Francesco Nava Microeconomic Principles II EC202 Lent Term 2010

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1 Answer Key Problem Set 1 Francesco Nava Microeconomic Principles II EC202 Lent Term 2010 Please give your answers to your class teacher by Friday of week 6 LT. If you not to hand in at your class, make arrangements with your class teacher about where to bring it. Thank you! 1. Consider the following complete information strategic form game: 1n2L L C R T 3; 4 1; 1 6; 2 M 2; 1 6; 4 0; 2 B 2; 2 2; 3 4; 2 (a) Find the pure strategy Nash equilibria. Answer: PNE of this game are: (T; L) ; (M; C) i) (T; L) is a PNE since: ii) (M; C) is a PNE since: u 1 (T; L) = 3 > 2 = u 1 (M; L) = u 1 (B; L) u 2 (T; L) = 4 > 2 = u 2 (T; R) > 1 = u 2 (T; C) u 1 (M; C) = 6 > 2 = u 1 (B; C) > 1 = u 1 (T; C) u 2 (M; C) = 4 > 2 = u 2 (M; R) > 1 = u 2 (M; L) (b) Are there any strictly dominated strategies if players can only play pure strategies? Answer: NO. For player 1: T is not dominated since it is best when player 2 chooses R; M is not dominated since it is best when player 2 chooses C; and B is not dominated since it beats T when player 2 chooses C and it beats M when player 2 chooses R. For player 2: L is not dominated since it is best when player 1 chooses T ; C is not dominated since it is best when player 1 chooses M; and R is not dominated since it beats L when player 1 chooses M and it beats C when player 1 chooses T. (c) Are there any strictly dominated strategies if players can employ a mixed strategies? Answer: Yes, (1=2)L + (1=2)C strictly dominates R and (3=4)T + (1=4)M strictly dominates B. In particular (1=2)L + (1=2)C strictly dominates R, since: 1=2u 2 (T; L) + 1=2u 2 (T; C) = 5=2 > 2 = u 2 (T; R) 1=2u 2 (M; L) + 1=2u 2 (M; C) = 5=2 > 2 = u 2 (M; R) 1=2u 2 (B; L) + 1=2u 2 (B; C) = 5=2 > 2 = u 2 (B; R) and (3=4)T + (1=4)M strictly dominates B, since: 3=4u 1 (T; L) + 1=4u 1 (M; L) = 11=4 > 2 = u 1 (B; L) 3=4u 1 (T; C) + 1=4u 1 (M; C) = 9=4 > 2 = u 1 (B; C) 3=4u 1 (T; R) + 1=4u 1 (M; R) = 9=2 > 4 = u 1 (B; R) (d) Find the mixed strategy Nash Equilibrium. Answer: 1 (T ) = 1 (M) = 1=2, 2 (L) = 5=6, 2 (C) = 1=6 & 1 (B) = 2 (R) = 0.

2 Microeconomic Principles II F. Nava Because strategies B and R are dominated we know that they cannot be played with positive probability in a Nash equilibrium or else a pro table deviation would exist (see slides). Thus we can focus on the two player two action remaining game. To prove that the proposed strategies constitute a MNE and to compute them simply verify that: u 1 (T; 2 ) = 3 2 (L) + (1 2 (L)) = 2 2 (L) + 6(1 2 (L)) = u 1 (M; 2 ) > u 1 (B; 2 ) u 2 ( 1 ; L) = 4 1 (T ) + (1 1 (T )) = 1 (T ) + 4(1 1 (T )) = u 2 ( 1 ; C) > u 2 ( 1 ; R) 2. Consider an economy with two producers competing to supply a market. Suppose that the cost function of the rst rm displays a constant marginal costs, while the second rm displays increasing marginal costs. In particular assume that: c 1 (q 1 ) = 2q 1 c 2 (q 2 ) = (q 2 + 1=2) 2 Suppose that the inverse demand in this market is linear and satis es: p(q) = 4 Assume that the two rms compete à la Cournot. 2q (a) Derive the Cournot production levels, pro ts and the equilibrium price. Answer: q 1 = 3=10, q 2 = 4=10, p = 13=5, 1 = 9=50, 2 = 23=100. To nd the Cournot outputs, let s nd the best response functions by solving the problem of each rm. Firm 1 s problem is: Firm 2 s problem is: max(4 2q 1 2q 2 )q 1 2q 1 q 1 4 4q 1 2q 2 = 2 [FOC] q 1 = 1 q 2 [BR1] 2 2 max(4 2q 1 2q 2 )q 2 (q 2 + 1=2) 2 q 2 4 2q 1 4q 2 = 2q [FOC] q 2 = 1 q 1 [BR2] 2 3 The solution q 1 = 3=10, q 2 = 4=10 is obtained by solving the system of the two best response functions (2 equations & 2 unknowns). The pro ts and prices are obtained by substituting the equilibrium quantities in the relevant expressions. (b) Assume that rms do not account for their market power, but simply equalize marginal costs to prices. Derive the competitive production levels, pro ts and the equilibrium price. Compare them to the Cournot outcomes. Answer: q 1 = 1=2, q 2 = 1=2, p = 2, 1 = 0, 2 = 0. The competitive production levels are obtained by equating prices to marginal costs: 4 2q 1 2q 2 = 2 4 2q 1 2q 2 = 2q The solution q 1 = 1=2, q 2 = 1=2 is obtained by solving the system of the two best response functions (2 equations & 2 unknowns). The pro ts and prices are obtained by substituting the equilibrium quantities in the relevant expressions. 2

3 Microeconomic Principles II F. Nava (c) Assume that rms form a cartel to sell their output as a monopolist. Derive the cartel production levels, pro ts and the equilibrium price. Compare them to the competitive and Cournot outcomes. Answer: q 1 = 0, q 2 = 1=2, p = 3, 1 = 0, 2 = 1=2. The cartel production levels are obtained by maximizing the joint pro ts: max(4 2q 1 2q 2 )(q 1 + q 2 ) 2q 1 (q 2 + 1=2) 2 q 1 ;q 2 4 4q 1 4q 2 = 2 [FOC1] 4 4q 1 4q 2 = 2q [FOC2] The solution q 1 = 0, q 2 = 1=2 is obtained by solving the system of the two best response functions (2 equations & 2 unknowns). The pro ts and prices are obtained by substituting the equilibrium quantities in the relevant expressions. Notice that the redistribution of pro ts among the two producers in a cartel is undetermined and that I have simply computed the factory pro ts. 3. Consider an auction with two buyers participating and a single object for sale. Suppose that each buyer knows the values of all the other bidders. Order players so that values decrease, x 1 > x 2. Consider a 2 nd price sealed bid auction. In such auction: all players simultaneously submit a bid b i ; the object is awarded to the highest bidder; the winner pays the second highest submitted bid to the auctioneer; the losers pay nothing. Suppose ties are broken in favor of player 1. That is: if b 1 = b 2 then 1 is awarded the object. (a) Characterize the best response correspondence of each player. Answer: Best responses are found simply by looking at the optimal bid holding xed the bid of the other player, as we have done in the slides for the rst price auction. In a second price auction with our tie-braking assumption they satisfy: 1 (b 2 ) = 2 (b 1 ) = 8 < : 8 < : b 1 < b 2 if b 2 > x 1 b 1 2 [0; 1) if b 2 = x 1 b 1 b 2 if b 2 < x 1 b 2 b 1 if b 1 > x 2 b 2 2 [0; 1) if b 1 = x 2 b 2 > b 1 if b 1 < x 2 (b) Characterize all the Nash equilibria for a given pro le (x 1 ; x 2 ). Answer: All Nash Equilibria of the game satisfy one of the following: b 2 x 1 and b 1 x 2 b 1 x 2 and b 2 minfx 1 ; b 1 g To nd the PNE simply intersect the two best response correspondences in a plot. You may verify that for each of the proposed strategies both players are best responding to their competitor. (c) Is bidding above the value weakly dominated? and bidding below the value? Answer: Yes both bidding above and below the value are weakly dominated, by bidding one s value. Bidding below your value is dominated by bidding your value. Consider player 1 bidding b 1 < x 1 and notice that: u 1 (x 1 ; b 2 ) = (x 1 b 2 ) = u 1 (b 1 ; b 2 ) if b 2 b 1 u 1 (x 1 ; b 2 ) = (x 1 b 2 ) > 0 = u 1 (b 1 ; b 2 ) if b 1 < b 2 < x 1 u 1 (x 1 ; b 2 ) = 0 = u 1 (b 1 ; b 2 ) if b 2 x 1 3

4 Microeconomic Principles II F. Nava Bidding above your value is dominated by bidding your value. Consider player 1 bidding b 1 > v 1 and notice that: u 1 (x 1 ; b 2 ) = (x 1 b 2 ) = u 1 (b 1 ; b 2 ) if b 2 x 1 u 1 (x 1 ; b 2 ) = 0 > (x 1 b 2 ) = u 1 (b 1 ; b 2 ) if x 1 < b 2 b 1 u 1 (x 1 ; b 2 ) = 0 = u 1 (b 1 ; b 2 ) if b 2 > b 1 A similar argument works for player 2. Note that for player 2 inequalities need to be adjusted to account for the tie-braking rule. (d) Are there any dominant strategy equilibria Answer: Yes, b i = x i for any i 2 f1; 2g. This is an immediate consequence of part (c) since it has been shown that bidding the valuation is a dominant strategy for both players. (e) If so, what are the seller s revenues in such equilibrium? Answer: The revenues are x 2. If both bid their value the highest value player wins the object and he pays the second highest value, namely x Consider two rms competing on prices to supply a market. Each rm can sell goods to the market at one of three prices f2; 4; 6g. Suppose that rm 1 s marginal cost of production is zero. While the cost of production of rm 2 is only known by rm 2 and is either 2 or 4 with equal probability. If the lowest price charged is p, then assume that market demand for goods is 7 p. The rm o ering the lowest price captures the entire market. To keep payo s simple establish the following convention when rms choose the same price: if the rms marginal costs are less than or equal to the common price, then the market is split evenly amongst them. Otherwise rm 1 captures the entire market at the common price. Therefore payo s in the game satisfy: 1n2L p L = 6 p L = 4 p L = 2 1n2H p H = 6 p H = 4 p H = 2 p 1 = 6 3; 2 0; 6 0; 0 p 1 = 6 3; 1 0; 0 0; 10 p 1 = 4 12; 0 6; 3 0; 0 p 1 = 4 12; 0 6; 0 0; 10 p 1 = 2 10; 0 10; 0 5; 0 p 1 = 2 10; 0 10; 0 10; 0 For this game of incomplete information: (a) Characterize the set of possible pure strategies for each player. Answer: S 1 = f2; 4; 6g & S 2 = fp 2 : fl; Hg! f2; 4; 6gg Equivalently S 2 can be de ne as a list of pairs of actions: S 2 = f(2; 2); (2; 4); (2; 6); (4; 2); (4; 4); (4; 6); (6; 2); (6; 4); (6; 6)g (b) Find the dominated strategies of each player. Answer: Strategies p 2 (L) 2 f2; 6g are weakly dominated for player 2L; strategies p 2 (H) 2 f2; 4g are weakly dominated for player 2H; strategy p 1 = 6 is strictly dominated for player 1. To check these notice that: u 2 (4; p 1 jl) u 2 (6; p 1 jl) u 2 (2; p 1 jl) for any p 1 (somewhere strict) u 2 (6; p 1 jh) u 2 (4; p 1 jh) u 2 (2; p 1 jh) for any p 1 (somewhere strict) u 1 (2; p 2 jx 2 ) > u 1 (6; p 2 jx 2 ) for any p 2 2 f2; 4; 6g & x 2 2 ft; Lg 4

5 Microeconomic Principles II F. Nava (c) Find a pure strategy Bayes Nash equilibrium of this game. Answer: A BNE satis es p 2 (L) = 4, p 2 (H) = 6, p 1 = 2. The other pure strategy BN equilibria satisfy p 1 = 2, for any (p 2 (L); p 2 (H)) 6= (6; 6). This is extra, but mixed strategy BNE satisfy 1 (2) = 1 and: 10(1 (1=2) 2L (2)) + 5(1=2) 2L (2) 12(1=2)( 2H (6) + 2L (6)) + 6(1=2)( 2H (4) + 2L (4)) Let s check that no player bene ts from a deviation in the rst BNE. Notice that the strategy used by player two of either type is dominant and that therefore he cannot bene t from a deviation. All we have to check is the optimality of the strategy of player 1 which holds since: U 1 (2; p 2 ) = 10 > (1=2)6 + (1=2)12 = U 1 (4; p 2 ) > U 1 (2; p 2 ) Similarly for the other pure strategy BN equilibria just notice that player 1 always prefers to play p 1 = 2 so long as p 2 6= (6; 6) and that player 2 s strategy is always optimal as he is always indi erent: where the rst inequality holds since: U 1 (2; p 2 ) > U 1 (4; p 2 ) > U 1 (2; p 2 ) u 2 (4; p 1 jl) = u 2 (6; p 1 jl) = u 2 (2; p 1 jl) u 2 (4; p 1 jh) = u 2 (6; p 1 jh) = u 2 (2; p 1 jh) 10 > (1=2)6 + (1=2)12 (1=2)5 + (1=2)10 > (1=2)12 Finally to test for the mixed BNE notice that if the above condition holds, player 1 does not bene t from a deviation since: U 1 (2; 2 ) U 1 (4; 2 ) U 1 (2; 2 ) (d) Is it a dominant strategy equilibrium? Answer: No, p 1 = 2 is not weakly dominant, since: 5. Consider the following extensive form game: u 1 (4; 6jL) = 12 > 10 = u 1 (2; 6jL) 5

6 Microeconomic Principles II F. Nava (a) Find the unique Subgame Perfect equilibrium of this game Answer: 1:1 (b) = 1, 2:1 (A) = 1, 2:2 (C) = 1 & 1:2 (e) = 1. The proposed strategies constitute a SPE since: u 1:2 (e) = 2 > u 1:2 (d) = 0 u 2:2 (C) = 1 > u 2:2 (D) = 0 u 2:1 (A; 1:2 ) = 4 > u 2:1 (B; 1:2 ) = 2 u 1:1 (b; 1:1 ) = 5 > u 1:1 (c; 1:1 ) = 3 > u 1:1 (a; 1:1 ) = 2 (b) Find a pure strategy Nash equilibrium with payo s (3; 5) Answer: 1:1 (c) = 1, 2:1 (A) = 1, 2:2 (D) = 1 & 1:2 (e) = 1. The proposed strategies constitute a NE since: u 1 (c; e; 2 ) = 3 = u 1 (c; d; 2 ) > u 1 (a; e; 2 ) = 2 > > u 1 (a; d; 2 ) = u 1 (b; e; 2 ) = u 1 (b; d; 2 ) = 0 u 2 (A; D; 1 ) = 5 = u 2 (A; C; 1 ) = u 2 (B; C; 1 ) = u 2 (B; D; 1 ) (c) Find a pure strategy Nash equilibrium with payo s (4; 2) Answer: 1:1 (a) = 1, 2:1 (B) = 1, 2:2 (D) = 1 & 1:2 (d) = 1. The proposed strategies constitute a NE since: u 1 (a; d; 2 ) = 4 = u 1 (a; e; 2 ) > u 1 (c; d; 2 ) = u 1 (c; e; 2 ) = 3 > > u 1 (b; d; 2 ) = u 1 (b; e; 2 ) = 0 u 2 (B; D; 1 ) = 2 = u 2 (B; C; 1 ) > u 2 (A; C; 1 ) = u 2 (A; D; 1 ) = 0 6

7 Solutions with notes for Hand-in Exercise 1 of Lent term Giovanni Ko February 22, 2010 How to read these solutions The main text provides the bare minimum that would be considered sufficient to answer the questions. Naturally, in an exam one would also supply the algebra required to progress from one step to the other. Text in sections formatted like this consists of supplementary explanatory material that is not part of the answer to the question. These notes contain insights obtained from observing mistakes made by students who submitted answers for the exercise. These solutions are complementary to those provided by Francesco and should therefore be used in conjunction with the latter. Any discrepancies in the answers between the two sets of solutions should be resolved in favour of Francesco s. 1 Question 1 The notation σ = al + bc + cr denotes a mixed strategy σ where L, C and R are played with probabilities a, b and c, respectively, i.e., σ(l) = a, σ(c) = b and σ(r) = c. 1.1 Part (a) L C R T 3, 4 1, 1 6, 2 M 2, 1 6, 4 0, 2 B 2, 2 2, 3 4, 2 Best reponses are shown by underlines in the payoff matrix. Reading off pairs of strategies that are best responses to each other, T, L and M, C are the only pure strategy Nash Equilibria. 1

8 1.2 Part (b) For either player, no pure strategy dominates any other, hence there are no dominated strategies. To get a better appreciation for why we cannot say that a several strategies together dominate another consider for example following game: L C R L 0, 10 10, 0 1, 9 C 10, 0 0, 10 1, 9 R 9, 1 9, 1 0, 0 In this game, we might be tempted to say that L and C together dominate R since either one of L or C gives a strictly higher payoff than R for any action of the other player. But it turns out that R is undominated in mixed strategies as there is no mixed strategy involving L and C alone that dominates R. 1 In fact, R is the best response to many mixed strategies, e.g., L + 1 C, 2 2 even though it is not a best response to any pure strategy. Furthermore, the MSNE of this game is for both to play 1 L + 1 C + 8 R: had we wrongly eliminated R from the game because it s dominated by L and C we would have ruled out a perfectly valid NE. 1.3 Part (c) The mixed strategy σ = 1 2 L + 1 2C strictly dominates the strategy R since u 2 (σ, T ) = = 2.5 > 2 = u 2 (R, T ) u 2 (σ, M) = = 2.5 > 2 = u 2 (R, M) u 2 (σ, B) = = 2.5 > 2 = u 2 (R, B). The mixed strategy σ = 3 4 T + 1 4M strictly dominates the strategy B since u 1 (σ, L) = = 2.75 > 2 = u 1 (B, L) u 1 (σ, C) = = 2.25 > 2 = u 1 (B, C) u 1 (σ, R) = = 4.5 > 2 = u 1 (B, R). Note how it suffices to provide one example of a mixed strategy that dominates R and another that dominates B. It is not necessary to provide all possible mixed strategies that dominate B and R. 1.4 Part (d) Since B and R are dominated in mixed strategies, they will never be played with positive probability in a mixed strategy Nash equilibrium. Therefore, we only need to consider mixed strategies σ 1 and σ 2 for P1 and P2 of the form, σ 1 (T ) = p, σ 1 (M) = 1 p, σ 2 (L) = q, σ 2 (C) = 1 q. In order for T and M to be played with positive probability it must be that P1 is indifferent between the two, so that u 1 (T, σ 2 ) = 3q + (1 q) = 2q + 6(1 q) = u 1 (M, σ 2 ) 2

9 which implies that q = 5 6. Likewise, in order for L and C to be played with positive probability it must be that P2 is indifferent between the two, so that u 2 (L, σ 1 ) = 4p + (1 p) = p + 4(1 p) = u 2 (C, σ 1 ) which implies that p = 1 2. Hence, the mixed strategy Nash equilibrium strategies are 1 2 T M for P1 and 5 6 L + 1 6C for P2. When computing mixed strategy equilibria, it is not necessary to write down the expected payoff from a mixed strategy and then maximise it. It simply suffices to apply the correct indifference conditions, as above. Note that in a mixed strategy Nash equilibrium, the player is indifferent only between those actions which are played with positive (non-zero) probability. This means that for P1, only the payoffs from T and M need to be equal to each other, whereas the payoff from B will be strictly less in equilibrium. Similarly, for P2, in equilibrium, only the payoffs from L and C are equal to each other, with R giving a strictly lower payoff than these. What would happen if we tried to find the MSNE of the 3x3 game without first noticing that B and R are strictly dominated and hence never part of a MSNE? First, assign probabilities a, b, 1 a b to T, M, B and α, β, 1 α β to L, M, R. Suppose that P1 plays T, M, B with positive probability so that he s indifferent between all of them. Then the payoffs must satisfy 3α + β + 6(1 α β) = 2α + 6β = 2α + 2β + 4(1 α β), but this system of two equations in two unknowns is inconsistent, as there are no values of α and β that satisfy it. Hence there is no MSNE where T, M and B are all played with positive probability. Similarly for P2, if we were to impose indifference between L, C and R we would obtain 4a + b + 2(1 a b) = a + 4b + 3(1 a b) = 2, which is also inconsistent. Hence, there is no MSNE where L, C and R are all played with positive probability. 2 Question Part (a) When choosing output simultaneously, profits of F1 and F2 are given by π 1 (q 1, q 2 ) = (4 2(q 1 + q 2 )) q 1 2q 1 π 2 (q 1, q 2 ) = (4 2(q 1 + q 2 )) q 2 (q 2 + 1/2) 2. Maximising π 1 with respect to q 1 whilst treating q 2 as fixed, we obtain the FOC 2 4q 1 2q 2 = 0 (1) which leads to the best response function for F1 b 1 (q 2 ) = q 2. 3

10 Similarly, maximising π 2 with respect to q 2 whilst treating q 1 as fixed, we obtain the FOC which leads to the best response function for F2 3 2q 1 6q 2 = 0 (2) b 2 (q 1 ) = q 1. The Nash equilibrium quantities satisfy q 1 = b 1 (q 2 ) and q 2 = b 2 (q 1 ), so that q 1 = 3 10 and q 2 = Total quantity is , equilibrium price is 2 5 and profits for F1 and F2 are and 100, respectively, with total profits 100. If the question does not ask to find the best response function and merely asks to find the Nash equilibrium, it is acceptable (and often easier) to solve the first order conditions (1) and (2) simultaneously rather than going through the best response functions first. 2.2 Part (b) Under perfect competition, all firms produce where price equals marginal cost, i.e., outputs q 1 and q 2 must satisfy the system of simultaneous equations MC 1 (q 1 ) = 2 = 4 2(q 1 + q 2 ) = p MC 2 (q 2 ) = 2q = 4 2(q 1 + q 2 ) = p with solution q 1 = q 2 = 1 2. The equilibrium price is 2 and profits are 0, both of which are lower than in the Cournot case. In this particular case profits are 0 for both firms, which seems natural enough. But note that the standard explanation for zero profits in perfect competition, namely free entry and exit in the industry, does not apply to this case. If the cost function of F2 were (q 2 + 1/3) 2 then q 2 = 2/3 and π 2 = 1/3 > Part (c) If F1 and F2 form a cartel, then they maximise joint profits π(q 1, q 2 ) = (4 2(q 1 + q 2 ))(q 1 + q 2 ) 2q 1 (q 2 + 1/2) 2 with respect to q 1 and q 2. The optimum is characterised by the FOCs 2 4q 1 4q 2 = 0 3 4q 1 6q 2 = 0 which are satisfied by q 1 = 0 and q 2 = 1 2. The price is 3 and joint profits are 1 2. Comparing all three cases we have P. comp. Cournot Cartel q 1 > 0.7 > 0.5 p 2 < 0.26 < 3 π 0 < 0.41 < 0.5. Since they are in a cartel, F1 and F2 would share profits in some way that is left unspecified by the question. Presumably, F1 would get at least the Cournot profits as it would otherwise have no reason to agree to the cartel production level of 0. 4

11 3 Question Part (a) The best response correspondences are b 1 < b 2 if b 2 > x 1 b 2 b 1 if b 1 > x 2 β 1 (b 2 ) = b 1 0 if b 2 = x 1 and β 2 (b 1 ) = b 2 0 if b 1 = x 2 b 1 b 2 if b 2 < x 1 b 2 > b 1 if b 1 < x 2 for player 1 and 2, respectively. The logic is simple: if the other player bids more than your value you want to lose the auction by bidding less; if the other player bids exactly your value you are indifferent between winning or losing so any bid is optimal; if the other player bids less than your own value you want to win the auction by bidding more. Note that if the other player bids your value or less, even bidding above your value is part of the best response since you pay his bid, not your own. 3.2 Part (b) The following diagram shows player 1 s best response correspondence in blue and player 2 s in red. 1 x 1 b x 2 1 b 1 From this we can see there are two regions where the two players strategies are best responses to each other: b 1 x 2 and b 2 x 1 ; b 1 x 2 and b 2 x 1 and b 2 b 1. The main difficulty is in correctly depicting the two best response correspondences on the same graph: drawing (and correctly shading) the one for the player whose bids are on the y-axis is particularly tricky. 5

12 In this case a picture is worth a hundred words if not perhaps a thousand. It is possible to pick out the two regions by testing all possible inequalities involving b 1, b 2, x 1 and x 2 but one has to be extremely careful not to omit cases. Furthermore, an examiner will be more likely to award marks for wrong answers if he s able to see which parts are correct; this is easier in a diagram than in a lengthy verbal explanation. 3.3 Part (c) For player 1 bidding b 1 > x 1 is weakly dominated by bidding x 1 since if x 1 < b 1 < b 2, he loses and gets 0 with either bid; if b 2 x 1 < b 1, he wins and gets x 1 b 2 with either bid; if x 1 < b 2 b 1, he wins and gets x 1 b 2 < 0 if he bids b 1, whereas he loses and gets 0 if he bids x 1. For player 2 bidding b 2 > x 2 is weakly dominated by bidding x 2 since if x 2 < b 2 b 1, he loses and gets 0 with either bid; if b 1 x 2 < b 2, he wins and gets x 2 b 1 with either bid; if x 2 < b 1 < b 2, he wins and gets x 2 b 1 < 0 if he bids b 2, whereas he loses and gets 0 if he bids x 2. For player 1 bidding b 1 < x 1 is weakly dominated by bidding x 1 since if b 1 < x 1 < b 2, he loses and gets 0 with either bid; if b 2 b 1 < x 1, he wins and gets x 1 b 2 with either bid; if b 1 < b 2 x 1, he loses and gets 0 if he bids b 1, whereas he wins and gets x 1 b 2 0 if he bids x 1. For player 2 bidding b 2 < x 2 is weakly dominated by bidding x 2 since if b 2 < x 2 b 1, he loses and gets 0 with either bid; if b 1 < b 2 < x 2, he wins and gets x 2 b 1 with either bid; if b 2 b 1 < x 2, he loses and gets 0 if he bids b 2, whereas he wins and gets x 2 b 1 > 0 if he bids x 1. Note that the arguments for player 1 and player 2 differ only in the cases where b 1 = b Part (d) Part (c) shows that bidding one s own value is a weakly dominant strategy. Therefore the dominant strategy equilibrium is b 1 = x 1 and b 2 = x Part (e) In this dominant strategy equilibrium player 1 wins by bidding x 1 and pays x 2 to the seller. 6

13 4 Question Part (a) The set S 1 of strategies of player 1 is merely the set of actions, i.e., S 1 = {2, 4, 6}. The set S 2 of strategies of player 2 is the set of maps from the set of states/types to the set of actions, i.e., S 2 = {p 2 : {L, H} {2, 4, 6}}. We can also denote every strategy of player 2 by the pair of actions played in each state/type, i.e., S 2 = {(2, 2), (2, 4), (2, 6), (4, 2), (4, 4), (4, 6), (6, 2), (6, 4), (6, 6)}, where p 2 = (4, 6) means p 2 (L) = 4 and p 2 (H) = 6. A common mistake was to write down the best response functions instead of the strategies. Read the question properly. You simply cannot be awarded any marks whatsoever if you answer the wrong question. 4.2 Part (b) Directly from the payoff matrices we see that 1\2L , 2 0, 6 0, , 0 6, 3 0, , 0 10, 0 5, 0 1\2H , 1 0, 0 0, , 0 6, 0 0, , 0 10, 0 5, 0 u 2 (4, p 1 L) u 2 (6, p 1 L) u 2 (2, p 1 L) for all p 1, u 2 (6, p 1 H) u 2 (4, p 1 H) u 2 (2, p 1 H) for all p 1, with strict inequalities if p 1 = 6, u 1 (2, p 2 x 2 ) > u 1 (6, p 2 x 2 ) for all p 2 S 2 and x 2 {L, H} so that all strategies for player 2 are weakly dominated by p 2 = (4, 6); also p 1 = 6 is strictly dominated by p 1 = Part (c) The best responses for player 2 are b 2 (p 1 L) = { 4 if p 1 {6, 4} 6 or 4 or 2 if p 1 = 2 6 if p 1 = 6 b 2 (p 1 H) = 6 or 4 if p 1 = 4 6 or 4 or 2 if p 1 = 2. The best responses for player 2 can be simply highlighted on the individual matrices and read off from there. The best responses for player 1 are { 6 if p 2 (L) = p 2 (H) = 6 b 1 (p 2 ) = 2 otherwise. The intersection of the best responses is then p 1 = 2 and any p 2 that does not satisfy p 2 (L) = p 2 (H) = 6. 7

14 In general, the best responses of player 1 cannot be read off from the individual matrices, but must be found instead by computing expected payoffs Eu 1(p 1, p 2) for every combination of p 1 and p 2 and then comparing. In this particular question, this is not necessary as it is clear that p 1 = 2 guarantees the highest payoffs except when p 2 = (6, 6). When in doubt, compute expected payoffs. Here, we could have also tried to guess a BNE by using the weakly dominant strategy for player 2, viz., p 2 = (4, 6), and either p 1 = 4 or p 1 = 2. We can see that p 1 = 4 is not a best response to (4, 6) as p 1 = 2 gives an expected payoff of 10 rather than = One could have also used iterated elimination of weakly dominated strategies to reduce the game. Note that if we eliminate weakly dominated strategies then we end up with the equilibrium p 1 = 2 and p 2 = (4, 6), but this does not mean that it is the unique Nash equilibrium, as we have shown that there are 7 other equilibria that involve weakly dominated strategies for player 2. On the other hand, if we use iterated elimination of strictly dominated strategies we are guaranteed to find all Nash equilibria, but it wouldn t have helped much in this particular game. 4.4 Part (d) Player 1 s strategy p 1 = 2 is not even weakly dominant as u 1 (4, (6, 6)) = 12 > 10 = u 1 (2, (6, 6)) but u 1 (2, (4, 4)) = 10 > 6 = u 1 (4, (4, 4)). It is incorrect to say that the Nash equilibrium that survives iterated elimination of dominated strategies must be a dominant strategy equilibrium. A strategy is said to be dominant if it dominates all others from the outset, before deleting strategies. 4.5 Ex-ante payoffs The whole of this question could have done by computing the expected payoffs of both players under all combinations of strategies to obtain the following payoff matrix 1\2 6, 6 6, 4 6, 2 4, 6 4, 4 4, 2 2, 6 2, 4 2, 2 6 3, , 1 1.5, 4 1.5, 3.5 0, 3 0, 2 1.5, 0.5 0, 0 0, , 0 9, 0 6, 5 9, 1.5 6, 1.5 3, 3.5 6, 0 3, 0 0, , 0 10, 0 10, 0 10, 0 10, 0 10, 0 7.5, 0 7.5, 0 7.5, 0 from which we can find all dominated/dominant strategies and all Bayesian Nash equilibria. 8

15 5 Question 5 The set of pure strategies of player 1 is S 1 = {a, b, c} {d, e} and the set of pure strategies of player 2 is S 2 = {A, B} {C, D}. Recall that it is necessary to specify all actions to be played, even ones which are in decision nodes that are not reached, so that for example c is not a strategy of player 1, whereas (c, e) is. 5.1 Part (a) By backward induction, e is played at 1 : 2, so that A is played at 2 : 1; C is played at 2 : 2 and so b is played at 1 : 1. The SPNE is therefore (b, e), (A, C). A very common mistake is to simply write (b, C) as the SPNE, since those are the actions that are actually played out in equilibrium. As mentioned above, a strategy must specify an action for every decision node, even those that are not reached in equilibrium. 5.2 Part (b) To find a PSNE with payoffs (3, 5), note that for c to be optimal, it must be that player 2 is playing A at 2 : 1 and D at 2 : 2, as otherwise a or b would give higher payoffs. Therefore the PSNE must be one of (c, d), (A, D) or (c, e), (A, D). 5.3 Part (c) To find a PSNE with payoffs (4, 2), note that for B to be optimal, it must be that player 1 is playing d at 1 : 2. Furthermore, for a to be optimal, it must be that player 2 is playing D at 2 : 2 as otherwise b would give a higher payoff. Therefore the PSNE must be (a, d), (B, D). 5.4 Strategic form Parts (b) and (c) could have been done also by converting the extensive form into the strategic form 1\2 A, C A, D B, C B, D a, d 0, 0 0, 0 4, 2 4, 2 a, e 2, 4 2, 4 4, 2 4, 2 b, d 5, 1 0, 0 5, 1 0, 0 b, e 5, 1 0, 0 5, 1 0, 0 c, d 3, 5 3, 5 3, 5 3, 5 c, e 3, 5 3, 5 3, 5 3, 5 from which we can read off all the PSNE by highlighting best responses. The downside of this method is that when computing MSNE, it is difficult to back out the probabilities with which individual actions are played from the probabilities of strategies. 9