# Epsilon-equilibrium

Epsilon-equilibrium
Epsilon-equilibrium
A solution concept in game theory
Relationships
Superset of Nash Equilibrium
Significance
Used for stochastic games

In game theory, an epsilon-equilibrium, or near-Nash equilibrium, is a strategy profile that approximately satisfies the condition of Nash equilibrium.

## Definition

Given a game and a real non-negative parameter ε, a strategy profile is said to be an ε-equilibrium if it is not possible for any player to gain more than ε in expected payoff by unilaterally deviating from his strategy. Every Nash Equilibrium is equivalent to a ε-equilibrium where ε = 0.

Formally, let $G=(N,A=A_1\times\cdots\times A_N, u: A \rightarrow \reals^N)$ be a N-player game with action sets Ai for each player and utility function u. A vector of strategies $\sigma \in \Delta = \Delta_1 \times \cdots \times \Delta_N$ is an $\epsilon$-Nash Equilibrium for G if

$u_i(\sigma)\geq u_i(\sigma_i^',\sigma_{-i})-\epsilon$ for all $\ \sigma_i^' \in \Delta_i, i \in N$

## Example

The notion of ε-equilibria is important in the theory of stochastic games of potentially infinite duration. There are simple examples of stochastic games with no Nash equilibrium but with an ε-equilibrium for any ε strictly bigger than 0.

Perhaps the simplest such example is the following variant of Matching Pennies, suggested by Everett. Player 1 hides a penny and Player 2 must guess if it is heads up or tails up. If Player 2 guesses correctly, he wins the penny from Player 1 and the game ends. If Player 2 incorrectly guesses that the penny is heads up, the game ends with payoff zero to both players. If he incorrectly guesses that it is tails up, the game repeats. If the play continues forever, the payoff to both players is zero.

Given a parameter ε > 0, any strategy profile where Player 2 guesses heads up with probability ε and tails up with probability 1-ε (at every stage of the game, and independently from previous stages) is an ε-equilibrium for the game. The expected payoff of Player 2 in such a strategy profile is at least 1-ε. However, it is easy to see that there is no strategy for Player 2 that can guarantee an expected payoff of exactly 1. Therefore, the game has no Nash equilibrium.

Another simple example is the finitely repeated prisoner's dilemma for T periods, where the payoff is averaged over the T periods. The only Nash equilibrium of this game is to choose Defect in each period. Now consider the two strategies tit-for-tat and grim trigger. Although neither tit-for-tat nor grim trigger are Nash equilibria for the game, both of them are $\epsilon$-equilibria for some positive $\epsilon$. The acceptable values of $\epsilon$ depend on the payoffs of the consituent game and on the number T of periods.

## References

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