# On Correlated Equilibria in Marinatto-Weber Type Quantum Games
## 4 Relationship between the Correlated Equilibrium and the Nash Equilibrium in the MW Scheme  
It is intuitively clear from the construction of the MW scheme that the initial quantum state correlates payoff vectors. After the players choose their own strategies \(V_{k}\), the payoff vectors are distributed according to the measurement on the resulting quantum state. On the other hand, a correlated equilibrium plays a similar role in the coordination of strategy profiles. An interesting question arises whether there is a connection between the notion of correlated equilibrium and the MW scheme. The following example confirms this fact. **Example 1**. : _Let us consider the following \(2\times 3\) game_  
\[ \begin{pmatrix}(3,2)&(2,0)&(1,3)\\ (4,2)&(1,3)&(2,1)\end{pmatrix} . \tag{22}\]  
_Conditions (5) and (6) for \(m=2\) and \(n=3\) are of the form:_  
\[ \begin{array}{l}(a_{00}-a_{10})p_{00}+(a_{01}-a_{11})p_{01}+(a_{02}-a_{12})p_ {02}\geq 0,\\ (a_{10}-a_{00})p_{10}+(a_{11}-a_{01})p_{11}+(a_{12}-a_{02})p_{12}\geq 0\\ (b_{00}-b_{01})p_{00}+(b_{10}-b_{11})p_{10}\geq 0\\ (b_{00}-b_{02})p_{00}+(b_{10}-b_{12})p_{10}\geq 0\\ (b_{01}-b_{00})p_{01}+(b_{11}-b_{10})p_{11}\geq 0\\ (b_{02}-b_{00})p_{02}+(b_{12}-b_{10})p_{12}\geq 0\\ (b_{02}-b_{01})p_{02}+(b_{12}-b_{11})p_{12}\geq 0.\end{array}  \tag{23}\]Substituting payoffs of (22) to (23) implies the system of inequalities  
\[ \begin{array}{c}-p_{00}+p_{01}-p_{02}\geq 0,\\ p_{10}-p_{11}+p_{12}\geq 0,\\ 2p_{00}-p_{10}\geq 0,\\ -p_{00}+p_{10}\geq 0,\\ -2p_{01}+p_{11}\geq 0,\\ -3p_{01}+2p_{11}\geq 0,\\ p_{02}-p_{12}\geq 0,\\ 3p_{02}-2p_{12}\geq 0.\end{array}  \tag{24}\]  
Inequalities (24) together with \(i=1,2\) and \(j=1,2,3\) reduce to  
\[p_{10}>0,\ p_{00}=\frac{p_{10}}{2},\ p_{12}=p_{02}=0,\ p_{11}=p_{10},\ p_{01}= \frac{p_{11}}{2}. \tag{25}\]  
Combining (25) with \(\sum_{ij}p_{ij}=1\) yields the unique correlated equilibrium  
\[(p_{00},p_{01},p_{02},p_{10},p_{11},p_{12})=\left(\frac{1}{6},\frac{1}{6},0, \frac{1}{3},\frac{1}{3},0\right). \tag{26}\]  
To see how correlated equilibrium (26) in game (22) implies a Nash equilibrium in the MW approach to that game, consider the initial state \(|\Psi\rangle\) in (14) such that \(\alpha_{ij}=\sqrt{p_{ij}}\) for \(i=0,1\) and \(j=0,1,2\). Then  
\[|\Psi\rangle=\sqrt{\frac{1}{6}}|00\rangle+\sqrt{\frac{1}{6}}|01\rangle+\sqrt{ \frac{2}{6}}|10\rangle+\sqrt{\frac{2}{6}}|11\rangle. \tag{27}\]  
Taking into account the strategy sets and the payoffs of (22), the other components of (14) are  
\[S_{1}=\{V_{0},V_{1}\},\quad S_{2}=\{V_{0},V_{1},V_{2}\}, \tag{28}\]  
and  
\[ \begin{array}{c}M_{1}=3|00\rangle\langle 00|+2|01\rangle\langle 01|+|02\rangle \langle 02|+4|10\rangle\langle 10|+|11\rangle\langle 11|+2|12\rangle\langle 12|,\\ M_{2}=2|00\rangle\langle 00|+3|02\rangle\langle 02|+2|10\rangle\langle 10|+3|11 \rangle\langle 11|+|12\rangle\langle 12|.\end{array}  \tag{29}\]  
Now, consider the strategy profile \((V_{0},V_{0})\). Payoffs of player 1 and 2 corresponding to that profile are  
\[u_{1}((V_{0},V_{0}))=\text{tr}\left(|\Psi\rangle\langle\Psi|M_{1}\right)= \frac{5}{2},\quad u_{2}((V_{0},V_{0}))=\text{tr}\left(|\Psi\rangle\langle\Psi |M_{2}\right)=2. \tag{30}\]  
Interestingly, no player has a profitable unilateral deviation from \((V_{0},V_{0})\). In the case of player 1, if he deviates to \(V_{1}\), then his payoff is still 5/2; i.e.,  
\[u_{1}((V_{1},V_{0}))=\text{tr}\left((V_{1}\otimes V_{0})|\Psi\rangle\langle \Psi|(V_{1}\otimes V_{0})^{\dagger}M_{1}\right)=\frac{5}{2}. \tag{31}\]  
Similarly, player 2 will not benefit from playing the strategies \(V_{1}\) and \(V_{2}\),  
\[u_{2}(V_{0},V_{j})=\text{tr}\left((V_{0}\otimes V_{j})|\Psi\rangle\langle \Psi|(V_{0}\otimes V_{j})^{\dagger}M_{2}\right)=\frac{11}{6}\ \ \text{for}\ \ j=1,2. \tag{32}\]  
Since  
\[ \begin{array}{c}u_{1}((V_{0},V_{0}))\geq u_{1}((V_{1},V_{0})),\\ u_{2}((V_{0},V_{0}))\geq u_{2}((V_{0},V_{j}))\ \ \text{for}\ \ j=1,2,\end{array}  \tag{33}\]  
the strategy profile \((V_{0},V_{0})\) is a Nash equilibrium (see Definition 2).