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### PS1solutions

Course: ECON 171, Fall 2009
School: UCSB
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171 Econ Spring 2010 Problem Set 1 Solutions to the two-star Problems In some cases Ive provided more explanation than was asked of you. ** Problem 3 Find the rationalizable strategies for this game. For each iteration, list the eliminated A B C W 5, 6 4, 3 6, 2 X 4, 0 3, 4 2, 0 Y 2, 1 6, 0 0, 5 Z 1, 2 2, 1 5, 7 strategy and which strategy dominates it. The unique rationalizable strategy is (C, Z ). The logic...

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171 Econ Spring 2010 Problem Set 1 Solutions to the two-star Problems In some cases Ive provided more explanation than was asked of you. ** Problem 3 Find the rationalizable strategies for this game. For each iteration, list the eliminated A B C W 5, 6 4, 3 6, 2 X 4, 0 3, 4 2, 0 Y 2, 1 6, 0 0, 5 Z 1, 2 2, 1 5, 7 strategy and which strategy dominates it. The unique rationalizable strategy is (C, Z ). The logic is as follows: All actions are rational for Player 1. A is optimal if she believes 2 will play X with certainty, B is optimal if she believes Y , and C is optimal if she believes W will be played. Only W , X and Z are rational for Player 2 because Y is dominated by Z . Y is eliminated. If Player 1 knows that 2 is rational, she will never play B . She knows Y wont be played. A is optimal if she thinks X will be played and C is optimal given W . Is B ever optimal? Well, if 1 believes that Pr(Z ) > 1/4 then she prefers C to B and if she believes that Pr(Z ) < 1/2 then she prefers A to B . This means that B is never optimal, which in this type of game tells us that it is dominated. Any 1 strategy that places probability q on A and 1 q on C , with q ( 2 , 3 ) dominates 4 B. If Player 2 only knows that 1 is rational, then she cant be certain that 1 will eliminate any action, but if she knows that 1 knows that she is rational, then she knows 1 wont play B . This means that X is dominated by W and can be eliminated. If Player 1 knows this, then she knows that 2 will only play W or Z , in which case A is dominated by C and is eliminated. Knowing this, W is dominated by Z and is eliminated. ** Problem 4 Find all Nash equilibria for these two games. (a) There is no pure strategy Nash Equilibrium. Any mixed strategy NE has to involve Player 2 mixing, because otherwise Player 1 would not mix either. Let pA , pB , and pC be the probabilities that 1 plays A, B , and C , respectively. Let q and 1 q be the probabilities with which 2 plays L and R, respectively. For Player 2 to be willing to mix, it is necessary that pB = pC , because 2 strictly prefers L if C is more likely and prefers R if B is more likely. 1 A B C L 2, 0 3, 0 0, 1 R 2, 0 0, 1 3, 0 A B C a) L 2, 0 5, 0 0, 2 R 2, 0 0, 3 5, 0 b) Case 1 Suppose pB = pC > 0. Player 1 will only be willing to mix B and C if L and R are equally likely (q = 1/2), but this would mean that A is preferred to both. There is no NE in this case. Case 2 Suppose pB = pC = 0. A is preferred to both B and C for 1/3 q 2/3. In this case, pA = 1, to all which q [0, 1] is a best response for 2. Thus 2 (A, q ) is a NE for all q [ 1 , 3 ]. These are all of the NE of the game. 3 (b) Again, there is no pure strategy NE. One way to begin is to note that A is not rationalizable. It is dominated by an even mix between B and C . Let p be the probability with which Player 1 plays B and let q be the probability that 2 plays L. For mixing to be sustained in equilibrium, each players strategy must make the other indierent between the two pure strategies in that persons mix. Player 2 is indierent when 2(1 p) = 3p, or p = 2/5. Player 1 is indierent (between B and C ) when q = 1/2. So the unique NE involves 1 playing (0, 2/5, 3/5) and 2 playing (1/2, 1/2). ** Problem 5 Watson 9.7 That (B, X ) is a Nash equilibrium does not put any restrictions on x. (A, Z ) is 1 ecient as long as x 4. For Y to be a best response to 1 = ( 2 , 1 ), we need 2 u2 (1 , Y ) = 3 x/2 + 1 = u2 (1 , Z ), which is satised for x 4. Thus, for all three statements to be true it must be the case that x = 4. ** Problem 6 Watson 9.9 (a) The Nash equilibria occur where the curves intersect, i.e. (2, 1), (5/2, 2), and (3, 3). (b) The set of rationalizable strategy proles is [2, 3] [1, 3]. Explanation (not required): Looking at BR1 we see that no matter Player 1s beliefs, [0, 1) and (4, 5] are never best responses, and thus not rationalizable. For Player 2, anything in the interval (4, 5] is never a best response. Player 1, knowing Player 2 is rational, knows that s2 4 which eliminates s1 > 3.5. Player 2, knowing 1 is rational, knows that 1 s1 4, which eliminates s2 > 3.5 and s2 less than some value that looks to be around 0.8. We see from here that each successive iteration will further restrict the remaining set of proles until it converges to the one stated above. ** Problem 7 Watson 11.3 (a) (N, L) and (L, N ). 2 (b) Firm Y chooses q so that Firm X is indierent between L and N . This yields 25x 5q +(x 15)(1 q ) = 10 10q . Rearranging yields q = 20x . Firm X chooses p so that Firm Y is indierent between L and N . This yields 5p +15 15p = 10 10p. Rearranging yields p = 1/2. (c) The probability of (L, N ) equals p(1 q ) = (1/2) 20x25+x 20x = (1/2) 5 x20 . (d) As x increases, (L, N ) becomes a better outcomes for Firm X, and no worse for Firm Y, but it becomes less likely to be played. This may seem counterintuitive, but cases like these force us to rene/relearn our intuition, and draw more insights in the future. Can you nd the intuition for this result? 3
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Econ 171 Spring 2010Problem Set 3Due Tuesday, June 1Important: hand in only the two-star problems. There are no one-star problems on thisproblem set. The notation a.b denotes problem number b from Chapter a in Watson.* Problem 1Consider a two-player
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Econ 171 Spring 2010Problem Set 3 - SolutionsDue Tuesday, June 1Important: hand in only the two-star problems. There are no one-star problems on thisproblem set. The notation a.b denotes problem number b from Chapter a in Watson.* Problem 1Consider
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Project 11MAC 2233In this project, we explore the average and instantaneous rates of change of total cost, total revenue, and total prot with respect to the number x of items produced/sold. In particular, note theinterpretations of the calculations, an
University of Florida - MAC - 2233
Project 12MAC 22331. Examine the function f (x) = x3 3x. From precalculus, what should be the two distinctpossibilities for the number of turning points of any cubic (degree 3) polynomial function?In order for a smooth graph to turn around, what value
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Econ 171 (Grossman) Spring 2010Exam 1April 20You have 75 minutes to take this exam. Please answer all 6 questions, each of which is worth5 points. Point subtotals are indicated. Show your work to obtain full credit.Part 1 4 questions, 20 points (5 ea
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Project 13/14MAC 22331. An underwater pipe begins to leak oil that spreads outward in the shape of an expandingsphere. Recall that the volume of a sphere is V = 4 r3 , where r is the length of its radius.3Suppose the company that owns the pipe takes
UCSB - ECON - 171
Econ 171 (Grossman) Spring 2010Exam 1April 20You have 75 minutes to take this exam. Please answer all 6 questions, each of which is worth5 points. Point subtotals are indicated. Show your work to obtain full credit.Part 1 4 questions, 20 points (5 ea
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Project 15MAC 22331. One model that is used for an agricultural yield Y bushels (in thousands) as a function of soilNnitrogen level N points is Y (N ) =.1 + N2Calculate Y (N ) and Y (N ). Give the units for each.Y (N ) =Y (N ) =1 N2(1 + N 2 )2
University of Florida - MAC - 2233
Project 16MAC 22331. Suppose the daily supply x of a certain product in an open-air market is related to the unitprice p (in dollars) by the equation 100p2 + 10xp = x2 + 500.Calculatedximplicitly. What rate of change does it represent?dpFind all t
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University of Florida - MAC - 2233
Project 24MAC 22331. (From lecture) You are required to construct a closed rectangular box with a surface area of 48square feet so that the length is twice the width. What dimensions will maximize the volumeof the box?Max volume when width is 2, leng
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UCSB - ECON - 177
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UCSB - ECON - 177
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UCSB - ECON - 177
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University of Florida - MAC - 2233
MAC 2233 Review Worksheet Answers1. a. 5/8 grams1. b. 20 years2. m = 7/23.x 8 x 4 ln xC4.23/ 2 x 1 2 x 1 C35.1C3x3 e 3x 6.13/ 2ln4x 2 x 2C217. e x 2C8.ln 229. 838 fish10. 56/9
UCSB - ECON - 177
Discrete BidsAverage Bid if saw 0: 1.364Average Bid if saw 3: 2.73Average Profit = .462 millionNumber positive = 6, Number negative = 4, Number zero = 3Continuous BidsAverage Bid if saw 0: 1.316Average Bid if saw 3: 2.605Average Profit = .649 mill