University of Hong Kong ECON6036 Stephen Chiu. Extensive Games with Perfect Information II. Outline

Transcription

1 University of Hong Kong ECON6036 Stephen Chiu Extensive Games with Perfect Information II 1 Outline Interpretation of strategy Backward induction One stage deviation principle Rubinstein alternative bargaining game Excess optimum Two paradoxes Chain store paradox Centipede game 2 1

2 An extensive game and its strategic form CE CF DE DF AGI 1,2 1,2 AGJ 1,2 1,2 AHI 3,4 3,4 AHJ 3,4 3,4 BGI BGJ BHI BHJ The Nash equilibrium of the strategic form is called the Nash equilibrium of the EGPI. 3 Interpretation of a strategy Consider 1 s strategy AGJ. If the strategy is followed, there is no chance that he will be called upon to move in history (B,E) A strategy needs to specify actions at histories, even at those that occur with zero probabilities! Rationale: people make mistakes! You need to specify what to do in case you have made mistakes in the past! 4 2

3 Interpretation of a strategy Unique SPE: (AHJ,CF) Given AHJ played by 1, 2 finds it optimal to choose C given history A (which is reached with probability 1). Why we need to specify 2 s action under history B (reached with 0 probability)? Reason 1: 1 may make mistake and so choose B by accident. We have to specify what 2 will do in this case. Reason 2: in general 1 needs to know what 2 will do under any circumstance before making up his own mind 5 Subgame perfect equilibrium SPE of an EGPI Γ is a NE which induces a NE in every subgame of Γ. SPE is found out by a method called backward induction. 6 3

4 Backward induction Unique subgame perfect equilibrium: 1 using AHJ and 2 using CF Verify that (AHJ,CF) induces NE in each subgame verify this by constructing a strategic game for each subgame 7 Extensive game with perfect information with simultaneous moves There is only one subgame game the whole game itself A subgame must start with a history about which the beginning player exactly knows. 8 4

5 Extensive game with perfect information with simultaneous moves Only two subgames The whole game and the subgame starts with the history M 9 One-deviation property Definition: One-Deviation property is satisfied if no player can increase her payoff by changing her action at the start of any subgame in which she is the first-mover, given the other player's strategies and the rest of her own strategy. Proposition (One-Deviation Property). Let Γ be a finite EGPI (or an infinite horizon EGPI under some qualification). The strategy profile s* is a subgame perfect equilibrium if and only if it satisfies the onedeviation property. proof see Fudenberg and Tirole Implication: Given a strategic profile for an EGPI, we wonder if it is an SPE. The proposition says we need to check only those alternative strategies with one deviation. 10 5

6 Infinite horizon For the proposition to hold for infinite horizon games, we need the following qualification: Continuity at infinity: A game is continuous at infinity if for each player i the utility function u i satisfies sup ui h ui h ~ ~ ( ) ( ) 0 ~ t h, h s.t. h = h t as t This condition says events in the distant future are relatively unimportant. It will be satisfied if the overall payoffs are a discounted sum of per-period payoffs g it (a t ) and the per-period payoffs are uniformly bounded, i.e., there exists a finite B such that max, t at g t i ( a t ) < B. 11 One Stage Deviation Principle (OSDP): an illustration Consider a one-player EGPI. Is s*=(a,c,f) is an SPE? We should check s* against ALL other strategies. However, according to the one deviation principle, we just need to fare s* against three of them: (B,C,F), (A,D,F), and (A,C,E). Since each such strategy does not give a strictly higher payoff for the subgame in which the deviation occurs, s* is an SPE. A B C D E F

7 One deviation property Note there are 7 alternative strategies to AHJ, and 3 to CF. A lot to check to ensure SPE! You just need to check AHJ against BHJ, AGJ, and AHI and to check CF against DF and CE only! Why? One deviation property! 13 Rubinstein alternative offer game One unit of cake is to be divided (x₁,x₂) be an offer Player 1 makes offers in periods t=0,2,4,..., and Player 2 makes offers in periods t=1,3,5,... An agreement (x₁,x₂) reached t periods later gives discounted payoffs of δ t x₁ and δ t x₂ to the two players. The players are risk neutral; the objective of each player is to maximize his discounted expected payoff. 14 7

8 An SPE Let x*=(x₁*,x₂*)=((1/(1+δ)),(δ/(1+δ))) and y*=(y₁*,y₂*)=((δ/(1+δ)),(1/(1+δ))). Player 1 always proposes x* and accepts any offer in which he is paid (δ/(1+δ)) or more. Player 2 always proposes y* and accepts any offer in which he is paid (δ/(1+δ)) or more. Proof: Check that nobody can benefit from unilaterally deviating once. Then by one deviation property proposition, nobody can benefit from multiple deviations. This indeed is an SPE. 15 Unique SPE Let G i be a subgame in which i is the proposer. Different SPEs give different discounted payoffs to player i. Let M i and m i be the supremum and infimum of them. We claim that M₁=m₁=(1/(1+δ)) and M₂=m₂=(1/(1+δ)). If this claim is true, then there is a unique SPE (not only unique SPE outcome). (verify this) 16 8

9 Unique SPE (cont.) We now turn to show the claim. It takes a few steps. Step 1: m₂ 1-δM₁. Suppose it is 2's turn to make an offer. If 1 gets a chance to make an offer in the next period, 1 will get at most M₁. This most optimistic outcome for 1 is worth δm₁ now after discounting. Hence any offer to 1 giving him δm₁+ɛ now must be accepted by 1, leaving 1-δM₁-ɛ for player 2. Hence it cannot be the case that m₂<1-δm₁. 17 Unique SPE (cont.) Step 2: M₁ 1-δm₂. Suppose it is 1's turn to make an offer. If 2 gets a chance to make an offer in the next period, he will get at least m₂. Then 2 will not accept an offer giving him δm₂-ɛ right now. Hence, if agreement is to be reached immediately, 1 cannot make an offer so that x₂<δm₂ or x₁>1- δm₂; 1's offer has to satisfy x₁ 1- δm₂. Suppose 1's offer is rejected, the supremum of his present value will become δ(1- m₂) (1-m₂) 1- δm₂. Whether 1's offer is accepted now or not, we have M₁ 1-δm₂. 18 9

10 Unique SPE Step 1: m₂ 1-δM₁; and Step 2: M₁ 1-δm₂ Also, Step 3: m₁ 1-δM₂; and Step 4: M₂ 1- δm₁. Combining Steps 1 and 2, we have m₂ 1-δM₁ 1-δ(1-δm₂) = (1-δ)+δ²m₂. Hence, we have m₂ ((1-δ)/(1-δ²))=(1/(1+δ)). Substituting this back to step 2, we have M₁ (1/(1+δ)).Taking into account of symmetry and the existence of SPE, the claim is shown. 19 Extension: Will Excess Optimum always lead to delay? Consider the case where two risk-neutral players are trying to divide a dollar, which is worth 1 at t=0, δ (1/2,1) at t=1, and zero afterwards. It is also common knowledge that each player believes with probability one that he will be picked to make an offer at t=1 so long as no agreement is made at t=0. Find the unique SPE

11 Extension: Will Excess Optimum always lead to Delay? (cont) Now consider a four-period version of the previous game. The dollar is worth 1, δ, δ², and δ³ at dates 0, 1, 2, and 3, respectively, where δ (1/2,1/ 2). The dollar is worth 0 afterwards. Assume that each player is always sure that he will make all the remaining offers. Find the unique SPE. See Yildiz (Econometrica, 2003) 21 Chain-store game A chain-store (CS) has branches in K cities, numbered 1,...,K. In each city k there is a single potential competitor, player k. In period k, competitor k chooses to enter or not; if so CS either fights or cooperates in that period. at every point in the game all players know all the actions previously chosen. The payoff of competitor k is its payoff in period k; the payoff of the chain-store in the game is the sum of its payoffs in the K cities. Fight CS In 0,0 2,2 Competitor k Cooperate Out 5,1 Chainstore s payoff 22 11

12 Chain store paradox unique SPE k always enters CS always cooperate after entry of the competitor Paradox: Suppose every time a competitor entered in the past, it was fought by the chainstore. Shouldn't the competitor who has the turn decides not to enter? The SPE says it should not!! 23 Centipede game Two players 1 and 2 play a 6 stage centipede game

13 Centipede game Using backward induction, we find that the game will end immediately. unless the number of stages is very small it seems unlikely that player 1 would immediately choose S at the start of the game. After a history in which both a player and his opponent have chosen to continue many times in the past, how should a player form a belief about his opponent's action in the next period? It is far from clear. 25 Summary Nash equilibrium is not particularly useful for the study of EGPI. Subgame perfect equilibrium is introduced to solve this problem. SPE presumes that whenever a history with zero is actually reached, some mistake has occurred. The prediction of SPE leads to famous paradoxes that call for re-examination of assumptions of the model. One deviation property theorem Bargaining model as a building block in the modeling of many interesting issues 26 13