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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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the other only if the winner can also beat the loser in a majority vote with sincere voting or there is a third alternative that can defeat the loser and itself be defeated by the winner in a majority vote with sincere voting (this is due to Shepsle and Weingast). In our situation y can beat z on its own with sincere voting, and y can beat x through z because z can defeat x in sincere voting and y in turn defeats z. Hence, there is an agenda that ensures y is reachable.11 Agenda-setting give player 2 the ability to impose her most preferred outcome and there is nothing (in this instance) that the others can do. For instance, player 1 and player 3 cannot collude to defeat her obvious intent. To see this, suppose player 3 proposed a deal to player 1: if player 1 would vote sincerely for x in the first round, then player 3 would reward him by voting for x on the 2nd round. Since x will then beat y in the first round, player 3’s insincere vote in the second round would ensure that x will defeat z as well. This would benefit both players: player 1 would get his most preferred outcome and player 3 would avoid the worst outcome y and get his second-best. Unfortunately (for player 3), he cannot make a credible promise to cast an insincere vote. If x defeats y in the first round, then player 3 can get his most preferred outcome by voting sincerely in (x, z) in the second round. Therefore, he would renege on his pledge, so player 1 has no incentive to believe him. But since this reneging would saddle player 1 with his worst outcome, player 1 would strictly prefer to cast his sophisticated vote in the first round even though he is perfectly aware of how player 2 has manipulated the agenda to her advantage. The inability to make credible promises, like the inability to make credible threats, can seriously hurt players. In this instance, player 3 gets the worst of it.

4.3.2 The Ultimatum Game

–  –  –

In this game, player 1 has a continuum of action available at the initial node, while player 2 has only two actions. (The continuum of actions ranging from offering 0 to offering the entire pie is represented by the dotted curve connecting the two extremes.) When player 1 makes some offer, player 2 can only accept or reject it. There is an infinite number of subgames following a history of length 1 (i.e. following a proposal by player 1). Each history is uniquely identified by the proposal, x. In all subgames with x π, player 2’s optimal action is to In our game, each of the three outcome is possible with an appropriate agenda. Sophisticated voting does not reduce the chaos.

accept because doing so yields a strictly positive payoff which is higher than 0, which is what she would get by rejecting. In the subgame following the history x = π, however, player 2 is indifferent between accepting and rejecting. So in a subgame perfect equilibrium player 2’s strategy either accepts all offers (including x = π ) or accepts all offers x π and rejects x = π.

Given these strategies, consider player 1’s optimal strategy. We have to find player 1’s optimal offer for every SPE strategy of player 2. If player 2 accepts all offers, then player 1’s optimal offer is x = π because this yields the highest payoff. If player 2 rejects x = π but accepts all other offers, there is no optimal offer for player 1! To see this, suppose player 1 offered some x π, which player 2 accepts. But because player 2 accepts all x π, player 1 can improve his payoff by offering some x such that x x π, which player 2 will also accept but which yields player 1 a strictly better payoff.

Therefore, the ultimatum game has a unique subgame perfect equilibrium, in which player 1 offers x = π and player 2 accepts all offers. The outcome is that player 1 gets to keep the entire pie, while player 2’s payoff is zero.

This one-sided result comes for two reasons. First, player 2 is not allowed to make any counteroffers. If we relaxed this assumption, the SPE will be different. (In fact, in the next section we shall analyze a very general bargaining model.) Second, the reason player 1 does not have an optimal proposal when player 2 accepts all offers has to do with him being able to always do a little better by offering to keep slightly more. Because the pie is perfectly divisible, there is nothing to pin the offers. However, making the pie discrete (e.g. by slicing it into n equal pieces and then bargaining over the number of pieces each player gets to keep) will change this as well. (You will do this in your homework.)

4.3.3 The Holdup Game

We now analyze a three-stage game. Before playing the Ultimatum Game from the previous section, player 2 can determine the size of the pie by exerting a small effort, eS 0 resulting in a small pie of size πS, or a large effort, eL eS, resulting in a larger pie of size πL πS.

Since player 2 hates exerting any efforts, her payoff from obtaining a share of size x is x − e, where e is the amount of effort expended. The extensive form of this game is presented in Fig. 19 (p. 32).

–  –  –

We have already analyzed the Ultimatum Game, so each subgame that follows player 2’s effort has a unique SPE where player 1 proposes x = π and player 2 accepts all offers (note that the difference between this version and the one we saw above is that player 2 gets a strictly negative payoff if she rejects an offer instead of 0). So, in the subgame following eS, player 1 offers πS and in the subgame following eL he offers πL. In both cases player 2 accepts these proposals, resulting in payoffs of −eS and −eL respectively. Given these SPE strategies, player 2’s optimal action at the initial node is to expend little effort, or eS because doing so yields a strictly better payoff.

We conclude that the SPE of the Holdup Game is as follows. Player 1’s strategy is (πS, πL ) and player 2’s strategy is (eS, Y, Y ), where Y means “accept all offers.” The outcome of the game is that player 2 invests little effort, eS, and player 1 obtains the entire small pie πS.

Note that this equilibrium does not depend on the values of sS, eL, πS, πL as long as eS eL.

Even if πL is much larger than πS and eL is only slightly higher than eS, player 2 would still exert little effort in SPE although it would be better for both players if player 2 exerted eL (remember, only slightly larger than eS ) and obtained a slice of the larger pie. The problem is that player 1 cannot credibly promise to give that slice tho player 2. Once player 2 expends the effort, she can be “held up” for the entire pie by player 1.

This result holds for similar games where the bargaining procedure yields a more equitable distribution. If player 2 must expend more effort to generate a larger pie and if the procedure is such that some of this surplus pie goes to the other player, then for some values of player 2’s cost of exerting this effort, she would strictly prefer to exert little effort. Although there are many outcomes where both players would be strictly better off if player 2 exerted more effort, these cannot be sustained in equilibrium because of player 1’s incentives. In the example above, player 1 would have liked to be able to commit credibly to offering some of the extra pie to induce player 2 to exert the larger effort. Just like the problem with noncredible threats, the problem of non-credible promises means that this cannot happen in subgame perfect equilibrium.

4.3.4 A Two-Stage Game with Several Static Equilibria

Consider the multi-stage game corresponding to two repetitions of the symmetric normal form game depicted in Fig. 20 (p. 33). In the first stage of the game, the two players simultaneously choose among their actions, observe the outcome, and then in the second stage play the static game again. The payoffs are simply the discounted average from the payoffs in each stage. That is, let pi represent player i’s payoff at stage 1 and pi represent his payoff at stage 2. Then player i’s payoff from the multi-stage game is ui = pi + δpi, where δ ∈ (0, 1) is the discount factor.

–  –  –

If the game in Fig. 20 (p. 33) is played once, there are three Nash equilibria, two asymmetric ones in pure strategies: B, A, A, B, and one symmetric in mixed strategies with σ (A) = 3/7, and σ (B) = 4/7.

How do we find the MSNE? Suppose σ1 (C) 0 in a MSNE. We have several cases to consider.

First, suppose that σ1 (B) 0 as well; that is, player 1 puts positive weight on both B and C (and possibly A). Since he is willing to mix, it follows that 4σ2 (A) = 5σ2 (C) ⇒ σ2 (C) = 4/ σ (A). There are two ways to satisfy this requirement. Suppose σ (C) = σ (A) = 0, but in this case we obtain A, B. Suppose σ2 (C) 0, which implies σ2 (A) 0 too; that is, player 2 must be willing to mix between A and C (and possibly B). This now implies that 3σ1 (B) + 6σ1 (C) = 5σ1 (C) ⇒ 3σ1 (B) + σ1 (C) = 0, a contradiction because σ1 (C) 0.

Therefore, if σ1 (C) 0, then σ1 (B) = 0 as well. Second, suppose that σ1 (A) 0 as well; that is, player 1 puts positive weight on both A and C. Since he is willing to mix, it follows that 3σ2 (B) + 6σ2 (C) = 5σ2 (C) ⇒ 3σ2 (B) + σ2 (C) = 0, which implies σ2 (B) = σ2 (C) = 0, which means σ2 (A) = 1, but this means that player 1 will not mix and we have B, A. From all this, we conclude that σ1 (C) = 0 in MSNE. Analogously, symmetry gives us σ2 (C) = 0 too. This now reduces the game to the 2 × 2 variant in Fig. 21 (p. 34).

–  –  –

This is now very easy to deal with. Since player 1 is willing to mix, it follows that 3σ2 (B) = 4σ2 (A) and since σ1 (B) = 1 − σ2 (A), this gives us σ2 (A) = 3/7. Analogously, we obtain σ1 (A) = 3/7 and σ1 (B) = 4/7. The last thing we need to do is check that the players will not want to use C given the mixtures (we already know this from the argument above, but it does not hurt to recall the MSNE requirement). It suffices to check for player 1: if he plays C, his payoff will be 0 given player 2’s strategy of playing only A and B with positive probability, which is strictly worse than the expected payoff from either A or B. Hence, we do have our MSNE indeed. The payoffs in the three equilibria are (4, 3), (3, 4), and ( 12/7, 12/7) respectively.

The efficient payoff (5, 5) is not attainable in equilibrium if the game is played once.12

However, consider the following strategy profile for the two-stage game:

• Player 1: play C at the first stage. If the outcome is (C, C), play B at the second stage, otherwise play σ1 (A) = 3/7, σ1 (B) = 4/7 at the second stage;

• Player 2: play C at the first stage. If the outcome is (C, C), play A at the second stage, otherwise play σ1 (A) = 3/7, σ2 (B) = 4/7 at the second stage.

Is it subgame perfect? Since the strategies at the second stage specify playing Nash equilibrium profiles for all possible second stages, the strategies are optimal there. At the first stage players can deviate and increase their payoffs by 1 from 5 to 6 (either player can choose A). However, doing so results in playing the mixed strategy Nash equilibrium at the second stage, which lowers their payoffs to 12/7 from 4 for player 1 and from 3 for player 2. Thus, An outcome is efficient if it is not possible to make some player better off without making the other one worse off. The outcomes with payoffs (0, 0) are all inefficient, as are the outcomes with payoffs (4, 3) and (3, 4). However, the outcomes (6, 0) and (0, 6) are also efficient.

player 1 will not deviate if:

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We conclude that the strategy profile specified above is a subgame perfect equilibrium if δ ≥ 7/9. In effect, players can attain the non-Nash efficient outcome at stage 1 by threatening to revert to the worst possible Nash equilibrium at stage 2. This technique will be very useful when analyzing infinitely repeated games, where we shall see analogous results.

4.4 The Principle of Optimality

This is an important result that you will make heavy use of both for multi-stage games and for infinitely repeated games that we shall look at next time. The principle states that to check whether a strategy profile of a multi-stage game with observed actions is subgame perfect, it suffices to check whether there are any histories ht where some player i can profit by deviating only from the actions prescribed by si (ht ) and conforming to si thereafter. In other words, for games with arbitrarily long (but finite) histories,13 it suffices to check if some player can profit by deviating only at a single point in the game and then continuing to play his equilibrium strategy. That is, we do not have to check deviations that involve actions at several points in the game. You should be able to see how this simplifies matters considerably.

The following theorem is variously called “The One-Shot (or One-Stage) Deviation Principle,” and is essentially the principle of optimality in dynamic programming. Since this is such a nice result and because it may not be obvious why it holds, we shall go through the proof.

–  –  –

The proof works as follows. You start from the last deviation in a sequence of multiple deviations and argue that it cannot be profitable by itself, or else the OSDP would be violated.

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