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Nash Equilibrium
A solution concept in game theory
Relationships
Subset of: Rationalizability, Correlated equilibrium
Superset of: Evolutionary stable strategy, Subgame perfect equilibrium, Perfect Bayesian equilibrium, Trembling hand perfect equilibrium
Significance
Proposed by: John Forbes Nash
Used for: All non-cooperative games
Example: Prisoner's dilemma

In game theory, the Nash equilibrium (named after John Forbes Nash, who proposed it) is a kind of solution concept of a game involving two or more players, where no player has anything to gain by changing only his or her own strategy unilaterally. If each player has chosen a strategy and no player can benefit by changing his or her strategy while the other players keep theirs unchanged, then the current set of strategy choices and the corresponding payoffs constitute a Nash equilibrium.

The concept of the Nash equilibrium (NE) is not exactly original to Nash (e.g., Antoine Augustin Cournot showed how to find what we now call the Nash equilibrium of the Cournot duopoly game). Consequently, some authors refer to it as a “Cournot-Nash equilibrium” (or as a “Nash-Cournot equilibrium”). However, Nash showed for the first time in his dissertation, Non-cooperative games (1950), that Nash equilibria must exist for all finite games with any number of players. Until Nash, this had only been proven for 2-player zero-sum games by John von Neumann and Oskar Morgenstern (1947).

Formal definition Edit

Let (S, f) be a game, where S is the set of strategy profiles and f is the set of payoff profiles. Let \sigma_{-i} be a strategy profile of all players except for player i. When each player i \in \{1, ..., n\} chooses strategy x_i resulting in strategy profile x = (x_1, ..., x_n) then player i obtains payoff f_i(x). Note that the payoff depends on the strategy profile chosen, i.e. on the strategy chosen by player i as well as the strategies chosen by all the other players. A strategy profile x^* \in S is a Nash equilibrium (NE) if no deviation in strategy by any single player is profitable, that is, if for all i

f_i(x^*_{i}, x^*_{-i}) \geq f_i(x_{i},x^*_{-i}) \forall i

A game can have a pure strategy NE or an NE in its mixed extension (that of choosing a pure strategy stochastically with a fixed frequency). Nash proved that, if we allow mixed strategies (players choose strategies randomly according to pre-assigned probabilities), then every n-player game in which every player can choose from finitely many strategies admits at least one Nash equilibrium.

Proof sketchEdit

As above, let \sigma_{-i} be a mixed strategy profile of all players except for player i. We can define a best response correspondence for player i, b_i. b_i is a relation from the set of all probability distributions over opponent player profiles to a set of player i's strategies, such that each element of

b_i(\sigma_{-i})

is a best response to \sigma_{-i}. Define

b(\sigma) = b_1(\sigma_{-1}) \times b_2(\sigma_{-2}) \times \cdots \times b_n(\sigma_{-n}).

One can use the Kakutani fixed point theorem to prove that b has a fixed point. That is, there is a \sigma^* such that \sigma^* \in b(\sigma^*). Since b(\sigma^*) represents the best response for all players to \sigma^*, the existence of the fixed point proves that there is some strategy set which is a best response to itself. No player could do any better by deviating, and it is therefore a Nash equilibrium.

Examples Edit

Competition game Edit

A competition game
Player 2 chooses '0' Player 2 chooses '1' Player 2 chooses '2' Player 2 chooses '3'
Player 1 chooses '0' 0, 0 2, -2 2, -2 2, -2
Player 1 chooses '1' -2, 2 1, 1 3, -1 3, -1
Player 1 chooses '2' -2, 2 -1, 3 2, 2 4, 0
Player 1 chooses '3' -2, 2 -1, 3 0, 4 3, 3

Consider the following two-player game: both players simultaneously choose a whole number from 0 to 3. Both players then win the smaller of the two numbers in points. In addition, if one player chooses a larger number than the other, then s/he has to give up two points to the other. This game has a unique Nash equilibrium: both players choosing 0. Any other choice of strategies can be improved if one of the players lowers his number to one less than the other player's number. In the table to the left, for example, when starting at the green square it is in player 1's interest to move to the purple square by choosing a smaller number, and it is in player 2's interest to move to the blue square by choosing a smaller number. 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 3 Nash equilibria.

Coordination game Edit

Main article: Coordination game
A coordination game
Player 2 adopts strategy 1 Player 2 adopts strategy 2
Player 1 adopts strategy 1 A, A B, C
Player 1 adopts strategy 2 C, B D, D

The coordination game is a classic (symmetric) two player, two strategy game, with the payoff matrix shown to the right, where the payoffs are according to A>C and D>B. The players should thus cooperate on either of the two strategies to receive a high payoff. Players in the game have to agree on one of the two strategies in order to receive a high payoff. If the players do not agree, a lower payoff is rewarded. An example of a coordination game is the setting where two technologies are available to two firms with compatible 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, and having to choose either to drive on the left or to drive on the right of the road, is also a coordination game. For example, with payoffs 100 meaning no crash and 0 meaning a crash, the coordination game can be defined with the following payoff matrix:

The driving game
Drive on the Left Drive on the Right
Drive on the Left 100, 100 0, 0
Drive on the Right 0, 0 100, 100

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 is (50%, 50%).

Prisoner's dilemma Edit

Main article: Prisoner's dilemma (but watch out for differences in the orientation of the payoff matrix)

The Prisoner's Dilemma has the same payoff matrix as depicted for the Coordination Game, but now C > A > D > B. Because C > A and D > B, each player improves his situation by switching from strategy #1 to strategy #2, no matter what the other player decides. The Prisoner's Dilemma thus has a single Nash Equilibrium: both players choosing strategy #2 ("betraying"). What has long made this an interesting case to study is the fact that D < A ("both betray" is globally inferior to "both remain loyal"). The globally optimal strategy is unstable; it is not an equilibrium.

As Ian Stewart put it, "sometimes rational decisions aren't sensible!"

Nash equilibria in a payoff matrix Edit

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 duplet 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.

We can apply this rule to a 3x3 matrix:

A Payoff Matrix
Option A Option B Option C
Option A 0, 0 25, 40 5, 10
Option B 40, 25 0, 0 5, 15
Option C 10, 5 15, 5 10, 10

Using the rule, we can very quickly (much faster than with formal analysis) see that the Nash Equlibria 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.

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 tuple has maximum of the row. If yes - you've got a Nash Equilibrium. Check all columns this way to find all NE cells. An NxN matrix may have between 0 and N pure strategy Nash equilibria.

Stability Edit

The concept of stability, useful in the analysis of many kinds of equilibrium, 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:

  1. the player who did not change has no better strategy in the new circumstance
  2. 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 his 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. John Nash showed that the latter situation could not arise in a range of well-defined games.

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 his probabilities slightly, they will be both at a disadvantage, and his opponent will have no reason to change his strategy in turn. The (50%,50%) equilibrium is instability. If either player changes his probabilities, then the other player immediately has a better strategy at either (0%, 100%) or (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 his 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.

Note that stability of the equilibrium is related to, but distinct from, stability of a strategy.

Occurrence Edit

If a game has a unique Nash equilibrium and is played among players with certain characteristics, then it is true (by definition of these characteristics) that the NE strategy set will be adopted. Sufficient conditions to be met by the players are:

  1. The players all will do their utmost to maximize their expected payoff as described by the game.
  2. The players are flawless in execution.
  3. The players have sufficient intelligence to deduce the solution.
  4. 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.

Where the conditions are not met Edit

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

  1. 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.
  2. Pong has an equilibrium which can be played perfectly by a computer, but to make human vs. computer games interesting the programmers add small errors in execution, violating the second condition.
  3. 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[1]. 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).
  4. The fourth 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.

Where the conditions are met Edit

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 money, and in evolutionary biology gene transmission, both are the fundamental bottom line of survival. Agents 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.[verification needed]

See also Edit

References Edit

  • Fudenberg, Drew and Jean Tirole (1991) Game Theory MIT Press.
  • Mehlmann, A. The Game's Afoot! Game Theory in Myth and Paradox, American Mathematical Society (2000).
  • Morgenstern, Oskar and John von Neumann (1947) The Theory of Games and Economic Behavior Princeton University Press
  • Nash, John (1950) "Equilibrium points in n-person games" Proceedings of the National Academy of the USA 36(1):48-49.
  • Nash, John (1951) "Non-Cooperative Games" The Annals of Mathematics 54(2):286-295.

Notes Edit

  1. Nash has proven that a perfect NE exists for this type of finite extensive form game – it can be represented as a strategy complying with his original conditions for a game with a NE. Such games may not have unique NE, but at least one of the many equilibrium strategies would be played by hypothetical players having perfect knowledge of all 10150 game trees.


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