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Must an isolated completely mixed Nash equilibrium (i.e., all strategies for all players receive positive weight) be essential?

(By essential, I mean the equilibrium z of the game G that for every eps>0 there is delta>0 s.t. if the payoffs of the game are perturbed at most delta to G', then G' has an equilibrium in an eps-neighborhood of z.)

If the answer is no in general - what if the players are binary (i.e., have two actions each)?

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The answer is negative: there is a 3-player binary game with a nonessential isolated completely mixed equilibrium. Here it is. Let $G(e)$ be the following game, whose payoff function depends on a parameter $e$ (only the payoffs of player 3 are affected):

\begin{array}{|c|c|c|c|c|} 1,1,1+e & -5,0,3+e & \ \ \ \ \ \ \ \ \ \ \ \ \ \ & 3,-5,0 & 1,0,0\\ 0,3,-5+e & 0,0,1+e & \ \ \ \ \ & 0,1,0 & 0,0,0 \end{array}

It requires some calculations to show that for $e=0$, the unique equilibrium is $(1/2,1/2,1/2)$. Few more calculations show that for $e > 0$, there is no completely mixed equilibrium.

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