# Can Pareto Optimality be compared to Nash Equilibrium?

Given a state $$s$$, and a value function $$v^i$$ that determines the expected payoff for the i-th agent in that state, can the two definitions below, one of Nash equilibrium and another of Pareto optimality be compared to one another?

Nash equilibrium: $$v^i(s, \pi^i_*, \pi^{-i}_*) \ge v^i(s, \pi^i, \pi^{-i}_*)\ \ \ \forall\ \ i \in \{1, 2, \cdots,N\}$$

Pareto optimality: $$v^i(s, \Pi_\#) \gt v^i(s, \Pi)$$ for atleast 1 agent $$i$$, and,

$$v^j(s, \Pi_\#) \ge v^j(s, \Pi)\ \ \ \forall\ \ j \in \{1, 2, \cdots,N\}$$

### Notation Used

Nash Equlibrium = $$(\pi^i_*, \pi^{-i}_*)$$

Pareto Optimal stragey = $$\Pi_\#$$

• I think your notation is somewhat confusing. What is $\Pi, \Pi_\ast, \Pi_\#$? That both concepts are not the same is easy to see, look e.g. at the Prisoner's Dilemma and find the Pareto-optimal point and the Nash equilibrium. However, you theoretical can compare optimal points and equilibria. There is still a lot of research going on that goes under the euphonic name "Price of Anarchy". – ttnick Jan 31 '19 at 10:03
• Hey, @ttnick I have updated the question, please have a look. – sayan Feb 1 '19 at 9:12
• I understand the notation now, but what are you asking? I do not know what to add to my first comment unless, you have a more specific question. – ttnick Feb 3 '19 at 9:44