ECN611 Handout1

ECN611 Handout1 - Economics 611 Handout #1 G a m e th e o r...

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Unformatted text preview: Economics 611 Handout #1 G a m e th e o r y - a general theory of strategic interaction Games of strategy, not games of skill - Equally intelligent players Game form: players / rules (e.g., admissibility of strategies) / outcomes Game: Game form plus preferences Games: Cooperative vs. noncooperative S im u lta n e o u s -M ov e G a m e s w ith P e r f e c t In f o r m a tio n In pure strategies game form: ( I, {Si}, g(￿) ) game à = ( I, {Si}, g(￿), {ui}) or ( ￿, {Si}, {ui}) The game form is common knowledge. Common knowledge: If everyone knows x, then x is known up to level 0. If everyone knows x, and (everyone knows that everyone knows x), then x is known up to level 1. If everyone knows x, and (everyone knows that everyone knows x), and (everyone knows that (everyone knows that everyone knows x)), then x is known up to level 2. .... Then x is common knowledge if it is known up to level t for all t ￿ 0. Solution Concept: What we might reasonably expect to observe. Profile: s = (s1, s2, ..., sn) with sj ￿ Sj for all j s–i = (s1, s2, ... , si–1, si+1, ... , sn) S–i = {(s1, s2, ... , si–1, si+1, ... , sn) : sj ￿ Sj for all j ￿ i} 611.01 - 1 Strategy si￿ strictly dominates strategy si in game à = ( I, {Si}, {ui} ) if for all s–i in S–i, ui(si￿, s–i) > ui(si, s–i). Strategy si is strictly dominated if there exists a strategy si￿ that strictly dominates si. Strategy si is strictly dominant if for all si￿ ￿ si, si strictly dominates si￿. "If a player has a strictly dominant strategy, ... we should expect him to play it. ... It is compelling that players should play strictly dominant strategies if they have them." Profile (s1, s2, ..., sn) is a strictly dominant strategy equilibrium if each si is strictly dominant in Si. Strictly dominant strategy equilibria need not exist. In outcome space X, x is Pareto superior to y, if x Ri y for all i, and x Pi y for at least one i. x is Pareto optimal in X if there is no x* in x such that x* is Pareto superior to x. [Notice this is NOT the same as saying that x is Pareto superior to every x* in X such that x* ￿ x.] Important!! It is possible that there is a strictly dominant strategy equilibrium (s1, s2, ..., sn) and yet g(s1, s2, ..., sn) is NOT Pareto optimal, i.e., g(s1, s2, ..., sn) is Pareto dominated by some other feasible alternative. Prisoner’s dilemma. 611.01 - 2 N a s h e q u ilib r iu m In pure strategies: First approach: Player i's best response correspondence, bi: S–i ￿ Si in game ( I, {Si}, {ui(￿)} ) is defined by bi(s–i) = {si ￿ Si : ui(si, s–i) ￿ ui(si￿, s–i) for all si￿ ￿ Si }. Then (s1, s2, ..., sI) is a Nash equilibrium of game ( I, {Si}, {ui(￿)} ) iff si ￿ bi(s–i) for all i ￿ I. l r L (0, 2) (0, 2) R (–3, –1) (2, 1) Second approach: i-variant profile: A strategy profile (s1￿, s2￿, ... , sn￿) is an i-variant of strategy profile (s1, s2, ... , sn) if sj￿ = sj for all j ￿ i A strategy profile (s1, s2, ... , sn) is a Nash equilibrium if there does not exist an i and an i-variant profile (s1￿, s2￿, ... , sn￿) with: ui(g(s1￿, s2￿, ... , sn￿)) > ui(g(s1, s2, ... , sn)). The Cournot Game Two firms (1,2): duopoly Demand: P = a – b(Q1 + Q2) [ Actually if we were being very careful, we’d preclude negative prices and say: P = a – b(Q1 + Q2) if Q1 + Q2 < a/b and P = 0 if Q1 + Q2 ￿ a/b. ] ði = P￿Qi – cQi for i = 1, 2. Here we only treat the economically interesting case: a, b, and c are all positive and a > c. ði = [a – b(Q1 + Q2)]￿Qi – cQi For a best response function we want to find the global maximizing value of Q1 for the function 611.01 - 3 ð1(Q1) = [a – b(Q1 + Q2)]￿Q1 – cQ1 = aQ1 – bQ12 – bQ2Q1 – cQ1 = (a – bQ2 – c)Q1 – bQ12 over the domain ￿+, the non-negative real numbers. Existence of a global maximum: Note that ð1(Q1) < 0 if (a – bQ2 – c)Q1 – bQ12 < 0 (a – bQ2 – c)Q1 < bQ12 (a – bQ2 – c) < bQ1 if Q1 ￿ 0 (a – bQ2 – c)/b < Q1 So consider the subdomain of ð1(Q1) that is the interval D = [0, (a – bQ2 – c)/b]. This is a compact (closed and bounded) set on which ð1(Q1) = [a – b(Q1 + Q2)]￿Q1 – cQ1 is a continuous function. By the Weierstrass theorem (Extreme value theorem), there exists a Q1* in this interval that maximizes ð1(Q1) over the interval, i.e., ð1(Q1*) ￿ ð1(Q1) for all Q1 ￿ [0, (a – bQ2 – c)/b]. In particular, ð1(Q1*) ￿ ð1(0) = 0. Since ð1(Q1) < 0 for all Q1 > (a – bQ2 – c)/b, we see that Q1* maximizes ð1(Q1) not just over the interval, but over all of ￿+, i.e., Q1* is a global maximizing value. Candidates: In general, candidates for a global maximum must be either (1) Boundary points of the domain ￿+, i.e., Q1 = 0; (2) Interior points of the domain where ð1(Q1) fails to be differentiable - and there are none here; or (3) Interior points of the domain where the function is differentiable and the derivative takes on the value 0. We turn to this category: DQ1ð1 = (a – bQ2 – c) – 2bQ1 (a – bQ2 – c) – 2bQ1* = 0. So there is a unique critical point: 611.01 - 4 Best response function: If (a – bQ2 – c) ￿ 0, i.e., Q2 ￿ (a – c)/b, then we have just a single candidate, Q1 = 0, which must then be the global maximum. If (a – bQ2 – c) > 0, i.e., Q2 < (a – c)/b, then we have two candidates, Q1 = 0, and Q1 = (a – bQ2 – c)/2b, and we must compare ð1(Q1) at those two values. Of course, ð1(0) = 0. What is ð1((a – bQ2 – c)/2b) ? Since ð1(Q1) = (a – bQ2 – c)Q1 – bQ12, ð1((a – bQ2 – c)/2b) = (a – bQ2 – c)(a – bQ2 – c)/2b – b[(a – bQ2 – c)/2b] 2, = (a – bQ2 – c)2{(1/2b) - (1/4b)} = (a – bQ2 – c)2(1/4b), which is positive since (a – bQ2 – c) > 0. Therefore ð1((a – bQ2 – c)/2b) > ð1(0). Putting all this together, firm #1's best response function is Similarly for firm #2: 611.01 - 5 Nash equilibria: These two candidates for each of the two players lead to just four apparently possible Nash equilibria: (1) Simultaneous solutions of Q1 = (a – bQ2 – c)/2b and Q2 = 0: ((a – c)/2b, 0) (2) Simultaneous solutions of Q1 = 0 and Q2 = (a – bQ2 – c)/2b: (0, (a – c)/2b) (3) Simultaneous solutions of Q1 = 0 and Q2 = 0: (0,0) (4) Simultaneous solutions of Q1 = (a – bQ2 – c)/2b and Q2 = (a – bQ2 – c)/2b: ((a – c)/3b, (a – c)/3b) [Do the algebra to get this.] But none of the first three are Nash equilibria. Consider the first, where #2 chooses 0. As we’ve seen, that is #2's best response just when (a – bQ1 – c) ￿ 0. But if Q1 = (a – c)/2b, a – bQ1 – c = a – b[(a – c)/2b] – c = (a – c) – (a – c)/2 = (a – c)/2 > 0. So #2's 0 is NOT a best response to #1's (a – c)/2b. Similarly, (2) and (3) can be eliminated. What remains is the strategy profile ((a – c)/3b, (a – c)/3b), but we have to check that, too: Is #1's strategy (a – c)/3b a best response to #2's strategy (a – c)/3b? As we have seen, the answer is “Yes” if and only if (a – bQ2 – c) > 0, i.e., if Q2 < (a – c)/b. But obviously, (a – c)/3b is less than (a – c)/b, so #1's strategy (a – c)/3b is a best response to #2's strategy (a – c)/3b and #2's strategy (a – c)/3b is a best response to #1's strategy (a – c)/3b, i.e., case (4) IS a Nash equilibrium. 611.01 - 6 Common knowledge / The game of red hats Once upon a time, many years ago, in a Kingdom far, far away, a cruel King, who was known by all to always tell the truth, decided to have sport with the n prisoners in his castle jail. The prisoners were all brought to the circular Great Hall where they were placed, evenly spread out with their backs to the wall. Blindfolded, each had a hat placed on their head. Before the blindfolds were removed, each was told that if they looked up at their own hat or if they tried to communicate with any other prisoner, they would be immediately beheaded. “All of the hats are either blue or red,” the King said. “We are going to play a game in a sequence of steps. At the first step, a gong will sound and each of you has the chance to raise your hand or not. If you actually have a red hat and you raise your hand, you immediately will be set free. If you have a blue hat and you raise your hand, you will be immediately beheaded. Then a bell will ring to signal the end of the first step. Then a gong will signal the start of the second round and the same rules apply: If you actually have a red hat and you raise your hand, you immediately will be set free. If you have a blue hat and you raise your hand, you will be immediately beheaded. Then a bell will signal the end of the second step. This will continue indefinitely.” What happens depends on (1) what the situation is (i.e., how many are wearing red hats), (2) what is known about the situation, and (3) the level of commonality of what is known about the situation. Case 1. Exactly one hat is red. Subcase 1A. That at least one hat is red is known up to level 0. Subcase 1B. That at least one hat is red is known up to level 1. Subcase 1C. That at least one hat is red is known up to level 2. Case 2. Exactly two hats are red. Subcase 2A. Subcase 2B. Subcase 2C. Subcase 2D. Subcase 2E. That at least one hat is red is known up to level 0. That at least one hat is red is known up to level 1. That at least one hat is red is known up to level 2. That at least two hats are red is known up to level 0. That at least two hats are red is known up to level 1. 611.01 - 7 Case 3. Exactly three hats are red. Subcase 3A. Subcase 3B. Subcase 3C. Subcase 2D. Subcase 2E. Subcase 2F. Subcase 2G. That at least one hat is red is known up to level 0. That at least one hat is red is known up to level 1. That at least one hat is red is known up to level 2. That at least two hats are red is known up to level 0. That at least two hats are red is known up to level 1. That at least three hats are red is known up to level 0. That at least three hats are red is known up to level 1. Exercises: 1. There are two players. Player 1's strategy space is {0,1,2,...,20} (numbers of nickels offered to #2); player 2's strategy space is all 21-entry lists of A (accept) and R (reject). The payoff at (x, arrarrr...ar) is a) 100-5x for #1 and 5x for #2 if #2 has a in the x+1st position b) (0,0) otherwise. What are all the Nash equilbria? 2. A. (Autarky) Firm A in the US is the only seller of wine there. Let QAUS be Firm A’s sales in the US where demand is given by 6 – (1/8)QAUS. Firm A’s costs are 1 per unit. ðA(QAUS) = (6 – (1/8)QAUS)￿ QAUS – QAUS What value of QAUS maximizes profit? What is the sum of consumer surplus and firm profit? B. (Free Trade) Now trade is opened between the US and France. Firm A is still the only producer in the US, while firm B is the only producer in France. Both can sell in either country: QAUS is A’s sales in the US; QAFr is A’s sales in France; QBUS is B’s sales in the US; QBFr is B’s sales in France. If a firm sells in the other country, their costs include not only their production costs but also a transportation cost of 1 per unit. Demand in the US is 6 – (1/8)(QAUS + QBUS) while demand in France is 6 – (1/8)(QAFr + QBFr). There is an important separability here; e.g., 611.01 - 8 ðA(QAUS, QAFr, QBUS, QBFr) = PUS(QAUS + QBUS)￿ QAUS + PFr(QAFr + QBFr)￿ QAFr – (QAUS + QAFr) – QAFr = [6 – (1/8)(QAUS + QBUS)]QAUS + [6 – (1/8)(QAFr + QBFr)]QAFr – QAUS – 2QAFr = {[6 – (1/8)(QAUS + QBUS)]QAUS – QAUS} + {[6 – (1/8)(QAFr + QBFr)]QAFr – 2QAFr} = ðA1(QAUS, QBUS) + ðA2(QAFr, QBFr) Assume A and B act as Cournot duopolists in each of the US and French markets. What are the Nash equilibrium values of QAUS, QAFr, QBUS, and QBFr? What is the value of the sum (US consumer surplus + Firm A’s profit)? 3. (One-sided tariff) Now assume the US imposes on Firm B a tariff of .75 on each unit Firm B exports to the US. France has no subsidy and no tariff. What are the Nash equilibrium values of QAUS, QAFr, QBUS, and QBFr? What is the value of the sum (US consumer surplus + Firm A’s profit + US tariff revenues}? [Assume the same demand curves, same production costs and same transportation costs as in the previous problem. Also assume A and B act as Cournot duopolists in both the US and Fr markets.] 4. Analyze the red hat question for Case 3. 5. Player 1 has 5 strategies and player 2 has 6 strategies. Are there payoffs such that there are no Nash equilibria? Are there payoffs such that there are 30 Nash equilibria? Are there payoffs such that there are 17 Nash equilibria? What numbers of Nash equilibria are possible? Generalize to k strategies for player 1 and ￿ strategies for player 2. 611.01 - 9 ...
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