微观经济学作业代写 Economics 3382A-001 

Economics 3382A-001
Advanced Microeconomics I Department of Economics

微观经济学作业代写  The entire game NE are colored in red. All of them are subgame perfect NE as they are just combination of NEs in each stage.

Problem 1  微观经济学作业代写

If 1 plays A, best response (BR) for 2 is b. if 1 plays B, BR for 2 is a. If 1 plays C, BR for 2 is c. Also, if 2 plays a, BR for 1 is B. If 2 plays b, BR for 1 is A. If 2 plays c, BR for 1 is C. So all pure NE are: (A,b), (B,a), (C,c).

To ensure subgame-perfect, use backward induction. In the second period, neither player has a profitable deviation from (B,a) as it is stage NE regardless of what is played in period 1. Then in the first stage, as its outcome does not affect the second stage, no profitable deviation from (B,a) as well given (B,a) is a stage NE.

Suppose there is a punishment NE in the second stage, arising after not observing (A,a) and a reward NE in the second stage, arising after observing (A,a). Notice that the punishment NE still has to be NE in the stage game.




    1. If the punishment NE is (B,a), player 1 has a profitable deviation in period 1 by playing B.

    2. If the punishment NE is (A,b), player 2 has a profitable deviation in period 1 by playing b.

    3. If the punishment NE is (C,c)

      1. If the reward NE is (A,b), player 1 has a profitable deviation in period 1 by playing B.

      2. If the reward NE is (B,a), player 2 has a profitable deviation in period 1 by playing b.







  • Draw the normal matrix of the entire game.






























































































































1\2aaabacbabbbccacbcc
AA8,85,94,45,92,101,54,41,50,0
AB9,54,44,46,61,51,55,10,00,0
AC4,44,47,71,51,54,80,00,03,3
BA9,56,65,14,41,50,04,41,50,0
BB10,25,15,15,10,00,05,10,00,0
BC5,15,18,40,00,03,30,00,03,3
CA4,41,50,04,41,50,07,74,83,3
CB5,10,00,05,10,00,08,43,33,3
CC0,00,03,30,00,03,33,33,36,6

The entire game NE are colored in red. All of them are subgame perfect NE as they are just combination of NEs in each stage.

 

Problem 2 微观经济学作业代写

Discuss all strategy profiles.

Player 1 splits, player 2 always steals. The expected payoff is (10,10), no profitable deviation for any player. So this is NE.

Player 1 splits, player 2 always splits. Player 2 can profitably deviate to steal if type S, so this is not NE.

Player 1 splits. Player 2 steals if type S and splits if type A. The expected payoff is (10,10), no profitable deviation for any player. So this is NE.

Player 1 splits. Player 2 splits if type S and steals if type A. Player 2 can profitably deviate to steal if type S, so this is not NE.

Player 1 steals, player 2 always steals. Player 1 is better off if he splits (expected payoff 10 vs. 0). So this is not NE.

Player 1 steals, player 2 always splits. Expected payoff for player 1 is 10, no profitable deviation. For player 2, if type S, he cannot be better off by choosing steal as the payoff is still 0; if type A, player 2 will be worse off if he chooses steal. So this is NE.

Player 1 steals, Player 2 steals if type S and splits if type A. Player 1 can be better off by choosing split (expected payoff 10 vs. 5). So this is not NE.

Player 1 steals, player 2 steals if type A and splits if type S. Player 1 can be better off by choosing split (expected payoff 10 vs. 5). So this is not NE.

Conclusion: NE of this game is (split, split if S and split if A), (split, split if A and steal if S), (steal, split if S and split if A).

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