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More Problems from Chapter 12

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Problem 1, Chapter 12

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Find a separating equilibrium Trial-and-error. Two possible separating strategies for Player 1: – Choose A if type s, B if type t. – Choose B if type s, A if type t. Player 2 sees action A or B, does not see type. If Player 2 believes that Player 1 will choose A if s and B if t, then Player 2’s best response strategy will be y if A, x if B. Given Player 2’s strategy, – Type s would rather have Player 2 do y than x and so would choose A – Type t would rather have Player 2 do x than y and so would do B. So the strategies A if type s, B if type t for Player 1 and y if A, x if B constitute a separating Bayes-Nash equilibrium.

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Beef or horse? In a restaurant, some customers prefer beef, some prefer horse. Type s likes beef, type t likes horse Waitress asks if you want beef or horse. – For Player 1, Action A—Say beef, Action B—say horse – For Player 2, Action x—Bring horse, Acton y, bring beef If she brings you the kind you like, you are happier. If you are happier, you leave a bigger tip and waitress is happier. Simplicity of this case because payoff to Player 1, given action of Player 2 depends on one’s type, but not on action A or B.

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Complications What if the customer is a foreigner and doesn’t know the words for beef and horse? Babbling equilibrum?

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Bayesian waitress Customer is an American in Japan. Waitress believes that only ¼ of American customers know the words for horsemeat and beef. She believes that 9/10 of Americans like beef and don’t like horse. Suppose that Americans who know the words order what they prefer. Those who don’t are equally likely to say the word for horse as that for beef. What should she do when an American orders beef? orders horse?

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If Americans who know the words order the meat they prefer and those who don’t order randomly, what is the probability that an American who orders horse wants horse? Try Bayes’ law. Let H mean likes horse and B mean that he prefers beef. Let h mean orders horse. Then P(H|h)=P(H and h)/P(h). The event H and h occurs if American likes horse and either knows the words or guesses correctly. P(H and h)=(1/10)(1/4)+(1/10)(3/4)(1/2)=5/80 Probability that customer prefers beef and orders horse is P(B and h)= (9/10)(3/4)1/2=27/80. Then P(h)=P(H and h)+P(B and h)=32/8 So P(H|h)=5/32.

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Babble and Horsemeat Since P(H|h)<1/2, the waitress will have a higher expected payoff from serving the customer beef even if he orders horsemeat. In equilibrium, waitress always brings beef for Americans, no matter what they say. Americans therefore would be indifferent about what they say.

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Problem 5, Page 384

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Find a separating Bayes-Nash equilibrium Candidate for equilibrium behavior by Sender: Say m1 if type 1, m2 if Type 2. If Receiver believes that this is Sender’s strategy, best response is B if m1, C if m2. If Receiver plays B if m1, C if m2, then Sender’s best response is m1 if type 1, m2 if Type 2. This is a Bayes-Nash equilibrium. The beliefs of each player about what the other player will do are confirmed by the responses.

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Find a pooling equilibrium Suppose that Player 1 says m1 if type 1 and m1 if type 2. What is the best response for Player 2? – Expected payoff from A is.6*3+.4*4=3.2 – Expected payoff from B is.6*4+.4*1=2.8 – Expected payoff from C is.6*0+.4*5=2 So A if m1, A if m2, is a best response for 2. If Player 2 always goes to A, then it doesn’t matter what Player 1 says, so m1 if type 1 and m1 if type 2 is a best response. – So is flip a coin about what to say This is a pooling equilibrium. Beliefs are confirmed by outcomes.

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Answer to part c Suppose probability that Sender is type 1 is p. When is there a pooling Bayes-Nash equilibrium where receiver always plays B? Suppose Sender plays m1 if type 1 and m1 if type 2. Expected payoffs to Receiver from strategies: – From A 3p+4(1-p) – From B 4p+(1-p) – From C 0p+5(1-p) For what p’s is payoff from B the largest? – B beats A, 4p+1-p>3p+4(1-p) if p>3/4. – B beats C, 4p+1-p>0+5(1-p) if p>1/2 So Always play B is a best response to Sender’s strategy of m1 if type 1 and m1 if the probability of type 1 if p>3/4. If Receiver always plays B, Sender is indifferent about what signal to send and might always say m1 or always say m2 or just “babble”.

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