14.12 Economic Application of Game Theory Prof. Muhamet Yildiz TA: Youngjin Hwang Problem Set 5 Solutions 1. Let’s denote each player’s horizontal and vertical actions by A and D respectively, and Player 1’s additional action at bottom-right node by E. And let’s write the strategies in the form },;,,{3212312111ttttt for Player 1, where ijt indicates the action taken by type i at the j-th move, and }{x for the strategy of Player 2 who has only one information set and is to move only once. We can easily see that Dt=12, Et=32, and At=31. Out of several possibilities, only the following case can be an equilibrium - Player 1 with the top type is mixing, and middle type is choosing D. Suppose that Player 1 (top) is choosing A with probability p, then by Bayes’ rule Player 2’s beliefs are from top to bottom )181,0,188(++ ppp. Since she will be indifferent between A and D, it must be that 4/1=p . Also Player 1 (top) will be indifferent when Player 2 is playing A with probability 1/3. In this case, there is no incentive to deviate. Therefore, PBE is },;,,4/34/1{ EDADDA+, }3/23/1{ DA + with belief (2/3, 0, 1/3) For other cases (pooling, separating, other hybrid..), you will end up with contradiction, or impossible probabilities. Hence, they cannot be an equilibrium. 2. In the second stage of the game, it is clear that each type will always play the static best response: 2=a type will fight and 1−=a type will accommodate. Then for the first stage, with 9.0=π, the entrant will always enters since 0)(8.012)(=>=−=outEUEnterEUπ, and 2=a type fights and 1−=a type accommodates. The PBE is )},,(),,;,{( EOEAFAF - here we are denoting the strategy for the incumbent as ),;,(22122111tttt , where ijt indicates the action taken by type i in the stage j, and for the entrant as ),,(22211aaa , where 1a refers to the first information set at 1=t , and the other two refer to the information sets at 2=t . Player 2’s beliefs are updated such that we have 1)|1(=−= Aaµ at 22a , and 1)|2(==Faµ at 21a .3. (a) At 1=t , the buyer of type 1 (2) will accept if and only if 11≤p ( 21≤p ). And the seller will offer 01≥p . Let’s write the seller’s belief as historyv |2(=µat t = 1). Then if 2/1>µ, the seller sets 21=p ; if 2/1<µ, he sets 11=p ; if 2/1=µ, he is indifferent between 1 and 2. If the seller observes 10>p , then 1=µ and he will accept 29.00⋅≥p . And if he observes 10≤p , then 8.0=µ (the same as his prior) and this offer is rejected since only 8.029.00⋅⋅≥p will be accepted. Then, type 1 (2) would set 10≤p (29.00⋅≥p ). In this case, if the seller observes 10≤p , he will set 11=p . But then type 2 buyer will not want to set 29.00⋅≥p and deviate. Therefore, there is no pure strategy equilibrium. Now consider mixed strategy equilibrium: type 1 buyer sets 9.00=p and type 2 buyer is mixing 8.10=p with probability 1- q and 9.00=p with probability q, and the seller is accepting with probability p. Then, type 2 buyer is indifferent between 0.2 (when sets )8.10=p and )9.02( −p (when )9.00=p only if 11/2=p . Also we must have type 2 buyer is mixing such that the seller is indifferent between accepting and rejecting – that is, qq⋅+⋅⋅=8.02.08.08.19.0, or q = ¼. Hence, in equilibrium - type 1 buyer sets 0.9, and type 2 buyer is mixing 1.8 and 0.9 with probability ¾ and ¼ at 0=t . And the seller accepts if he observes 8.10=p , and if he observes 9.00=p , he accepts with probability 2/11 and rejects with probability 9/11 then sets 21=p . (b) At 1=t , both types of buyer will offer 01=p , and type 1 (2) will get 1 (2). At 0=t , type 1 (2) buyer will accept offer by the seller if 19.0)1(0⋅≥−p , or 1.00≤p (if 29.0)2(0⋅≥− p , or 2.00≤p ). If the seller offers 1.000≤≤p , then both types of buyer will accept, and his expected payoff is 000)1( pppEUS=⋅+⋅−=ππ, or 0.1, which is the optimal value for the seller. If the seller offers 2.01.00≤< p , only type 2 buyer will accept, and his expected payoff is 008.00)1( ppEUS=⋅+⋅−=ππ, or 0.8*0.2 = 0.16, which is the optimal value for the seller. Hence in equilibrium, the seller sets 2.00=p at 0=t , and accepts offer 01=p at 1=t .4. As proposed in the question, suppose there is an equilibrium in which the Sender’s strategy is (R, R), where the first letter is type 1t ’s strategy and the second is 2t ’s. Then the Receiver’s information set corresponding to R is on the equilibrium path, and her belief at this information set is determined by Bayes’ rule and the Sender’s strategy - )5.0,5.0( . Note that this is the same as the prior distribution since there is no updating in belief for pooling eqilibrium. Given this belief, the Receiver’s best response following R is to play d since 5.015.015.0)|(125.005.0)|( =⋅+⋅=>=⋅+⋅= RuEURdEURR. Now consider the Receiver’s beliefs and best response at information set following L. To determine whether both types of Sender are willing to choose R, we need to specify how the Receiver would react to L. In particular, we need to pin down her beliefs such that )|()|( LdEULuEURR> in order for both types of Sender to play R. In order for u to be the best response for the Receiver, her belief at the information set following L should be 3/1≥q (where q is the probability for 1t ). Thus, if there is an equilibrium in which the Sender’s strategy is (R, R), then the Receiver’s strategy must be such that he plays d following R, and a following L. Concluding, pooling Perfect Bayesian Nash equilibrium is }3/1,5.0);,(),,{(≥=qpudRR 5. (a) We can consider the following several cases. * Pooling on R: - the Receiver’s belief on the equilibrium path is 5.0)|()|(21==RtRtµµ - the Receiver’s best response is u since 5.015.005.0)|(105.025.0)|( =⋅+⋅=>=⋅+⋅= RdEURuEURR - each type of Sender’s payoff is 2)(1=RUt and 1)(2=RUt. - any play by the Receiver after observing L sustains (R, R). Hence PBE in this case are: }2/1);,(),,{( >quuRR , }2/1);,(),,{(<qduRR , }2/1;)1(,(),,{(=−+qduuRRαα. * Pooling on L:This cannot be an equilibrium since the Sender )(2t will always want to play R. * Separating equilibrium in which the Sender (1t ) plays R and the Sender (2t ) plays L: - both of the Receiver’s information sets are on the equilibrium path and beliefs are 1)|(1=Rtµ and