Question

Is every Nash equilibrium of a normal form game Pareto optimal? If yes, provide a proof. Otherwise, provide a counterexample.

Answer 1

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Nash Equilibrium:

Informally, a strategy profile is a Nash equilibrium if no player can do better by unilaterally changing her/his 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?"

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]

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

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.

Strict/Weak Equilibrium[edit]

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?"

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 the exactly same payout (i.e. this player is indifferent between switching and not), then the equilibrium is classified as a weak Nash equilibrium.

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's Existence Theorem[edit]

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).

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 the other 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.

Examples[edit]

Coordination game[edit]

Main article: Coordination game

A sample coordination game showing relative payoff for player 1 (row) / player 2 (column) with each combination

Player 2 Player 1	Player 2 adopts strategy A	Player 2 adopts strategy B
Player 1 adopts strategy A	4 4	3 1
Player 1 adopts strategy B	1 3	2 2

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).

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 payload (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.

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:

The driving game

Driver 2 Driver 1	Drive on the Left	Drive on the Right
Drive on the Left	10 10	0 0
Drive on the Right	0 0	10 10

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%).

Prisoner's dilemma[edit]

Main article: Prisoner's dilemma

Example PD payoff matrix

Prisoner 2 Prisoner 1	Cooperate (with other)	Defect (betray other)
Cooperate (with other)	−1, −1	−3, 0
Defect (betray other)	0, −3	−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).

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.

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.

Network traffic[edit]

See also: Braess's paradox

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.

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 x "cars" traveling from A to D, what is the expected distribution of traffic in the network?

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 ABD, ABCD, or ACD). The "payoff" of each strategy is the travel time of each route. In the graph on the right, a car travelling via ABD experiences travel time of (1+x/100)+2, where x is the number of cars traveling on edge AB. 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 ABD, 50 via ABCD, and 25 via ACD. Every driver now has a total travel time of 3.75 (to see this, note that a total of 75 cars take the AB edge, and likewise, 75 cars take the CD edge).

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.

Competition game[edit]

A competition game

Player 2 Player 1	Choose '0'	Choose '1'	Choose '2'	Choose '3'
Choose '0'	0, 0	2, −2	2, −2	2, −2
Choose '1'	−2, 2	1, 1	3, −1	3, −1
Choose '2'	−2, 2	−1, 3	2, 2	4, 0
Choose '3'	−2, 2	−1, 3	0, 4	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).

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 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.

A payoff matrix – Nash equilibria in bold

Player 2 Player 1	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

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.

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.

Stability[edit]

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:

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 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%).

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.

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.

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.

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. 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).
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.^[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).
4. 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.

Where the conditions are met[edit]

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

(...) can be connected with observable phenomenon. One interpretation is rationalistic: if we assume that players are rational, know the full structure of the game, the game is played just once, and there is just one Nash equilibrium, then players will play according to that equilibrium. This idea was formalized by Aumann, R. and A. Brandenburger, 1995, Epistemic Conditions for Nash Equilibrium, Econometrica, 63, 1161-1180 who interpreted each player's mixed strategy as a conjecture about the behaviour of other players and have shown that if the game and the rationality of players is mutually known and these conjectures are commonly know, then the conjectures must be a Nash equilibrium (a common prior assumption is needed for this result in general, but not in the case of two players. In this case, the conjectures need only be mutually known).

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

[i]t is unnecessary to assume that the participants have full knowledge of the total structure of the game, or the ability and inclination to go through any complex reasoning processes. What is assumed is that there is a population of participants for each position in the game, which will be played throughout time by participants drawn at random from the different populations. If there is a stable average frequency with which each pure strategy is employed by the average member of the appropriate population, then this stable average frequency constitutes a mixed strategy Nash equilibrium.

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.

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]