ECO 5341 (Section 2) Spring 2016 Midterm March 24th 2016 Total Points: 100

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1 Name:... ECO 5341 (Section 2) Spring 2016 Midterm March 24th 2016 Total Points: 100 For full credit, please be formal, precise, concise and tidy. If your answer is illegible and not well organized, if your arguments are informal and imprecise, you will lose points. Show all your algebra! Question 1 (20 points): Consider the following two person game. Player 1 is row player, Player 2 is column player. L C R T 3,3 1,2 5,1 M 2,3 2,4 6,0 B 1,2 0,5 1,5 a) (5 points) Does player 1 have a strictly dominated strategy? Explain formally by writing player 1 s best response to every possible strategy of player 2. Once you eliminate player 1 s strictly dominated strategy, does player 2 now have a strictly dominated strategy? Again explain formally. Answer: Player 1 s best reponses are BR 1 (L) = T BR 1 (C) = M BR 1 (R) = M Note that for player 1, Strategy B is strictly dominated by Strategy M. After eliminating B, we have Player 2 s best reponses are L C R T 3,3 1,2 5,1 M 2,3 2,4 6,0 BR 2 (T) = L BR 2 (M) = C Note that for player 2, Strategy R is strictly dominated by both Strategy C and Strategy L. 1

2 b) (5 points) After eliminating the strictly dominated strategies for each player, state all pure strategy NE of this game. Note: Just stating the pure strategy NE is suffi cient. Answer: (T,L) and (M,C) are pure NE. 2

3 b) (10 points) Find all mixed strategy NE of this game. We have L C T 3,3 1,2 M 2,3 2,4 3q + (1 q) = 2q + 2(1 q) 2q + 1 = 2 q = 1 2 and 3p + 3(1 p) = 2p + 4(1 p) 3 = 4 2p p = 1 2 The Mixed Strategy NE is Player 1 plays T with probability 1 2 and plays M with probability 1 2. Player 2 plays L with probability 1 2 and plays C with probability 1 2 3

4 Question 2 (20 points) Two research firms, Firm 1 and Firm 2 simultaneously choose how much time x i to spend on research to develop a new drug. Firm 1 chooses x 1 0 and Firm 2 chooses x 2 0. If Firm i spends x i > x j, that is, spends strictly more time on research than its rival, then Firm i gets 10 x i. If Firm i spends x i < x j, that is, spends strictly less time than its rival, all time spent on research is lost and Firm i receives a payoff of x i. If the two firms spend the exact same amount x i = x j, then each firm gets 5 x i. The two firms payoff functions are given by u 1 (x 1, x 2 ) = u 2 (x 1, x 2 ) = 10 x 1 if x 1 > x 2 5 x 1 if x 1 = x 2 x 1 if x 1 < x 2 10 x 2 if x 2 > x 1 5 x 2 if x 1 = x 2 x 1 if x 1 < x 2 a) (5 points) Is the strategy pair (x 1 = 5, x 2 = 5) a NE? Your argument needs to be formal. Answer: The strategy pair (x 1 = 5, x 2 = 5) is NOT a NE. In this proposed NE, player 1 s payoff is u 1 (x 1 = 5, x 2 = 5) = 5 5 = 0 Instead of playing x 1 = 5, player 1 can set any x 1 = 5 + ε where ε < 5 and get u 1 (x 1 = 5 + ε, x 2 = 5) = 10 (5 + ε) = 5 ε > 0 which is strictly better than his current payoff of 0. Since player 1 has a profitable deviation, this is not a NE. 4

5 b) (5 points) Is there any NE in which x 1 < x 2 < 5, that is Firm 1 spends strictly less time on research than Firm 2 and also both firms spend strictly less than 5? To receive any credit your argument needs to be formal, that is, you need to formally show if there is or there is not any such NE. Answer: Suppose there is a strategy profile in which x 1 = a < x 2 = b < 5 and this strategy profile is a NE. In this proposed NE, player 1 s payoff is u 1 (x 1 = a, x 2 = b) = a < 0 since a < b. But player 1 can instead play x 1 = b and get u 1 (x 1 = b, x 2 = b) = 5 b > 0. which is strictly better than his current payoff of a < 0. Since player 1 has a profitable deviation, there cannot be a NE where x 1 < x 2 < 5. 5

6 c) (10 points) Is any strategy pair with x 1 = x 2 where 0 x 1 = x 2 < 5 a NE? (That is, both firms spend the same time on research and both firms spend strictly less than 5). Your argument needs to be formal to receive any credit. Answer: First, suppose there is a strategy profile in which 0 < x 1 = x 2 = a < 5 and this strategy profile is a NE. In this proposed NE, player 1 s payoff is u 1 (x 1 = a, x 2 = a) = 5 a < 5. But player 1 can instead play x 1 = 5 and get u 1 (x 1 = 5, x 2 = a) = 10 5 = 5 > 5 a. which is strictly better than his current payoff of 5 a. Since player 1 has a profitable deviation, there cannot be a NE where 0 < x 1 = x 2 = a < 5. The only remaining possibility is the profile x 1 = x 2 = 0. In this proposed profile, player 1 s payoff is u 1 (x 1 = 0, x 2 = 0) = 5. But note that player 1 can instead play x 1 = ε > 0 and get u 1 (x 1 = ε, x 2 = 0) = 10 ε > 5 as long as ε < 5. Since player 1 has a profitable deviation, the profile x 1 = x 2 = 0 is not a NE. Therefore, there is no NE in which 0 x 1 = x 2 < 5. 6

7 Question 3 (30 points) Consider two farmers, Farmer 1 and Farmer 2 who choose the number of palm trees x i to plant in a common green field. If Farmer 1 plants x 1 trees and Farmer 2 plants x 2 trees, then there will a total number of X = x 1 + x 2 trees. If a total number of X trees are planted, then each palm tree can fetch a price P given by P = 600 X where X = x 1 + x 2 is the total number of palm trees. Note that the more palm trees planted, the lower the price each tree can fetch. Each palm tree costs the farmer c = $60. Therefore the profit of each farmer is given by π 1 (x 1, x 2 ) = (600 x 1 x 2 )x 1 60x 1 π 2 (x 1, x 2 ) = (600 x 1 x 2 )x 2 60x 2 Each farmer i simultaneusly chooses his own number of trees x i to maximize own profits π i. That is, Farmer 1 chooses x 1 to maximize π 1 (x 1, x 2 ) and Farmer 2 chooses x 2 to maximize π 2 (x 1, x 2 ). 7

8 a) (15 points) Derive Farmer 1 s best response x1 (x 2) and Farmer 2 s best response x2 (x 1).Find the equilibrium number of trees (x1, x 2 ) planted by the farmers. How many palm trees in total are planted, that is, what is X = x1 + x 2? What is the equilibrium profit of each farmer in this equilibrium? You need to show all your algebra. Answer: Choose x 1 to Maximize π 1 (x 1, x 2 ) = (600 x 1 x 2 )x 1 60g 1 FOC is 600 2x 1 x 2 60 = 0 which gives us x 1(x 2 ) = x 2 Similarly, choosing x 2 to Maximize π 2 (x 1, x 2 ) = (600 x 1 x 2 )x 2 60g 2 FOC is 600 2x 2 x 1 60 = 0 which gives us x 2(x 1 ) = x 1 The NE pair (x 1, x 2 ) solves x1 = x 2 x 1 = [ ] 2 2 x x 1 = x 1 = 180 Hence x 2 = x 2 = 180 We have P = = 240 π 1 = π 2 = (240 60) 180 = = 32, 400 8

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10 b) (10 points) What is the total number of palm trees X M that would maximize the joint profits of the two farmers. Notice that the joint profits of the two farmers are given by π (X) = (600 X)X 60X How does the optimal number of total goats X M that maximizes π (X) above compare with the total number X = x1 + x 2 you found in part a? Do the farmers overplant or underplant when they choose x 1 and x 2 individually to maximize own profits. Note: To find X M, you need to maximize π (X) = (600 X)X 60X with respect to X. yields Answer: Maximizing π (X) = (600 X)X 60X 600 2X M 60 = 0 X M = 270 Since x 1 + x 2 = = 360 > X M = 270 farmers overuse the green when maximizing individually. 10

11 c) (5 points) Suppose the two farmers try to cooperate and each decide to plant exactly half of the joint profit maximizing number of trees X M you found in part (b). Is this cooperative outcome ) (x 1 = x 2 = XM 2 a Nash Equilibrium? Explain why or why not. Answer: No, strategy profile x 1 = x 2 = XM 2 = = 135 is not NE. When Farmer 2 sets x 2 = 135, Farmer 1 s best response is x 1 = x 2 = = >

12 Question 4 (15 points) Consider the following game between a homeowner and a contractor. In the first stage of the game, the contractor chooses his hourly wage w to maximize own payoff u c (w, L) = (w 20)L where L > 0 is the amount of service hours provided. L is chosen by the homeowner in the second stage of the game. After observing w chosen by the contractor, in the second stage the homeowner chooses the number of service hours L to maximize own payoff u h (w, H) = 100βL βl 2 wl where β > 0 is the skill level of the contractor. Find the backward induction equilibrium (w, L ) as a function of β. How does the contractor s equilibrium hourly wage w and homeowner s demand for contractor s service L depend on the contractor s skill level β? Answer: Choosing L to maximize yields the FOC u h (w, H) = 100βL βl 2 wl 100β 2βL w = 0 L (w) = 50 w 2β In the first stage the contractor chooses w to maximize u c (w, L) = (w 20)L (w) = 50w w2 2β + 10w β 1000 FOC is given by Hence 50 w β + 10 β = 0 w = 50β + 10 ( ) L (w 50β + 10 ) = 50 L (w ) = β β 12

13 Question 5 (10 points) Consider the following sequential move game. First, Player 1 chooses between L and R. If Player 1 chhoses L, the game ends. If Player 1 chooses R, in the second stage Player 2 chooses between In and Out. If Player 2 chooses Out, the game ends. If Player 2 plays In, it is again Player 1 s turn to move at the last stage and Player 1 chooses between T and B. The first payoff is Player 1 s payoff, the second payoff is Player 2 s payoff. Player 1 L R 4,2 Player 2 Out In 3,3 T Player 1 B 5,2 4,3 Describe the "backward induction equilibrium" of the game by stating the equilibrium strategy of each player at every decision node that the player has to choose an action. What is the backward induction equilibrium "outcome" of this game. Answer: The backward induction equilibrium is Player 1 plays L in the first stage If called to action Player 2 plays Out in the second stage If called to action Player 1 plays T in the last stage. The backward induction equilibrium outcome is Player 1 plays L in the first stage and the game ends with players getting a payoff (4,2). 13

14 Question 6 (5 points) Consider the game below and find the unique outcome that survives the Iterated Elimination of Strictly Dominated Strategies by filling in the blanks below. You just need to fill in the blanks in the statements below. Player 1 is the row player, Player 2 is the column player. L C R T 3,1 2,3 0,4 M 1,2 2,2 3,1 B 4,2 3,1 2,1 Step 1: Strategy T is strictly dominated by Strategy B for Player 1 hence can be eliminated. L C R M 1,2 2,2 3,1 B 4,2 3,1 2,1 Step 2: Having eliminated Strategy T in the first step, now Strategy R is strictly dominated by Strategy L for Player.2 and hence can be eliminated. Step 3: Having eliminated Strategy R L C M 1,2 2,2 B 4,2 3,1 in the second step, now Strategy M is strictly dominated by Strategy B for Player 1 and hence can be eliminated. L C B 4,2 3,1 Step 4: The only pair of strategies that survive the Iterated Elimination of Strictly dominated Strategies is Strategy B for Player 1 and Strategy L for Player 2. 14