MIDTERM 2 Question 1. (a-5 points) Define a perfect equilibrium. (b-5 points) Define a subgame perfect equilibrium. Solution: See class notes. Question 2. For this question, consider the game below. (a-8 points) Find the unique subgame perfect Nash equilibrium of this game. This game has two subgames: the whole game and the game beginning at the node where Player 1 chooses between L and M. In the game beginning at this node, Player 1 has a dominant strategy (L). The unique equilibrium of the subgame is (L, l). The unique SPNE is therefore ((N R, L), l). (b-8 points) Show carefully that this equilibrium is also −perfect. (There exist sequences such that...) Note: if you only found the −perfect equilibrium of the subgame, you lost 4 points. The normal form version of the game is: 1 NR,L NR,M R,L R,M l r (4,1) (1,0) (0,0) (0,1) (2,2) (2,2) (2,2) (2,2) We need to show that there exist sequences {(n)} and {s(n)} such that the following conditions are satisfied: 1. ij (n) > 0 ∀i ∈ {1, ..., N }, j ∈ {1, ..., #Ai }, n ∈ {1, 2, 3, 4, ...}. P i 2. j j (n) < 1 ∀i, n. 3. limn→∞ ij (n) = 0 ∀i, j. 4. s(n) ∈ N E(G(n) ) ∀n. 5. limn→∞ s(n) = ((1, 0, 0, 0), (1, 0)). Let ij (n) = satisfied. 1 n ∀i, j, n. For a large enough n, conditions 1 and 2 will be Let s(n) = ((1 − 3/n, 1/n, 1/n, 1/n), (1 − 1/n, 1/n)). We need to shown that s(n) is a Nash Equilibrium of the (n)−perturbed game. 1 If player 1 plays (1 − 3/n, 1/n, 1/n, 1/n), player 2 gets (1 − 3n ) + n2 + n2 from playing l and n1 + n2 + n2 from playing r. Since the payoff from l is strictly higher, player 2 wants to put as large a probability on left as possible. He therefore plays (1 − 1/n, 1/n). If player 2 plays (1 − 1/n, 1/n), player 1 gets 4(1 − 1/n) + n1 from N R, L. For a large enough n, this payoff is strictly greater than player 1’s payoff from any other pure strategy. Player 1 therefore puts the largest possible probability on N R, L. He therefore plays (1 − 3/n, 1/n, 1/n, 1/n). (c-8 points) Are there other −perfect equilibria? If yes, find one and show carefully that it satisfies the definition. Yes, ((R, L), r) and ((R, M ), r) are both −perfect. To see this, let’s consider ((R, L), r). Let {(n)} be any sequence such that the perturbation on (N R, M ) is greater than the perturbation on (N R, L). In any equilibrium of the (n)-perturbed 2 game, Player 2 will best respond by putting the highest probability possible on r. If Player 2 puts the highest probability possible on r, it is a best response for Player 1 to put the highest probability possible on (R, L). The s(n) in which Player 1 puts the highest probability on (R, L) and Player 2 puts the highest probability on r converges to ((R, L), r) as n → ∞. This shows that ((R, L), r) is a perfect equilibrium. Question 3. Consider the following game, sometimes called the ultimatum game. Player 1 decides how to split 100 cents between himself and Player 2. Thus, Player 1 proposes an offer y to Player 2. The proposal can be anything from 1 cent up to 100 cents in 1 cent increments. Thus, 1 cent, 2 cents, 3 cents, ..., 99 cents, 100 cents are all possible proposals. Player 2 decides whether to accept or reject the proposal. If the proposal is accepted, Player 2 gets y and Player 1 gets 100 − y. If the proposal is rejected, both players get 0. (a-8 points) What are the pure strategy Nash equilibria of this game? The pure strategies of Player 1 are indexed by the offer: (1,2,3,...,100). A pure strategy of Player 2 specifies a decision (Accept/Reject) at any information set. Thus, one pure strategy is (Accept 1, accept 2, accept 3,..., accept 100). Another is to reject any offer, etc. For any offer, if player 2’s pure strategy is willing to accept a lower offer, it is not an equilibrium for player 1 to make it because he will want to switch to a lower offer. Any offer is an equilibrium together with any strategy of player 2 that is not willing to accept a lower offer. For example, it is an equilibrium for player 1 to offer 2 and for player 2 to (reject 1, accept 2, ... ). What player 2 does when the offer is 3, 4, 5, 6,... is irrelevant in this example. Thus, (offer 2, (reject 1, accept 2, accept 3, accept 4, accept 5,...)) is an equilibrium, but so is (offer 2, (reject 1, accept 2, reject 3, reject 4, reject 5,...)). (b-8 points) What is the unique subgame perfect Nash equilibrium? There is a subgame of the game for any offer of player 1. In each subgame, player 1 gets more by accepting than by rejecting. Therefore, the SPNE has player 1 accepting everywhere. If player 1 accepts everywhere, player 2 offers the smallest possible amount: 1 cent. Thus, the unique SPNE is for player 1 to offer 1 cent and for player 2 to accept everywhere. 3
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