# On Correlated Equilibria in Marinatto-Weber Type Quantum Games
## 4 Relationship between the Correlated Equilibrium and the Nash Equilibrium in the MW Scheme
Example 1 shows that the MW approach to (22) in which the initial state is based on correlated equilibrium (26) results in the payoff-equivalent Nash equilibrium. As the following proposition states, this property is true in general. The MW approach in which the amplitudes of the initial state are the square roots of the respective probabilities of the correlated equilibrium always implies the Nash equilibrium \((V_{0},V_{0})\). Proposition 1: _Let \((p_{ij})\) be a correlated equilibrium in (1), and let \((|\Psi\rangle,(S_{1},S_{2}),M)\) be the MW approach associated with (1). If \(|\Psi\rangle=\sum_{ij}\sqrt{p_{ij}}\mathrm{e}^{\mathrm{i}\theta_{ij}}|ij\rangle\) then \((V_{0},V_{0})\) is a Nash equilibrium in \((|\Psi\rangle,(S_{1},S_{2}),M)\) with the equilibrium outcome \(\sum_{ij}p_{ij}(a_{ij},b_{ij})\)._  
Proof: First note that indicies \(k\) and \(l\) in (5) and (6) go from \(0\) to \(m-1\) and from \(0\) to \(n-1\), respectively. Thus, (5) and (6) can be written as  
\[\sum_{j=0}^{n-1}(a_{ij}-a_{i+n,k_{j}})p_{ij}\geq 0\quad\text{for all} \quad i,k=0,\ldots,m-1, \tag{34}\] \[\sum_{i=0}^{m-1}(b_{ij}-b_{i,j+n})p_{ij}\geq 0\quad\text{for all} \quad j,l=0,\ldots,n-1. \tag{35}\]  
Let us determine \(u_{1}(V_{k},V_{0})\). First we obtain  
\[V_{k}\otimes V_{0}|\Psi\rangle=\sum_{i=0}^{m-1}\sum_{j=0}^{n-1}\alpha_{ij}|i+ _{m}k,j\rangle. \tag{36}\]  
Then  
\[u_{1}(V_{k},V_{0})=\text{tr}\left(V_{k}\otimes V_{0}|\Psi\rangle\langle\Psi|V _{k}^{\dagger}\otimes V_{0}^{\dagger}M_{1}\right)=\sum_{i=0}^{m-1}\sum_{j=0}^{ n-1}|\alpha_{ij}|^{2}a_{i+n,k_{j}}. \tag{37}\]  
It follows that  
\[u_{1}(V_{0},V_{0})\geq u_{1}(V_{k},V_{0})\Leftrightarrow\sum_{i=0}^{m-1}\sum_ {j=0}^{n-1}|\alpha_{ij}|^{2}(a_{ij}-a_{i+n,k,j})\geq 0 \tag{38}\]  
for each \(k=0,1,\ldots,m\). Using the fact that \(\alpha_{ij}=\sqrt{p_{ij}}\mathrm{e}^{\mathrm{i}\theta_{ij}}\) for all \(i,j\) we can rewrite (38) as  
\[\sum_{i=0}^{m-1}\sum_{j=0}^{n-1}p_{ij}(a_{ij}-a_{i+_{m}k,j})\geq 0. \tag{39}\]  
Inequality (39) is a consequence of the correlated equilibrium conditions (34). In a similar manner we can prove that  
\[u_{2}(V_{0},V_{0})\geq u_{2}(V_{0},V_{l})\Leftrightarrow\sum_{i=0}^{m-1}\sum_ {j=0}^{n-1}|\alpha_{ij}|^{2}(b_{ij}-b_{i,j+n}l)\geq 0. \tag{40}\]  
It is worth noting that the converse of Proposition 1 may not hold. Given the Nash equilibrium \((V_{0},V_{0})\) in the MW game, the corresponding initial state may not determine a correlated equilibrium in the classical game. The following example illustrates this point. **Example 2**. : _Consider again the game given by bimatrix (22) and the MW approach to that game with the initial state_  
\[|\Psi\rangle=\sqrt{\frac{1}{6}}\sum_{i=0}^{1}\sum_{j=0}^{2}|ij\rangle. \tag{41}\]  
_The state \(|\Psi\rangle\) in the equal superposition means that the players cannot change the state \(|\Psi\rangle\); i.e.,_  
\[(V_{k}\otimes V_{l})|\Psi\rangle=|\Psi\rangle\;\;\text{for all}\;\;k=0,1\; \;\text{all}\;\;l=0,1,2. \tag{42}\]  
_As a result, each strategy profile \((V_{k}\otimes V_{l})\) implies the same payoff vector,_  
\[u_{1}((V_{k},V_{l}))=\text{tr}\,(|\Psi\rangle\langle\Psi|M_{1})=\frac{13}{6}, \tag{43}\]  
\[\sum_{i=0}^{m-1}(b_{ij}-b_{i,j+n})p_{ij}\geq 0\quad\text{for all} \quad j,l=0,\ldots,n-1. \tag{35}\]0  
_for \(k=0,1\) all \(l=0,1,2\). The same payoff vectors for each strategy profile imply that \((V_{0},V_{0})\) satisfies the Nash equilibrium conditions (2). However, a probability distribution \(P=(p_{ij})\) such that \(p_{ij}=1/6\) for each \(i=0,1\) and \(j=0,1,2\) is not a correlated equilibrium in (22)--the left hand side of the first condition of (23) is equal to \(-1\)/6._