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Median Voter Theorem In this final problem, we'll look at a game with political candidates who
decide where to locate in an ideological policy space, before voters cast their ballots. Voters are
not strategic actors here. Our focus is on the game played between candidates, who want to
choose policies which maximize their chances of winning the election. Formally, let a large
number of voters be distributed uniformly along the unit interval. Voters always cast their ballot
for the candidate "closest" to them in the policy space, breaking evenly in the event of a tie.
Voters at 0 are more liberal than everyone else; voters at 1 are more conservative than everyone
else. Plurality rule determines who wins: the candidate who receives the most votes receives a
payoff of 1 , while other candidates receive 0 . If candidates tie, they each win with equal
probability. Let's look at what happens in a couple of examples to make this concrete. I highly
suggest drawing this all out! Say that candidate A locates at x A = 5 1 so that they are pretty far
left. Candidate B locates at x B = 5 3 so that they are just right of center. Clearly, all voters to the
left of A - i.e., 5 1 of all voters - cast their ballot for A . Similarly, all voters to the right of B -i.e.,
5 2 of all voters, from [ 5 3 , 1 ] - cast their ballot for B . What happens with the remainder
between the candidates? They break evenly. So since 5 2 of all voters lie between A and B on the
interval [ 5 1 , 5 3 ] , A receives 5 1 additional votes, and so does B . Since candidate A only
receives 5 2 of all votes and candidate B receives 5 3 , B wins the election and receives the
payoff of 1 . Now say that candidate A locates at x A = 4 1 and B locates at x B = 4 3 . Here, A
receives 2 1 of all votes and so does B . Then the expected payoffs are 2 1 for each player. The
same result obtains whenever the candidates are equidistant from the center, including when they
both locate at x A = x B = 2 1 . Let's finally consider two examples with three candidates. Say
that candidate A locates at x A = 5 1 , candidate B locates at x B = 5 3 , and candidate C locates
at x C = 5 4 . Candidate A wins all voters in the interval [ 0 , 5 2 ] , candidate B wins all voters in
the interval [ 5 2 , 10 7 ] , and candidate C wins all voters in the interval [ 10 7 , 1 ] . Candidate A
receives 5 2 of all votes, candidate B receives 10 3 , and candidate C receives 10 3 . Candidate A
wins the election decisively. Now say that candidate A locates at x A = 5 1 , while B and C each
locate at x B = x C = 5 2 . Candidate A wins all voters in the interval [ 0 , 10 3 ] and candidates B
and C split the remaining voters, for 20 7 each. Since 20 7 > 10 3 , B and C win with equal
probability. Solve for the unique pure strategy Nash equilibrium in the game with two
candidates, A and B . You should not need any calculus to find the correct answer. (b) The game
with three candidates has multiple pure strategy Nash equilibria. Provide an example of one. No
calculus is required here, either.

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Median Voter Theorem In this final problem- we'll look at a game with.docx

  • 1. Median Voter Theorem In this final problem, we'll look at a game with political candidates who decide where to locate in an ideological policy space, before voters cast their ballots. Voters are not strategic actors here. Our focus is on the game played between candidates, who want to choose policies which maximize their chances of winning the election. Formally, let a large number of voters be distributed uniformly along the unit interval. Voters always cast their ballot for the candidate "closest" to them in the policy space, breaking evenly in the event of a tie. Voters at 0 are more liberal than everyone else; voters at 1 are more conservative than everyone else. Plurality rule determines who wins: the candidate who receives the most votes receives a payoff of 1 , while other candidates receive 0 . If candidates tie, they each win with equal probability. Let's look at what happens in a couple of examples to make this concrete. I highly suggest drawing this all out! Say that candidate A locates at x A = 5 1 so that they are pretty far left. Candidate B locates at x B = 5 3 so that they are just right of center. Clearly, all voters to the left of A - i.e., 5 1 of all voters - cast their ballot for A . Similarly, all voters to the right of B -i.e., 5 2 of all voters, from [ 5 3 , 1 ] - cast their ballot for B . What happens with the remainder between the candidates? They break evenly. So since 5 2 of all voters lie between A and B on the interval [ 5 1 , 5 3 ] , A receives 5 1 additional votes, and so does B . Since candidate A only receives 5 2 of all votes and candidate B receives 5 3 , B wins the election and receives the payoff of 1 . Now say that candidate A locates at x A = 4 1 and B locates at x B = 4 3 . Here, A receives 2 1 of all votes and so does B . Then the expected payoffs are 2 1 for each player. The same result obtains whenever the candidates are equidistant from the center, including when they both locate at x A = x B = 2 1 . Let's finally consider two examples with three candidates. Say that candidate A locates at x A = 5 1 , candidate B locates at x B = 5 3 , and candidate C locates at x C = 5 4 . Candidate A wins all voters in the interval [ 0 , 5 2 ] , candidate B wins all voters in the interval [ 5 2 , 10 7 ] , and candidate C wins all voters in the interval [ 10 7 , 1 ] . Candidate A receives 5 2 of all votes, candidate B receives 10 3 , and candidate C receives 10 3 . Candidate A wins the election decisively. Now say that candidate A locates at x A = 5 1 , while B and C each locate at x B = x C = 5 2 . Candidate A wins all voters in the interval [ 0 , 10 3 ] and candidates B and C split the remaining voters, for 20 7 each. Since 20 7 > 10 3 , B and C win with equal probability. Solve for the unique pure strategy Nash equilibrium in the game with two candidates, A and B . You should not need any calculus to find the correct answer. (b) The game with three candidates has multiple pure strategy Nash equilibria. Provide an example of one. No calculus is required here, either.