Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Most references seem to state that Nash showed every symmetric game has a symmetric equilibrium point, but as far as I can tell from Nash's paper, he actually showed the much more general statement that every finite game has a symmetric equilibrium point.

I realise the references wouldn't be wrong, but why don't they state the more general result?

share|improve this question
    
As far as I remember, Nash's paper only talks about equilibrium, there is no reference to any symmetry whatsoever. Sorry if I am wrong. –  GH from MO Apr 14 at 16:13
4  
To be clear, I'm referring to his PhD thesis, a copy of which is located at princeton.edu/mudd/news/faq/topics/… –  Nick Ham Apr 14 at 16:15
    
Thanks. I did not read his Ph.D. thesis. –  GH from MO Apr 14 at 16:16
1  
Hi Nick! Asymmetric finite games need not have symmetric equilibria -- the players need not even have e.g. the same number of strategies. Did you ask what you meant to ask? –  Noah Stein Apr 14 at 16:43
1  
@GHfromMO There are two papers. Equilibrium Points in n-Person Games from 1950 only contained a proof that an equilibrium exists. Non-Cooperative Games from 1951 (essentially the published thesis) contains a proof that every game has a symmetric equilibrium. –  Michael Greinecker Apr 14 at 17:01

2 Answers 2

Nash defined symmetries of finite games and proved existence of an equilibrium point that is invariant under all symmetries. He called such an equilibrium a symmetric equilibrium. For this to make a difference, a game needs to have nontrivial symmetries. Generic games don't.

A clearer source for his result would be the published version in the Annals of Mathematics. Theorem 2 there says: Any finite game has a symmetric equilibrium point.

share|improve this answer
    
> For this to make a difference, a game needs to have nontrivial symmetries. Generic games don't. >Theorem 2 there says: Any finite game has a symmetric equilibrium point. –  Nick Ham Apr 14 at 20:52
    
Don't those two statements contradict each other? –  Nick Ham Apr 14 at 20:54
2  
@NickHam No. A symmetric equilibrium is an equilibrium invariant under all symmetries of the game. In a generic game, the only such symmetry is the identity (there is a open dense, full measure set of games with no nontrivial symmetry). Nash did not define symmetric games. –  Michael Greinecker Apr 14 at 20:58

I'd like to complete the answer by giving reference to this work by Fey which shows that in infinite symmetric games, symmetric equilibrium doesn't always exist.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.