ECON 803: MICROECONOMIC THEORY II Arthur J. Robson Fall 2016 Assignment 9 (due in class on November 22)

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1 ECON 803: MICROECONOMIC THEORY II Arthur J. Robson all 2016 Assignment 9 (due in class on November 22) 1. Critique of subgame perfection. 1 Consider the following three-player sequential game. In the first stage, can either play L, ending the game with payoffs (6, 0, 6), or play R, which gives the move to. can then play either r, ending the game with payoffs (8, 6, 8), or play l, thereby moving the game to the third stage. In stage 3, and P3 (but not ) play a simultaneous-move coordination game: they each choose or. If their choices differ, they each receive 7 and gets 10; if their choices coincide, all three players get 0. (a) Draw the extensive form game in question. Solution: L R (6, 0, 6) l r P3 (8, 6, 8) (0, 0, 0) (7, 10, 7) (7, 10, 7) (0, 0, 0) (b) ind ALL subgame perfect equilibria in this game. Solution: Recall that a SPE induces a NE in every subgame. There are three subgames in this game (each stage initiates a subgame). In the smallest subgame, the coordination game in stage 3, there are three NEs. Two of these NEs are in pure strategies: (i) chooses and P3 chooses, and (ii) P3 chooses and chooses ; both equilibria lead to the payoff vector (7, 10, 7). There is also a mixed strategy NE, in which both and P3 randomize between and with equal probability. This mixed NE leads to an expected payoff vector (3.5, 5, 3.5). 1 rom udenberg and Tirole (1991) and Rabin (1988). Page 1 of 5

2 In stage 2, chooses l if the equilibrium in stage 3 is either of the two pure strategy NEs; otherwise, he chooses r if the equilibrium in stage 3 is in mixed strategies. In stage 1, chooses R, anticipating a payoff of either 7 (when she can successfully coordinate with P3 in stage 3) or 8 (when she fails to coordinate with P3 and hence playing the mixed NE in stage 3). To summarize, there are three SPEs: two in pure strategies, (R; l; ) and (R; l; ), and one involving mixing ( R( ); r; ( )). (c) Take note of s choice in stage 1 in all of the subgame perfect equilibria. Discuss carefully whether it could be rational for to choose a different action in stage 1. Solution: A subgame perfect equilibrium requires not only that and anticipate NEs in all subgames, but also that the NEs they anticipate are the same. This is a point that Rabin (1988) makes in critiquing the notion of subgame perfection. In our example, suppose anticipates the mixed strategy NE to be played in stage 3 (if reached). But she also believes that would expect the stage 3 outcome to be one of the pure strategy NEs (and hence would choose l). Then would choose L to secure a payoff of 6, as opposed to the expected payoff of 3.5 if he chooses R as dictated by her equilibrium strategy in any SPE. 2. Consider the following extensive form game. A (1, 1) L (0, 2) B a R b (2, 0) (3, 3) (a) ind all subgame perfect equilibria of this game. Solution: Using backward induction, we can find (Bb, R) to be the unique SPE. (b) Convert the game to its normal form. ind all Nash equilibria in pure strategies. Page 2 of 5

3 Solution: The normal form is Aa Ab Ba Bb L 1, 1 1, 1 2, 0 2, 0 R 1, 1 1, 1 0, 2 3, 3 In addition to the SPE (Bb, R), there are two more NEs that are not subgame perfect: (Aa, L) and (Ab, L). (c) Verify whether these Nash equilibria are trembling hand perfect. Solution: We first argue that the SPE (Bb, R) is THP. Suppose for, the sequence of totally mixed strategy is σ k 2 = ɛk L + (1 ɛ k )R, for k N and some appropriate ɛ (0, 1). Clearly, σ k 2 R as k We need to show that Bb is a best response to σk 2, for all k N. Observe that u 1 (Aa, σ k 2) = 1 u 1 (Ab, σ k 2) = 1 u 1 (Ba, σ k 2) = 2(1 ɛ k ) u 1 (Bb, σ k 2) = 3(1 ɛ k ). Thus for Bb to be a best response to σ k 2, it suffices to let 3(1 ɛ) 1, or ɛ 2 3. We also need R to be a best response to s sequence of completely mixed strategy σ k 1. Suppose σ k 1 = ɛk Aa + ɛ k Ab + ɛ k Ba + (1 3ɛ k )Bb, where ɛ k (0, 1) and 3ɛ k < 1 for each k. Clearly, σ k 1 Bb as k. We need to find ɛk such that u 2 (σ k 1, L) u 2(σ k 1, R) ɛ k + ɛ k + 2ɛ k + 2(1 3ɛ k ) ɛ k + ɛ k + 3(1 3ɛ k ). It suffices to pick ɛ 1 5. Hence, there exists a sequence of completely mixed strategy profiles σ k = (σ1 k, σk 2 ) as described above, such that σk (Bb, R) and each player s equilibrium strategy is a best response to the other player s trembled strategy in σ k for every k. Therefore, (Bb, R) is a THP NE. Next, we argue that (Aa, L) (and (Ab, L)) is THP. Let σ k 2 = (1 ɛk )L + ɛ k R, and we want to find an ɛ (0, 1) such that Aa is a best response to σ2 k. Observe that u 1 (Aa, σ k 2) = 1 u 1 (Ab, σ k 2) = 1 u 1 (Ba, σ k 2) = 2ɛ k u 1 (Bb, σ k 2) = 3ɛ k. Page 3 of 5

4 It suffices to pick ɛ 1 3. We also need to make L a best response to σk 1 = (1 3ɛk )Aa + ɛ k Ab + ɛ k Ba + ɛ k Bb. In fact, given this σ1 k, any ɛ (0, 1) will make L a best response: (1 3ɛ k ) + ɛ k + 2ɛ k + 2ɛ k > (1 3ɛ k ) + ɛ k + 3ɛ k, ɛ (0, 1), k N. Hence, we conclude that (Aa, L) is a THP NE. The argument for (Ab, L) being THP is similar. This example shows that trembling hand perfection, as we have studied in class (which is defined based on normal form games), is a weaker criterion than subgame perfection in ruling out implausible NEs. A stronger version of THP is the extensive form trembling hand perfection, which requires a player s strategy to be robust not only to other players mistakes, but also to one s own mistakes made at different information sets. See p of MW for a discussion. 3. Consider the following three-player normal form game, where S 1 = {U, D}, S 2 = {L, R}, and S 3 = {X, Y}: L R U 1, 1, 1 1, 0, 1 D 1, 1, 1 0, 0, 1 X L R U 1, 1, 0 0, 0, 0 D 0, 1, 0 1, 0, 0 Y (a) Perform iterated deletion of weakly dominated strategies. What are the strategies that survive this process? Solution: L and X are the strictly dominant strategy for players 2 and 3, respectively. However, there is no domination between U and D, either before or after removing the other players dominated strategies. So the strategies that survive IDWDS are {{U, D}, {L}, {X}}. (b) ind all Nash equilibria (both pure and mixed). Does any of the Nash equilibrium involve weakly dominated strategies? Solution: In a NE, both players 2 and 3 would play their respective dominant strategy. Player 1 is then indifferent between U and D. So the NEs of this game can be described by (pu + (1 p)d, L, X), for any p [0, 1]. None of the NEs involves weakly dominated strategies. (c) ocus on the pure strategy Nash equilibria. Are they trembling hand perfect? Page 4 of 5

5 Solution: The two pure strategy NEs are (U, L, X) and (D, L, X). We first argue that (D, L, X) is not THP. Let the profile of completely mixed strategies be σ(ɛ), where σ 1 (ɛ 1 ) = ɛ 1 U + (1 ɛ 1 )D σ 2 (ɛ 2 ) = (1 ɛ 2 )L + ɛ 2 R σ 3 (ɛ 3 ) = (1 ɛ 3 )X + ɛ 3 Y, so that as ɛ (0, 0, 0), σ(ɛ) (D, L, X). It is straightforward to verify that, for player 2, L is the best response to players 1 and 3 s completely mixed strategies. Similarly, X is player 3 s best response to players 1 and 2 s completely mixed strategies. Now consider player 1 s best response. u 1 (U, σ 2 (ɛ 2 ), σ 3 (ɛ 3 )) = (1 ɛ 2 )(1 ɛ 3 ) + ɛ 2 (1 ɛ 3 ) + (1 ɛ 2 )ɛ 3 = 1 ɛ 2 ɛ 3 (1) u 1 (D, σ 2 (ɛ 2 ), σ 3 (ɛ 3 )) = (1 ɛ 2 )(1 ɛ 3 ) + ɛ 2 ɛ 3 = 1 ɛ 2 ɛ 3 + 2ɛ 2 ɛ 3. (2) or D to be a best response, we need 1 ɛ 2 ɛ 3 + 2ɛ 2 ɛ 3 1 ɛ 2 ɛ 3, or 3ɛ 2 ɛ 3 ɛ 2 + ɛ 3. But as ɛ i 0 (i = 2, 3), this inequality will eventually be violated. or example, when ɛ 2 (0, 1 3 ), the inequality is reversed regardless of the value of ɛ 3. Therefore, D cannot be a best response to the entire sequence of totally mixed strategies of players 2 and 3. Hence, (D, L, X) is not THP. Using (1) and (2), we can also verify that (U, L, X) is THP, since all we need is to find a sequence of ɛ 2 and ɛ 3 converging to zero that makes U a best response for player 1. The condition for U being a best response is 3ɛ 2 ɛ 3 ɛ 2 + ɛ 3, or 3ɛ ɛ 2 ɛ 3. It suffices to let ɛ 2 be any sequence bounded above by 1 3 and converging to 0, and ɛ 3 can be any sequence that converges to 0. In contrast to two-player normal form games, where THP is equivalent to the requirement that strategies must not be weakly dominated, this example shows that in the case of more than two players, there may be NEs with undominated strategies that are not THP. Page 5 of 5