Simon Fraser University Fall 2014

Transcription

1 Simon Fraser University Fall 2014 Econ 302 D100 Final Exam Solution Instructor: Songzi Du Monday December 8, 2014, 12 3 PM This brief solution guide may not have the explanations necessary for full marks. NE = Nash equilibrium, SPE = subgame perfect equilibrium, PBE = perfect Bayesian equilibrium 1. (10 points) Consider the following simultaneous-move game (Table 1). W X Y Z A 9, 2 4, 3 1, 2 4, 2 B 3, 2 3, 6 2, 5 7, 5 C 4, 0 7, 1 3, 0 3, 2 D 4, 3 4, 4 9, 3 3, 3 Table 1 Part i: Find the strategies that survive iterative deletion of strictly dominated strategies (ISD). For each strategy that you delete, write down the strategy that strictly dominates it. Part ii: Find all NE (pure and mixed) in this game, and for each NE that you find, calculate the two players expected payoffs in the equilibrium. Part i, ISD: Round 1: both W and Y are strictly dominated by X. Delete W and Y. Round 2: both A and D are strictly dominated by 1/2B + 1/2C. Delete A and D. The strategies that survive: B, C for player 1, and X, Z for player 2. Part ii, NE: For NE we can look at the smaller game given by ISD. Clearly, there is no pure-strategy NE. 1

2 Suppose player 1 uses mixed strategy pb + (1 p)c and player 2 uses mixed strategy qx + (1 q)z. For both B and C to be player 1 s best responses, we must have: 3q + 7(1 q) = 7q + 3(1 q), i.e., q = 1/2. For both X and Z to be player 2 s best responses, we must have: 6p+(1 p) = 5p+2(1 p), i.e., p = 1/2. Thus, the unique mixed-strategy NE is (1/2B + 1/2C, 1/2X + 1/2Z). In this equilibrium player 1 gets an expected payoff of 5, and player 2 gets an expected payoff of (10 points) Three firms are considering entering a new market. The payoff for each firm that enters is 150, where n is the number of firms that enter. The cost of entering is 62. n (So the net payoff for a firm that enters is , and 0 for a firm that does not enter.) n Part i: Find all the pure-strategy NE, and calculate the firms payoffs in the equilibrium. Part ii: Find the symmetric mixed-strategy NE in which all three firms enter with the same probability, and calculate the firms payoffs in the equilibrium. Part i: Clearly, it cannot be an equilibrium for only one firm to enter, since 150/2 62 > 0. Moreover, it cannot be an equilibrium for all three firms to enter, since 150/3 62 < 0. Thus, the pure-strategy NE is for two firms to enter (each with payoff 150/2 62 = 13) and one firm to not enter (with payoff 0). So there are three pure-strategy NE, in which firm i (and only firm i) does not enter, for each i = 1, 2, 3. Part ii: In the symmetric mixed-strategy NE, suppose that each firm enters with probability 0 < p < 1. For both entering and not entering to be best responses, we must have: p 2 (150/3 62) + (1 p) 2 (150 62) + 2p(1 p)(150/2 62) = 0, i.e., 50p 2 150p + 88 = 0. The solutions to the above quadratic equation are p = 4/5 and p = 11/5, so we take the first solution since it must be a number between 0 and 1. Thus, in the symmetric mixed- 2

3 Figure 1 strategy NE, each firm enter with probability 4/5. In this NE, each firm gets an expected payoff of (10 points) Find and report all pure-strategy SPE in Figure 1. Explain your answers. Clearly, in any SPE player 2 must use b and c, and player 1 must use D. In the subgame following B, there are two pure-strategy NE: (E, e) and (F, f). Thus there are two pure-strategy SPE: SPE #1: player 1 plays (A, D, E) and player 2 plays (b, c, e). SPE #2: player 1 plays (A, D, F ) and player 2 plays (b, c, f). 4. (10 points) Two players must choose among three alternatives, a, b, and c. Their payoffs from each alternative are as follows: a b c player player The rules are that player 1 moves first and vetoes one of the three alternatives. Then player 2 chooses one of the remaining two alternatives. 3

4 Figure 2 Part i: Draw the game tree. Part ii: Find and report the pure-strategy SPE. Be sure to report a complete strategy for each player. Part i: see Figure 2. Part ii: the SPE is player 2 chooses b if a is vetoed, chooses a if b is vetoed, and chooses b if c is vetoed; and player 1 chooses to veto b. (You may also write the SPE by clearly marking the strategies on the tree.) 5. (10 points) There are two players. Player 1 is either a high (H) or low (L) type worker, with probability 0.3 and 0.7 respectively. Player 1 knows his type and chooses to get an MBA degree (D) or be content with his undergraduate degree (U). Player 2 who is an employer observes player 1 s degree but not his type, and must decide to assign player 1 to be a manager (M) or a blue-colar worker (B). Getting an MBA degree has a cost of c H = 3 for a high type worker and c L = 5 for a low type worker. The market wage for a manager is w M = 10, and the market wage for a blue-collar worker is w B = 6. The payoff of a high type worker who gets a blue-collar assignment is w B c H if he has gotten an MBA and is w B if he has not, and likewise in the other cases. The employer does not care about the MBA degree per se. The employer s net 4

5 payoff (after paying the wage) depends on the worker s type and assignment and is given by the following table: M B H 10 5 L 0 4 Part i: Draw the game tree. Part ii: Find all pure-strategy PBE (if there is any), and explain. Include conditional beliefs (such as P(H D) = 1 and P(H U) 1/2.) in your description of PBE. Part i: see Figure 3. Figure 3 Part ii: After observing MBA degree D, player 2 gets 10P(H D) from assigning manager job (D-M) and 5P(H D)+4(1 P(H D)) = 4+P(H D) from assigning blue collar job (D-B); so D-M is a best response if P(H D) 4/9 and D-B is a best response if P(H D) 4/9. Likewise, after observing undergraduate degree U, U-M is a best response if P(H U) 4/9 and U-B is a best response if P(H U) 4/9. We now check four cases: 5

6 1. If both types of player 1 get MBA (H-D, L-D), then player 2 would assign blue collar job given MBA degree (D-B), since P(H D) = 3/10 < 4/9. This means a payoff of 3 for the high type of player 1, but he can get at least 6 from not getting MBA (H-U). So (H-D, L-D) cannot be a part of a PBE. 2. If both types of player 1 are do not get MBA (H-U, L-U), then player 2 would assign blue collar job given undergraduate degree (U-B), since P(H U) = 3/10 < 4/9. For both types of player 1 to not pursue MBA, player 2 must also assign blue collar job given MBA degree (D-B), which is possible because P(H D) is arbitrary (D happens with probability 0). Therefore, it is a PBE that player 1 plays (H-U, L-U), and player 2 plays (U-B, D-B) with beliefs P(H U) = 3/10 and P(H D) 4/9. 3. If high type of player 1 gets MBA and low type does not (H-D, L-U), then player 2 would assign manager job given MBA degree and assign blue collar job given undergraduate degree (D-M, U-B), since P(H D) = 1 and P(H U) = 0. Given (D-M, U-B), it is easy to check that (H-D, L-U) are best responses for player 1 of both types. Therefore, it is a PBE that player 1 plays (H-D, L-U), and player 2 plays (D-M, U-B) with beliefs P(H D) = 1 and P(H U) = If high type of player 1 does not get MBA and low type gets MBA (H-U, L-D), then player 2 would assign manager job given undergraduate degree and assign blue collar job given MBA degree (D-B, U-M), since P(H D) = 0 and P(H U) = 1. Given (D-B, U-M), it is easy to check that (H-U, L-D) is not a best response for player 1 of either type. So (H-U, L-D) cannot be a part of a PBE. 6. (15 points) Players 1 and 2 put a dollar each in a pot, and player 1 is dealt a card which is either a king (K) or an ace (A) (with equal probability). Player 1 observes his card and then decides whether to fold, forfeiting his dollar to player 2, or to bid, proceeding with the game. If player 1 bids, then without knowing the card of player 1 player 2 can fold and forfeit his dollar to player 1, or bid, in which case each player must add another dollar to the pot. After bidding by both players, if player 1 has a king then player 2 wins the pot, while if player 1 has an ace then player 1 wins the pot. See the game tree in Figure 4. Part i: Find all pure-strategy PBE, if there is any. Explain your answers. Part ii: Find a mixed-strategy PBE, with the following steps: 6

7 Figure 4 (0) Notice that if player 1 has an ace, then he never folds (i.e., A-Bid); suppose player 1 with a king bids with probability p, and player 2 bids with probability q, where 0 < p < 1 and 0 < q < 1. We need to calculate p and q that make a PBE. (1) For player 1 with a king to mix between bidding and folding, he must be indifferent; what probability q makes player 1 with a king indifferent? (2) For player 2 to mix between bidding and folding, he must be indifferent; what belief about player 1 s card (conditional on player 1 bidding) makes player 2 indifferent? (3) What probability p gives (via Bayes rule) the conditional belief that makes player 2 indifferent? (4) Steps 0 3 determine the mixed-strategy PBE. What are players 1 and 2 s expected payoffs in this equilibrium? (For player 1, calculate his expected payoff before he is dealt the card.) Is the game rigged in a player s favor? Part i: Clearly, player 1 with an ace will bid (A-Bid) in any PBE. 7

8 1. If player 1 with a king bids (K-Bid), then player 2 believes that P(A Bid) = 1/2 so his best response is to bid if player 1 bids. If player 2 bids, then player 1 with a king prefers to fold. 2. If player 1 with a king folds (K-Fold), then player 2 believes that P(A Bid) = 1 so his best response is to fold if player 1 bids. If player 2 folds, then player 1 with a king prefers to bid. Therefore, there is no pure-strategy PBE. Part ii: (1) 2q + 1(1 q) = 1, i.e., q = 2/3. (2) 2P(A Bid) + 2(1 P(A Bid)) = 1, i.e., P(A Bid) = 3/4. (3) 1/2 1/2+p/2 = P(A Bid) = 3/4, i.e., p = 1/3. (4) In this equilibrium, player 1 with an ace gets 2 2/ /3 = 5/3, and player 1 with a king gets 1 (since he is indifferent between bidding and folding). So overall (before player 1 getting his card), player 1 gets 5/3 1/2 1 1/2 = 1/3. If player 1 bids, then player 2 gets 1 (since he is indifferent between bidding and folding), and if player 1 does not bid, then player 2 gets 1. Since player 1 bids with probability 1/2 + 1/2 1/3 = 2/3, player 2 gets 2/3 ( 1) + 1/3 1 = 1/3. The game is rigged in player 1 s favor, since his expected payoff is positive. If the game were fair, then the expected payoffs of both players would be zero. 8