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{{Infobox equilibrium

{{Infobox equilibrium

模板:Infobox equilibrium



In game theory, the Nash equilibrium, named after the mathematician John Forbes Nash Jr., is a proposed solution of a non-cooperative game involving two or more players in which each player is assumed to know the equilibrium strategies of the other players, and no player has anything to gain by changing only their own strategy.[1]

在博弈论中, 纳什均衡(Nash equilibrium)由数学家(约翰·福布斯·纳什 John Forbes Nash Jr.)提出并以其名字命名,是一种非合作博弈的解的概念。这种博弈模型假定每个局中人都知道其他局中人的均衡策略。在纳什均衡下,没有一个局中人仅仅通过改变自己的策略就能得到额外任何好处。[1]

If each player has chosen a strategy—an action plan choosing its own action based on what it has seen happen so far in the game—and no player can increase its own expected payoff by changing its strategy while the other players keep theirs unchanged, then the current set of strategy choices constitutes a Nash equilibrium.

如果每位局中人都选择了一个策略——即一个根据博弈中所有已知信息选择行动的行动计划,并且没有局中人可以在其他局中人则保持自己的策略不变的情况下,通过改变策略来提高自己的预期收益,那么当前的一组策略就构成了一个 纳什均衡点

If two players Alice and Bob choose strategies A and B, (A, B) is a Nash equilibrium if Alice has no other strategy available that does better than A at maximizing her payoff in response to Bob choosing B, and Bob has no other strategy available that does better than B at maximizing his payoff in response to Alice choosing A. In a game in which Carol and Dan are also players, (A, B, C, D) is a Nash equilibrium if A is Alice's best response to (B, C, D), B is Bob's best response to (A, C, D), and so forth.

如果两位局中人 Alice 和 Bob 选择策略A 和 B,如果 Bob 采取 B策略时Alice 没有比 A 更好的策略来增加收益,且Alice 采取 A策略时Bob 没有比 B 更好的策略来增加收益,那么(A, B) 是一个纳什均衡点 。在一个 Carol 和 Dan 也是局中人的博弈中,如果 A 是 Alice 对(B, C, D)的最佳对策,B 是 Bob 对(A, C, D)的最佳对策,那么(A, B, C, D)就是纳什均衡点,如此等等。

Nash showed that there is a Nash equilibrium for every finite game: see further the article on strategy.



Game theorists use Nash equilibrium to analyze the outcome of the strategic interaction of several decision makers. In a strategic interaction, the outcome for each decision-maker depends on the decisions of the others as well as their own. The simple insight underlying Nash's idea is that one cannot predict the choices of multiple decision makers if one analyzes those decisions in isolation. Instead, one must ask what each player would do taking into account what she/he expects the others to do. Nash equilibrium requires that their choices be consistent: no player wishes to undo their decision given what the others are deciding.

博弈论学者使用纳什均衡点分析多个决策者策略互动的结果。在策略互动中,每个决策者的决策结果既取决于其他决策者的决策,也取决于他们自己的决策。纳什对此的一个简单的见解是,如果一个人对每个决策者单独地分析他们的决策,那么他就无法正确地预测多个决策者的决策。也就是说,决策者必须考虑到每个局中人在期望其他人做什么。 纳什均衡点要求他们的决策选择是一致的,即在考虑到其他人的决策时,没有局中人希望取消自己的决策。

The concept has been used to analyze hostile situations like wars and arms races[2] (see prisoner's dilemma), and also how conflict may be mitigated by repeated interaction (see tit-for-tat). It has also been used to study to what extent people with different preferences can cooperate (see battle of the sexes), and whether they will take risks to achieve a cooperative outcome (see stag hunt). It has been used to study the adoption of technical standards,[citation needed] and also the occurrence of bank runs and currency crises (see coordination game). Other applications include traffic flow (see Wardrop's principle), how to organize auctions (see auction theory), the outcome of efforts exerted by multiple parties in the education process,[3] regulatory legislation such as environmental regulations (see tragedy of the commons),[4] natural resource management,[5] analysing strategies in marketing,[6] even penalty kicks in football (see matching pennies),[7] energy systems, transportation systems, evacuation problems[8] and wireless communications.[9]

这个概念已经被用来分析战争和军备竞赛等敌对情况[2] (参见囚徒困境) ,以及如何通过反复互动来缓和冲突(参见以牙还牙)。它也被用来研究不同偏好的人可以在多大程度上合作(参见性别大战),以及他们是否愿意冒着风险实现合作成果(参见猎鹿博弈)。它被用来研究怎样采用技术标准[citation needed],以及银行挤兑和货币危机(Currency crisis)的发生机制(参见协调博弈)。其他应用领域包括交通流量(参见 Wardrop 原理) ,组织拍卖(参见拍卖理论) ,多方努力下的教育结果[3],监管立法——如环境法规(参见公共地悲剧[4] ,自然资源管理[5] ,市场营销策略分析[6] ,甚至包括足球点球(参见匹配便士[7] ,能源系统,交通系统,疏散问题[8] 和无线通信[9]


Nash equilibrium is named after American mathematician John Forbes Nash, Jr. The same idea was used in a particular application in 1838 by Antoine Augustin Cournot in his theory of oligopoly.[10] In Cournot's theory, each of several firms choose how much output to produce to maximize its profit. The best output for one firm depends on the outputs of the others. A Cournot equilibrium occurs when each firm's output maximizes its profits given the output of the other firms, which is a pure-strategy Nash equilibrium. Cournot also introduced the concept of best response dynamics in his analysis of the stability of equilibrium. Cournot did not use the idea in any other applications, however, or define it generally.

纳什均衡是以美国数学家纳什(John Forbes Nash Jr.)命名的。1838年, 安托万·奥古斯丁·古诺(Antoine Augustin Cournot)在他的寡头垄断理论[10]中体现了同样的思想。根据古诺的理论,几家公司中的每一家都致力于实现利润最大化。一个公司的最佳决策取决于其他公司的产出情况。给定其他企业的产出情况,若企业实现利益最大化,则达到了古诺均衡,这是一个纯策略的纳什均衡点。古诺还在其均衡稳定性分析中引入了最佳响应动力学的概念。然而,古诺并没有在其他任何应用领域中使用这个概念,或者给出一般性定义。

The modern game-theoretic concept of Nash equilibrium is instead defined in terms of mixed strategies, where players choose a probability distribution over possible actions (rather than choosing a deterministic action to be played with certainty). The concept of a mixed-strategy equilibrium was introduced by John von Neumann and Oskar Morgenstern in their 1944 book The Theory of Games and Economic Behavior. However, their analysis was restricted to the special case of zero-sum games. They showed that a mixed-strategy Nash equilibrium will exist for any zero-sum game with a finite set of actions.[11] The contribution of Nash in his 1951 article "Non-Cooperative Games" was to define a mixed-strategy Nash equilibrium for any game with a finite set of actions and prove that at least one (mixed-strategy) Nash equilibrium must exist in such a game. The key to Nash's ability to prove existence far more generally than von Neumann lay in his definition of equilibrium. According to Nash, "an equilibrium point is an n-tuple such that each player's mixed strategy maximizes his payoff if the strategies of the others are held fixed. Thus each player's strategy is optimal against those of the others." Just putting the problem in this framework allowed Nash to employ the Kakutani fixed-point theorem in his 1950 paper, and a variant upon it in his 1951 paper used the Brouwer fixed-point theorem to prove that there had to exist at least one mixed strategy profile that mapped back into itself for finite-player (not necessarily zero-sum) games; namely, a strategy profile that did not call for a shift in strategies that could improve payoffs.[12]

现代博弈理论中,纳什均衡点是用混合策略来定义的,局中人选择的是可能行为的概率分布(而不是选择一个确定性的行动)。混合策略均衡的概念是由约翰·冯·诺依曼(John von Neumann)和摩根·斯特恩(Oskar Morgenstern)在他们1944年出版的《博弈论与经济行为》一书中提出的。然而,他们的分析仅限于零和博弈的特殊情况。他们证明了一个混合策略的纳什均衡点对于任何零和博弈都是存在的[11] 。纳什在1951年的文章《非合作博弈》中定义了一个混合策略纳什均衡点,并证明了在任何有限博弈中至少存在一个混合策略纳什均衡点。与冯 · 诺依曼相比,纳什能够证明存在性的关键在于他对均衡的定义。按照纳什的说法,“如果其他局中人的策略保持不变,那么每个局中人使他的收益最大化的混合策略的n元数组即为一个均衡点。这样一来,每个局中人的策略都是相对于其他局中人的最优策略。”把这个问题放在这个框架中,纳什就可以在他1950年的论文中使用角谷静夫不动点定理,以及在他1951年的论文中使用布劳威尔不动点定理来证明至少存在一个混合策略组合,可以映射回有限局中人的博弈(不限于零和博弈)中。也就是说,存在一个不能依靠改变其中某个策略来提高收益的策略组合[12]

Since the development of the Nash equilibrium concept, game theorists have discovered that it makes misleading predictions (or fails to make a unique prediction) in certain circumstances. They have proposed many related solution concepts (also called 'refinements' of Nash equilibria) designed to overcome perceived flaws in the Nash concept. One particularly important issue is that some Nash equilibria may be based on threats that are not 'credible'. In 1965 Reinhard Selten proposed subgame perfect equilibrium as a refinement that eliminates equilibria which depend on non-credible threats. Other extensions of the Nash equilibrium concept have addressed what happens if a game is repeated, or what happens if a game is played in the absence of complete information. However, subsequent refinements and extensions of Nash equilibrium share the main insight on which Nash's concept rests: the equilibrium is a set of strategies such that each player's strategy is optimal given the choices of the others.

随着纳什均衡概念的发展,博弈论学者发现它在某些情况下会做出误导性的预测,或者不能做出特殊的预测。他们提出了许多相关的解决性的概念,也被称为纳什均衡的”精炼“ ,旨在克服纳什概念中的缺陷。一个特别重要的问题是,某些纳什均衡可能是基于“不可信”的威胁。1965年,赖因哈德·泽尔腾(Reinhard Selten)提出了一种改进后的 子博弈精炼均衡(Subgame perfect equilibrium),它消除了依赖于不可信威胁的均衡。其他纳什均衡概念的扩展解决了如果重复进行博弈,或者如果一个博弈在没有完整信息的情况下会发生什么的问题。不管怎样,后续对纳什均衡点的改进和扩展与纳什均衡概念所倚赖的主要观点是一致的: 均衡是这样一组策略,在这组策略下,每个局中人的策略在其他人的策略下都是最优的。



Informally, a strategy profile is a Nash equilibrium if no player can do better by unilaterally changing their strategy. To see what this means, imagine that each player is told the strategies of the others. Suppose then that each player asks themselves: "Knowing the strategies of the other players, and treating the strategies of the other players as set in stone, can I benefit by changing my strategy?"

通俗地说,如果没有局中人能够单方面改变他们的策略来获得更多收益,那么这种情况下的策略组合就是一个 纳什均衡点。要理解这点,需要设想一下如下场景,如果每个局中人都被告知其他局中人的策略,然后每个局中人都扪心自问: “知道其他参与者的策略,并且把其他参与者的策略当作是固定不变的,我能通过改变策略而受益吗? ”

If any player could answer "Yes", then that set of strategies is not a Nash equilibrium. But if every player prefers not to switch (or is indifferent between switching and not) then the strategy profile is a Nash equilibrium. Thus, each strategy in a Nash equilibrium is a best response to all other strategies in that equilibrium.[13]

如果任何一个参与者都能回答“是” ,那么这套策略就不是 纳什均衡点。但是如果每个局中人都不愿意切换(或者对切换和不切换无所谓)策略 ,那么这个策略就是一个纳什均衡点。因此,纳什均衡点中的每个策略都是对均衡中所有其他策略的最佳回应[13]

The Nash equilibrium may sometimes appear non-rational in a third-person perspective. This is because a Nash equilibrium is not necessarily Pareto optimal.

从第三者的角度来看,纳什均衡点有时可能显得非理性。这是因为 纳什均衡点不一定是 帕累托最优的。

The Nash equilibrium may also have non-rational consequences in sequential games because players may "threaten" each other with non-rational moves. For such games the subgame perfect Nash equilibrium may be more meaningful as a tool of analysis.



Suppose that in the Nash equilibrium, each player asks themselves: "Knowing the strategies of the other players, and treating the strategies of the other players as set in stone, would I suffer a loss by changing my strategy?"

假设在纳什均衡中,每个局中人都问自己: “如果知道其他局中人的策略,并且把其他们的策略当作是固定不变的,我会因为改变策略而遭受损失吗? ”

If every player's answer is "Yes", then the equilibrium is classified as a strict Nash equilibrium.[14]


If instead, for some player, there is exact equality between the strategy in Nash equilibrium and some other strategy that gives exactly the same payout (i.e. this player is indifferent between switching and not), then the equilibrium is classified as a weak Nash equilibrium.

如果相反,对于某些局中人来说,纳什均衡点的策略和其他策略给出的回报完全相同(例如:这个参与人在切换和不切换之间是无关紧要的) ,那么这个均衡被归类为弱纳什均衡点。

A game can have a pure-strategy or a mixed-strategy Nash equilibrium. (In the latter a pure strategy is chosen stochastically with a fixed probability).



Nash proved that if we allow mixed strategies (where a player chooses probabilities of using various pure strategies), then every game with a finite number of players in which each player can choose from finitely many pure strategies has at least one Nash equilibrium (which might be a pure strategy for each player or might be a probability distribution over strategies for each player).

纳什证明,如果我们允许混合策略(局中人选择使用各种纯策略的概率) ,那么局中人数量有限且每个局中人可以从有限多个纯策略中选择的博弈都至少有一个纳什均衡点(这可能是每个局中人的纯策略,也可能是每个局中人的纯策略概率分布)。

Nash equilibria need not exist if the set of choices is infinite and noncompact. An example is a game where two players simultaneously name a number and the player naming the larger number wins. Another example is where each of two players chooses a real number strictly less than 5 and the winner is whoever has the biggest number; no biggest number strictly less than 5 exists (if the number could equal 5, the Nash equilibrium would have both players choosing 5 and tying the game). However, a Nash equilibrium exists if the set of choices is compact with each player's payoff continuous in the strategies of all the players.[15] An example, in which the equilibrium is a mixture of continuously many pure strategies, is a game where two players simultaneously pick a real number between 0 and 1 (inclusive) and player one's winnings (paid by the second player) equal the square root of the distance between the two numbers.

如果策略集是无限的且非紧凑的,那么纳什均衡就不存在。比如,两位局中人同时说出一个数字,而说出较大数字的一方获胜。又比如,每位局中人选择一个严格小于5的实数,获胜者是选择了最大数字的人(没有严格小于5的最大实数存在,如果数字可以等于5,纳什均衡点将有两个参与者选择5并平局)。然而,如果对于所有局中人而言,其策略集是紧凑的,且自身的收益是连续的,则存在一个纳什均衡点[15] 。这样的示例为,两位局中人同时选择一个介于0和1之间的实数(包括0和1在内) ,局中人一的奖金(由局中人二支付)等于两个数字之间距离的平方根。



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A sample coordination game showing relative payoff for player 1 (row) / player 2 (column) with each combination
模板:Diagonal split header Player 2 adopts strategy A 局中人2采用策略 a Player 2 adopts strategy A Player 2 adopts strategy B 局中人2采用策略 b Player 2 adopts strategy B
Player 1 adopts strategy A 局中人1采用策略 a Player 1 adopts strategy A 模板:Diagonal split header 模板:Diagonal split header
Player 1 adopts strategy B 局中人1采用策略 b Player 1 adopts strategy B 模板:Diagonal split header 模板:Diagonal split header

The coordination game is a classic (symmetric) two player, two strategy game, with an example payoff matrix shown to the right. The players should thus coordinate, both adopting strategy A, to receive the highest payoff; i.e., 4. If both players chose strategy B though, there is still a Nash equilibrium. Although each player is awarded less than optimal payoff, neither player has incentive to change strategy due to a reduction in the immediate payoff (from 2 to 1).

协调博弈是一个经典的两位(对称的)局中人、两个策略的博弈,右边显示了一个收益矩阵的例子。因此,局中人应该相互协调,都采用策略 A,以获得最高的收益。如果两位局中人都选择了 B 策略,那么仍然有一个纳什均衡点。虽然每位局中人的收益都低于最优收益,但是由于直接收益减少(从2减少到1) ,于是两位局中人都没有改变策略的动机。

A famous example of this type of game was called the stag hunt; in the game two players may choose to hunt a stag or a rabbit, the former providing more meat (4 utility units) than the latter (1 utility unit). The caveat is that the stag must be cooperatively hunted, so if one player attempts to hunt the stag, while the other hunts the rabbit, she/he will fail in hunting (0 utility units), whereas if they both hunt it they will split the payoff (2, 2). The game hence exhibits two equilibria at (stag, stag) and (rabbit, rabbit) and hence the players' optimal strategy depend on their expectation on what the other player may do. If one hunter trusts that the other will hunt the stag, they should hunt the stag; however if they suspect that the other will hunt the rabbit, they should hunt the rabbit. This game was used as an analogy for social cooperation, since much of the benefit that people gain in society depends upon people cooperating and implicitly trusting one another to act in a manner corresponding with cooperation.

一个著名的例子是猎鹿博弈,在博弈中,两位局中人可以选择猎鹿或猎兔,前者(4个效用单位)提供比后者(1个效用单位)更多的肉。需要注意的是,这只鹿必须被合作猎杀,所以如果一位局中人试图猎杀鹿,而另一位局中人猎杀兔子,那么猎鹿者将在猎杀中失败(0个实用单位),猎兔者获得兔子(1个效用单位) ,而如果他们都猎杀鹿,他们将分享收益(2,2)。这个博弈因此展现了两个均衡(雄鹿,雄鹿)和(兔子,兔子) ,因此局中人的最佳策略取决于他们对另一位局中人可能做什么的期望。如果一个猎人相信另一个会猎鹿,他们就应该猎鹿; 但是如果他们怀疑另一个会猎兔,他们就应该猎兔。这个博弈被用来作为社会合作的类比,因为人们在社会中获得的大部分利益取决于人们之间的合作和相互间的隐性信任,并且以与合作相对应的方式行事。

Another example of a coordination game is the setting where two technologies are available to two firms with comparable products, and they have to elect a strategy to become the market standard. If both firms agree on the chosen technology, high sales are expected for both firms. If the firms do not agree on the standard technology, few sales result. Both strategies are Nash equilibria of the game.


Driving on a road against an oncoming car, and having to choose either to swerve on the left or to swerve on the right of the road, is also a coordination game. For example, with payoffs 10 meaning no crash and 0 meaning a crash, the coordination game can be defined with the following payoff matrix:


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The driving game
The driving game
模板:Diagonal split header Drive on the Left 向左开 Drive on the Left Drive on the Right 向右开 Drive on the Right
Drive on the Left 向左开 Drive on the Left 模板:Diagonal split header 模板:Diagonal split header
Drive on the Right 向右开 Drive on the Right 模板:Diagonal split header 模板:Diagonal split header

In this case there are two pure-strategy Nash equilibria, when both choose to either drive on the left or on the right. If we admit mixed strategies (where a pure strategy is chosen at random, subject to some fixed probability), then there are three Nash equilibria for the same case: two we have seen from the pure-strategy form, where the probabilities are (0%, 100%) for player one, (0%, 100%) for player two; and (100%, 0%) for player one, (100%, 0%) for player two respectively. We add another where the probabilities for each player are (50%, 50%).

在这种情况下,存在两个纯策略纳什均衡,即当两者都选择左侧或右侧开。如果我们承认混合策略(一个纯策略是随机选择的,且服从某个固定概率) ,那么对于同样的情况有三个纳什均衡: 从纯策略形式中存在两个纳什均衡,其中局中人1的策略的概率组合为(0% ,100%) ,局中人2为(0% ,100%) ;又或者局中人1的概率组合为(100% ,0%) ,局中人2为(100% ,0%)。除此之外,可以再加上一个每位局中人的策略概率组合是(50% ,50%)的情况。

模板:Clear left


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Example PD payoff matrix
示例 PD 支付矩阵
Example PD payoff matrix
模板:Diagonal split header Cooperate (with other) 合作 Cooperate (with other) Defect (betray other) 背叛 Defect (betray other)
Cooperate (with other) 合作 Cooperate (with other) −1, −1 −1, −1 −1, −1 −3, 0 −3, 0 −3, 0
Defect (betray other) 背叛 Defect (betray other) 0, −3 0, −3 0, −3 −2, −2 −2, −2 −2, −2

Imagine two prisoners held in separate cells, interrogated simultaneously, and offered deals (lighter jail sentences) for betraying their fellow criminal. They can "cooperate" (with the other prisoner) by not snitching, or "defect" by betraying the other. However, there is a catch; if both players defect, then they both serve a longer sentence than if neither said anything. Lower jail sentences are interpreted as higher payoffs (shown in the table).

想象一下,两个犯人被关在不同的牢房里,同时接受审讯,并且提供只要背叛了同伙就可以减轻刑罚的交易 。他们可以通过不告密来达成“合作” ,或者通过告密而“背叛”同伙。然而,这里有一个陷阱: 如果两个囚犯都叛变了,那么他们服刑的时间都会比双方都不告密的时间要长。较低的监禁刑罚被表示为较高的收益(如表所示)。

The prisoner's dilemma has a similar matrix as depicted for the coordination game, but the maximum reward for each player (in this case, a minimum loss of 0) is obtained only when the players' decisions are different. Each player improves their own situation by switching from "cooperating" to "defecting", given knowledge that the other player's best decision is to "defect". The prisoner's dilemma thus has a single Nash equilibrium: both players choosing to defect.

囚徒困境(Prisoner's dilemma)具有与协调博弈相似的矩阵,但是只有当局中人的决策不同时,才能得到对每个局中人的最大报酬(在这种情况下,最小损失为0)。因为知道另一位局中人的最佳决定是“背叛”,所以每位局中人会考虑到改善自身处境,由“合作”转变到“背叛”。因此,囚徒困境只有一个纳什均衡点:双方都选择背叛。

What has long made this an interesting case to study is the fact that this scenario is globally inferior to "both cooperating". That is, both players would be better off if they both chose to "cooperate" instead of both choosing to defect. However, each player could improve their own situation by breaking the mutual cooperation, no matter how the other player possibly (or certainly) changes their decision.

长期以来,这一直是一个值得研究的有趣案例,因为这种场景在全局范围内来看不如”双方合作”。也就是说,如果双方都选择合作 ,而不是双方都选择背叛,那么双方都会过得更好。然而,每位局中人都可以通过打破相互合作来改善自身处境,不管其他局中人可能(或肯定)如何改变他们的决定。


Sample network graph. Values on edges are the travel time experienced by a 'car' traveling down that edge. is the number of cars traveling via that edge.

文件:Nash graph equilibrium.png
Sample network graph. Values on edges are the travel time experienced by a 'car' traveling down that edge. x is the number of cars traveling via that edge.

thumb | 250px |网络图示例。边上的值是“汽车”沿该边行驶所经历的行驶时间。x 是沿该边行驶的汽车数量。

An application of Nash equilibria is in determining the expected flow of traffic in a network. Consider the graph on the right. If we assume that there are "cars" traveling from A to D, what is the expected distribution of traffic in the network?

纳什均衡的一个应用是确定网络中的预期流量。看看右边的图表。如果我们假设有从 A 到 D 的“汽车” ,网络中预期的流量分布是什么?

This situation can be modeled as a "game" where every traveler has a choice of 3 strategies, where each strategy is a route from A to D (either , , or ). The "payoff" of each strategy is the travel time of each route. In the graph on the right, a car travelling via experiences travel time of , where is the number of cars traveling on edge . Thus, payoffs for any given strategy depend on the choices of the other players, as is usual. However, the goal, in this case, is to minimize travel time, not maximize it. Equilibrium will occur when the time on all paths is exactly the same. When that happens, no single driver has any incentive to switch routes, since it can only add to their travel time. For the graph on the right, if, for example, 100 cars are travelling from A to D, then equilibrium will occur when 25 drivers travel via , 50 via , and 25 via . Every driver now has a total travel time of 3.75 (to see this, note that a total of 75 cars take the edge, and likewise, 75 cars take the edge).

针对这种情况可以建立一种“博弈”模型 ,每个旅行者都有3个策略可供选择,每个策略都是从 A 到 D 的路线。每个策略的“收益”是每条路线的行驶时间。在右边的图表中,一辆汽车经历的旅行时间取决于在边上旅行的汽车的数量。因此,通常情况下,任何给定策略的收益都取决于其他局中人的选择。然而,在这种情况下,旅行时间的目标是最小化,而不是最大化。当所有路径上的时间完全相同时,即达到平衡。当这种情况发生时,没有一个司机有任何动机改变路线,因为这只能增加他们的旅行时间。例如,对于右边的图表,如果100辆汽车从 A 到 D,那么平衡将发生在线路上分别是25个司机,50个司机,和25个司机的情况下。每个司机的总旅行时间为3.75(注意,总共有75辆车在边AB上,同样地,有75辆汽车在边CD上)。<!——25名司机选择 ABD,50名司机选择 ABCD,因此有75辆汽车在边AB上行驶 。同样,75辆汽车行驶在边 CD 上。通过 ABD 旅行需要(1 + 75/100) + 2 = 3.75小时。其他路线的行程时间是相同的。-->

Notice that this distribution is not, actually, socially optimal. If the 100 cars agreed that 50 travel via ABD and the other 50 through ACD, then travel time for any single car would actually be 3.5, which is less than 3.75. This is also the Nash equilibrium if the path between B and C is removed, which means that adding another possible route can decrease the efficiency of the system, a phenomenon known as Braess's paradox.

注意,这种分布实际上并不是社会最优的。如果100辆车中50辆经过ABD,其他50辆经过ACD,那么任何一辆车的旅行时间实际上是3.5,小于3.75。如果 B 和 C之间的路径被移除,这也是纳什均衡点,这意味着增加另一条可能的路径会降低系统的效率,这种现象被称为布雷斯悖论(Braess's paradox )。


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A competition game
a competition game
模板:Diagonal split header Choose '0' 选择「0」 Choose '0' Choose '1' 选择「1」 Choose '1' Choose '2' 选择「2」 Choose '2' Choose '3' 选择“3” Choose '3'
Choose '0' 选择「0」 Choose '0' 0, 0 0,0 0, 0 2, −2 2, −2 2, −2 2, −2 2, −2 2, −2 2, −2 2, −2 2, −2
Choose '1' 选择「1」 Choose '1' −2, 2 −2, 2 −2, 2 1, 1


1, 1 3, −1 3, −1 3, −1 3, −1 3, −1 3, −1
Choose '2' 选择「2」 Choose '2' −2, 2 −2, 2 −2, 2 −1, 3 −1, 3 −1, 3 2, 2 2,2 2, 2 4, 0


4, 0
Choose '3' 选择“3” Choose '3' −2, 2 −2, 2 −2, 2 −1, 3 −1, 3 −1, 3 0, 4


0, 4 3, 3


3, 3

This can be illustrated by a two-player game in which both players simultaneously choose an integer from 0 to 3 and they both win the smaller of the two numbers in points. In addition, if one player chooses a larger number than the other, then they have to give up two points to the other.


This game has a unique pure-strategy Nash equilibrium: both players choosing 0 (highlighted in light red). Any other strategy can be improved by a player switching their number to one less than that of the other player. In the adjacent table, if the game begins at the green square, it is in player 1's interest to move to the purple square and it is in player 2's interest to move to the blue square. Although it would not fit the definition of a competition game, if the game is modified so that the two players win the named amount if they both choose the same number, and otherwise win nothing, then there are 4 Nash equilibria: (0,0), (1,1), (2,2), and (3,3).

这个博弈模型有一个独特的纯策略纳什均衡点: 两位局中人都选择0(用淡红色突出显示)。任何其他的策略都可以通过一位局中人把他们的数字换成比另一位局中人更小的数字来改进。在相邻的表格中,如果游戏从绿色的方块开始,那么局中人1的最佳收益会转移到紫色的方块上,而局中人2的最佳收益会转移到蓝色的方块上。虽然它不符合竞争博弈的定义,但是如果对模型进行修改,使得两位局中人在选择相同数字的情况下都赢得指定的金额,那么就有4个纳什均衡: (0,0) ,(1,1) ,(2,2)和(3,3)。


There is an easy numerical way to identify Nash equilibria on a payoff matrix. It is especially helpful in two-person games where players have more than two strategies. In this case formal analysis may become too long. This rule does not apply to the case where mixed (stochastic) strategies are of interest. The rule goes as follows: if the first payoff number, in the payoff pair of the cell, is the maximum of the column of the cell and if the second number is the maximum of the row of the cell - then the cell represents a Nash equilibrium.

有一种简单的数值方法可以确定收益矩阵中的纳什均衡。在局中人有两个以上策略的双人博弈中,这种方法尤其有用。在这种情况下,形式上的分析可能会显得冗长。这条规则并不适用于混合(随机)策略的情况,它可以被描述为: 如果单元格的收益对中的第一个数是该单元格列的最大值,且第二个数是该单元格行的最大值,那么该单元格表示一个纳什均衡点。

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A payoff matrix – Nash equilibria in bold
A payoff matrix – Nash equilibria in bold
模板:Diagonal split header Option A 选项 a Option A Option B 选项 b Option B Option C 选项 c Option C
Option A 选项 a Option A 0, 0


0, 0 25, 40


25, 40 5, 10


5, 10
Option B 选项 b Option B 40, 25


40, 25 0, 0


0, 0 5, 15


5, 15
Option C 选项 c Option C 10, 5


10, 5 15, 5


15, 5 10, 10


10, 10

We can apply this rule to a 3×3 matrix:

我们可以把这个规则应用到3 × 3矩阵上:

We can apply this rule to a 3×3 matrix:

Using the rule, we can very quickly (much faster than with formal analysis) see that the Nash equilibria cells are (B,A), (A,B), and (C,C). Indeed, for cell (B,A) 40 is the maximum of the first column and 25 is the maximum of the second row. For (A,B) 25 is the maximum of the second column and 40 is the maximum of the first row. Same for cell (C,C). For other cells, either one or both of the duplet members are not the maximum of the corresponding rows and columns.

使用这个规则,我们可以很快(比正式分析快得多)看到纳什均衡的单元格是(B,A), (A,B), 和 (C,C)。实际上,对于单元格(B,A),40是第一列的最大值,25是第二行的最大值。对于(A,B),25是第二列的最大值,40是第一行的最大值。单元格(C,C)也是如此。对于其他单元格,收益对中任何一个数都不是相应行和列的最大值。

This said, the actual mechanics of finding equilibrium cells is obvious: find the maximum of a column and check if the second member of the pair is the maximum of the row. If these conditions are met, the cell represents a Nash equilibrium. Check all columns this way to find all NE cells. An N×N matrix may have between 0 and N×N pure-strategy Nash equilibria.

也就是说,找到平衡单元的实际机制是显而易见的: 找到一列的最大值,然后检查这一对的第二个数是否是行的最大值。如果满足这些条件,则该单元格表示一个纳什均衡点。检查所有单元格后即可找到所有的纳均衡点。一个 n × n 矩阵可能存在0到 n × n 个纯策略纳什均衡。

< ! -- 宽屏布局修正 -- >


The concept of stability, useful in the analysis of many kinds of equilibria, can also be applied to Nash equilibria.


A Nash equilibrium for a mixed-strategy game is stable if a small change (specifically, an infinitesimal change) in probabilities for one player leads to a situation where two conditions hold:

一个混合策略博弈的纳什均衡点是稳定的,如果一位局中人采取的混合策略的概率发生了微小的变化(具体地说,是一个无穷小的变化) ,将导致出现两种情况:

the player who did not change has no better strategy in the new circumstance


  1. the player who did not change has no better strategy in the new circumstance
the player who did change is now playing with a strictly worse strategy.


  1. the player who did change is now playing with a strictly worse strategy.

If these cases are both met, then a player with the small change in their mixed strategy will return immediately to the Nash equilibrium. The equilibrium is said to be stable. If condition one does not hold then the equilibrium is unstable. If only condition one holds then there are likely to be an infinite number of optimal strategies for the player who changed.


In the "driving game" example above there are both stable and unstable equilibria. The equilibria involving mixed strategies with 100% probabilities are stable. If either player changes their probabilities slightly, they will be both at a disadvantage, and their opponent will have no reason to change their strategy in turn. The (50%,50%) equilibrium is unstable. If either player changes their probabilities (which would neither benefit or damage the expectation of the player who did the change, if the other player's mixed strategy is still (50%,50%)), then the other player immediately has a better strategy at either (0%, 100%) or (100%, 0%).

在上面的“撞车博弈”的例子中,既有稳定的均衡,也有不稳定的均衡。含有100% 概率的混合策略的均衡是稳定的,如果任何一位局中人稍微改变他们的概率,他们都将处于劣势,而他们的对手将没有理由依次改变他们的策略。(50% ,50%)平衡是不稳定的,当任何一位局中人改变了他的概率,如果另一位局中人的混合策略仍然是(50% ,50%) ,这既不会有利于也不会损害改变策略的玩家的期望值,但他立即有一个更好的策略是(0% ,100%)或(100% ,0%)。

Stability is crucial in practical applications of Nash equilibria, since the mixed strategy of each player is not perfectly known, but has to be inferred from statistical distribution of their actions in the game. In this case unstable equilibria are very unlikely to arise in practice, since any minute change in the proportions of each strategy seen will lead to a change in strategy and the breakdown of the equilibrium.


The Nash equilibrium defines stability only in terms of unilateral deviations. In cooperative games such a concept is not convincing enough. Strong Nash equilibrium allows for deviations by every conceivable coalition.[16] Formally, a strong Nash equilibrium is a Nash equilibrium in which no coalition, taking the actions of its complements as given, can cooperatively deviate in a way that benefits all of its members.[17] However, the strong Nash concept is sometimes perceived as too "strong" in that the environment allows for unlimited private communication. In fact, strong Nash equilibrium has to be Pareto efficient. As a result of these requirements, strong Nash is too rare to be useful in many branches of game theory. However, in games such as elections with many more players than possible outcomes, it can be more common than a stable equilibrium.

纳什均衡仅仅用单边偏离来定义稳定性。在合作博弈中,这样的概念是不够令人信服的。强纳什均衡可以允许任何可以想象的联盟出现偏差[16] 。从形式上讲,强纳什均衡是指没有任何一个联盟,在其补集的行动被给定的情况下,能够合作地偏离一个有利于所有成员的方向[17]。然而,强纳什概念有时被认为过于“强” ,因为环境允许无限的私人交流。事实上,强纳什均衡必须是帕累托最优的。由于这些要求,强纳什均衡在博弈论的许多分支中实在是太稀有了。然而,在诸如选举这样的博弈中,参与者比可能的结果多得多,强纳什均衡可能比稳定的均衡更常见。

A refined Nash equilibrium known as coalition-proof Nash equilibrium (CPNE)[16] occurs when players cannot do better even if they are allowed to communicate and make "self-enforcing" agreement to deviate. Every correlated strategy supported by iterated strict dominance and on the Pareto frontier is a CPNE.[18] Further, it is possible for a game to have a Nash equilibrium that is resilient against coalitions less than a specified size, k. CPNE is related to the theory of the core.

一种被称为“防联盟的纳什均衡”(CPNE[16] )的精炼纳什均衡,其发生在——即使允许局中人进行交流并达成“自我执行”协议,也无法做得更好的情况下。在帕累托边界上,每一种由迭代严格优势度支持的相关策略都是一个CPNE[18] 。另外,对于一个博弈来说,有可能有一个对小于一定规模的联盟具有弹性的纳什均衡点,这个纳什均衡点与核心论(theory of the core)有关。

Finally in the eighties, building with great depth on such ideas Mertens-stable equilibria were introduced as a solution concept. Mertens stable equilibria satisfy both forward induction and backward induction. In a game theory context stable equilibria now usually refer to Mertens stable equilibria.

最后,在八十年代,作为解决方案,引入了对这种思想进行深入研究的默斯顿稳定均衡(Mertens-stable equilibria)的概念。默顿斯稳定均衡同时满足正向归纳和逆向归纳法。在博弈论中,稳定均衡现在通常指的是默斯顿稳定均衡。


If a game has a unique Nash equilibrium and is played among players under certain conditions, then the NE strategy set will be adopted. Sufficient conditions to guarantee that the Nash equilibrium is played are:


The players all will do their utmost to maximize their expected payoff as described by the game.


The players are flawless in execution.


The players have sufficient intelligence to deduce the solution.


The players know the planned equilibrium strategy of all of the other players.


The players believe that a deviation in their own strategy will not cause deviations by any other players.


There is common knowledge that all players meet these conditions, including this one. So, not only must each player know the other players meet the conditions, but also they must know that they all know that they meet them, and know that they know that they know that they meet them, and so on.



Examples of game theory problems in which these conditions are not met:


The first condition is not met if the game does not correctly describe the quantities a player wishes to maximize. In this case there is no particular reason for that player to adopt an equilibrium strategy. For instance, the prisoner's dilemma is not a dilemma if either player is happy to be jailed indefinitely.


Intentional or accidental imperfection in execution. For example, a computer capable of flawless logical play facing a second flawless computer will result in equilibrium. Introduction of imperfection will lead to its disruption either through loss to the player who makes the mistake, or through negation of the common knowledge criterion leading to possible victory for the player. (An example would be a player suddenly putting the car into reverse in the game of chicken, ensuring a no-loss no-win scenario).


译者注:胆小鬼博弈(game of chicken),亦称之为斗鸡博弈,该模型可以被描述为,两名车手向对方驱车而行,谁最先让开的一方被耻笑为“胆小鬼”(chicken),而另一方胜出,但如果两人拒绝收掣,任由两车相撞,则谁都无法胜出。这里的举例是指,若参与者做出了倒车这种非规则内行为,则无法由规则来界定其本身的输赢。

In many cases, the third condition is not met because, even though the equilibrium must exist, it is unknown due to the complexity of the game, for instance in Chinese chess[19] . Or, if known, it may not be known to all players, as when playing tic-tac-toe with a small child who desperately wants to win (meeting the other criteria).

在许多情况下,第三个条件是不能满足的,因为即使均衡一定存在,但由于博弈的复杂性,它是未知的,例如在下中国象棋的时候[19] 。又或者,即使知道的话,也许不是所有的玩家都知道,比如和一个非常想赢的小孩玩井字棋(假设其他标准都满足)。


The criterion of common knowledge may not be met even if all players do, in fact, meet all the other criteria. Players wrongly distrusting each other's rationality may adopt counter-strategies to expected irrational play on their opponents’ behalf. This is a major consideration in "chicken" or an arms race, for example.



In his Ph.D. dissertation, John Nash proposed two interpretations of his equilibrium concept, with the objective of showing how equilibrium points

约翰 · 纳什在他的博士论文中,为了描述平衡点是怎么达到的,对他的平衡概念给出了两种解释

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A second interpretation, that Nash referred to by the mass action interpretation, is less demanding on players:


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For a formal result along these lines, see Kuhn, H. and et al., 1996, "The Work of John Nash in Game Theory," Journal of Economic Theory, 69, 153–185.

关于这些方面的正式结果,参见 Kuhn, H. and et al., 1996, "The Work of John Nash in Game Theory," Journal of Economic Theory, 69, 153–185.

Due to the limited conditions in which NE can actually be observed, they are rarely treated as a guide to day-to-day behaviour, or observed in practice in human negotiations. However, as a theoretical concept in economics and evolutionary biology, the NE has explanatory power. The payoff in economics is utility (or sometimes money), and in evolutionary biology is gene transmission; both are the fundamental bottom line of survival. Researchers who apply games theory in these fields claim that strategies failing to maximize these for whatever reason will be competed out of the market or environment, which are ascribed the ability to test all strategies. This conclusion is drawn from the "stability" theory above. In these situations the assumption that the strategy observed is actually a NE has often been borne out by research.[20]

由于能够实际观察到纳什均衡的条件有限,因此它很少被作为日常行为的指南,或在人类谈判实践中被观察到。然而,作为经济学和进化生物学中的一个理论上的概念,纳什均衡仍具有解释力。经济学的收益是效用(有时是金钱) ,而进化生物学的收益是基因传递:两者都是生存的基本底线。将博弈论应用于这些领域的研究人员断言,无论出于什么原因,由于市场或环境具有测试所有策略的能力,未能实现效用最大化的那些策略都会在竞争中被市场或环境排除在外。这个结论是基于如上的“稳定性”理论得出的。在这些情况下的研究中,经常可以证实所观察到的策略实际上是一个纳什均衡[20]


Extensive and Normal form illustrations that show the difference between SPNE and other NE. The blue equilibrium is not subgame perfect because player two makes a non-credible threat at 2(2) to be unkind (U).

大量标准形式的图例显示了 SPNE 和其他 NE 的区别。蓝色均衡不是子博弈完美的,因为局中人二在2(2)时做出不可信的威胁是不友善的(u)。

文件:SGPNEandPlainNE explainingexample.svg
Extensive and Normal form illustrations that show the difference between SPNE and other NE. The blue equilibrium is not subgame perfect because player two makes a non-credible threat at 2(2) to be unkind (U).

The Nash equilibrium is a superset of the subgame perfect Nash equilibrium. The subgame perfect equilibrium in addition to the Nash equilibrium requires that the strategy also is a Nash equilibrium in every subgame of that game. This eliminates all non-credible threats, that is, strategies that contain non-rational moves in order to make the counter-player change their strategy.


The image to the right shows a simple sequential game that illustrates the issue with subgame imperfect Nash equilibria. In this game player one chooses left(L) or right(R), which is followed by player two being called upon to be kind (K) or unkind (U) to player one, However, player two only stands to gain from being unkind if player one goes left. If player one goes right the rational player two would de facto be kind to her/him in that subgame. However, The non-credible threat of being unkind at 2(2) is still part of the blue (L, (U,U)) Nash equilibrium. Therefore, if rational behavior can be expected by both parties the subgame perfect Nash equilibrium may be a more meaningful solution concept when such dynamic inconsistencies arise.

右边的图片展示了一个简单的序贯博弈,说明了子博弈不完美纳什均衡的问题。在这个博弈中,一号局中人选择左(l)或右(r) ,然后二号局中人被要求对一号局中人表示友好(k)或不友好(u) ,然而,如果一号局中人选择左,二号局中人只能从不友好中获益。如果一号局中人选择右,理性的二号局中人在子博弈中实际上对她/他很友好。然而,2(2)这个不可信的不友好的威胁仍然是蓝色(l,(u,u))纳什均衡点的一部分。因此,如果双方都能预期到理性行为,那么当这种动态不一致性出现时,子博弈完美纳什均衡点可能是一个更有意义的解决方案概念。


利用角谷静夫不动点定理(Kakutani fixed-point theorem)的证明

Nash's original proof (in his thesis) used Brouwer's fixed-point theorem (e.g., see below for a variant). We give a simpler proof via the Kakutani fixed-point theorem, following Nash's 1950 paper (he credits David Gale with the observation that such a simplification is possible).

纳什在他的论文中的原始证明使用了布劳威尔不动点定理(Brouwer's fixed-point theorem)。在1950年纳什的论文之后,我们可以通过角谷静夫不动点定理给出一个更简单的证明,纳什本人称赞了大卫·盖尔(David Gale)对这一简化所展现出的洞察力。

To prove the existence of a Nash equilibrium, let [math]\displaystyle{ r_i(\sigma_{-i}) }[/math] be the best response of player i to the strategies of all other players.

为了证明纳什均衡点的存在性,让策略[math]\displaystyle{ r_i(\sigma_{-i}) }[/math]是局中人 i 对其他局中人策略的最佳对策。

[math]\displaystyle{ r_i(\sigma_{-i}) = \mathop{\underset{\sigma_i}{\operatorname{arg\,max}}} u_i (\sigma_i,\sigma_{-i}) }[/math]

Here, [math]\displaystyle{ \sigma \in \Sigma }[/math], where [math]\displaystyle{ \Sigma = \Sigma_i \times \Sigma_{-i} }[/math], is a mixed-strategy profile in the set of all mixed strategies and [math]\displaystyle{ u_i }[/math] is the payoff function for player i. Define a set-valued function [math]\displaystyle{ r\colon \Sigma \rightarrow 2^\Sigma }[/math] such that [math]\displaystyle{ r = r_i(\sigma_{-i})\times r_{-i}(\sigma_{i}) }[/math]. The existence of a Nash equilibrium is equivalent to [math]\displaystyle{ r }[/math] having a fixed point.

这里,[math]\displaystyle{ \sigma \in \Sigma }[/math] ,其中[math]\displaystyle{ \Sigma = \Sigma_i \times \Sigma_{-i} }[/math] ,是所有混合策略集合中的混合策略轮廓,[math]\displaystyle{ u_i }[/math]是局中人 i 的收益函数。定义一个集值函数[math]\displaystyle{ r\colon \Sigma \rightarrow 2^\Sigma }[/math],使得[math]\displaystyle{ r = r_i(\sigma_{-i})\times r_{-i}(\sigma_{i}) }[/math]。纳什均衡点的存在相当于[math]\displaystyle{ r }[/math]有一个不动点。

Kakutani's fixed point theorem guarantees the existence of a fixed point if the following four conditions are satisfied.


[math]\displaystyle{  \Sigma }[/math] is compact, convex, and nonempty.

[math]\displaystyle{ \Sigma }[/math]是紧的,凸的,非空的。

[math]\displaystyle{ r(\sigma) }[/math] is nonempty.

[math]\displaystyle{ r(\sigma) }[/math]是非空的。

[math]\displaystyle{ r(\sigma) }[/math] is upper hemicontinuous

[math]\displaystyle{ r(\sigma) }[/math]是上半连续的

[math]\displaystyle{ r(\sigma) }[/math] is convex.

[math]\displaystyle{ r(\sigma) }[/math]是凸的。

Condition 1. is satisfied from the fact that [math]\displaystyle{ \Sigma }[/math] is a simplex and thus compact. Convexity follows from players' ability to mix strategies. [math]\displaystyle{ \Sigma }[/math] is nonempty as long as players have strategies.

对于条件一,事实上,[math]\displaystyle{ \Sigma }[/math]是一个单形,因此是紧的。而其凸性取决于玩家混合策略的决策能力。只要玩家有策略,[math]\displaystyle{ \Sigma }[/math]就不是空的。

Condition 2. and 3. are satisfied by way of Berge's maximum theorem. Because [math]\displaystyle{ u_i }[/math] is continuous and compact, [math]\displaystyle{ r(\sigma_i) }[/math] is non-empty and upper hemicontinuous.

条件2和3可通过 Berge 最大值定理满足其成立。因为[math]\displaystyle{ u_i }[/math]是连续的,并且也是紧的,从而[math]\displaystyle{ r(\sigma_i) }[/math]是非空的和上半连续的。

Condition 4. is satisfied as a result of mixed strategies. Suppose [math]\displaystyle{ \sigma_i, \sigma'_i \in r(\sigma_{-i}) }[/math], then [math]\displaystyle{ \lambda \sigma_i + (1-\lambda) \sigma'_i \in r(\sigma_{-i}) }[/math]. i.e. if two strategies maximize payoffs, then a mix between the two strategies will yield the same payoff.

由于采用的是混合策略,则满足条件4。假设 [math]\displaystyle{ \sigma_i, \sigma'_i \in r(\sigma_{-i}) }[/math],那么有[math]\displaystyle{ \lambda \sigma_i + (1-\lambda) \sigma'_i \in r(\sigma_{-i}) }[/math]。也就是说,如果两个策略的收益最大化,那么两个策略的混合会产生相同的收益。

Therefore, there exists a fixed point in [math]\displaystyle{ r }[/math] and a Nash equilibrium.[21]

因此,在 [math]\displaystyle{ r }[/math]中存在一个不动点和一个纳什均衡点[21]

When Nash made this point to John von Neumann in 1949, von Neumann famously dismissed it with the words, "That's trivial, you know. That's just a fixed-point theorem." (See Nasar, 1998, p. 94.)

1949年,当纳什向约翰·冯·诺伊曼(John von Neumann)提出这个观点时,冯·诺伊曼用一句名言驳斥了他: “你知道,这是微不足道的。这只是一个不动点定理。”(见 Nasar,1998,p. 94。)

利用布劳威尔不动点定理(Brouwer fixed-point theorem)的替代证明

We have a game [math]\displaystyle{ G=(N,A,u) }[/math] where [math]\displaystyle{ N }[/math] is the number of players and [math]\displaystyle{ A = A_1 \times \cdots \times A_N }[/math] is the action set for the players. All of the action sets [math]\displaystyle{ A_i }[/math] are finite. Let [math]\displaystyle{ \Delta = \Delta_1 \times \cdots \times \Delta_N }[/math] denote the set of mixed strategies for the players. The finiteness of the [math]\displaystyle{ A_i }[/math]s ensures the compactness of [math]\displaystyle{ \Delta }[/math].

对于博弈[math]\displaystyle{ G=(N,A,u) }[/math],其中[math]\displaystyle{ N }[/math] 代表局中人的数目,[math]\displaystyle{ A = A_1 \times \cdots \times A_N }[/math]代表局中人的动作集合。所有的动作集[math]\displaystyle{ A_i }[/math]都是有限的。设 [math]\displaystyle{ \Delta = \Delta_1 \times \cdots \times \Delta_N }[/math]代表局中人的混合策略集合。[math]\displaystyle{ A_i }[/math]的有限性保证了[math]\displaystyle{ \Delta }[/math]的紧性。

We can now define the gain functions. For a mixed strategy [math]\displaystyle{ \sigma \in \Delta }[/math], we let the gain for player [math]\displaystyle{ i }[/math] on action [math]\displaystyle{ a \in A_i }[/math] be

我们现在可以定义增益函数了。对于混合策略 [math]\displaystyle{ \sigma \in \Delta }[/math],我们令局中人[math]\displaystyle{ i }[/math][math]\displaystyle{ a \in A_i }[/math]动作下的增益为

[math]\displaystyle{ \text{Gain}_i(\sigma,a) = \max \{0, u_i(a, \sigma_{-i}) - u_i(\sigma_{i}, \sigma_{-i})\}. }[/math]

The gain function represents the benefit a player gets by unilaterally changing their strategy. We now define [math]\displaystyle{ g = (g_1,\dotsc,g_N) }[/math] where

增益函数代表局中人通过单方面改变策略而获得的收益。我们现在定义了 [math]\displaystyle{ g = (g_1,\dotsc,g_N) }[/math],其中

[math]\displaystyle{ g_i(\sigma)(a) = \sigma_i(a) + \text{Gain}_i(\sigma,a) }[/math]

for [math]\displaystyle{ \sigma \in \Delta, a \in A_i }[/math]. We see that

且有[math]\displaystyle{ \sigma \in \Delta, a \in A_i }[/math],我们看到

[math]\displaystyle{ \sum_{a \in A_i} g_i(\sigma)(a) = \sum_{a \in A_i} \sigma_i(a) + \text{Gain}_i(\sigma,a) = 1 + \sum_{a \in A_i} \text{Gain}_i(\sigma,a) \gt 0. }[/math]

Next we define:


[math]\displaystyle{ \begin{cases} f = (f_1, \cdots, f_N) : \Delta \to \Delta \\ f_i(\sigma)(a) = \frac{g_i(\sigma)(a)}{\sum_{b \in A_i} g_i(\sigma)(b)} & a \in A_i \end{cases} }[/math]

It is easy to see that each [math]\displaystyle{ f_i }[/math] is a valid mixed strategy in [math]\displaystyle{ \Delta_i }[/math]. It is also easy to check that each [math]\displaystyle{ f_i }[/math] is a continuous function of [math]\displaystyle{ \sigma }[/math], and hence [math]\displaystyle{ f }[/math] is a continuous function. As the cross product of a finite number of compact convex sets, [math]\displaystyle{ \Delta }[/math] is also compact and convex. Applying the Brouwer fixed point theorem to [math]\displaystyle{ f }[/math] and [math]\displaystyle{ \Delta }[/math] we conclude that [math]\displaystyle{ f }[/math] has a fixed point in [math]\displaystyle{ \Delta }[/math], call it [math]\displaystyle{ \sigma^* }[/math]. We claim that [math]\displaystyle{ \sigma^* }[/math] is a Nash equilibrium in [math]\displaystyle{ G }[/math]. For this purpose, it suffices to show that

很容易看出,每一个 [math]\displaystyle{ f_i }[/math]都是 [math]\displaystyle{ \Delta_i }[/math]中有效的混合策略。检查每个 [math]\displaystyle{ f_i }[/math][math]\displaystyle{ \sigma }[/math]中的连续函数也很容易,因此 [math]\displaystyle{ f }[/math]是一个连续函数。作为有限个紧凸集的叉积,[math]\displaystyle{ \Delta }[/math] 也是紧的和凸的。对[math]\displaystyle{ f }[/math][math]\displaystyle{ \Delta }[/math]应用布劳威尔不动点定理,我们得出结论: [math]\displaystyle{ f }[/math][math]\displaystyle{ \Delta }[/math]有一个不动点,记为[math]\displaystyle{ \sigma^* }[/math]。我们称 [math]\displaystyle{ \sigma^* }[/math][math]\displaystyle{ G }[/math]中的纳什均衡点。为了达到这个目的,只需要说明

[math]\displaystyle{ \forall i \in \{1, \cdots, N\}, \forall a \in A_i: \quad \text{Gain}_i(\sigma^*,a) = 0. }[/math]

This simply states that each player gains no benefit by unilaterally changing their strategy, which is exactly the necessary condition for a Nash equilibrium.


Now assume that the gains are not all zero. Therefore, [math]\displaystyle{ \exists i \in \{1, \cdots, N\}, }[/math] and [math]\displaystyle{ a \in A_i }[/math] such that [math]\displaystyle{ \text{Gain}_i(\sigma^*, a) \gt 0 }[/math]. Note then that

现在假设收益不全为零。因此,[math]\displaystyle{ \exists i \in \{1, \cdots, N\}, }[/math][math]\displaystyle{ a \in A_i }[/math],使得[math]\displaystyle{ \text{Gain}_i(\sigma^*, a) \gt 0 }[/math]成立。可以注意到

[math]\displaystyle{ \sum_{a \in A_i} g_i(\sigma^*, a) = 1 + \sum_{a \in A_i} \text{Gain}_i(\sigma^*,a) \gt 1. }[/math]


[math]\displaystyle{ C = \sum_{a \in A_i} g_i(\sigma^*, a). }[/math]

Also we shall denote [math]\displaystyle{ \text{Gain}(i,\cdot) }[/math] as the gain vector indexed by actions in [math]\displaystyle{ A_i }[/math]. Since [math]\displaystyle{ \sigma^* }[/math] is the fixed point we have:

此外,我们还可以用[math]\displaystyle{ \text{Gain}(i,\cdot) }[/math] 表示[math]\displaystyle{ A_i }[/math] 中动作对应的增益向量。于是[math]\displaystyle{ \sigma^* }[/math]是我们得出的不动点:

[math]\displaystyle{ \begin{align} :\lt math\gt \begin{align} \sigma^* = f(\sigma^*) &\Rightarrow \sigma^*_i = f_i(\sigma^*) \\ Sigma ^ * = f (sigma ^ *) & right tarrow sigma ^ * i = f _ i (sigma ^ *) \sigma^* = f(\sigma^*) &\Rightarrow \sigma^*_i = f_i(\sigma^*) \\ &\Rightarrow \sigma^*_i = \frac{g_i(\sigma^*)}{\sum_{a \in A_i} g_i(\sigma^*)(a)} \\ [6pt] & right tarrow sigma ^ * i = frac { g _ i (sigma ^ *)}{ sum _ { a in a _ i } g _ i (sigma ^ *)(a)}[6 pt ] &\Rightarrow \sigma^*_i = \frac{g_i(\sigma^*)}{\sum_{a \in A_i} g_i(\sigma^*)(a)} \\ [6pt] &\Rightarrow \sigma^*_i = \frac{1}{C} \left (\sigma^*_i + \text{Gain}_i(\sigma^*,\cdot) \right ) \\ [6pt] & right tarrow sigma ^ * i = frac {1}{ c } left (sigma ^ * i + text { Gain } i (sigma ^ * ,cdot) right)[6 pt ] &\Rightarrow \sigma^*_i = \frac{1}{C} \left (\sigma^*_i + \text{Gain}_i(\sigma^*,\cdot) \right ) \\ [6pt] &\Rightarrow C\sigma^*_i = \sigma^*_i + \text{Gain}_i(\sigma^*,\cdot) \\ & right tarrow c sigma ^ * _ i = sigma ^ * _ i + text { Gain } _ i (sigma ^ * ,cdot) &\Rightarrow C\sigma^*_i = \sigma^*_i + \text{Gain}_i(\sigma^*,\cdot) \\ &\Rightarrow \left(C-1\right)\sigma^*_i = \text{Gain}_i(\sigma^*,\cdot) \\ & right tarrow left (C-1 right) sigma ^ * _ i = text { Gain } _ i (sigma ^ * ,cdot) &\Rightarrow \left(C-1\right)\sigma^*_i = \text{Gain}_i(\sigma^*,\cdot) \\ &\Rightarrow \sigma^*_i = \left(\frac{1}{C-1}\right)\text{Gain}_i(\sigma^*,\cdot). & right tarrow sigma ^ * _ i = left (frac {1}{ C-1} right) text { Gain } _ i (sigma ^ * ,cdot). &\Rightarrow \sigma^*_i = \left(\frac{1}{C-1}\right)\text{Gain}_i(\sigma^*,\cdot). \end{align} }[/math]

Since [math]\displaystyle{ C \gt 1 }[/math] we have that [math]\displaystyle{ \sigma^*_i }[/math] is some positive scaling of the vector [math]\displaystyle{ \text{Gain}_i(\sigma^*,\cdot) }[/math]. Now we claim that

由于[math]\displaystyle{ C \gt 1 }[/math] ,我们知道[math]\displaystyle{ \sigma^*_i }[/math] 是向量[math]\displaystyle{ \text{Gain}_i(\sigma^*,\cdot) }[/math]的某种正比例缩放。现在我们可以断言

[math]\displaystyle{ \forall a \in A_i: \quad \sigma^*_i(a)(u_i(a_i, \sigma^*_{-i}) - u_i(\sigma^*_i, \sigma^*_{-i})) = \sigma^*_i(a)\text{Gain}_i(\sigma^*, a) }[/math]

To see this, we first note that if [math]\displaystyle{ \text{Gain}_i(\sigma^*, a) \gt 0 }[/math] then this is true by definition of the gain function. Now assume that [math]\displaystyle{ \text{Gain}_i(\sigma^*, a) = 0 }[/math]. By our previous statements we have that

为了看到这一点,我们首先注意到,如果 [math]\displaystyle{ \text{Gain}_i(\sigma^*, a) \gt 0 }[/math],那么根据增益函数的定义,这是成立的。现在假设 [math]\displaystyle{ \text{Gain}_i(\sigma^*, a) = 0 }[/math]。通过前面的表述,我们得到了这个式子

[math]\displaystyle{ \sigma^*_i(a) = \left(\frac{1}{C-1}\right)\text{Gain}_i(\sigma^*, a) = 0 }[/math]

and so the left term is zero, giving us that the entire expression is [math]\displaystyle{ 0 }[/math] as needed.

所以左边的项是零,即我们需要整个表达式是[math]\displaystyle{ 0 }[/math]

So we finally have that


[math]\displaystyle{ \begin{align} :\lt math\gt \begin{align} 0 &= u_i(\sigma^*_i, \sigma^*_{-i}) - u_i(\sigma^*_i, \sigma^*_{-i}) \\ 0 & = u _ i (sigma ^ * _ i,sigma ^ * _ {-i })-u _ i (sigma ^ * _ i,sigma ^ * _ {-i }) 0 &= u_i(\sigma^*_i, \sigma^*_{-i}) - u_i(\sigma^*_i, \sigma^*_{-i}) \\ &= \left(\sum_{a \in A_i} \sigma^*_i(a)u_i(a_i, \sigma^*_{-i})\right) - u_i(\sigma^*_i, \sigma^*_{-i}) \\ & = left (sum _ { a in a _ i } sigma ^ * _ i (a _ i,sigma ^ * _ {-i }) right)-u _ i (sigma ^ * _ i,sigma ^ * _ {-i }) &= \left(\sum_{a \in A_i} \sigma^*_i(a)u_i(a_i, \sigma^*_{-i})\right) - u_i(\sigma^*_i, \sigma^*_{-i}) \\ & = \sum_{a \in A_i} \sigma^*_i(a) (u_i(a_i, \sigma^*_{-i}) - u_i(\sigma^*_i, \sigma^*_{-i})) \\ & = sum _ { a in a _ i } sigma ^ * _ i (u _ i (a _ i,sigma ^ * _ {-i })-u _ i (sigma ^ * _ i,sigma ^ * _ {-i })) & = \sum_{a \in A_i} \sigma^*_i(a) (u_i(a_i, \sigma^*_{-i}) - u_i(\sigma^*_i, \sigma^*_{-i})) \\ & = \sum_{a \in A_i} \sigma^*_i(a) \text{Gain}_i(\sigma^*, a) && \text{ by the previous statements } \\ 和 = sum _ { a in a _ i } sigma ^ * _ i (a) text { Gain } _ i (sigma ^ * ,a) & text { by the previous statements } & = \sum_{a \in A_i} \sigma^*_i(a) \text{Gain}_i(\sigma^*, a) && \text{ by the previous statements } \\ &= \sum_{a \in A_i} \left( C -1 \right) \sigma^*_i(a)^2 \gt 0 & = sum _ { a in a _ i } left (c-1 right) sigma ^ * _ i (a) ^ 2 \gt 0 &= \sum_{a \in A_i} \left( C -1 \right) \sigma^*_i(a)^2 \gt 0 \end{align} }[/math]

where the last inequality follows since [math]\displaystyle{ \sigma^*_i }[/math] is a non-zero vector. But this is a clear contradiction, so all the gains must indeed be zero. Therefore, [math]\displaystyle{ \sigma^* }[/math] is a Nash equilibrium for [math]\displaystyle{ G }[/math] as needed.

从最后一个不等式可以看出,[math]\displaystyle{ \sigma^*_i }[/math]是一个非零向量。但这是一个明显的矛盾,因此所有的增益必然是零。因此,可以说,[math]\displaystyle{ \sigma^* }[/math][math]\displaystyle{ G }[/math] 的纳什均衡点。


If a player A has a dominant strategy [math]\displaystyle{ s_A }[/math] then there exists a Nash equilibrium in which A plays [math]\displaystyle{ s_A }[/math]. In the case of two players A and B, there exists a Nash equilibrium in which A plays [math]\displaystyle{ s_A }[/math] and B plays a best response to [math]\displaystyle{ s_A }[/math]. If [math]\displaystyle{ s_A }[/math] is a strictly dominant strategy, A plays [math]\displaystyle{ s_A }[/math] in all Nash equilibria. If both A and B have strictly dominant strategies, there exists a unique Nash equilibrium in which each plays their strictly dominant strategy.

如果局中人A有一个占优策略[math]\displaystyle{ s_A }[/math] ,那么存在一个纳什均衡,在这个均衡中A选择策略[math]\displaystyle{ s_A }[/math]。在两个博弈者A和B的情况下,存在一个纳什均衡,其中A选择策略[math]\displaystyle{ s_A }[/math],而B有对[math]\displaystyle{ s_A }[/math]的最佳对策。如果 [math]\displaystyle{ s_A }[/math]是严格占优策略,则A在所有的纳什均衡中都会采用策略[math]\displaystyle{ s_A }[/math] 。如果A和B都有严格的占优策略,则存在一个唯一的纳什均衡,在这个均衡中,每个人都发挥各自的严格占优策略。

In games with mixed-strategy Nash equilibria, the probability of a player choosing any particular (so pure) strategy can be computed by assigning a variable to each strategy that represents a fixed probability for choosing that strategy. In order for a player to be willing to randomize, their expected payoff for each (pure) strategy should be the same. In addition, the sum of the probabilities for each strategy of a particular player should be 1. This creates a system of equations from which the probabilities of choosing each strategy can be derived.



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Matching pennies
Matching penny
模板:Diagonal split header Player B plays H 局中人 B 采取 策略 H Player B plays H Player B plays T 局中人 B 采取 策略 T Player B plays T
Player A plays H 局中人 A 采取 策略 H Player A plays H −1, +1 −1, +1 −1, +1 +1, −1 +1, −1 +1, −1
Player A plays T 局中人 A 采取 策略 T Player A plays T +1, −1 +1, −1 +1, −1 −1, +1 −1, +1 −1, +1

In the matching pennies game, player A loses a point to B if A and B play the same strategy and wins a point from B if they play different strategies. To compute the mixed-strategy Nash equilibrium, assign A the probability p of playing H and (1−p) of playing T, and assign B the probability q of playing H and (1−q) of playing T.

在匹配便士游戏中,如果A和B采用相同的策略,玩家 A会输给 B 1分,如果A和B采用不同的策略,玩家A会从B那里赢1分。为了计算混合策略纳什均衡点,记 A 采取 策略T 的概率为p,采取策略H的概率为(1-p),记 B 采取 策略T 的概率为q,采取策略H的概率为(1-q)。

E[payoff for A playing H] = (−1)q + (+1)(1−q) = 1−2q

E [ A 采取 H 的收益] = (- 1) q + (+ 1)(1-q) = 1-2q

E[payoff for A playing T] = (+1)q + (−1)(1−q) = 2q−1

E [ A 采取 T 的收益] = (+ 1) q + (- 1)(1-q) = 2q-1

E[payoff for A playing H] = E[payoff for A playing T] ⇒ 1−2q = 2q−1 ⇒ q = 1/2

E [ A 采取 H 的收益] = E[ A 采取 T 的收益]⇒ 1−2q = 2q−1 ⇒ q = 1/2

E[payoff for B playing H] = (+1)p + (−1)(1−p) = 2p−1

E [ B 采取 H 的收益] = (+ 1) p + (- 1)(1-p) = 2p-1

E[payoff for B playing T] = (−1)p + (+1)(1−p) = 1−2p

E [ B 参与 T 的收益] = (- 1) p + (+ 1)(1-p) = 1-2p

E[payoff for B playing H] = E[payoff for B playing T] ⇒ 2p−1 = 1−2p ⇒ p = 1/2

E [ B 采取 H 的收益] = E [ B 参与 T 的收益] = ⇒ 2p−1 = 1−2p ⇒ p = 1/2

Thus a mixed-strategy Nash equilibrium, in this game, is for each player to randomly choose H or T with p = 1/2 and q = 1/2.

因此在这个博弈中,混合策略的纳什均衡点是让每个参与者随机选择 p = 1/2和 q = 1/2的 H 或 T 。



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模板:Game theory

Category:Game theory equilibrium concepts

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Category:1951 in economics

分类: 1951年经济学

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