9.2 完全信息静态博弈
9.2.1 博弈的战略式表述
Definition A normal (strategic) form game G consists of: (1) a finite set of agents D?{1,2,?,n}. (2) strategy sets S1,S2,?,Sn.
(3) payoff functions ui:S1?S2???Sn?R(i?1,2,?,n).
囚徒A
完全信息静态博弈是一种最简单的博弈，在这种博弈中，战略和行动是一回事。
博弈分析的目的是预测博弈的均衡结果，即给定每个参与人都是理性的，什么是每个参与人的最优战略？什么是所有参与人的最优战略组合？
纳什均衡是完全信息静态博弈解的一般概念，也是所有其他类型博弈解的基本要求。 下面，我们先讨论纳什均衡的特殊情况，然后讨论其一般概念。
坦白 抵赖 囚徒B 坦白 -8,-8 -10,0 抵赖 0,-10 -1,-1
9.2.2 占优战略（Dominated Strategies）均衡
一般说来，由于每个参与人的效用（支付）是博弈中所有参与人的战略的函数，因此，
每个参与人的最优战略选择依赖于所有其他参与人的战略选择。但是在一些特殊的博弈中，一个参与人的最优战略可能并不依赖于其他参与人的战略选择。也就是说，不管其他参与人选择什么战略，他的最优战略是唯一的，这样的最优战略被称为“占优战略”。
Definition Strategy si is strictly dominated for player i if there is some si??Si such that
Proposition a rational player will not play a strictly dominated strategy.
抵赖 is a dominated strategy. A rational player would therefore never 抵赖. This solves the game since every player will 坦白. Notice that I don't have to know anything about the other player. 囚徒困境：个人理性与集体理性之间的矛盾。
This result highlights the value of commitment in the Prisoner's dilemma – commitment consists of credibly playing strategy 抵赖.
囚徒困境的广泛应用：军备竞赛、卡特尔、公共品的供给。
9.2.3 Iterated Deletion of Dominated Strategies (重复剔除劣战略)
ui(si?,s?i)?ui(si,s?i) for al s?i?S?i.
智猪博弈（boxed pigs）
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大猪
按 等待
按 3,1 7,-1 小猪
等待 2,4 0,0
此博弈没有占优战略均衡。因为尽管“等待”是小猪的占优战略，但是大猪没有占优战略。大猪的最优战略依赖于小猪的占略: ---。
大猪会正确地预测到小猪会选择“等待”；给定此预测，大猪的最优选择只能是“按”。这样，（按，等待）就是唯一的均衡。
重复剔除的占优均衡：先剔除某个参与人的劣策略，重新构造新的博弈，再剔除，---。
应用：大股东监督经理，小股东搭便车；大企业研发，小企业模仿。
9.2.4 Nash equilibrium
性别战博弈（battle of the sexes）:
女
足球赛 演唱会
男
足球赛 演唱会
2,1 0,0 0,0 1,2
在上面的博弈中，两个参与者都没有占优策略，每个参与者的最优策略都依赖于另一个参与人的战略。所以，没有重复剔除的占优均衡。
Definition A strategy profile s* is a pure strategy Nash equilibrium of G if and only if
ui(si*,s?i*)?ui(si,s?i*) for all players i and all si?Si
求解Nash均衡的方法。
A Nash equilibrium captures the idea of equilibrium: Both players know what strategy the other player is going to choose, and no player has an incentive to deviate from equilibrium play because her strategy is a best response to her belief about the other player's strategy.
对纳什均衡的理解：设想所有参与者在博弈之前达成一个（没有约束力的）协议，规定每个参与人选择一个特定的战略。那么，给定其他参与人都遵守此协议，是否有人不愿意遵守此协议？如果没有参与人有积极性单方面背离此协议，我们说这个协议是可以自动实施的（self-enforcing），这个协议就构成一个纳什均衡。否则，它就不是一个纳什均衡。
问题：纳什均衡与重复剔除（严格）劣战略均衡之间的关系。
9.2.5 Cournot Competition （古诺竞争）
This game has an infinite strategy space.
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Two firms choose output levels qi，cost function ci (qi) = cqi.
market demand determines a price p?f(q1?q2)????(q1?q2)：the products of both firms are perfect substitutes, i.e. they are homogenous products. D = {1; 2} S1 = S2 = R+
u1 (q1, q2) = q1 f (q1 + q2) -c1 (q1) u2 (q1, q2) = q2 f (q1 + q2) - c2 (q2)
the 'best-response' function BR(qj) of each firm i to the quantity choice qj of the other firm: 由?1?q1[???(q1?q2)]?cq1，得FOC: ???(q1?q2)??q1?c?0 ? q? 1??cq2??c。因 ?；又q1?0 ? q2?2?2????cqj??c?, if qj??BRi(qj)??2?2?
?0, otherwise? q2 BR1(q2) (q1*,q2* )BR2(q1)(??c)(2?)q1(??c)?
The best-response function is decreasing in my belief of the other firm's action.
Using our new result it is easy to see that the unique Nash equilibrium of the Cournot game is the intersection of the two BR functions.
Because of symmetry we know that q1 = q2 = q*.
2(??c)??cq*q*?q??Hence we obtain, This gives us the solution.
3??2*问题：将寡头竞争的古诺均衡与垄断企业的最优产量和利润进行比较。
9.2.6 Bertrand Competition （伯特兰竞争）
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Firms compete in a homogenous product market but they set prices. Consumers buy from the lowest cost firm. demand curve q = D(p)
Therefore, each firm faces demand
?D(pi) if pi?pj?Di(p1,p2)??D(pi)2 if pi=pj
? 0 if p?p ij?
We also assume that D(c) > 0, i.e. firms can sell a positive quantity if they price at marginal cost.
Lemma The Bertrand game has the unique NE (p1,p2)= (c; c).
Proof: First we must show that (c,c) is a NE. It is easy to see that each firm makes zero profits. Deviating to a price below c would cause losses to the deviating firm. If any firm sets a higher price it does not sell any output and also makes zero profits. Therefore, there is no incentive to deviate.
To show uniqueness we must show that any other strategy profile (p1; p2) is not a NE. It's easiest to distinguish lots of cases.
Case I: p1 < c or p2 < c. In this case one (or both players) makes negative losses. This player should set a price above his rival's price and cut his losses by not selling any output.
Case II: c < p1 < p2 or c < p2 < p1. In this case the firm with the higher price makes zero profits. It could profitably deviate by setting a price equal to the rival's price and thus capture at least half of his market, and make strictly positive profits.
Case III: c = p1 < p2 or c = p2 < p1. Now the lower price firm can charge a price slightly above marginal cost (but still below the price of the rival) and make strictly positive profits.
Case IV: c < p1 = p2. Firm 1 could profitably deviate by setting a price p1?p2???c. The firm's profits before and after the deviation are:
**?B?D(p2)(p2??) 2?A?D(p2??)(p2???c)
Note that the demand function is decreasing, soD(p2??)?D(p2). We can therefore deduce:
????A??B?Therefore, (p1; p2) cannot be a NE.
9.2.7 Mixed Strategies （混合战略）
D(p2)(p2?c)??D(p2??) 2This expression (the gain from deviating) is strictly positive for sufficiently small?.
猜谜游戏 matching pennies game:
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儿童A
H(正面) T(反面)
儿童B
H(正面) 1,-1 -1,1 T(反面) -1,1 1,-1
每一个参与者都想猜透对方的战略，而又不能让对方猜透自己的战略。This game has no pure-strategy Nash equilibrium. Whatever pure strategy player 1 chooses, player 2 can beat him. 如果一个参与人采用混合战略（以一定的概率选择某种概率），他的对手就不能准确地猜出他实际上会选择的战略，尽管在均衡点上，每个人都知道其他参与人在不同战略上的概率分布。
Intuitively, games in which the participants have a large number of strtegies will often offer suffifient flexibility to ensure that at least one Nash Equilibrium must exist. If we permit the players to use “mixed ” strategies, the above game will be converted into one with an infinite number of (mixed ) strategies and, again, the existence of a Nash equilibrium is ensured.
Suppose that the players decide to randomize amongst his strategies and play a mixed
strategy. Player A could flip a coin and play H with probability r and T with probability 1-r , and player B flip a coin and play H with probability s and T with probability 1-s.
Given these probabilities, the outcomes of the game occur with the following probabilities: H-H , rs; H-T, r(1-s); T-H, (1-r)s; T-T,(1-r)(1-s). Player A’s expected utility is then given by
E(uA)?rs(1)?r(1?s)(?1)?(1?r)s(?1)?(1?r)(1?s)(1)
?4rs?2r?2s?1?2r(2s?1)?2s?1
Oviously, A’s optimal choice of r depends on B’s probability, s. If s?12, utility is maximized by choosing r?0. If s?12, A should opt for r?1. And when s?12, A’s expected utility
is 0 no matter what value of r is choosen.
For player B, expected utility is given by
E(uB)?rs(?1)?r(1?s)(1)?(1?r)s(1)?(1?r)(1?s)(?1) ??(4rs?2r?2s?1)?2s(1?2r)?(1?2r)
Now, when r?12, B’s expected utility is maximized by choosing s?0. If r?12, A should opt for s?1. And when r?12, A’s expected utility is independent of what s is choosen.
sresponse curve of B1response curve of A1/21/21
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博弈论(2)—讲义
