Question 1: Basic Concepts (20 points) Economics 171: Final Exam 1. Is it true that every strategy is either strictly dominated or is a dominant strategy? Explain. (5) No, some strategies are neither dominated nor dominant. For example, see the Battle of the Sexes game below no strategy is either dominant or dominated. 2. What is the difference (in terms of beliefs) between the set of rationalizable outcomes and the set of Nash equilibrium outcomes? Is one set contained within the other? (5) For rationalizable outcomes, players must have possible beliefs about the other player s strategies; these beliefs need not be correct. On the other hand, in Nash equilibrium, beliefs must be correct. The set of Nash equilibrium outcomes is contained in the set of rationalizable outcomes. 3. According to the Folk Theorem, what is the set of per-period outcomes that can be supported in an infinitely-repeated game with sufficiently high discount factors for the players (note: allow for the possibility of mixed strategies)? Apply this to the Battle of the Sexes game below: (5) Player 1 Player 2 L R U 3, 1 0, 0 D 0, 0 1, 3 The Folk Theorem (p. 222, footnote 9) says that any outcome where the payoffs are just slightly greater than those in a stage Nash payoffs can be sustained with discount factors close to 1. The equilibrium in which the players get the worst payoffs is the mixed-strategy outcome, where each person earns an average of 3/4. Thus, any outcomes in which both players make slightly more than 3/4 can be sustained, up until the upper boundaries of the set [(3,1) can be sustained, (2,2) can be sustained, and (1,3) can be sustained], which will resemble Figure 22.7, but with the maximum on the axes being 3 and no negative payoffs. 4. Suppose Player 1 could either choose to play the game above or to instead choose an outside option in which he or she would get a utility of 2. What outcome(s) might you expect in this case? Explain. (5) If player 1 foregoes the outside option of 2, s/he must be planning to make 3 by choosing U. If player 1 believes that player 2 will understand this forwardinduction argument, the outcome should be (3,1). However, some player 1 s wouldn t trust player 2 to make this inference, and so would take the outside option
Question 2: Technology adoption (15 points) During the adoption of a new technology a CEO (player 1) can design a new task for a division manager (player 2). The new task can be a high level (H) or a low level (L). The manager simultaneously chooses to invest in good training (G) or bad training (B). The payoffs from this interaction are given by the following bi-matrix: Player 1 Player 2 G B H 5, 4-5, 2 L 2, -2 0, 0 (a) Present the game in extensive form (a game tree). How many proper subgames (subgames that aren t the whole game) does it have? (4) 1 H L 2 ---------------------------- 2 G B G B 5-5 2 0 4 2-2 0 There are no proper subgames. (b) Solve for all the Nash equilibria and subgame-perfect Nash equilibria (SPNE), whether in pure or mixed strategies. (4) Both (H, G) and (L, B) are pure-strategy Nash equilibria. There is also a mixedstrategy Nash equilibrium in which H is played 1/2 of the time and G is played 5/8 of the time. Since there are no proper subgames, all of these Nash equilibria are subgame-perfect Nash equilibria. Now assume that before the game is played the CEO can choose not to adopt this new technology, in which case the payoffs are (1,1), or to adopt it and then the game above is played. (c) Present the entire game in extensive form. How many proper subgames does it have? (4)
N 1 A 2 1 1 G B 1 ---------------------------- 1 H L H L 5 2-5 0 4-2 2 0 There is one proper subgame. This extensive form could also be drawn differently. (d) Solve for all the pure-strategy Nash equilibria and SPNE. (3) Player 1 Player 2 G B NH 1, 1 1, 1 NL 1, 1 1, 1 AH 5, 4-5, 2 AL 2, -2 0, 0 There are three pure-strategy Nash equilibria: (NH, B), (NL, B), and (AH, G). But in the proper subgame, only (H, G) and (L, B) are equilibria, so we can rule out (NH, B) as a SPNE. Thus, the two SPNE here are (NL, B) and (AH, G). Question 3: Best response graphs (10 points) For each of these games, draw the best-response functions for Player 1 and Player 2, assigning p as the probability that Player 1 chooses U and q as the probability that Player 2 chooses L. Then use these best-response functions to find all of the equilibria (the intersections), whether in pure of mixed strategies. (5 points each game) L R L R U 4, 2-5, 6 U 3, 2 0, 1 D -1, 5 0, -2 D 1, 0 4, 6 Game A Game B
p = 0, q = 1 p = 7/11 p = 1, q = 1 q = 1/2 Game A p = 0, q = 0 p = 1, q = 0 There is only one Nash, in mixed strategies, with p = 7/11 and q = 1/2. p = 0, q = 1 p = 1, q = 1 BR for player 2, given p BR for player 1, given q Game B q = 2/3 p = 0, q = 0 p = 6/7 p = 1, q = 0 There are two Nash equilibria in pure strategies: (U, L) and (D, R). There is also one Nash in mixed strategies, with p = 6/7 and q = 2/3. Question 4: Rationalizability and iterated deletion (20 pts.) E F G L R A 8, 0 8, 1 1, 8 U 1, 0 4, 1 0, 8 B 0, 1 6, 8 2, 0 D 1, 1 0, 8 4, 0 C 12, 8 5, 0 3, 1 0, 8 2, 0 2, 1 Game X Game Y i) In Game X, what are the set of rationalizable outcomes? (3)
(C, E) or (12, 8) ii) In Game X, what are the set of outcomes that survive the iterated deletion of those pure strategies that are strictly dominated by other pure strategies? (3) All outcomes survive. iii) In Game X, what are the set of outcomes that survive the iterated deletion of those pure strategies that are strictly dominated by some mixed strategy and pure strategies? (4) Same answer as in (i). First, B is dominated by a mix of 2/5 A and 3/5 C. After eliminating B, then F is dominated by G. After eliminating F, then A is dominated by C. Since player 2 prefers (C, E) to (C, G), only (C, E) survives. iv) In Game Y, what are the set of rationalizable outcomes? (3) Every outcome is rationalizable. v) In Game y, what are the set of outcomes that survive the iterated deletion of those pure strategies that are strictly dominated by other pure strategies? (3) Every outcome is rationalizable. vi) In Game Y, what are the set of outcomes that survive the iterated deletion of those pure strategies that are strictly dominated by some mixed strategy and pure strategies? (4) Every outcome is rationalizable. Question 5: Two-period repeated game (20 points) Suppose that two players play the following normal-form game: L C R U 5, 5 0, 6 0, 0 M 6, 0 2, 2 0, 0 D 0,0 0, 0 0, 0 a) Find all of the game s pure-strategy Nash equilibria. (3) (M,C) and (D, R) Now suppose that the players play this game twice in a row. They observe what each other did in the first stage before they decide what to do in the second stage. Each player s payoff is the (undiscounted) sum of the payoffs in the first and second stages.
b) Find a SPNE in which the players decisions in the second stage do not depend on their decisions in the first stage. Be sure to specify players strategies clearly, recalling that a strategy must be a complete contingent plan for playing the game. Play (M,C) in the first stage and play (M,C) in the second stage. Are there any such equilibria in which players do anything other than play one of the equilibria identified in part a) above in each stage? (5) No, there aren t (except for possible mixed strategies, but part a) specifies we are only speaking about pure strategies). c) Now find a pure-strategy SPNE in which the players play U, L in the first stage, and therefore do better than by repeating the best symmetric Nash equilibrium in the onestage version of the game? Specify the full strategies. (7) Play (U, L) in stage 1. If no one deviates, play (M, C) in stage 2. If either player (or both) deviates, then play (D, R) in stage 2. The utility from not deviating is 7, while the utility from deviating is 6. d) Why is it possible to support a desirable but non-equilibrium outcome like U, L in the first-stage of this two-stage game, but not the desirable but non-equilibrium outcome Cooperate, Cooperate in the first stage of a two-stage Prisoner s Dilemma? (5) This is because there is no credible threat that can sustain cooperation in the finite Prisoner s Dilemma. The unique Nash equilibrium will be played in the last stage. Knowing this, they must play the unique Nash in the penultimate stage, etc. Essentially, there must be at least two Nash equilibria in the stage game to make it possible to sustain a desirable non-equilibrium in the first stage of a finite (twoperiod here) game. Question 6: Stackelberg competition and collusion in an infinitely-repeated game (30 points) Stackelberg competition is similar to Cournot competition, except that one of the firms chooses a quantity to produce before the other firm does, the second firm observes the first firm s production choice and then makes a production choice, the first firm knows that the second firm observes the first firm s production choice and then makes a production choice, etc. Let firm 1 be the firm that produces first (the Stackelberg leader) and firm 2 be the firm that produces last (the Stackelberg follower). Let q 1 denote firm 1 s quantity and q 2 denote firm 2 s quantity, where each of q 1 and q 2 is non-negative; thus, the industry s total output is q 1 + q 2. Suppose the price is given by the function p = 300 - q 1 - q 2, and that there is no cost of production. Note that the profit for each firm is given by that firm s production multiplied by
the industry price given by the formula above, and assume that each firm is interested only in maximizing its own profit. a) What quantity does each firm produce in equilibrium in a one-shot game? (Hint: Think about the firms best-response functions; in particular, realize that firm 1 can derive firm 2 s best-response function and will choose a quantity that maximizes its own profits given this best-response function). (5) 150 and 75 Now suppose that this game is played an infinite number of times and that each player has the same discount factor d. Suppose further that the firms can collude to together produce the monopoly level of production, 150, with firm 1 producing 100 and receiving 2/3 of the total profits and firm 2 producing 50 and receiving 1/3 of the total profits. b) What is the profit for each firm in the equilibrium in part a) above? What is the profit for each firm in the monopoly-sharing equilibrium? (5) Fir m 1 makes 11,250 and firm 2 makes 5,625 in he part a) equilibrium. Firm 1 makes 15,000 and firm 2 makes 7,500in the monopoly-sharing equilibrium. Suppose each firm 1 has the following strategy in the infinitely-repeated game: produce its monopoly-sharing output (100) as long as firm 2 produces its monopoly-sharing output (50); however, if firm 2 ever deviates from this level of production (i.e., produces the output level that is its best response in the stage game), then firm 1 produces its Stackelberg output forever. c) In the stage game, what is firm 2 s best response to firm 1 producing 100 units? What profits does it make in the stage game by doing so? (5) The best response is to produce100. It makes 10,000 by doing so. d) Given the discount factor d, what is the sum of the profits over time that firm 2 would get by deviating in the first period and producing this best response? (7) 10,000 + 5625d + 5625d 2 + = 10,000 + 5625d/(1-d) e) Given the discount factor d, what is the sum of the profits over time that firm 2 would get by always producing the monopoly-sharing output level? [Note: If you don t remember the summation formulas, at least set up the equation for partial credit] 7500 + 7500d + 7500d 2 + = 7500/(1-d) f) How large does d have to be in order to sustain collusion in the infinitely-repeated game? (8) 7500/(1-d) >= 10,000 + 5625d/(1-d) d >=4/7
Question 7: Perfect Bayesian equilibria in a signaling game (10 points) Consider the following game of incomplete information. For what values of a and b is the following a PBE? t 1 and t 2 both go Left (L or L ). Player 2 goes u if Player 1 goes Left, and goes d if Player 1 goes Right. 0, 0 u u a, b [p] L t 1 R [q] d Prob =.5 d -1, 0 0, 0 Nature 0, 0 u Prob =.5 u 1, 1 [1-p] L t 2 R [1-q] -1, 0 0, 0 d d t 1 gets 0, would get -1 by deviating. No profitable deviation here t 2 gets 0, would get -1 by deviating. No profitable deviation here But if b > 0, then u strictly dominates d. So player 2 could only play this strategy if b <= 0. There is no restriction on a. Question 8: Perfect Bayesian equilibria in a signaling game (25 points) Consider the following game of incomplete information. 1, 1 u u 2, 2 [p] L t 1 R [q] d Prob =.5 d -1, 0 2, 0 Nature 0, 0 u Prob =.5 u 1, 0 [1-p] L t 2 R [1-q] 1, 1 0, 1 d d
The four possible pure-strategy Bayesian equilibria in this two-type, two-message game are: (1) pooling on LL ; (2) pooling on RR ; (3) separation with t 1 playing L and t 2 playing R ; and (4) separation with t 1 playing R and t 2 playing L. Compute whether there is a pure-strategy PBE in each of these four cases. If not, explain why not. If so, list the full strategy for each player and also specify the values of p and q that obtain in the equilibrium. Separating Try RL. Then q = 1, p = 0, so player 2 plays du ; t 1 would get 2, would also get 2 by deviating. Thus, there is no profitable deviation for t 1. t 2 gets 0, but would get 1 by deviating. So s/he would deviate, and so this is not a PBE. (5) Try R L. Then q = 0, p = 1, so player 2 plays ud ; t 1 would get 1, would get -1 by deviating. Thus, there is no profitable deviation for t 1. t 2 gets 1, would get 0 by deviating. So there is no profitable deviation for t 2. Therefore R L, q = 0, p = 1, ud is a PBE. (5) Pooling Try RR. Then p is undetermined, q = 1/2. If p > 1/2, player 2 plays uu ; t 1 would get 2, would get 1 by deviating. Thus, there is no profitable deviation for t 1. t 2 gets 1, would get 0 by deviating. So there is no profitable deviation for t 2. Therefore RR, q = 1/2, p > 1/2, uu is a PBE. (3.5) If p < 1/2, player 2 plays du ; t 1 would get 2, would get 2 by deviating. Thus, there is no profitable deviation for t 1. t 2 gets 1, would get 0 by deviating. So there is no profitable deviation for t 2. Therefore RR, q = 1/2, p < 1/2, du is a PBE. (3.5) Try LL. Then p = 1/2, q is undetermined. If q > 1/3, then player 2 plays either ud or dd, as s/he is indifferent between u and d when p = 1/2. If s/he plays ud, then t 2 gets 0, but would get 1 by deviating. So this is not a PBE. If s/he plays dd, then t 2 gets 0, but would get 1 by deviating. So this is not a PBE. (4) If q < 1/3, then player 2 plays either uu or du, as s/he is indifferent between u and d when p = 1/2. If s/he plays uu t 1 would get 1, but would get 2 by deviating. So s/he would deviate, and so this is not a PBE. If s/he plays du t 2 would get 0, but would get 1 by deviating. So s/he would deviate, and so this is not a PBE. (4)