POLS 8000 Dougherty Spring 2018 Homework Assignment 4 (answers) A. The questions in this section use the following stage game: CD C -1,-1 -10,0 C 0,-10 -5,-5 1. (5 points) Suppose the game is repeated for exactly 100 finite rounds, δ = 1, and player 2 plays TFT. Which would be more rational for player 1: TFT or always defect? Why? What is the difference between the two payoffs for player 1? For 100 finite rounds with δ = 1, (TFT;TFT ) will produce 100 C's for player 1. ... Hence, U1(TFT;TFT ) = 100(−1) = −100. For player 1, (AllD,TFT) will produce 0 for t = 1 and -5 for the remaining 99 rounds. ... Hence, U1(AllD; T F T ) = 99(−5) = −495. Therefore, TFT is more rational for player 1 than AllD, because it produces a greater payoff. 2. (5 points) Compare your answer in the previous question to the subgame perfect equilibrium prediction for finite games. What is the difference be- tween the two? Why? The distinction is based on the difference between Nash Equilibrium (which can miss some sequential rationality) and subgame perfect equilibrium, which is based on antic- ipation. Although player 1 receives greater utility by playing TFT against TFT than by playing AllD against TFT, this does not change the fact that TFT would not be a rational strategy in the final period as noted by backward induction. Because it is not rational to cooperate in the last round, it's not rational to cooperate in the second to last round, and by similar reasoning anticipation makes cooperation unravel. 1 B. The questions in this section use the following infinitely repeated stage game CD C 2,2 0,3 C 3,0 1,1 3 Delayed Grim Trigger. In the delayed grim trigger, players wait one period before implementing the punishment part of the strategy. Specifically, the strategy says to choose C in period 1 and any history in which (C,C) was played in every previous period. If (C,C) is not played in period t, then choose C in period t + 1, and then play D in period t + 2 and every subsequent period. (a) (5 points) Compute the discounted sum of payoffs for two players playing (DGT, DGT) infinitely. (DGT,DGT) Player 1: C,C,C,C... Player 2: C,C,C,C... EU1(DGT; DGT ): 2 + 2δ + 2δ2 + ::: 2 (1−δ) . (b) (5 points) What is the best possible deviation for player 1? What would be the discounted sum of the payoffs for player 1 if he/she plays the deviation instead? The best deviation against the delayed grim trigger (DGT) is AllD. To see this, sup- pose Player 1 deviates in the first period. In the second period, Player 2 will still play C then plays D from period 3 onward. Given that the punishment was triggered in period 1, the best thing for Player 1 to do is to stick with D in every subsequent period. Player 1's discounted sum of payoffs from (AllD, DGT) Player 1: D,D,D,D... Player 2: C,C,D,D... EU1(AllD; DGT ): 3 + 3δ + δ2 + δ3 + ::: δ2 3 + 3δ + (1−δ) . (c) (5 points) Set up the Nash condition. In other words, solve for the values of (between 0 and 1) where (DGT, DGT) can be sustained as an equilibrium. 2 This deviation is rational if: δ2 3 + 3δ + (1−δ) > 2=(1 − δ) 3 − 3δ + 3δ − 3δ2 + δ2 > 2 1 − 2δ2 > 0 1 2 2 > δ q 1 2 > δ. Hence, (DGT,DGT) is a Nash equilibrium if δ ≥ p1=2. 4 Pavlov. In this strategy, except for the first period, the choice of action in period t + 1 depends only on the actions in the last period (period t). The strategy is to play C in the first period, then play C if the actions in the last period were (C,C) or (D,D) and to play D if the actions in the last period were either (C,D) or (D,C). Consider two possible deviations: 1) always defect or 2) play D for the first two periods and then choose C in period 3 onward. (a) (5 points) Compute the discounted sum of payoffs for two players playing (pavlov, pavlov) infinitely. (Pavlov,Pavlov) Player 1: C,C,C,C... Player 2: C,C,C,C... EU1(P avlov; P avlov): 2 + 2δ + 2δ2 + ::: 2 (1−δ) . (b) (5 points) What is the best possible deviation for player 1? What would be the discounted sum of the payoffs for player 1 if he/she plays the deviation instead? Consider the first deviation where Player 1 plays AllD. Player 1's discounted sum of payoffs from (AllD, Pavlov): Player 1: D,D,D,D... Player 2: C,D,C,D... EU1(AllD; P avlov): 3 + δ + 3δ2 + δ3 + ::: 3(1 + δ2 + δ4 + :::) + δ(1 + δ2 + δ4 + :::) Let w = δ2, then 3(1 + w + w2 + w3 + :::) + δ(1 + w + w2 + w3 + :::) 3 δ (1−w) + (1−w) 3+δ (1−δ2) . 3 Next, consider the second deviation where Player 1 plays DDthenC. Player 1's discounted sum of payoffs from (DDthenC, Pavlov) Player 1: D,D,C,C... Player 2: C,D,C,C... EU1(DDthenC; P avlov): 3 + δ + 2δ2 + 2δ3 + ::: 3 + δ + 2δ2(1 + δ + δ2 + :::) 2δ2 3 + δ + (1−δ) . (c) (5 points) Set up the Nash condition. In other words, solve for the values of (between 0 and 1) where (pavlov, pavlov) can be sustained as an equi- librium. The deviation to AllD is rational if: EU1(AllD; P avlov) > EU1(P avlov; P avlov) 3+δ 2 (1−δ2) > (1−δ) 3 − 3δ + δ − δ2 > 2 − 2δ2 1 − 2δ + δ2 > 0 (1 − δ)2 > 0 (1 − δ) > 0 1 > δ This will always be true because for infinite games we assume δ < 1. Hence, the deviation is always rational and (pavlov,pavlov) is not an equilibrium. The deviation to DDthenC is rational if: EU1(DDthenC; P avlov) > EU1(P avlov; P avlov) 2δ2 2 3 + δ + (1−δ) > (1−δ) 3 − 3δ + δ − δ2 + 2δ2 > 2 1 − 2δ + δ2 > 0 (1 − δ)2 > 0 (1 − δ) > 0 1 > δ. This will always be true because for infinite games we assume δ < 1. Hence, the deviation is always rational and (Pavlov,Pavlov) is not an equilibrium. 4 C. Median Voter Theorem 5. Several people sit on a committee (designated by letters A ... H). Each committee member has an ideal point marked by a number. Numbers represent alternatives on a single-dimension. Given this information, find the median of each committee. Medians are determined by ordering numeric values from left to right and canceling the same number of alternatives to the left as you do to the right. The median alternative is the number(s) that remain. a. D=16, E=22, C=57, G=57, F=63, B=100, A=200 median is C=G=57. b. D=16, E=22, C=25, F=63, B=100, A=200 median is the interval from 25 to 63, inclusive, C to F, or [25, 63]. c. D=16, E=22, H=22, C=25, G=57, F=63, B=100, A=200 median is the interval from 25 to 57, inclusive, C to G, or [25, 57]. 6. Assume that each committee member from the previous problem has single peaked and symmetric utility. Describe the win set for each part of question 1 assuming that the status quo equals 50 [hint: you should have a separate answer for parts a-d of the previous question. Your answer should be in a range of numbers. You “might” use >, $, or interval notation to denote your answer]. Single peaked and symmetric utility allows us to use the corollary to the median voter theorem. This corollary shows that the alternative closer to the median voter wins in pair-wise majority rule. Hence the win set is the set of alternatives closer to the median voter than the status quo is to the median voter. These are marked with vertical lines and indicated below. a. Q <–-16—22-----------------|------57-----63-|----------------------------100-----------------------200---> 50 < W(Q) < 64 (50, 64) b. Q <–-16—22–25------------|--------------63------------------------------100-----------------------200---> Note that the median is 25 to 63 inclusive. Since 50 is in the median, nothing can beat it using pair-wise majority rule. Hence, the win set is empty. c. Q <–-16—22–25-------------|----57-----63------------------------------100-----------------------200---> The median is 25 to 57 inclusive. Since Q is in the median, nothing can beat it using pair-wise majority rule. Hence, the win set is empty. 5 D. Pivotal Politics 7. Pivotal Politics. Assume a game with one chamber, filibuster pivots, a Presidential veto, two-a potential two- thirds override, which proceeds as follows. 1) the median of the sole chamber (m) proposes bill b; 2) the filibuster pivots (fL and fR) successfully filibuster or cloture is invoked; 3) if the bill passes, the President (p) signs or vetoes; and 4) if the bill is vetoed, the veto pivots (vL and vR) override or sustain (as shown in the figure to the right). All alternatives are on a single dimension without bounds. The status quo (q) is one of those alternatives. The actors are in the following locations: p = 0, vL = 5, fL = 10, m = 12, fR = 15, and vR = 17. a. (10 points) Assume the status quo q = 8. What is the policy outcome of this game in sub-game perfect equilibria? In other words, what number results? Why? Explain all of the steps in the backward induction and state the pure strategy, sub-game perfect equilibria for the game.