Nash existence theorem
Every game with finitely many players and finitely many actions has at least one equilibrium once players are allowed to randomise.
Every game with finitely many players and finitely many actions has at least one equilibrium once players are allowed to randomise.
Often misread asTaking the theorem to promise a unique or sensible prediction, so existence is treated as though it also settled which equilibrium gets played.
How to see it
There is always one, and that is all it promises
Every game with finitely many players and finitely many actions has at least one equilibrium, provided players are allowed to randomise. The proof is a fixed point argument: map each profile of strategies to the best responses it provokes and show that some profile maps to itself.
The checkIt guarantees at least one. It says nothing about how many there are, which one occurs, or whether anybody could compute it.
Try itName the two finiteness conditions, then say what would fail in matching pennies if randomising were banned. Then ask what the theorem tells you about which equilibrium of the stag hunt gets played. The honest answer is nothing at all.
Know these first
When it stops being true
- finite-action-sets
- randomisation-permitted
Easily confused with
Sources
- Nash, J. (1950). PNAS 36(1), 48-49.
- Nash, J. (1951). Annals of Mathematics 54(2), 286-295.
- Osborne, M. & Rubinstein, A. (1994). MIT Press.