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«Contents. 1 Nash Equilibrium in Extensive Form Games 2 1.1 Selten’s Game......... 1.2 The Little Horsey....... 1.3 Giving Gifts.. ...»

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This now means that if you use the multiple-deviation strategy up to that point and follow the original OSDP strategy from that point on, you would get at least as good a payoff (again, because the last deviation could not have been the profitable one, so the original OSDP strategy will do at least as good in that subgame). You then go up one step to the new “last” deviation and argue that this deviation cannot be profitable either: since we are comparing a subgame with this deviation and the original OSDP strategy to follow with the OSDP strategy itself, the fact that the original strategy satisfies OSDP implies that this particular deviation cannot be profitable. Hence, we can replace this deviation with the action from the OSDP strategy too and obtain at least as good a payoff as the multi-deviation strategy. You repeat this process until you reach the first stage with a deviation and you reach the contradiction because this deviation cannot be profitable by itself either. In other words, if a strategy satisfies OSDP, it must be subgame perfect.

An example here may be helpful. Since in equilibrium we hold all other players strategies constant when we check for profitable deviations, the diagram in Fig. 22 (p. 36) omits the strategies for the other players and shows only player 1’s moves at his information sets.

Label the information sets consecutively with small Roman numerals for ease of exposition.

Suppose that the strategy (adegi) satisfies OSDP. We want to show that there will be no more profitable other strategies even if they involve multiple deviations from this one. To make the illustration even more helpful, I have bolded the actions specified by the OSDP strategy.

i

–  –  –

Because (adegi) satisfies OSDP, we can infer certain things about the ordering of the payoffs. For example, OSDP implies that changing from g to h at (iv) cannot be profitable, which implies u ≥ v. Also, at (v), changing from i to j cannot be profitable, so y ≥ z. At (ii), changing to c cannot be profitable; since the strategy specifies playing g at (iv), this deviation leads to w ≥ u. At (iii), changing to f cannot be profitable. Since the original strategy specifies i at (v), this deviation will lead to x ≥ y. Finally, at (i) changing to b cannot be profitable. Since the original strategy specifies e at (iii), this deviation will lead to w ≥ x. The implications of

OSDP are listed as follows:

–  –  –

These inequalities now imply some further relationships: from the first and the third, we get w ≥ y, and putting this together with the last yields w ≥ z as well. Furthermore, from the third and last we obtain x ≥ z, and from the second and fourth we obtain w ≥ v. Putting

everything together yields the following orderings of the payoffs:

w≥x≥y ≥z w ≥ u ≥ v.

and We can now check whether there exist any profitable multi-stage deviations. (Obviously, there will be no single-stage profitable deviations because the strategy satisfies OSDP.) Take, for example, an alternative strategy that deviates at (ii) and (iv); that is in the subgame starting at (ii), it specifies ch. This will lead to the outcome v, which cannot improve on w, the outcome from following the original strategy. Consider another alternative strategy which deviates twice in the subgame starting at (iii); i.e., it prescribes f j, which would lead to the outcome z. This cannot improve on x, the outcome the player would get from following the original strategy. Going up to (i), consider a strategy that deviates at (i) and (v). That is, it prescribes b and j at these information sets. Since (v) is still never reached, this actually boils down to a one-shot deviation with the outcome x, which (not surprisingly) cannot improve on w, which is what the player can get from following the original strategy. What if he deviated at (i) and (iii) instead? This would lead to y, which is also no better than w. What if he deviated at (i), (iii), and (v)? This would lead to z, which is also no better than w. Since all other deviations that start at (i) leave (ii) and (iv) off the path of play, there is no need to consider them. This example then shows how OSDP implies subgame-perfection. Intuitively, if a strategy satisfies OSDP, then it implies a certain preference ordering, which in turn ensures that no multi-stage deviations will be profitable.

To see how the proof would work here. Take the longest deviation, e.g., a strategy that deviates at (i), (iii), and (v). Since it leaves (ii) and (iv) off the path, let’s consider (bdf gj) as such a supposedly better alternative. Observe now that because (adegi) satisfies OSDP, the deviation to j at (v) cannot be improving. This means that the strategy (bdf gi) is at least as good as (bdf gj). Hence, if (bdf gj) is better than the original, then (bdf gi) must also be better. Consider now (bdf gi): since it matches the original at (v), OSDP implies that the deviation to f cannot be improving. Hence, the strategy (bdegi) is at least as good as (bdf gi), which implies it is also at least as good as (bdf gj). Hence, if (bdf gj) is better than the original, then (bdegi) must also be better. However, (bdegi) matches the

–  –  –

Let’s now check if player 1’s strategy is indeed subgame perfect. Recall that this requires that it is optimal for all subgames. This is easy to see for the subgame that begins with player 1’s second information set that follows history (b, d). How about player 1’s choice at the first information set? If we were to examine all possible deviations, we must check the alternative strategies (ae), (af ), and (be) because these are the other strategies available for that subgame. The one-shot deviation principle allows us to check just one thing: whether player 1 can benefit by deviating from b to a at his first information set! (This is why the OSDP is your friend.) In this case, deviating to a would get player 1 a payoff of 1 instead of 3, which he would get if he stuck to his equilibrium strategy. Therefore, this deviation is not profitable. We already saw that deviating to e in the subgame that begins with player 1’s second information set is not profitable either. Therefore, by the OSDP, the strategy is subgame perfect.





This proof does not work for games with infinite horizon because the essential step in it requires that there is a finite number of possible deviations. However, in a game with infinite horizon, there are strategies with an infinite number of deviations. Fortunately, if we assume that the payoff function takes the form of a discounted sum of per-period payoffs, we can get the result “back.” Again fortunately, all infinite-horizon games that we shall look at will have payoff functions that meet this requirement. The OSDP may not be that helpful in finding SPE. However, it’s extremely useful for checking whether some profile is SPE.

4.5 The Rubinstein Bargaining Model

There are at least two basic ways one can approach the bargaining problem. (The bargaining problem refers to how people would divide some finite benefit among themselves.) Nash initiated the axiomatic approach with his Nash Bargaining Solution (he did not call it that, of course). This involves postulating some desirable characteristics that the distribution must meet and then determining whether there is a solution that meets these requirements. This is very prominent in economics but we shall not deal with it here.

Instead, we shall look at strategic bargaining. Unlike the axiomatic solution, this approach involves specifying the bargaining protocol (i.e. who gets to make offers, who gets to respond to offers, and when) and then solving the resulting extensive form game.

People began analyzing simple two-stage games (e.g. ultimatum game where one player makes an offer and the other gets to accept or reject it) to gain insight into the dynamics of bargaining. Slowly they moved to more complicated settings where one player makes all the offers while the other accepts or rejects, with no limit to the number of offers that can be made. The most attractive protocol is the alternating-offers protocol where players take turns making offers and responding to the other player’s last offer.

The alternating-offers game was made famous by Ariel Rubinstein in 1982 when he published a paper showing that while this game has infinitely many Nash equilibria (with any division supportable in equilibrium), it had a unique subgame perfect equilibrium! Now this is a great result and since it is the foundation of most contemporary literature on strategic bargaining, we shall explore it in some detail.15

4.5.1 The Basic Alternating-Offers Model

Two players, A and B, bargain over a partition of a pie of size π 0 according to the following procedure. At time t = 0 player A makes an offer to player B about how to partition the pie.

If player B accepts the offer, then an agreement is made and they divide the pie accordingly, ending the game. If player B rejects the offer, then she makes a counteroffer at time t = 1. If the counteroffer is accepted by player A, the players divide the pie accordingly and the game ends. If player A rejects the offer, then he makes a counter-counteroffer at time t = 2. This process of alternating offers and counteroffers continues until some player accepts an offer.

To make the above a little more precise, we describe the model formally. Two players, A and B, make offers at discrete points in time indexed by t = (0, 1, 2,...). At time t when t is even (i.e. t = 0, 2, 4,...) player A offers x ∈ [0, π ] where x is the share of the pie A would keep and π − x is the share B would keep in case of an agreement. If B accepts the offer, the division of the pie is (x, π − x). If player B rejects the offer, then at time t + 1 she makes a counteroffer y ∈ [0, π ]. If player A accepts the offer, the division (π − y, y) obtains.

Generally, we shall specify a proposal as an ordered pair, with the first number representing player A’s share. Since this share uniquely determines player B’s share (and vice versa) each proposal can be uniquely characterized by the share the proposer offers to keep for himself.

The payoffs are as follows. While players disagree, neither receives anything (which means that if they perpetually disagree then each player’s payoff is zero). If some player agrees on a partition (x, π −x) at some time t, player A’s payoff is δt x and player B’s payoff is δt (π −x).

The players discount the future with a common discount factor δ ∈ (0, 1). This captures the time preferences of the players: it is better to obtain an agreement sooner than later.

Here’s how it works. Suppose agreement is reached on a partition (2, π − 2) at some time t. Player A’s payoff is 2δt. Since 0 δ 1, as t increases, δt decreases. Let δ =.9 (so a dollar tomorrow is only worth 90 cents today). If this agreement is reached at t = 0, player A’s payoff (2)(.9)0 = 2. If the agreement is reached at t = 1, player A’s payoff is (2)(.9)1 = 1.8. If it happens at t = 2, player A’s payoff is (2)(.9)2 = 1.62. At t = 10, the payoff is (2)(.9)10 ≈.697, at t = 100, it is only (2)(.9)100 =.000053, and so on and so forth.

The point is that the further in the future a player gets some share, the less attractive this same share is compared to getting it sooner.

The Rubinstein bargaining model is extremely attractive because it can be easily modified, adapted, and extended to various settings. There is significant cottage industry that does just that. The Muthoo (1999) book gives an excellent overview of the most important developments. The discussion that follows is taken almost verbatim from Muthoo’s book. If you are serious about studying bargaining, you should definitely get this book.

Yours truly also tends to use variations of the Rubinstein model in his own work on intrawar negotiations.

This completes the formal description of the game. You can draw the extensive form tree for several periods, but since the game is not finite (there’s an infinite number of possible offers at each information set and the longest terminal history is infinite—the one where players always reject offers), we cannot draw the entire tree.

4.5.2 Nash Equilibria

Let’s find the Nash equilibria in pure strategies for this game. Actually, we cannot find all Nash equilibria because there’s an infinite number of those. What we can do, however, is characterize the payoffs that players can get in equilibrium. It turns out that any division of the pie can be supported in some Nash equilibrium. To see this, consider the strategies where player A demands x ∈ (0, π ) in the first period, then π in each subsequent period where he gets to make an offer, and always rejects all offers. This is a valid strategy for the bargaining game. Given this strategy, player B does strictly better by accepting x instead of rejecting forever, so she accepts the initial offer and rejects all subsequent offers. Given that B accepts the offer, player A’s strategy is optimal.

The problem, of course, is that Nash equilibrium requires strategies to be mutually best responses only along the equilibrium path. It is just not reasonable to suppose that player A can credibly commit to rejecting all offers regardless of what player B does. To see this, suppose at some time t 0, player B offers y π to player A. According to the Nash equilibrium strategy, player A would reject this (and all subsequent offers) which yields a payoff of 0. But player A can do strictly better by accepting π − y 0! The Nash equilibrium is not subgame perfect because player A cannot credibly threaten to reject all offers.

4.5.3 The Unique Subgame Perfect Equilibrium

Since this is an infinite horizon game, we cannot use backward induction to solve it. However, since every subgame that begins with an offer by some player is structurally identical with all subgames that begin with an offer by that player, we shall look for an equilibrium with two intuitive properties: (1) no delay: whenever a player has to make an offer, the equilibrium offer is immediately accepted by the other player; and (2) stationarity: in equilibrium, a player always makes the same offer.



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