The Nash equilibrium may sometimes appear non-rational in a third-person perspective. This is because it may happen that a Nash equilibrium is not pareto optimal.
The Nash-equilibrium may also have non-rational consequences in sequential games because players may "threat" each other with non-rational moves. For such games the Subgame perfect Nash equilibrium may be more meaningful as a tool of analysis.
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.
When the inequality above holds strictly (with instead of ) for all players and all feasible alternative strategies, then the equilibrium is classified as a strict Nash equilibrium. If instead, for some player, there is exact equality between and some other strategy in the set , then the equilibrium is classified as a weak Nash equilibrium.
|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|
This can be illustrated by a two-player game in which both players simultaneously choose a whole number 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 he/she has 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 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 4 Nash equilibria (0,0...1,1...2,2...and 3,3).
|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 satisfy A>C and D>B. The players should thus coordinate, either on A or on D, to receive a high payoff. If the players' choices do not coincide, 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:
|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%).
We can apply this rule to a 3x3 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 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 NxN matrix may have between 0 and NxN pure strategy 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:
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 unstable. 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.
A Coalition-Proof Nash Equilibrium (CPNE) (similar to a Strong Nash Equilibrium) occurs when players cannot do better even if they are allowed to communicate and collaborate before the game. Every correlated strategy supported by iterated strict dominance and on the Pareto frontier is a CPNE. 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.
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. Researchers who apply games theory in these fields claim that 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.
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 his 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 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.
One can use the Kakutani fixed point theorem to prove that has a fixed point. That is, there is a such that . Since represents the best response for all players to , 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.
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.)
We have a game where is the number of players and is the action set for the players. All of the actions sets are finite. Let denote the set of mixed strategies for the players. The finiteness of the s insures the compactness of .
We can now define the gain functions. For a mixed strategy , we let the gain for player on action be
The gain function represents the benefit a player gets by unilaterally changing his strategy. We now define where
for . We see that
We now use to define as follows. Let
for . It is easy to see that each is a valid mixed strategy in . It is also easy to check that each is a continuous function of , and hence is a continuous function. Now is the cross product of a finite number of compact convex sets, and so we get that is also compact and convex. Therefore we may apply the Brouwer fixed point theorem to . So has a fixed point in , call it .
I claim that is a Nash Equilibrium in . For this purpose, it suffices to show that
Now assume that the gains are not all zero. Therefore, , , and such that . Note then that
Since we have that is some positive scaling of the vector . Now I claim that
. To see this, we first note that if then this is true by definition of the gain function. Now assume that . By our previous statements we have that
So we finally have that
where the last inequality follows since is a non-zero vector. But this is a clear contradiction, so all the gains must indeed be zero. Therefore is a Nash Equilibrium for as needed.
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