Apologies if there is anything wrong with my formatting, this is my first post.
I'm trying to use the linprog function in MatLab to calculate the correlated equilibrium in a normal form game represented by this $3\times 3$ matrix:
$$ \begin{array}{|c|c|c|c|} \hline & \mathrm{A} & \mathrm{B} & \mathrm{C} \\ \hline \mathrm{X} & 10,5 & 0,6 & 0,0 \\ \hline \mathrm{Y} & 7,7 & 0,0 & 6,0 \\ \hline \mathrm{Z} & 0,0 & 7,7 & 5,10 \\ \hline \end{array} $$
I know how to do this in a $2\times 2$ game, however in a $3\times 3$ I'm stumped. In a $2\times 2$ the number of coefficients match the inequalities used for the linprog function, whereas when calculating inequalities for a $3\times 3$ (unless I'm doing it wrong) you end up with $12$ inequalities ($2$ for each row/column) but $9$ coefficients.
Apologies if this is a silly question, I'm new to game theory and MatLab and I'm struggling to wrap my head around some of the concepts. I'd really appreciate some help.
For a simpler notation lets number the strategies. Let for Player 1 $X=s^1_1$, $Y=s^1_2$, and $Z=s^1_3$. Similarly for Player 2 $A=s^2_1$, $B=s^2_2$, and $C=s^2_3$. So you are looking for vector $\mu=(\mu(s^1_1,s^2_1),\ldots, \mu(s^1_3,s^1_3))$, such that $\sum_{i=1}^3\sum_{j=1}^3\mu(s^1_i,s^2_j)=1$, $\mu(s^1_i,s^2_j)\geq0$ for all $i,j$, and $A\mu\geq0$, for a 12x9 Matrix given by the incentive constraints. Or equivalently, you solve the linear problem $$\max \sum_{i=1}^3\sum_{j=1}^3\mu(s^1_i,s^2_j)$$ $$\text{s.t. }A\mu\geq0$$ $$\sum_{i=1}^3\sum_{j=1}^3\mu(s^1_i,s^2_j)\leq1$$ $$\mu(s^1_i,s^2_j)\geq0 \text{ for all } i,j\in\{1,2,3\}$$. Why is this equivalent? As any finite game has a mixed strategy equilibrium, we know that there exists a solution of the program with $\sum_{i=1}^3\sum_{j=1}^3\mu(s^1_i,s^2_j)=1$. So any solution has the value 1. I suppose your problem is with coming up with the matrix $A$. However, it is easy once you write down the incentive constraints, subtract the right-hand side from the left-hand side, and organize them by the variables $\mu(s^1_i,s^2_j)$. This should give you the following matrix. $$A= \begin{bmatrix} 3 & 0 & -6 & 0 & 0 & 0 & 0 & 0 & 0 \\ 10 & -7 & -5 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -3 & 0 & 6 & 0 & 0 & 0 \\ 0 & 0 & 0 & 7 & -7 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & -10 & 7 & 5 \\ 0 & 0 & 0 & 0 & 0 & 0 & -7 & 7 & -1 \\ -1 & 0 & 0 & 7 & 0 & 0 & -7 & 0 & 0 \\ 5 & 0 & 0 & 7 & 0 & 0 & -10 & 0 & 0 \\ 0 & 1 & 0 & 0 & -7 & 0 & 0 & 7 & 0 \\ 0 & 6 & 0 & 0 & 0 & 0 & 0 & -3 & 0 \\ 0 & 0 & -5 & 0 & 0 & -7 & 0 & 0 & 10 \\ 0 & 0 & -6 & 0 & 0 & 0 & 0 & 0 & 3 \end{bmatrix}. $$