2008 Midterm 2: Solutions

14.12 Game Theory Fall 2008 Midterm 2 solutions Problem 1 a) If we substitute in the value e = -1, the payof matrix becomes: X Y A B C 6, 0 0, 0 0, 1 0, 0 2, -1 9, 1 It is easy to see that strategy C for player 2 dominates all the other strategies. Once we eliminate A and B, then strategy X for player 1 will become conditionally dominated. Therefore, the unique Nash equilibrium of this game is (Y, C). In class we have seen a theorem which states that whenever a stage game has a unique Nash equilibrium and it is repeated a fnite number of times, then there is only one subgame perfect equilibrium with the players playing the Nash equilibrium in every period. b) Let's now substitute in the value e = 2. The stage game becomes: X Y A B C 6, 6 0, 0 0, -2 0, 0 2, 2 9, -2 We can immediately see that C is dominated for player 2, thus we can safely ignore it when computing the Nash equilibria of the game. There are 2 pure-strategy Nash equilibria (X, A) and (Y, B) and one mixed-strategy Nash equilibria (01 , 0 2 ) = 1 X + � Y, 1 A + � B . The payofs associated with the latter equilibrium are �2 , �2 , which are lower than those associated with the two pure-strategy Nash equilibria. Thus we can try to use the usual trick of rewarding the players when they comply with the proposed strategy and punish them in case of deviation. The reward will be the most proftable Nash equilibrium (X, A) while the punishment will be the mixed-strategy equilibrium (0 1 , 0 2 ). 1. Consider the strategies: 1 = play Y at t = 1 play X at t = 2 if (Y, C) play 0 1 otherwise = play C at t = 1 play A at t = 2 if (Y, C) play 0 2 otherwise and 2 1 The only player who may have an incentive to deviate is player 2 (player 1 is getting the highest payo¤ at t = 1). If player 2 deviates, he will play B and get 2 in the …rst period and 32 in the second period. His total payo¤ from deviating will be 72 which is lower than the total payo¤ of 4 he gets from not deviating. Thus, (Y; C) can be played in the …rst round of a SPE. 2. Consider the strategies: 8 < and play X at t = 1 play X at t = 2 if (X; B) s1 = : play 1 otherwise 8 < play B at t = 1 play A at t = 2 if (X; B) s2 = : play 2 otherwise Also in this case player 2 is the player with the most pro…table deviation. Indeed he can deviate by getting 6 instead of 0 in the …rst round. Therefore, his total payo¤ from deviating will be 15 which is higher than the total payo¤ of 6 he gets 2 from not deviating. Thus, (Y; C) cannot be played in the …rst round of a SPE. 3. Consider the strategies: s1 = and s2 = 8 > > > > < > > > > : 8 > > > > < > > > > : play X at t = 1 play X at t = 2 if (X; B) play 1 otherwise play X at t = 3 if (X; B) , (X; A) play 1 otherwise play C at t = 1 play A at t = 2 if (X; C) play 2 otherwise play A at t = 3 if (X; C) , (X; A) play 2 otherwise The player with the most pro…table deviation is now player 1. She can deviate in period 1 by getting 9 and then 32 in each of the following periods. Her total payo¤ from deviating will be 12 which is therefore equal to the total payo¤ of 12 she gets from not deviating. Thus, (X; C) can be played in the …rst round of a SPE. c) From part a) we know that whenever = 1, C is a dominant strategy for player 2. Thus, in any Bayesian Nash Equilibrium type = 1 of player 2 will always play his dominant strategy, that is, s2 ( 1) = C. Let’s now consider player 1. Given s2 ( 1) = C, player 1 will get at most 12 6 + 21 0 = 3 by playing X (which happens 1 1 when type = 2 plays A) and at least 2 0 + 2 9 = 4:5 by playing Y (which happens when type = 2 plays A). In other words, given s2 ( 1) = C, X is dominated by Y for player 1. 2 Let’s now consider type = 2 of player 2. As shown already above, for this type C is a dominated action. Thus type = 2 will choose either A or B (or he will randomize between the two). Given s1 = Y and s2 ( 1) = C, type = 2 will face the following choice: 0 if s2 = A u2 (s1 ; s2 ( 1); s2 ; = 2) = 2 if s2 = B Thus type = 2 will choose B. Therefore (s1 ; s2 ( 1); s2 (2)) = (Y; C; B) is the unique BNE of this game. 3 Problem 2 a) To be a subgame perfect equilibrium, neither of the players must have a single deviation that could make them better of, given any possible history. Therefore, to show that this is not SPE, we must fnd a history and a deviation that will make one of the player's better of. The key lies in Bob's acceptance strategy. The strategy says that Bob will accept if X 2 1. So he will reject anything less than 1. This is not SPE. Suppose Alice ofers X E (Ps + PR , 1). Let's check the single deviation principle. If he accepts, he will get X. If he rejects, with probability Ps + PR he will get 1 in the next round (whether he or Alice ofers), so he will get a payof of Ps + PR . Since X > Ps + PR , he will be strictly better of deviating by accepting Alice's ofer. b) To fnd the SPE, we will look at their acceptance strategies and apply the single deviation principle: Alice will accept if X 2 XA . Suppose Bob ofers some Xx < XA . If Alice were to deviate and to accept, she will get Xx. If she rejects (and then strategies are played as called for), there will be a Ps probability that Bob will ofer again and she'll get XA , there will be a PR probability that she will ofer and get 1 - XB , and there will be a PE probability that she will get 0. So her payof, if she rejects would be Ps XA +PR (1-XB ). Therefore, for her NOT to have an incentive to deviate, it must be the case that: Xx � Ps XA + PR (1 - XB ) �x X < XA Now consider an Xx > XA . If Alice deviates and rejects this ofer, she will get Ps XA + PR (1 - XB ). If she accepts it as she is supposed to in the strategy, she will get Xx. Therefore, for her NOT to have an incentive to deviate, it must be the case that: Xx 2 Ps XA + PR (1 - XB ) for all Xx 2 XA Therefore, the inequality must bind for Xx = XA . So, we get the equation: XA = Ps XA + PR (1 - XB ) Parallel analysis for Bob's acceptance strategy will yield the equation: XA = Ps XB + PR (1 - XA ) Solving these two equations and two unknowns will yield: XA = X B = PR 1 + PR - Ps Thus, the SPE of the game will be the strategies as listed in the problem where XA and XB are as in the above equation. We have calculated it by checking the single deviation principle for both of their acceptance strategies. (You should therefore get 15 points up to this point). The fnal thing that we need to do is to check, using the single deviation strategies, that neither of them have an incentive to deviate when ofering. Alice is suppose to ofer XB to Bob. If she does this, he will accept and she will get a payof of 1 - XB . If she deviates by ofering more than XB to Bob, he will accept (given his 4 strategies), and she will do strictly worse of. Suppose she deviates by ofering X < XB . Then Bob will reject, and she will get an expected payof of Ps (1 - XB ) + PR XA . She will not have an incentive to deviate as long as: (1 - XB ) 2 Ps (1 - XB ) + PR XA . Plug in the values of XA and XB to see that this is equivalent to the condition that j Ps -Psj +PR 1-Ps 2 . So we just need to check that: 1 - Ps 2 Ps - Ps2 + PR2 . 1+PR -Ps 1+PR -Ps Rearranging, we get that this would mean that 1 - 2Ps + Ps2 - PR2 2 0. Which can be changed to: (1 - Ps )2 - PR2 2 0, or (1 - Ps + PR )(1 - Ps - PR ) 2 0. Since the frst factor is clearly greater than 0, and the second factor is equal to PE > 0, this must be true. Therefore, Alice will not have an incentive to deviate when she is ofering XB to Bob. With the exact same analysis, Bob will not have an incentive to deviate when ofering XA to Alice. 5 Problem 3 1. (a) If the entrant exits, qI = 12 : (-2 points for not noting this as part of SPE). If the entrant enters, k is a sunk cost, so this is strategically equivalent to a standard Cournot game, and, from class, qI = qE = 13 : Entrant enters, since 13 (1 23 ) :1 = 19 :1 > 0, so higher pro…ts from entering than exiting. (b) There are several possible subgames we need to check. When I say “deviation” here, I mean a deviation away from q = 14 : – First, note that all subgames which occur after a deviation satisfy the single deviation principle, since, by (a), the strategies in these history are repeated SPE of the stage game. – Second, in no-deviation histories, note that the entrant never wants to exit: this doesn’t a¤ect the future, and it lowers his pro…t in the present period. – Third, in a no-deviation history where the entrant enters, the single deviation principle requires max qE (1 qE 1 4 qE ) k + 1 ( 1 9 k) 1 8 k + 1 ( 1 8 k) and, since the k’s fall out of this equation, the same condition has to hold for the incumbent. Solving the max gives qE = 3=8, so the condition reduces to 9 1 1 1 + 64 1 9 8 1 8 or 72 . 1 64 Solving for the critical gives b = 9 : (8 points for getting this far) 17 – Fourth, consider the nodes when there has been no deviation (away from q = 14 ) but after an “exit”by the Encumbent. According to the proposed strategies, the incumbent is supposed to produce qI = 14 . He’d prefer to produce qI = 12 , but this will lead to a deviation history. He’s willing to produce 14 precisely when: 1 + 4 1 (c) 1 ( ) 9 3 + 16 1 1 ( ) 8 9 9 Solving for the critical gives b = 11 :. Since this is bigger than 17 ; this is the critical value for these strategies to be an SPE. – Note that the nodes from the previous bullet aren’t very important –it would be easy to adjust the strategies so that this would be an SPE for b = 9 . This will be my “trick”for part (d). 17 Yes. There are many possibilities. Here’s one: 6 – Normal mode: always exit and produce qI = 12 . If enter, produce qI = 1 and qE = 0: – Trigger mode: play the SPE from (a) forever. – Stay in Normal mode unless Enter AND qI 6= 1: Using single deviation principle, note that the only place where anyone would potentially want to deviate is the incumbent producing qI = 1 after an entry. She is willing to do this whenever 0+ or when (d) 1 4 1 1 + 4 1 1 9 ; > 59 : Since :9 > 59 ;, we’re good. Something like “carrot and stick”from class would work. But here it’s easier if you got (b) right. Simply change the strategies from (b) from – if any producer has ever produced a quantity other than 14 , “trigger” to playing the subgame perfect equilibrium from part (a) in every period. TO – if any producer has ever produced a quantity other than 14 AFTER ENTRY, “trigger”to playing the subgame perfect equilibrium from part (a) in every period. AND from – Every producer produces q = 14 , so long as no producer has ever produced a quantity other than 14 : TO – After entry, every producer produces q = 14 , so long as no producer has ever produced a quantity other than 14 IN A ROUND WHERE THERE WAS ENTRY. After Exit, produce q = 12 : Then the same reasoning from (b) applies, and we get an equilibrium when9 or greater, which is lower than the correct answer from (b). ever b = 17 7