In this answer there is the offhand remark

Of course, a game with no Nash equilibria may have a correlated equilibrium, but I'm not aware of any simple examples where this is the case.

Can anyone provide an example of such a game?

The game should allow for mixed strategies, otherwise correlation would be meaningless. Because of this finite games are out, since their mixed extensions always have a Nash equilibrium.

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    $\begingroup$ Ha, I was wondering whether I should post exactly this question. You beat me to the punch. $\endgroup$ Apr 20, 2018 at 8:11

1 Answer 1


There are three players, 1,2,3, and the action spaces of players 1 and 2 are both $[0,1]\times\mathbb{N}$, the action space of player 3 is $[0,1]$. The generic action of player 1 is written as $(x,m)$, the generic action of player 2 is written as $(y,n)$, and the generic action of player 3 is written as $(x')$.

The payoff of player 3 is simply $1$ if $x'=x$ and $0$ otherwise.

The payoffs of players 1 and 2 are more complicated. The payoff of player 1 and 2 is both $2$ if $y=x\neq x'$, and both $-2$ if $x=x'$. In the case that $y\neq x\neq x'$, the player with the highest number in the second coordinate gets a payoff of $1$ and the one with the lower number gets $-1$. If they pick the same number, they get both $0$.

This game has no Nash equilibrium. Suppose there were one. If player 1 would not randomize in equilibrium, player 3 can match them for a lousy payoff of $-2$. By randomizing the payoff will be higher, but it will also be the case that with positive probability $y\neq x\neq x'$, since the behavior of all players is stochastically independent. But then both 1 and 2 have to play the game of picking a higher number and that game has no equilibrium even in mixed strategies (if mixed strategies must be countably additive).

But there is a correlated equilibrium in which the behavior of 1 and 2 is given by a distribution on the diagonal $D=\{(x,y)\mid x=y\}$ without mass points, and player 3 plays anything stochastically independent. Player 1 and 2 can clearly do no better and there is no way for 3 to match 1 with positive probability.


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