Is a steady state necessarily Nash?

Question

A steady state is Nash if it is strict and pure. And there can be multiple steady states, but not all of them satisfy the conditions of a Nash Equilibrium. Is this right?

Answer 1

Not quite. Let's look at the definitions.

1. Steady state: In the context of dynamic systems or games, a steady state refers to a point or state in which the system remains unchanged over time. It is a state where variables or strategies no longer change. In game theory, a steady state can correspond to an equilibrium.

2. Nash Equilibrium: In game theory, a Nash Equilibrium is a steady state where each player's strategy is optimal given the strategies chosen by all other players. In other words, no player has an incentive to unilaterally deviate from their chosen strategy.

Based on these definitions, we can say:

1. A steady state can be Nash if it satisfies the conditions of a Nash Equilibrium, where each player's strategy is optimal given the strategies of others.
2. A Nash Equilibrium can be strict if at least one player's strategy is strictly optimal, meaning no other strategy would yield a higher payoff.
3. A Nash Equilibrium can be pure if each player chooses a single strategy without any randomization or mixing of strategies.

It is possible to have multiple steady states in a game, but not all of them may satisfy the conditions of a Nash Equilibrium. In other words, not all steady states are Nash Equilibria. Only those steady states where players' strategies are mutually optimal and no player has an incentive to deviate form a Nash Equilibrium.

So, to summarize, a steady state can be Nash if it meets the conditions of a Nash Equilibrium, but not all steady states necessarily satisfy these conditions.