Simon Fraser University Fall Econ 302 D200 Final Exam Solution Instructor: Songzi Du Wednesday December 16, 2015, 8:30 11:30 AM

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1 Simon Fraser University Fall 2015 Econ 302 D200 Final Exam Solution Instructor: Songzi Du Wednesday December 16, 2015, 8:30 11:30 AM NE = Nash equilibrium, SPE = subgame perfect equilibrium, PBE = perfect Bayesian equilibrium 1. There is a shared internet connection which has a maximum capacity of 100. There are n = 10 users of this connection, and user i uses x i units of data, where 0 x i 100. (x i can be a real number, not necessarily an integer.) If n i=1 x i 100, then the connection is jammed and each user gets zero payoff. If n i=1 x i < 100, then each user i gets a payoff of x i (100 n i=1 x i). Suppose that each user chooses his/her x i simultaneously. (Recall that n i=1 x i = x 1 + x x n.) i. (5 points) The total payoff of all users is ( n i=1 x i)(100 n i=1 x i). Let y = n i=1 x i. What value of y maximizes the total payoff? ii. (10 points) Calculate the symmetric NE (in pure strategy) such that n i=1 x i < 100. Would everyone be better off (relative to the NE outcome) if x i = y/n, where y is determined in part (i), and why or why not? Part i: we maximize the total payoff y(100 y). The first order condition is 100 2y = 0, i.e., y = 50 maximizes the total payoff. Part ii: suppose x i = x is the symmetric NE. Let us pick user 1 and look at his best response: player 1 maximizes x 1 (100 x 1 9x) since all other users uses x. Player 1 s first order condition is then: i.e., x = 100/11 is the symmetric NE x 1 9x x1 =x = 0, 1

2 In this NE, each user gets payoff 100/11( /11) = 10000/ If x i = y/10 = 5 for everyone, then each user gets payoff 5(100 50) = 250. Clearly, everyone is much better off with x i = 5. Thus, this is a prisoner s dilemma. 2. (10 points) Consider a new version of rock-paper-scissors game which has the following payoff matrix (player 1 has a big advantage if he uses Rock and player 2 uses Scissors): Rock Scissors Paper Rock 0, 0 30, , 10 Scissors -10, 10 0, 0 10, -10 Paper 10, , 10 0, 0 As before each player chooses his action simultaneously. Find the mixed-strategy NE (this equilibrium is asymmetric: the two players use different mixed strategies). What are players expected payoff in this equilibrium? Suppose the mixed-strategy equilibrium is player 1 using pr + qs + (1 p q)p and player 2 using xr + ys + (1 x y)p. Player 1 must be indifferent between R, S and P : i.e., 30y 10(1 x y) = 10x + 10(1 x y), 30y 10(1 x y) = 10x 10y, i.e., x = 1/3, y = 1/5. Player 2 must be indifferent between R, S and P : 10q + 10(1 p q) = 30p 10(1 p q), 10q + 10(1 p q) = 10p + 10q, i.e., p = 1/5, q = 1/3. Notice that in equilibrium player 1 places less probability on Rock, relative to the symmetric version of the game. 3. There are 4 houses in a neighborhood. For a resident in a house, the cost of installing an alarm system is 13, and the cost of his/her house being burglarized is 50. If n (0 n 4) 2

3 houses have the alarm, then these n houses will not be burglarized (each gets a payoff of 13), while for the remaining 4 n houses each will be burglarized with probability 1/(4 n) (each gets a payoff of 50/(4 n)). Suppose the residents sequentially decide whether or not to install the alarm system: resident 1 moves first and either installs the alarm or does not; after observing what resident 1 did, resident 2 decides; after observing what resident 1 and 2 did, resident 3 decides; after observing what resident 1, 2 and 3 did, resident 4 decides. i. (7 points) Draw the game tree (write the first payoff for resident 1 s payoff, the second payoff for resident 2 s payoff, and so on). ii. (8 points) Find and describe the SPE in pure strategy. Part i: (I omit the game tree because it is straightforward.) Part ii: notice that 13 > 50/(4 n) if and only if n = 0. Here is the SPE which is solved by backward induction: Let 0 n 4 3 be the number of houses that have installed the alarm before resident 4. Resident 4 installs the alarm if n 4 > 0 and does not install the alarm if n 4 = 0. Let 0 n 3 2 be the number of houses that have installed the alarm before resident 3. Resident 3 installs the alarm if n 3 > 0 and does not install the alarm if n 3 = 0. Let 0 n 2 1 be the number of houses that have installed the alarm before resident 2. Resident 2 installs the alarm if n 2 > 0 and does not install the alarm if n 2 = 0. Resident 1 does not install the alarm in the beginning of the game. 4. A pharmaceutical company (player 1) introduces a new cold medicine. The medicine may either be highly effective (H type) or have little effect (L type). The company knows the type of the drug, but the consumer (player 2) does not know the type (he/she only knows the medicine is H type with probability 0.3 and is L type with probability 0.7). The company can choose either to advertise the drug (action A), at a cost of c, or not to advertise 3

4 (action N), which costs nothing. The consumer decides whether or not to buy the drug after observing whether the company advertised the drug. The consumer gets a payoff of 1 if he buys a highly effective drug, -1 if he buys the little-effect drug, 0 if he does not buy. If the drug is highly effective, then the company gets 100 if the consumer buys the drug (i.e., making money from repeated purchases); if the drug has little effect, then the company gets 10 if the consumer buys the drug (i.e., making money from one time purchase). The company gets 0 if the consumer does not buy. The net payoff of the company is the payoff from sales minus the advertisement cost. For example, if the drug has little effect but the company advertises it, and the consumer buys, then the net payoff of the company is 10 c. i. (4 points) Draw the game tree. ii. (6 points) Suppose c = 20, find all pure-strategy PBE (if any exists). iii. (10 points) Suppose c = 5, find all pure-strategy PBE (if any exists). Find a mixedstrategy PBE in which the L-type company randomizes between advertising and not advertising. In part (ii) and (iii), include conditional beliefs P(H A) and P(H N) in your description of PBE. Part ii: Clearly, the consumer s best response is to buy after advertisement (A-buy) if P(H A) 1/2, and not buy after advertisement (A-not) if P(H A) 1/2; to buy after no advertisement (N-buy) if P(H N) 1/2, and not buy after no advertisement (N-not) if P(H N) 1/2. The low-type company clearly always prefers no advertisement (L-N) to advertisement (L-A). Thus, we can safely ignore L-A. There are only two strategies of the company that are potentially PBE: Suppose the company uses the strategy (H-A, L-N). Then P(H A) = 1 > 1/2 and P(H N) = 0 < 1/2. So consumer s best response is (A-buy, N-not). And given (A-buy, N-not), (H-A, L-N) is clearly also a best response for the company. Thus, this is a PBE. 4

5 Figure 1 : part i when c = 20. Suppose the company uses the strategy (H-N, L-N). Then P(H A) can be anything, and P(H N) = 0.3. Consumer s best response must use N-not since 0.3 < 1/2. If the consumer uses A-buy, then H-N is not a best response for the high-type company, which is inconsistent with the initial conjecture. If the consumer uses A-not, then H-N is a best response for the high-type company. Thus, the PBE here is: the company uses (H-N, L-N); the consumer believes that P(H N) = 0.3 and P(H A) 1/2 and uses (A-not, N-not). Part iii: consumer s best response as a function of his beliefs is the same as part ii, because his payoff is not changed. Now both the high-type and the low-type company have no dominant strategy. So we go through all four strategies of the company: Suppose the company uses the strategy (H-A, L-A). Then P(H A) = 0.3 < 1/2 and P(H N) can be anything. So consumer s best response must include A-not. But given A-not by the consumer, H-N is always better for the high-type company than H-A, so (H-A, L-A) cannot be a PBE. Suppose the company uses the strategy (H-A, L-N). Then P(H A) = 1 > 1/2 and 5

6 P(H N) = 0 < 1/2. So consumer s best response is (A-buy, N-not). And given (Abuy, N-not), L-A is better for the low-type company than L-N, so (H-A, L-N) cannot be a PBE. Suppose the company uses the strategy (H-N, L-A). Then P(H N) = 1 > 1/2 and P(H A) = 0 < 1/2. So consumer s best response is (A-not, N-buy). And given (Anot, N-buy), L-N is better for the low-type company than L-A, so (H-N, L-A) cannot be a PBE. Suppose the company uses the strategy (H-N, L-N). Then P(H A) can be anything, and P(H N) = 0.3. Consumer s best response must use N-not since 0.3 < 1/2. If the consumer uses A-buy, then H-N is not a best response for the high-type company, which is inconsistent with the initial conjecture. If the consumer uses A-not, then H-N is a best response for the high-type company. And given N-not, L-N is also a best response for the low-type company. Thus, the PBE here is: the company uses (H-N, L-N); the consumer believes that P(H N) = 0.3 and P(H A) 1/2 and uses (A-not, N-not). Now for the mixed-strategy equilibrium in which the low-type company is indifferent between L-A and L-N and randomizes between the two actions. Then, the high-type company must strictly prefers H-A over H-N, because they have the same advertisement cost but the high type makes more money when the consumer buys. Suppose that the low type plays L-A with probability p and the high type plays H-A with probability 1. Then P(H A) = 0.3/( p) and P(H N) = 0. So the best response of the consumer must use N-not. For the low type to be indifferent between L-A and L-N, the consumer must randomize after advertisement; the only way for that to happen is P(H A) = 0.3/( p) = 1/2, i.e., p = 3/7. Suppose the consumer uses A-buy with probability q. For the low type to be indifferent we must have 5q 5(1 q) = 0, i.e., q = 1/2. Thus, the mixed-strategy PBE is: the company uses (H-A, 3/7 L-A + 4/7 L-N), the consumer believes that P(H A) = 1/2 and P(H N) = 0, and uses (1/2 A-buy + 1/2 A-not, N-not). This mixed-strategy equilibrium is more efficient than the pure-strategy equilibrium in which nobody buys. 6