I'm reading Computing Correlated Equilibria in Multi-Player Games (C.H. Papadimitriou, T. Roughgarden). Consider the following extracts
While the following extract is from here, it is also described in the paper mentioned above. It describes the constraints a distribution must satisfy to be a Correlated Equilibrium (CE).
I've written the constraints described above in full for the given game of chicken.
$\begin{align} -4 & x_A + x_B \geq 0 \\ 4 & x_C - x_D \geq 0 \\ -4 & x_A + x_C \geq 0 \\ 4 & x_B - x_D \geq 0 \\ & x_A + x_B + x_C + x_D = 1 \\ & x_A, x_B, x_C, x_D \geq 0 \end{align}$
I've named the different outcomes $A, B, C, D$ from left to right and from up to down. So $(stop, stop) = A; (stop, go) = B; (go, stop) = C; (go, go) = D$
I've tried to verify that the given CE are indeed CE for the given game (i.e. they satisfy the constraints).
$(x_A, x_B, x_C, x_D) = (0,1,0,0)$ is indeed a CE because it satisfies the constraints above. $(x_A, x_B, x_C, x_D) = (0,0,1,0)$ is indeed a CE because it satisfies the constraints above. $(x_A, x_B, x_C, x_D) = (0,\frac{1}{2},\frac{1}{2},0)$ is indeed a CE because it satisfies the constraints above. $(x_A, x_B, x_C, x_D) = (\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{4})$ is not a CE because it doesn't satisfy all constraints above. $(x_A, x_B, x_C, x_D) = (\frac{1}{3},\frac{1}{3},\frac{1}{3},0)$ is not a CE because it doesn't satisfy all constraints above.
As you can see, I fail to verify that the CE given in the paper are indeed CE for the given game of chicken. Moreover, I can find new CE not mentioned, such as $(x_A, x_B, x_C, x_D) = (0,\frac{1}{10},\frac{9}{10},0)$. This last distribution also satisfies the constraints. I'm not sure if this is indeed a CE or not, since the authors do not explicitly state that the given CE are the only CE of the game.
Obviously I'm missing something. I must've misinterpreted the constraints, but I cannot see how.
I use your notation and the notation from the paper you cited. Let's write the constraints for Player I (rows): \begin{align}\left(u^1(A)-u^1(C)\right)\cdot x_A+\left(u^1(B)-u^1(D)\right)\cdot x_B\ge 0 \\ \left(u^1(C)-u^1(A)\right)\cdot x_C+\left(u^1(D)-u^1(B)\right)\cdot x_D\ge 0\end{align}
The first corresponds to the recommendation for Player I (rows) to play "stop" and the second to play "go". Substituting his payoffs we obtain
$$\begin{align}-x_A+ x_B\ge 0 \implies x_B\ge x_A\\ x_C-x_D\ge 0 \implies x_C\ge x_D\end{align} \tag{*}$$
Similarly for Player II (columns) the constraints are \begin{align}\left(u^2(A)-u^2(B)\right)\cdot x_A+\left(u^2(C)-u^2(D)\right)\cdot x_C\ge 0 \\ \left(u^2(B)-u^2(A)\right)\cdot x_B+\left(u^2(D)-u^2(C)\right)\cdot x_D\ge 0\end{align}
The first corresponds to the recommendation for Player II (columns) to play "stop" and the second to play "go". Substituting his payoffs we obtain
$$\begin{align}-x_A+ x_C\ge 0 \implies x_C\ge x_A\\ x_B-x_D\ge 0 \implies x_B\ge x_D\end{align} \tag{**}$$ Bringing $(*)$ and $(**)$ together and taking into account that $x_i$ should be a probability distribution we obtain the set of constraints that a correlated has to satisfy: \begin{align}x_C &\ge x_A\\x_B&\ge x_D\\x_B&\ge x_A\\x_C&\ge x_D\\x_A+x_B+x_C+x_D&=1\\x_A,x_B,x_C,x_D&\ge0\end{align}
Now, it is straightforward to verify that all the given distributions are correlated equilibria. Moreover any convex combination of Nash equilibria (or correlated equilibria) is again a correlated equilibria, so the set of correlated equilibria is a convex set. Hence, it is not uncommon that there infinitely many correlated equilibria, as is the case in the game that you are studying. So, no, these (that the authors suggest) are not all equilibria, and yes the one you found is also one. There are many more as well.