Exercises Solutions: Game Theory

Transcription

1 Exercises Solutions: Game Theory Exercise. (U, R).. (U, L) and (D, R). 3. (D, R). 4. (U, L) and (D, R). 5. First, eliminate R as it is strictly dominated by M for player. Second, eliminate M as it is strictly dominated by U for player. Third, eliminate L as it is strictly dominated by M for player. Last, eliminate U as it is strictly dominated by D for player. The Nash equilibrium is (D, M). Note: Alternatively, you can show that this is the unique Nash equilibrium by applying the usual reasoning in terms of best responses. Exercise. For all A, B, C, D that satisfy C A and B D.. For all A, B, C, D that satisfy (A > C and B D) or (C A and D > B). 3. For all A, B, C, D that satisfy A > C and D > B. Exercise 3. There is a unique Nash equilibrium given by (A, H): player plays A and player plays H.. The outcome that maximizes the sum of the payoffs is (L, H): plays L and plays H. 3. Player benefits more from collusion. It suffices to propose 0 to firm so that it accepts this action profile. Exercise 4. There are two Nash equilibria. First equilibrium: Airbus produces and Boeing does not. Second equilibrium: Boeing produces and Airbus does not. There is a coordination problem on which equilibrium will be played in practice.. The new payoff matrix is: A\B Produce Not produce Produce 40 \ \ 0 Not produce 0 \ 00 0 \ 0 The subsidy resolves the coordination problem. Now, there is only one equilibrium: Airbus produces and Boeing does not.

2 Exercise 5. The payoff matrix is given by: 5 \ 5 3 \ 6 6 \ 3 4 \ 4. Producing units is strictly dominant for both firms. The unique Nash equilibrium is therefore given by both firms choosing to produce units. In the sequential game, Firm chooses to produce units and then the firm does the same. Exercise 6 Objective: One needs to show that if there are many equilibria, the payoffs received by a given player are identical. Suppose that (high, left) and (low, right) are the equilibria. The associated payoffs then are (a, a) and (b, b). What does this imply? Apply the standard reasoning for finding a Nash equilibrium. We get that a d, b c... This requires globally that b c a d b, which is impossible unless we have equalities everywhere. And a = b. Therefore the payoffs received by the agents should be the same in each equilibrium. You can redo this analysis by taking some other equilibria - the final conclusion has to be the same. Exercise 7 If y = y <, then each of the players wants to increase by a little bit his effort in the fight. They will win then with probability instead of 0,5. y y cannot be an equilibrium. The one who makes the highest effort would have an incentive to slightly decrease the effort in the fight. There is a unique Nash equilibrium in which y = y =. Clearly it is inefficient as both put maximum effort in fighting and the good is eventually not produced. Exercise 8. To find a Nash equilibrium of the game, we need to derive the best-response or reaction curves of both players to each other s actions c (c ) and c (c ). Hence, we need to maximize each neighbor s objective function. For neighbor, the maximization of his objective is (taking c as a given) max u (c, c ) = b (c + c ) + ( c ) + ( c )(c + c ). () c The first order conditions yield b + c c c = 0 Rearranging we get The problem of neighbor is c (c ) = b c () max c u (c, c ) = b (c + c ) + ( c ) + ( c )(c + c ). (3) After taking the first order conditions and rearranging as above we get c (c ) = b c (4)

3 If b = b = b then the Nash equilibrium values c and c have to solve the system of equations c = b c which yields c = b c c = c = b 3 As b, the weight they attach to having a clean walkway, increases both neighbors have an incentive to clean it longer.. We need to maximize now the sum of the two utilities, taking into account that b = b = b. max u (c, c ) + u (c, c ) c,c = b(c + c ) + ( c ) + ( c )(c + c ) + +b(c + c ) + ( c ) + ( c )(c + c ) = b(c + c ) + ( c c )( + c + c ) The first order conditions yield two identical relationships (notice that c and c enter symmetrically into the problem) b c c + = 0 b c c + = 0 Therefore, it must be true that c + c = b + for both neighbors to achieve the maximum of payoffs. Comparing with the Nash outcome we see that this is impossible to achieve if they are playing a noncooperative game as in the Nash outcome c + c = 3 b only. 3. (More difficult) We can use the maximization problems () and (3) and the respective best response curves () and (4) we found in question this time keeping the b and b in place. To find the Nash equilibrium, we need now to find the solution to equations c = b c (5) which yields c = b c (6) c = 3 b 3 b (7) c = 3 b 3 b (8) Now, b > b. Moreover, we know that b [0, ] and b [0, ]. Therefore, we know that c > 0. But if 3 b 3 b < 0 or b < b then we cannot use the formulas (7) and (8) since we no longer have a solution in the interior of the set since we would have from (8) c < 0 (you cannot spend negative time cleaning the pavement). Then (6) cannot hold and the closest to the optimal value that neighbor can be is when c = 0. Then from (5) we get c = b. Exercise 9 - Second price auctions See solution in slides 5-7 of Part. Exercise 0 - Committee and voting We will eliminate dominated strategies of players in an iterative manner, i.e, looking at one player at a time. 3

4 Starting with the committee member A (the President), we see that he will not vote 3 (his least preferred candidate), and this is his dominated strategy (the one that is always worse no matter what other players do). We can see this in the following way: suppose that player A would play 3. Then, if players B and C would play different strategies or, he obtains 3, his least preferred candidate. If players B and C happen to vote for the same candidate,, or 3, A clearly cannot change the outcome of the election, so in that outcome playing 3 is at least as good (bad) as other strategies. In the case that one of the players - B or C vote for 3, player A will get 3 voted in by using this strategy, his least preferred candidate. We can see thus that strategy 3 is weakly dominated for player A. A rational player A will not use therefore strategy 3 ; hence players B and C can eliminate this from the set of strategies played by A. Let s scrutinize the strategies of player B in the reduced game, i.e. when A does not play 3. Suppose we look at B s strategy of voting for. If A plays, no matter what C plays candidate is going to be voted in, the least preferred candidate by B. If A plays and C plays, there is no difference in what player B chooses (all strategies give the same payoff in that outcome). If A plays, and player C plays 3, candidate is voted in as the privilege of the president (tie-breaking) is exercised. Note that this is the case where strategy dominates playing 3! Hence, is not dominated in the reduced game for player B. We can see that neither playing or 3 is dominated either; player B does not have a dominated strategy even if the execution of strategy 3 by player A is eliminated from the strategy set. Let s turn to player C now; Suppose he would play. If A and B play or, the choice of C s strategy does not matter. But, if A plays and B or 3, playing is dominated for C by playing 3 or (i.e., it is weakly the worst strategy to follow). If A plays and B plays, playing is dominated by and 3 gives exactly the same payoff, and if A plays and B plays again it is better for C not to exercise - or 3 will make candidate being voted in. Therefore, is weakly dominated for C. Rational players know thus that A will not exercise 3 and C will not play. Suppose we turn now to player A and consider the strategy of playing in the reduced game. If either B or C play this strategy is weakly dominated by. If B plays and C plays 3, A can get his preferred candidate voted in by playing - the president s privilege prevails. When both B and C play 3, though, neither playing or gives A an advantage. We conclude that for A, strategy is dominated (weakly) by. A rational player A will thus never play either, and the only strategy he would follow is. This means, for B, that if A plays and C plays 3, the strategy is dominated for B, as is because then the least preferred candidate (by the presidents A privilege) is voted in. And if both A and C play, no strategy of B gives him better payoffs. Hence, playing or is weakly dominated for B and a rational player B will never use such strategies. It is easy to check that then playing is dominated for C in the remaining reduced form of the game. We have thus a solution: the only remaining non-dominated strategies are for A, 3 for B and 3 for C. Candidate C is thus voted in. Exercise - Simultaneous vs. sequential choices. (High, Right) is the only Nash equilibrium.. One needs to start the analysis from the decisions of the Column player for each decision taken by the Row player. By backward induction, we can therefore analyze the decision of the Row player. The Row player will play Down and the Column player plays Left. Exercise - The role of agendas. We use backward induction to solve the game. At the last possible node, a vote is made either for K or to keep the seat vacant. The outcome through majority voting is to vote for K. We can therefore move to the preceding node of the game tree, and we know that rational players 4

5 are going to ponder at that stage whether they will vote G in or continue voting (which means for rational players that then they will effectively vote for G or K if they continue). Therefore, through majority voting, senators decide for G. Moving back one node they will vote whether to vote for B or continuing the vote, but they know effectively that a vote for continuation is a vote for G. Therefore, they prefer B and this is the candidate that is voted in during the... first stage of voting.. All senators prefer V to B that was voted in through the previous procedure. Note this exercise shows that, in practice, agenda setting may matter. That is, the order by which decisions are taken (here voting) is important for the outcome if the preferences of the committee members differ. Exercise 3 - UN elections Applying backward induction, USA anticipates that Africa will choose its preferred candidate between the two who were not banned by USA. The best USA can do is to veto B. Africa will then choose between A and H, and will choose A. The outcome is 0 for both. Exercise 4 - The pirates game We solve the exercise by backward induction. In the last stage of the game, the 4th and 5th remain alive. If the 5th rejects the proposition of pirate #4, he will get the entire bounty. Hence, the 4th pirate must offer 00 to the 5th. With pirates #3, #4, #5 pirate #3 who makes the offer notices that pirate #4 is not going to get more than zero if the game continues. Hence, he could be offered ε > 0, a small amount and will vote in favor. Hence, if the game reaches this stage, pirate #3 is going to offer 00 ε to himself, ε to #4 and 0 to #5. In the preceding stage of the game, with pirate # making the offer, he notices that #3 is going to vote against unless he gets 00 ε, but #4 and #5 are going to vote in favor if offered ε. Hence pirate # gets 00 ε, #3 gets zero and #4 and #5 obtain both ε. We arrive at the first stage of the game. Pirate knows that if the game continues # is going to get 00 ε so this is what he would have to be offered to vote in favor. But, if #3,#4 and #5 each get ε they will be happy to vote for the proposal of pirate : they know that if players are rational they will not get anything more. Hence, pirate can offer 00 3ε to himself, zero to # and ε to #3,#4 and #5 each. Note that drawing the game tree here is complex. Exercise 5 - The survival game. Enstock.net has two options. Either it quits immediately with a profit of 0 or it waits that Jambes.com exits at k > and then capture all the market. In this case, Enstock.net s profit will be 80 3 () = 47. The last option is best.. Quitting generates zero profits. Waiting that Jambes.com exits generates profit equal to 80 3 (34) =. Thus, it is better to exit. 3. The losses made at the moment of taking the decision are irrecoverable costs. We should therefore not take them into account then. After 4 years we had 4 4 trimesters = 6. This means that from this point on, Jambes.com will stay 8 turns more in the market. Quitting gives zero profits. Waiting that Jambes.com exits brings profit equal to 80 3 (8) = 6. The firm has therefore an interest to remain in the market. 4. Only one firm can survive. The equilibrium (and therefore the best responses) is therefore such that one firm quits the market now and the other declares to remain on the market long enough - at least 7 trimesters, in order to encourage the other firm to exit. We have therefore k Enstock = 0 and k Jambes > 6 or k Enstock > 6 and k Jambes = 0. 5

6 5. The same reasoning applies but, now, the equilibrium is k Jambes = 0 and k Enstock >. The other one k Enstock = 0 and k Jambes > 6 also survives. Exercise 6 - Boeing-Airbus. The game tree is: B I N A A I N I N 4, Y X, 6 3, Y/3,. We need Y > 6 to incentivize Airbus to invest. And, since 4 > 3, Boeing has always an incentive to invest. 3. The game matrix is: Boeing\Airbus I N I 4, Y X, 6 N 3, Y 3, The Nash Equilibrium is still (I, I) and it is credible. Indeed, here there are no credibility problems. Exercise 7 - A sequential game. If n = or n =, the first player will take, respectively, either or tokens, and win the game.. For n = 3, the game is as follows: Pl. Pl. Pl. Pl. -, -,, - Hence, player knows that, whatever he plays, he will lose. Player is going to play if he takes one token and he is going to play if he takes two tokens. 3. For n = 4, the game tree is as follows: 6

7 Pl. Pl. Pl. Pl. Pl. Pl.,-,- Pl. -,, - -, By taking one token, player puts player in a situation in which player would be as if he was starting a game with n = 3 tokens on the table. Hence, player wins the game. 4. For n > 4, observe that if n = 5, the starting player can take two tokens and put player as if player were starting the game with n = 3. Hence, player would win. If n = 6, player will always be put in the position by player of playing a game with n = 3 when he makes the next move... Hence, whenever n is a multiple of 3, the starting player will lose. Otherwise, he will win. Exercise 8 - The hold-up problem. The game has two stages: an investment stage (when the firms are playing a simultaneous game) and the revenue partition stage (again a simultaneous game). At the revenue partition stage, the simultaneous game can be described by a matrix A \ B No decision Decision No decision, -, 3 Decision 3, - -, -. The equilibrium of this game is (D, D). Each player has a dominant strategy to hold-up the other one. This stage of the game where the firms partition their revenues resembles a prisoner s dilemma game. 3. Therefore, effectively, the first investment stage can be depicted as follows. A\ B Invest Not Invest Invest -, - -, 0 Not Invest 0, - 0, 0 The two firms therefore recognize that in the second stage of the game they are going to play a PD game, and that their payoffs are going to be negative from the project. Not investing in the first place is a dominant strategy for both of them, and is the outcome of the game. 7

8 The simultaneous game in the last table above can be represented in extensive form as follows: each player has two actions (Invest, Not Invest), and the decision nodes of the player moving second, say B, are joined with a dotted circle. This means that the second player does not observe the choice made by the first player. Therefore, it is as if the choices were made simultaneously. Note: This problem shows how a hold-up problem (for example, the threat of suing in courts) may prevent projects with positive net present value being pursued in joint-ventures. A solution to improve the situation may be integration of the two firms (i.e., a merger). Exercise 9 - Laptops (final exam Fall 007). The payoff matrix is:. The Nash equilibrium is (x = 4, y = 4). Alice\Bob y=4 y=5 y=6 x = 4 4, 4 6, 6, x = 5, 6 5, 5 7, 3 x = 6, 6 3, 7 6, 6 3. They would choose (x = 6, y = 6) which maximizes total well-being. 4. The subgame-perfect equilibrium (i.e., the outcome found by applying backward induction) is still (x = 4, y = 4). It does not matter whether Alice plays first or second. In this game there is no first-mover advantage. Exercise 0 - Joint contributions. The payoff matrix is: Player /Player S L S 0, 0 4, L, 4 3, 3. There are two Nash Equilibria: (L, S) and (S, L). 3. If they could find an agreement, they would choose (L, L) since these strategies collectively maximize total well-being. 4. The payoff matrix is: Player /Player S L S 0, 0, 3 L 3, 3, 3 L is now a strictly dominant strategy and (L, L) the only Nash equilibrium. 5. Through backward induction, the subgame-perfect equilibrium is (L, L). There is no gain from moving first or second. 8

9 Pl. L S Pl. Pl. L S L S 3,3 3,,3 0,0 6. The only Nash equilibrium is still (L, L) as we found in q.4. Exercise - Games and contracts (Final Exam 05). The payoffs in the matrix are justified by the following calculations for the producer (player ) and the consumer (player ), respectively. In (N, C), = 0 8 and 0 = In (N, R), = 0 8 and 0 = 0 0. In (O, C), 0 = 0 0 and 0 = 0 0. In (O, R), 0 = 0 0 and 0 = The unique Nash equilibrium is (O, R). O is strictly dominant for player. 3. The game tree is: Pl. O N Pl. Pl. R C R C 0,0 0,-0,0,0 By backward induction, player anticipates that if he chooses N, player s best response is C; whereas, if he chooses O, player s best response is R. Hence, player will choose N and the backward-induction equilibrium is (N, C). 4. If player moves first, the game tree is the following: Pl. R C Pl. Pl. O N O N 0,0,0 0,-0,0 By backward induction, player anticipates that if he chooses R, player s best response is O; whereas, if he chooses C, player s best response is still O. Hence, player will choose R and the backward-induction equilibrium is (R, O). 9

10 5. If the producer signs the contract, the profit for the producer will be 30 8 =, whereas the payoff for the customer will be = 0. If the producer does not sign the contract, we know from question that the payoffs will be (0, 0). Thus, the producer will sign the contract since he will get instead of just This contract is not optimal for the client. Indeed, any price p < 30 such that p 8 > 0 (thus, any price 30 > p > 8) would give the consumer a higher profit and, at the same time, would incentivize the producer to sign. 0