Econ 711 Homework 1 s January 4, 014 1. 1 Symmetric, not complete, not transitive. Not a game tree. Asymmetric, not complete, transitive. Game tree. 1
Asymmetric, not complete, transitive. Not a game tree. (both and preceed 4, but neither preceeds the other). 4 Asymmetric, not complete, not transitive. Not a game tree. 5 Asymmetric, not complete, vacuously transitive. Not a game tree 1 and both preceed 4, but neither preceeds the other.. 1) WTS Defn 1 Defn and ) Defn Defn 1 1) Suppose Prop A and not Prop B. There are two possibilities: a) no sequence exists or b) two or more sequences exist. First, we ll consider a): t k t by assumption. Then i, j, t i t and t j t implies t i t j or t j t i by Prop A. Asymmetry implies only one of those can hold, and adding transitivity implies there can be no cycles. Thus we can construct a sequence as in Defn. Now let s consider b): Suppose we can order these nodes two different ways. It must be the case, then, that at least one pair of nodes in the set of preceeding nodes switches order between the two sequences, and thus by transitivity that each node in this pair preceeds the other, violating asymmetry. Contradiction. ) Suppose Prop B holds and we have (t,t,t ) such that t t and t t. Prop B tells us that, since t and t preceed t, there is a unique sequence over this set of two nodes. Thus, either t t or t t, proving Prop A.. 1
Yes it is. Alice only moves at one node, so she has perfect information, and Bob always knows what node he is at because he knows how much money Alice has offered him the distinguishing feature between the nodes he plays. For the subgame starting with B 1, Bob is indifferent between accepting and rejecting, so either is valid. For all other subgames, Bob strictly prefers accepting. Thus, for Alice, there are two cases: 1) Bob accepts all. In this case, Alice s best response is to offer nothing, so the SPNE for (Alice,Bob) is (0,aaaaa). ) Bob rejects 0 and accepts all others. Then Alice s best response is to offer 1, the least she can give while ensuring Bob s acceptance. Then the SPNE is (1,raaaa). 4
Now Bob moves before Alice, but Alice doesn t observe Bob, so she only has one information set despite the fact that she has 5 nodes. Thus, in this case, SPNE doesn t impose any additional constraints beyond NE. There are now many equilibria. If Alice plays 0, all s 1 s s s 4 s 5 are best responses for Bob. If Alice plays i > 0, all strategies where s i = accept are best responses for Bob. For Alice s strategy i to be a best response, we must have i = min{j : s j = a, 0}. Thus the set of SPNE is {0,rrrrr} and {i, s 1 s s s 4 s 5 } where i = min{j : s j = a, 0}. 5 4. 1 Since Bob now cares about all possible actions Alice could chose, rather than just one that she s picked, his only two possibilities are raaaa and aaaaa, as in. However, Alice has no weakly dominant strategies, as her best response is contingent on Bob s strategy. Thus there are no equilibria in weakly dominant strategies. Actions: q 1 [0, ], q (q 1 ) [0, ]. Strategies: firm 1: q 1 [0, ], firm : the set of functions q : [0, ] [0, ] For backwards induction equilibria, first we solve the subgame starting at firm s turn, then we solve firm 1 s problem. Given q 1, firm s best response is to maximize Π (q ) = q ( 1 q 1 q ). Computing the FOC gives us 1 q 1 q = 0 or q = 1 q 1. Thus firm s best response is q (q 1 ) = 1 q 1. Consequently, firm 1 must maximize Π 1 (q 1 ) = q 1 ( 1 q 1 1 q 1 ) = q 1 (1/ q 1 /). Computing the FOC gives us 1/ q 1 = 0 or q 1 = 1. Plugging that into q (q 1 ),we have q = 1/4. Thus the equilibrium outcome is Π 1 = 1/(1 1/ 1/4) = 1/8 and Π = 1/4(1 1/ 1/4) = 1/16. The profit from Cournot is 1/(1/)=1/9, so firm 1 is better off and firm is worse off. 4
5 We ll consider 5 first because and 4 are special cases of it. Consider a strategy of q = 1 q1 for firm if q 1 = q 1and q = 1 otherwise. Essentially, we re making deviating from (q1, 1 q 1 ) as costly as possible for firm 1, to give them as strong incentives as we can to produce those quantities. If firm 1 produces anything else, firm will produce a quantity 1, making firm 1 s profit less than or equal to zero. Firm s behavior for contingencies other than q 1 =q 1 seems irrational, and wouldn t survive in a SPNE, but since we only require a Nash Equilibrium it doesn t matter because firm never has to play those irrational moves in the equilibrium. Thus, given that firm 1 s strategy is q 1 =q 1, firm is playing its best response, and so long as firm 1 s profit is nonnegative, playing q 1 is better than the next best response of producing 0 and getting 0 profit. So long as q 1 1, this condition will be satisfied. Thus the range of supportable q 1 s is [0,1] and the range of supportable q s is [0,1/]. Choosing q 1 = 1/ and using the strategies in part 5 gives us the desired equilibrium. 4 5. 1 Choosing q 1 = 1/4 and using the strategies in part 5 gives us the desired equilibrium, as it is the reverse of the SPNE Stackleberg outcome we derived before. u i (x i, x i x i ) = x i ( x i x i ). Notice this has the same functional form as the profit function in the cournot game, with price and quantity replaced with the public and private goods. The best responses are thus the same: x i = x i. Then the equilibrium x is are x i x i = = 1/ + x i /4 and x i = /. & 5
This is equivalent to Stackleberg, which was solved earlier. 4-1 As we saw in the last homework, then the set of possible Nash Equilibria is x such that x 1 x = 1/4. and x i 1. 4- Player 1 s strategies are simply x 1 [0, 1]. Player s strategies are the set of functions x : [0, 1] [0, 1]. 4- First, we ll solve player s problem. Given x 1, u (x, x 1 ) = x + x 1 x F OC : u (x, x 1 ) x = 1 1 x 1 x = 0 x = 7/4 x 1 This is our best response for player. Then player 1 solves u 1 (x 1, x ) = x 1 + 1/4,which has a corner solution of x 1 = 1. Essentially, since expenditures on the public good are the same no matter what player 1 does, since player has an incentive to make up for player 1 s shortchanging, player 1 might as well not contribute anything and leave it all to player to finance the public good. Thus we have a NE of (1, 7/4 x 1 ) and an equilibrium outcome of / for player 1 and 5/4 for player. Note that te SPNE is the element of the set of NE in the simultaneous move game that is most favorable to player 1 and least favorable to player. 6. 1 6
Notice that at the last node of the game tree (the last subgame), stopping is the unique best response. Given that player will stop in the last period, player 1 s best response at the second to last period (the second to last subgame) is to stop. Continuing inductively, we see that, given optimal play in all later periods, the best response at every node is to stop. Thus the only SPNE strategy profile is s 1 s...s 99 s 100 for both players. As in part, the game must stop in the first period. Suppose we had a NE where the first n actions were continue and the n+1th action was stop (any strategy profile with continue as the first move will be of this form, since the game stops after the last period). Then whoever plays at period n will have an incentive to deviate, as stopping n the nth period rather than the n+1th period will net them 1 additional dollar. Then the only possible NE strategy profiles have player 1 stopping in the first period, and player stopping in the second period. However, all other elements of the strategy that is, actions at every node 7
after the second can vary freely, since they are never reached and thus have no effect on the payoffs. Player must play stop at the second period so that player 1 doesn t have an incentive to deviate to continue. Thus the NE strategy profiles are s 1 a...a 99 a 100 for player 1 and s 1 a...a 99 a 100, where a can be either stop or continue. 4 There are no strictly dominated strategies. Every node after the first will never be reached under certain strategies specifically where player 1 stops in the first period, so there are situations where the actions at those nodes don t matter. Thus, we only need consider player 1 s first action. Continue is strictly better if player s first action is continue, and strictly worse otherwise, so there is no action in period 1 that is strictly better for player 1 under all contingencies. 7. 1 8
Strategies for player 1 are of the form a 1 a a a 4...a 8, where a 1 {(A, B), (B, C), (C, D)}and all other actions are chosen from the two options that are being voted for at that node. We end up with 16 information sets because player 1 chooses the first voting pair (1 info set), votes on the first pair ( info sets), and votes on the second pair (*8=4 info sets). Strategies for player are of the form a 1 a a a 4...a 54, where all actions are chosen from the two options that are being voted for at that node. We end up with 54 information sets because player votes on the first pair (*=6 info sets) and votes on the second pair (*8*=48 info sets). Strategies for player are of the form a 1 a a a 4...a 108, where all actions are chosen from the 9
two options that are being voted for at that node. We end up with 108 information sets because player votes on the first pair (*4=1 info sets) and votes on the second pair (*4*8=96 info sets). As you can see in the extensive form of the game tree, the second rounds of voting yield outcome A for (A,B), C for (A,C), and B for (B,C). Then, once we ve finished the first round of voting, we know the payoff, and can plug it in as in the figure. Now we see that, if player 1 chooses (A,C), the voting will proceed to (A,B), which is optimal for player 1. Thus player 1 will choose (A,C) and the final vote will be for A. There are many SPNE strategy profiles there are many information sets where players are indifferent, and each of those doubles the number of SPNE strategy profiles since either choice is consistent with best responding (a quick count gives me ˆ50 strategies, but I m obviously not going to write them out). As before, there are many, many SPNE strategy profiles, and they can be defined analogously to 7., though the reordering of moves will change the number of information sets each player has and the way their strategies must be written down. The set of SPNE outcomes will remain the same, however notice that, in a voting game with options and voters, a given voter is either not pivotal, in which case their vote doesn t matter, or they are pivotal, in which case they choose their most preferred option. Later voters observe the votes of earlier voters, but the only change this can cause in their strategy is to make them vote for a less preferred option if they know their vote isn t pivotal, which only happens in this case if the same outcome obtains regardless of their vote. Thus, changing the order of voting will not change the equilibrium payoffs. 10