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Simon Fraser University Fall 2012 Econ 302 Final Exam — solution Instructor: Songzi Du Section D200 Thursday Dec. 6, 2012, 3:30 – 6:30 PM This exam has two parts. Write your name, SFU ID number, and tutorial section number on each part. 1. This is a closed-book exam. 2. You may use a non-graphing calculator. 3. There is no separate exam booklet. Write your solution in the space following each question. 4. Show your work! Partial credits are given. 5. We will accept a request for regrade only if the solution is written with a pen. 6. The final has 135 points (45% of the class grade). 7. Stay in your seat if less than 20 minutes remain in the exam. You may leave early if you finish more than 20 minutes before the exam is over. 1 1. (15 points) The market (inverse) demand function for a good is P (Q) = 20 − 2Q. There is a monopolist with a constant marginal cost of 3 for producing each unit of the good. Assume that the government imposes a sales tax of 12%, and the sales tax does not impact the monopolist’s cost. Calculate the monopolist’s price and quantity, and calculate the corresponding consumer’s price. Answer: The monopolist solves max Q 1 (20 − 2Q)Q − 3Q. 1 + 0.12 The first order condition is: 1 (20 − 4Qm ) − 3 = 0. 1.12 Solving this gives the monopolist’s quantity Qm = (20 − 3.36)/4 = 4.16. The consumer’s 1 price is then P (4.16) = 20 − 8.32 = 11.68. The monopolist’s price is 1.12 P (4.16) ≈ 10.43. 2 2. (15 points) Part i: Find the strategies that survive iterative deletion of strictly dominated strategies (ISD). For each strategy that you delete, write down the (possibly mixed) strategy that dominates it. Hint: two strategies of each player survive. Part ii: Calculate all Nash equilibria (pure and mixed) in this game. W X Y Z A 3, 3 0, 1 3, 0 0, -1 B 0, 1 3, 3 0, -1 3, 0 C 1, -1 1, 0 1, 3 1, 1 D 2, 0 0, -1 2, 1 0, 3 Answer: Part i: First iteration of deletion: C is dominated by 1/2A + 1/2B; D is dominated by 0.9A + 0.1B. Second iteration of deletion: Y and Z are both dominated by W . ISD strategies: A and B for player 1; W and X for player 2. Part ii: Three Nash equilibria: (A, W ), (B, X), and a mixed-strategy Nash equilibrium (pA + (1 − p)B, qW + (1 − q)X) which is solved by: 3q = 3(1 − q), 3p + (1 − p) = p + 3(1 − p), i.e., q = 1/2, p = 1/2. 3 3. (15 points) Consider the battle of the sexes game, with a twist: player 1 (the girl) has an additional strategy X (say, X = going to ballet with Mom). Specify the payoffs for the strategy X (i.e., complete the third row) so that (B, B) is the unique Nash equilibrium in the game with the strategy X. Hint: ISD. B H B 3, 1 0, 0 H 0, 0 1, 3 X Answer: There are many possible answers to this question. Consider: B H B 3, 1 0, 0 H 0, 0 1, 3 X 1, 0 2, -1 Clearly, X dominates H for the girl; with H of the girl deleted, B dominates H for the guy; and with H of the guy deleted, B dominates X. Thus, ISD delivers a unique action profile (B, B), which is the unique Nash equilibrium. 4 4. (15 points) Calculate all Nash equilibria (pure and mixed) in this game. Hint: there are two. D E F A 4, 0 0, 4 3, 0 B 0, 4 4, 0 3, 0 C 0, 3 0, 3 3, 3 Answer: The two Nash equilibria: (1/2A + 1/2B, 1/2D + 1/2E), and (C, F ). 5 5. (15 points) Find all pure-strategy subgame perfect equilibria of the following game. Hint: there are two. 1 a b 0.5 1 2 c d 0 0 1 e f 2 2 g h i j 1 1 2 0 2 −2 1 −1 Answer: The two subgame perfect equilibria: (b, e; d, g, j), and (a, f ; c, g, j). 6 6. (15 points) The market (inverse) demand function for a good is P (Q) = 20 − Q, where Q is the total quantity of the good on the market. There are two firms, and each of them has a constant marginal cost of 1 for producing each unit of the good. Firm 1 is the industry leader, so it sets its quantity of production first. Firm 2 sets its quantity of production after observing firm 1’s production. Calculate the subgame perfect equilibrium in this game. Answer: In the subgame (in stage 2) in which firm 1 had produced q1 units, firm 2’s best response is to solve: max(20 − (q1 + q2 ))q2 − q2 . q2 The first-order condition is: 20 − q1 − 2q2 − 1 = 0, or 19 1 19 − q1 = − q1 . (1) 2 2 2 Equation (1) is the equilibrium strategy of firm 2; it specify firm 2’s best-responding action after every contingency (which is a quantity of firm 1, q1 ). Given that player 2 is playing according to the strategy (1), firm 1 in the beginning of the game solves: 19 1 − q1 q1 − q1 . max 20 − q1 + q1 2 2 q2∗ (q1 ) = The first order condition is: 20 − q1 − 19 − 1 = 0, 2 or 19 . (2) 2 Equation (2) is the equilibrium strategy of firm 1, and together with the strategy of firm 2 in (1) they form the subgame-perfect equilibrium of this game. q1∗ = 7 7. (15 points) You have utility over wealth u(w) = w0.3 and start with $10000. Consider a risky project with probability 0.5 of gaining $5000, and probability 0.5 of losing $3500. Find the expected value of this project. Find your expected utility, certainty equivalent, and risk premium for the this project. Would you take the project? Answers: i Expected value is 0.5 × 15000 + 0.5 × 6500 = 10750 dollars. ii Expected utility is 0.5 × (15000)0.3 + 0.5 × (6500)0.3 ≈ 15.91. iii (x)0.3 = 0.5 × (15000)0.3 + 0.5 × (6500)0.3 ⇒ x ≈ 10135.88. So the certainty equivalent is 10135.88 dollars. iv Risk premium = 10750 − 10135.88 = 614.12 dollars. v You will take this project, because the certainty equivalent is above your initial wealth of $10000, or equivalently 0.5 × (15000)0.3 + 0.5 × (6500)0.3 > (10000)0.3 . 8 8. (15 points) Recall the model of education from class. The worker is either skilled (s = 1) or unskilled (s = 0), and this is his private information (his type); the probability of a skilled worker is p. Worker of either type chooses to be educated (e = 1) or not (e = 0), and education does not change his skill; education has a cost of c to the skilled worker and k to the unskilled worker. The firm observes the education level (e) of the worker, and offers him a fair wage which is the probability that he is skilled conditional on his education level: we = µ(s = 1 | e), and this conditional belief µ is based on the strategy of worker (how worker of each type chooses education). The utility of a worker from choosing education level e (e is either 0 or 1) is we − e · c or we − e · k, depending on his type. Assume k > c. Calculate a semi-separating equilibrium where an unskilled worker gets education with probability q. What is the condition on k and p for this equilibrium to exist? Answer: Since an unskilled worker mixes between acquiring an education (with probability q) and acquiring no education (with probability 1 − q), we must have w1 − k = w0 . But this means that w1 − c > w0 , which means that a skilled worker acquires an education (with probability 1). Thus, we have w0 = µ(s = 1 | e = 0) = 0. Moreover, w1 = µ(s = 1 | e = 1) = Thus, p . (1 − p)q + p p − k = 0, (1 − p)q + p i.e., p = k(1 − p)q + kp, i.e., q= p(1 − k) . k(1 − p) We must have 0 ≤ q ≤ 1, i.e., p ≤ k ≤ 1. 9 9. (15 points) Consider the third price auction with 4 bidders and 2 identical goods: the top two bidders win the two goods and each pays the third highest bid; everyone else do not pay. Assume that each bidder wants only one good and has a value vi for it. Suppose that bidder 1 bids b1 = 7, bidder 3 bids b3 = 9, and bidder 4 bids b4 = 5. If v2 = 6, explain why in this situation it is optimal for bidder 2 to bid his true value of 6. And explain why it is optimal for bidder 2 to bid his true value if v2 = 8. Answer: Suppose that v2 = 6. By bidding more than 7 (b2 > 7), bidder 2 wins a good and pays 7, which gives a profit of −1. By bidding less than 7 (b2 < 7) bidder 2 does not win and gets 0 profit. And by bidding at exactly 7, bidder 2 wins a good and pays 7 with probability 0.5 (and gets nothing with probability 0.5), which gives a (expected) profit of −0.5. The second outcome is clearly better than the rest. In particular, bidding his value (b2 = 6) induces the second outcome. Suppose that v2 = 8. By bidding more than 7 (b2 > 7), bidder 2 wins a good and pays 7, which gives a profit of 1. By bidding less than 7 (b2 < 7) bidder 2 does not win and gets 0 profit. And by bidding at exactly 7, bidder 2 wins a good and pays 7 with probability 0.5 (and gets nothing with probability 0.5), which gives a (expected) profit of 0.5. The first outcome is clearly better than the rest. In particular, bidding his value (b2 = 8) induces the first outcome. 10