i prefers to all other policies onthe right and the left of a, and the further the alternative is from a, the less iwill like it. Give examples of how this translates on a political spectrum.In problem 1‘s set up, assuming candidates can be ranked from left to right on thepolitical spectrum: JM < BH < EM < FF < MP, an individual with single peakedpreference would have the following rankings: JM BH EM FF MP or EM FF MP and EM BH JM but could not have these preferences for instance: JM MP EM FF BH2. The median voter prefers the policy that has equally many alternative policiesto its right and to its left. Who would that be on a political spectrum.It would be the voter who prefers the middle ground. If we keep to the ranking weassumed on the French political spectrum, and have 5 voters, two voting MP, one FF,one EM and one JM, then the median voter is the one voting for FF (and not EM!).3. Why is it important that #I is odd?Assume there are only 4 voters, one voting MP, one FF, one EM and one JM. Then therecannot be as many voters to the left as to the right of any voter. Take the FF voter.There is only one voter to its right and two to his left. For the EM voter it is the opposite:there are two voters to her right but only one to her left.34. Draw an example of single peaked preferences P on a graph. Use it to show why policiesto the left and to the right of the median …
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