Extensive Games with Perfect Information II

1. Extensive Games with Perfect Information II

2. Outline • One stage deviation principle • Rubinstein alternative bargaining game • Excess optimum • Two paradoxes • Chain store paradox • Centipede game • Review of the mid term exam

3. 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 or infinite horizon) EGPI. The strategy profile s* is a subgame perfect equilibrium if and only if it satisfies the one-deviation property. • proof see Fudenberg and Tirole • Implication: Suppose we are given a strategic profile for an EGPI, and we wonder if it is indeed an SPE. The proposition says we need only check those alternative strategies with one deviation.

4. Infinite horizon • For the proposition to hold also for infinite horizon games, we need the following very mild qualification: • Continuity at infinity: A game is continuous at infinity if for each player i the utility function ui satisfies • 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 git(at) and the per-period payoffs are uniformly bounded, i.e., for is a B such that

5. Consider a one-player EGPI. You are told that s*=(A,C,F) is an SPE. To check this, according to the definition of SPE, you need to check s* against ALL other strategies. However, according to the one deviation principle, you just need to fare s* against three of them: (B,C,F), (A,D,F), and (A,C,E). So far as each of these alternative strategies does not give a strictly higher payoff for the subgame in which the deviation occurs, then s* is in an SPE. A B C D F E 1 2 0 0 One Stage Deviation Principle (OSDP): an illustration

6. 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 δtx₁ and δtx₂ to the two players. • The players are risk neutral; the objective of each player is to maximize his discounted expected payoff.

7. 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.

8. Unique SPE • Let Gi be a subgame in which i is the proposer. Different SPEs give different discounted payoffs to player i. Let Mi and mi 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)

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.

10. 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 .

11. 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.

12. 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.

13. 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, 2004??)

14. Competitor k In Out CS Fight Cooperate 0,0 2,2 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. That period's payoff is • 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.

15. Chain store paradox • unique SPE • k always enters • CS always cooperate after entry of the competitor • Paradox: Suppose upon every time a competitor entered in the past, it was fought by the chain-store. Shouldn't the competitor who has the turn decides not to enter? The SPE says it should not!!

16. Centipede game Two players 1 and 2 play a 6 stage centipede game.

17. 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.

18. Summary • The notion of Nash equilibrium is not particularly useful for the study of EGPI. • The notion of subgame perfect equilibrium is introduced to solve this problem. • SPE presumes that a history with zero probability according to the strategy profile is reached is viewed as an outcome of mistakes. • The prediction of SPE leads to famous paradoxes that call for re-examination of assumptions of the model. • One deviation property theorem • Alternative bargaining model as a building block in the modeling of many interesting issues (e.g., outside option, threat point, trade bargaining, bargaining between firm and union, the role of property rights, debt renegotiation, holdup, etc.)