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Brief Idea about Applicable Game Theory

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Lu Victoria

on 4 February 2013

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Transcript of Brief Idea about Applicable Game Theory

Brief Idea about Applicable Game Theory Dynamic Games with Complete Information Dynamic Games with Incomplete Information Static Games with Complete Information iterated elimination of dominated strategies A Game that Relies on a Noncredible Threat Robert Gibbons 1997 Created by Lu Zhao Nash Equilibrium Bayesian Nash Equilibrium perfect Baysian Equilibrium The Prisoners' Dilemma The Dating Game The Trust Game short-term gain from defection followed by the long-term loss from punishment Subgame-Perfect Nash Equilibrium backward induction Nash equilibrium Game Tree Static Games with Incomplete Information insert ti : player i's type p(t-i | ti)= The Dating Game with Incomplete Information The Auction: sealed bid, simultaneous move Spence's (1973) classic model signaling game the definition of equilibrium no longer consists of just a strategy for each player but now also includes a belief for each player whenever the player has the move but is uncertain bout the history of prior play. Prisoner Dilemma Bertrand Complete information:

Common knowledge of timing, feasible moves and payoffs Cournot P= 30-Q Q= Q1+Q2 MC1= MC2= 0 Price Competition D1:Q1 = 24 - 4P1 + 2P2    ①



    ② D2:Q2 = 24 −- 4P2 + 2P1 P=6 A Game that Relies on a Non-credible Threat The Trust Game Dating Game
with incomplete information Why Players' Beliefs are as Important as their Strategies My model: Tennis Game 1 0 0 Player 1 Player 2 Dynamic Game with Complete Information Discrete Model 1 Nash equilibrium Backward attack defend Discrete Model 2 2 (0,1) (0,1) (1,0) (1,0) attack defend p 1-p q 1-q 1-q q front back front front back back p= 1/2 q= 1 Continuous Model 1 2 2 (0,1) (0,1) (1,0) (1,0) attack defend p(x) q(y) front back front front back q(y) f1(x,y) , f2(x,y) ( ) Player 1 let Player 2 decide first So I use Iteration Method For a given y1, y2 which satisfied with 0<y1,y2< 1 if f2(x,y1)p(x)>f2(x,y2)p(x) then we can find Let q1(y1)=a, q1(y2)=b Then we can find another optional q2(y) q2(y)= { a+b 0 q1(y) y=y1 y=y2 y≠y1,y2 a> b Then it constant exists that Take iterative method as mentioned above
Then there must exist optimal q*(y) q*(y)= { ∞ 0 if f2(x,y)p(x)=Max, which y=yk y≠yk or q*(y)= 1 if f2(x,y)p(x)= constant Num so P1 will choose the second one for P2 Conclusion f2(x,y)p(x)= constant Num p(x)=Constant Number/f2(x,y) q(y)=1 for any y Continuous Model Continuous Model Continuous Model Continuous Model Behavior:

Rational Choices Made Brief Look of Game Theory f2 f1 cournot equilibrium Competive equi Monopoly equi conspiracy function P= MC MR==MC==0 MR1=MC1=0 MR2=MC2=0 Quantity Competition ------- p(t-i,ti) p(ti) = ------- p(t-i,ti) G={Ac, Ap, Bc, Bp; Tc, Tp; pc, pp; uc ,up} p/x (x-p)/x Tc=Tp=[0,x] uniform distribution c/x (x-c)/x Chris Pat Steak Chris's strategy is a rule specifying Chris's action for each possible value of tc; steak, if tc exceeds a critical value c
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