Microeconomic Theory Jamison / Kohlberg / Avery Problem Set 4 Solutions Spring (a) LEFT CENTER RIGHT TOP 8, 5 0, 0 6, 3 BOTTOM 0, 0 7, 6 6, 3

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1 Microeconomic Theory Jamison / Kohlberg / Avery Problem Set 4 Solutions Spring Subgame Perfect Equilibrium and Dominance (a) LEFT CENTER RIGHT TOP 8, 5 0, 0 6, 3 BOTTOM 0, 0 7, 6 6, 3 Highlighting the best responses to each pure strategy, we find that there are two pure strategy Nash equilibria: (Top, Left) and (Bottom, Center). Next consider the possibility of mixed strategy equilibria. Since there are just two pure strategies for player 1, it is easier to consider player 2 s best responses to player 1 s mixed strategies than the reverse. Suppose that player 1 plays Top with probability p and Bottom with probability 1-p. Then player 2 s expected payoffs are: EU 2 (Left) = 5p EU 2 (Center) = 6(1-p) EU 2 (Right) = 3 Comparing these payoffs, EU 2 (Left) > EU 2 (Center) if p > 6/11; EU 2 (Left) > EU 2 (Right) if p > 6/10; EU 2 (Center) > EU 2 (Right) if p < 1/2. Therefore, player 2 s best response to player 1 s mixed strategy with P(Up) = p is (1) Left is the unique best response if p > 6/10; (2) Any combination of Left and Right is a best response if p = 6/10; (3) Right is the unique best response if 1/2 < p < 6/10; (4) Any mixed strategy combination of Center and Right is a best response if p = ½; (5) Center is the unique best response if p < ½. The figure on the next page shows these best responses in a graph of player 2 s expected payoffs as a function of player 1 s mixing probability P(Up) = p. Following the procedure of Guess and Verify consider each of (1) through (5) in turn to determine which of these cases, if any, produce mixed strategy Nash equilibria.

2 Case 1: p > 6/10, player 2 plays Left. Player 1 has a pure strategy best response, Up or p = 1 to Left. But ( Up, Left ) is one of the pure strategy equilibria identified above. Case 2: p = 6/10, player 2 mixes between Left and Right. If player 2 does not play Center, then Top weakly dominates Bottom and so Player 1 has a pure strategy best response Top (p = 1) to any mixed strategy by player 2 that includes both Left and Right with positive probability. This best response p = 1 is not consistent with the conjecture that p = 6/10 so there is no Nash equilibrium with p = 6/10 and player 2 playing both Left and Right with positive probability. There is however, an equilibrium where player 2 plays Right with probability 1 and player 1 plays the mixed strategy with p = 6/10. Case 3: 1/2 < p < 6/10, player 2 plays Right. If player 2 plays Right, then player 1 is indifferent between Top and Bottom so any mixed strategy (i.e. any choice of p) is a best response for player 1. That is, every combination of a mixed strategy with ½ < p < 6/10 and player 2 playing Right is a Nash equilibrium in Case 3. We find separately in our analysis of Cases 2 and 4 that the boundary values for p in this range also produce Nash equilibria: (p = ½, Right) and (p = 6/10, Right).

3 Case 4: p = 1/2, player 2 mixes between Right and Center. Similar to Case 2, Bottom weakly dominates Top if player 2 does not play Left, so player 1 will have a unique pure strategy best response of Bottom if player 2 plays Center with positive probability. This rules out the possibility of an equilibrium in this case where each player plays 2 strategies with positive probability. The only remaining possibility is that player 2 plays the pure strategy Right and player 1 plays a mixed strategy with p = ½ -- this is a Nash equilibrium. Case 5: p < 1/2, player 2 plays Center. Bottom is player 1 s unique best response to Center, so the only Nash equilibrium in this case is p = 0 the pure strategy equilibrium ( Bottom, Center ) we identified in our analysis of pure strategies at the outset of the problem. Combining these results, we conclude that there are two pure strategy Nash equilibria (Top, Left) and (Bottom, Center) and a range of mixed strategy equilibria with ½ < P(Up) < 6/10 and player 2 playing the pure strategy Right. (b) LEFT CENTER RIGHT TOP (8, 5) (0, 0) (6, 3) BOTTOM (0, 0) (7, 6) (6, 3) STOP (5, 4) (5, 4) (5, 4) Highlighting the best responses to each pure strategy, we find that there are two pure strategy Nash equilibria: (Top, Left) and (Bottom, Center). These are the same pure strategy Nash equilibria as in the 2x3 game. Next consider the mixed strategy equilibria from (a). In these equilibria, player 2 plays Right, player 1 plays Up with probability p, player 1 plays Bottom with probability 1-p and ½ < p < 6/10. We ve added Stop as a possible strategy for player 1, but this does not affect player 2 s best response calculations if player 1 plays Stop with probability 0. Also, when player 2 plays Right, player 1 strictly prefers Top and Bottom to Stop so the addition of Stop does not affect player 1 s best response calculations. Therefore, the mixed strategy Nash equilibria from (a) remain mixed strategy equilibria of the 3x3 game. (c) If player 1 plays the pure strategy Stop, then player 2 is indifferent between Left and Center, so any strategy with P(Left) = q, P(Center) = 1-q is a best response for Player 2.

4 If player 2 mixes between Left and Center with P(Left) = q, then player 1 gets expected utility 8q from Left, expected utility 7(1-q) from Center and expected utility 5 from Stop. That is, Stop is a best response to P(Left) = q if 5 > 8q and 5 > 7(1-q) These inequalities both hold if 2/7 < q < 5/8, so there is a range of mixed strategy equilibria of the 3x3 game where player 2 mixes with P(Left) = q and 2/7 < q < 5/8, while player 1 plays Stop. (d) Each of the Nash equilibria from (a) produces a subgame perfect equilibrium of the whole game where player 1 plays continue in Stage 1 and then the players follow the prescribed Nash equilibrium in Stage 2. The key point is that player 1 gets expected payoff greater than 5 in every Stage 2 Nash equilibrium. Therefore, player 1 has no incentive to play Stop in Stage 1. However, this analysis also indicates that there is no possibility of a subgame perfect equilibrium where player 1 plays Stop in Stage 1, since any continuation Nash equilibrium in Stage 2 produces a higher expected payoff for player 1 than Stop. (e) There are no weakly dominated strategies in the 3x3 game. Therefore, this game demonstrates that there can be a Nash equilibrium that does not use weakly dominated strategies that is not a subgame perfect equilibrium.

5 2. The Median Voter Theorem (a) If both policies are to the left of x 2, then players 2 and 3 vote for the rightmost policy and it wins. Similarly if both policies are to the right of x 2, then players 1 and 2 vote for the leftmost policy and it wins. Finally, if one policy is to the left of x 2 and the other is to the right of x 2, then player 1 votes for the policy to the left, player 3 vote for the policy to the right, and player 2 votes for the policy that is closer to her ideal point. In every case, player 2 is the swing vote her vote causes her favorite policy to win (though there are some cases with both policies to the same side of x 2, where there is a unanimous vote). (b) Any policy to the left of x 2, will lose a majority vote to another policy that is also to the left, but closer to x 2. Similarly, any policy to the right of x 2, will lose a majority vote to another policy that is also to the right, but closer to x 2. So all policies other than x 2, are amendable. But x 2 wins every majority vote, so it is unamendable. (c) Player 1 would vote for an amendment that is farther away from the status quo point x 1 if that would amendment would lose a majority vote to x 1 and the other amendment would win a majority vote against x 1. That is, if z 2 - x 2 < x 1 x 2 and z 3 - x 2 > x 1 - x 2, (i.e. z 2 is closer to person 2 s ideal point than x 1 and x 1 is closer to person 2 s ideal point than z 2 ) then person 2 would vote for z 2 over x 1 and x 1 over z 3. So person 1 can anticipate that x 1 would defeat z 3, but x 1 would lose to z 2 in majority rule voting. In this case, person 1 should always vote for z 3 over z 2. (d) In any subgame perfect equilibrium, person 2 proposes z 2 = x 2 and this policy wins both majority rule votes. Person 3 could only propose an amendment z 3 that would defeat z 2 in the first stage of voting if this amendment would ultimately lose to x 1. But since person 3 prefers x 2 to x 1, person 3 does better by proposing an amendment that loses to x 2 (or alternately person 3 could also propose the same amendment as person 2: z 3 = x 2 ). (e) There are many other Nash equilibria. For instance, suppose that x 3 - x 2 < x 2 x 1, meaning that player 2 prefers x 3 to x 1. Then there is a Nash equilibrium enacting x 3 based on player 3 s (noncredible) threat to vote for x 1 in the final vote against any other policy other than x 3. That is, player 3 s strategy is to propose x 3, to vote for x 3 whenever possible and to vote for x 1 in the final vote if x 3 loses at the amendment stage. Player 2 s best response to this strategy by player 3 is to vote for x 3 in the amendment stage and also in the final vote (it doesn t matter what player 2 proposes). Given this set of strategies, player 3 s threat to vote for x 1 in the final stage is off the equilibrium path, so it is allowable in a Nash equilibrium. Extending this logic, there is even a Nash equilibrium where x 1 is sustained as the final policy, but it is not very sensible. Player 2 and player 3 both propose z > 2x 2 x 1, and both pledge to vote for x 1 in the final stage regardless of what amendment wins in the second stage. Fixing the strategies of players 1 and 3, x 1 will be enacted regardless of player 2 s strategy (and similarly for player 3), so player 2 s strategy is a best response. The strategy for these players to vote for

6 x 1 in the final stage is not optimal in most circumstances, but it is optimal on the equilibrium path, which pits x 1 against z. 3. High Low Poker with Varying Parameters (a) As in the problem from class, player 1 has four strategies: (Bet, Bet), (Bet, Pass), (Pass, Bet), and (Pass, Pass). Each strategy is a paired list of components, where the first element is player 1 s action with a High card and the second element is player 1 s action with a Low card. Given a strictly positive probability of a high card (λ > 0), we know that (Pass, Pass) is strictly dominated by (Bet, Pass) and that (Pass, Bet) is strictly dominated by (Bet, Bet) because player 1 gets a higher payoff by betting than by passing with a high card. So we focus on player 1 s undominated strategies (Bet, Bet) and (Bet, Pass) in combination with player 2 s two strategies of and. In addition, this is a constant sum game with total payoff to the two players of 20 in all instances. So if we identify a payoff π for player 1 under some circumstances, player 2 s payoff is simply 20 π. Table 1 lists the payoffs to player 1 from each combination of strategies given (A) a High Card and (B) a Low Card. TABLE 1: Panel A: Payoffs with a High Card. Bet, Bet 20 + B, -B 20, 0 Bet, Pass 20 + B, -B 20, 0 Panel B: Payoffs with a Low Card. Bet, Bet -B, 20 + B 20, 0 Bet, Pass 0, 20 0, 20 Table 2 takes a probability weighted average of the payoffs in Table 1 to produce the normal form for the full game. TABLE 2 Normal Form Payoffs for the Full Game Bet, Bet -B + 20λ + 2Bλ, 20, B - 20λ 2Bλ Bet, Pass (20 + B)λ, 20 - (20 + B)λ 20λ, 20-20λ (a) Table 3 shows the Normal Form for the game when B = 10: TABLE 3 Normal Form Payoffs for the Full Game with B = 10 Bet, Bet λ, 30-40λ 20, 0 Bet, Pass 30λ, 20-30λ 20λ, 20-20λ

7 We highlight three pure strategy best responses in Bold. For any value of λ, player 1 s best response to is (Bet, Pass) (avoiding a loss by betting with a low card) and player 1 s best response to is (Bet, Bet) (making maximum gains with both cards given that player 2 is going to ). For any value of λ, player 2 s best response to (Bet, Pass) is since then a bet by player 1 indicates a High card so that player 2 can only lose by choosing. However, player 2 s best response to Bet, Bet depends on the value of λ. will strictly dominate if 0 > 30-40λ of λ > 3/4. weakly dominates is λ = 3/4. That is, the game is solvable by iterative weak dominance if λ > 3/4, with predicted solution [(Bet, Bet), )]. Intuition for Dominated Strategy Solution: When λ is high, player 2 loses proportionally often enough by calling player 1 s bets that player 2 prefers to no matter what player 1 does with a Low card. In essence, the odds are so much against player 2 that it is preferable to simply give up. Then player 1 can take advantage of the advantageous odds by betting with both High and Low cards. In equilibrium, player 2 recognizes that player 1 will bluff by betting whenever the card is Low, but this doesn t happen often enough for player 2 to do anything other than. Importance of λ = 3/4: When B = 10, player 2 stands to gain 30 or lose 10 by calling a bet. This 3 to 1 ratio is exactly matched by the ratio of probability of high card to low card (3/4 to 1/4 is in proportions 3 to 1). As a result, if player 1 always bets and player 2 calls, player 2 s expected payoff when λ = ¾ is exactly equal to 0. Comment on Order of Elimination by Dominance: The order of elimination of strategies can affect the outcome of iterated weak dominance. In this case, however, player 1 has strictly dominated strategies, so eliminating them first cannot affect the outcome of the iterated dominance process. (b) Returning to the game in Table 3, (1) If λ > ¾ then since strictly dominates in the reduced Normal form, there is a unique Nash equilibrium [(Bet, Bet), ]. (2) If λ = ¾ [(Bet, Bet), ] remains a Nash equilibrium. If player 1 plays (Bet, Pass) with probability 1, then player 2 has a strict best response of, which can t be part of a Nash equilibrium. But player 2 can also mix between and so long as player 1 plays [(Bet, Bet)] with probability 1. That is, there is a range of Nash equilibria where player 1 plays [Bet, Bet] and player 2 mixes between and. To identify this range of equilibria, suppose that player 2 plays with probability q and

8 with probability 1-q. Then player 1 s payoffs are π 1 (Bet, Bet) = q( λ) + (1-q) 20 = 20 30q + 40λq π 1 (Bet, Pass) = q(30λ) + (1-q) 20λ = 20λ + 10λq Player 1 prefers (Bet, Bet) if 20 30q + 40λq > 20λ + 10λq OR 20-20λ > 30q - 30λq OR 20 (1- λ) > 30q (1 - λ) OR q < 2/3. That is, when λ = ¾, there is a range of Nash equilibria where player 1 plays ( Bet, Bet ) and player 2 mixes between and with probability q < 2/3 of. (3) If λ < ¾, then there are no pure strategy equilibria and there is a unique mixed strategy Nash equilibrium. Player 2 Mixed Strategy The calculations from (2) indicate that player 1 is indifferent between ( Bet, Bet ) and ( Bet, ) if player 2 plays with probability 2/3. Player 1 Mixed Strategy If player 1 plays ( Bet, Bet ) with probability p, then player 2 s payoffs are π 2() = p(30-40λ) + (1-p) (20-30λ) = p 30λ - 10λp π 2 () = p(0) + (1-p) (20-20λ) = 20 20p - 20λ +20λp Setting these equal gives p 30λ - 10λp = 20 20p - 20λ +20λp OR 30p 30 λp = 10 λ OR p = λ / [3(1-λ)]. That is, when λ < ¾, there is a range of Nash equilibria where player 1 plays ( Bet, Bet ) with probability λ / [3(1-λ)] and player 2 plays with probability 2/3. (c) Table 4 shows the Normal Form for the game when λ = 1/2: TABLE 4 Normal Form Payoffs for the Full Game Bet, Bet 10, 10 20, 0 Bet, Pass 10 + B/2, 10 B/2 10, 10 Highlighting the best responses, we see that since B > 0 and 0 < λ < 1, neither player has a dominated strategy. Therefore, the game cannot be solved by iterated dominance for any value of B. (d) Player 2 Mixed Strategy Suppose that player 2 plays with probability q and with probability 1-q. Then player

9 1 s payoffs are π 1 (Bet, Bet) = q(10) + (1-q) 20 = 20 10q π 1 (Bet, Pass) = q(10 + B/2) + (1-q) 10 = 10 + Bq / 2. Player 1 is indifferent between these two outcomes if 20 10q = 10 + Bq/2 OR 10 = q (10 + B/2) OR q = 20 / (20 + B). Player 1 Mixed Strategy If player 1 plays ( Bet, Bet ) with probability p, then player 2 s payoffs are π 2() = p(10) + (1-p) (10 B/2) = 10 + pb / 2 B/2 π 2 () = p(0) + (1-p) (10) = 10 10p Setting these equal gives 10 + pb/2 B/2 = 10 10p OR pb B = -20 p OR p(b + 20) = B OR p = B / (20 + B). That is, for λ = 1/2, there is a unique Nash equilibrium where player 1 plays ( Bet, Bet ) with probability B / [20 + B] and player 2 plays with probability 20 / (20 + B). (e) When λ is close to zero in (b), player 1 is very unlikely to have a High card. So player 2 would have incentive to any bet unless player 1 plays Bet with very small probability given a Low card. As λ increases, player 1 is more and more likely to have a high card (and always bets given a high card), so player 1 s probability of betting with a Low card increases as well to maintain player 2 s indifference between and in response to a bet. When λ = 1/2, player 1 wins and loses bets equally often given the strategy Bet, Bet. As a result, B does not appear in the payoffs for Bet, Bet. Then the only effect of B on the normal form is from the combination of strategies Bet, Pass and. As B increases, player 1 s payoffs in this cell increase and player 2 s payoffs decrease. To maintain indifference for both players, this cell has to occur with decreasing probability as B increases. That is q is decreasing and p is increasing in B.