<p>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 $(x')$. </p>

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

<p>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 $0$.</p>

<p>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).</p>

<p>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.</p>

CC BY-SA 3.0
2018-04-20T09:35:19.847Z
2018-04-20T09:35:19.847Z
Michael Greinecker
<p>In <a href="https://economics.stackexchange.com/a/21582/1601">this answer</a> there is the offhand remark</p>

<blockquote>
  <p>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.</p>
</blockquote>

<p>Can anyone provide an example of such a game?</p>

<p>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.</p>

CC BY-SA 3.0
2018-04-20T08:11:31.400Z
Ha, I was wondering whether I should post exactly this question. You beat me to the punch.
Theoretical Economist
<game-theory><nash-equilibrium>
Example of a game with no Nash equilibria but at least one correlated equilibrium
Giskard
# Example of a game with no Nash equilibria but at least one correlated equilibrium

(Asked by: username_0 on 2018-04-20)

In [this answer](https://economics.stackexchange.com/a/21582/1601) 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.

**Tags:** game-theory, nash-equilibrium
> **Comments:**
> **username_1**: Ha, I was wondering whether I should post exactly this question. You beat me to the punch.  

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## 1 answers

### Answer by username_2 on 2018-04-20. Score: 6 | Accepted answer

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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