A probabilistic approach to quantum Bayesian games of incomplete information

Abstract

A Bayesian game is a game of incomplete information in which the rules of the game are not fully known to all players. We consider the Bayesian game of Battle of Sexes that has several Bayesian Nash equilibria and investigate its outcome when the underlying probability set is obtained from generalized Einstein–Podolsky–Rosen experiments. We find that this probability set, which may become non-factorizable, results in a unique Bayesian Nash equilibrium of the game.

Appendix

Using the standard notation, the set of 16 probabilities in (17) is obtained from a pure state of two qubits

$$\begin{aligned} \left| \psi _{0}\right\rangle =\alpha \left| 00\right\rangle +\beta \left| 01\right\rangle +\gamma \left| 10\right\rangle +\delta \left| 11\right\rangle \end{aligned}$$

(33)

where $\alpha ,\beta ,\gamma ,\delta \in \varvec{C}$ and $\left| \alpha \right| ^{2}+\left| \beta \right| ^{2}+\left| \gamma \right| ^{2}+\left| \delta \right| ^{2}=1$. We assume that observer 1’s directions $D_{1}$ and $D_{2}$ are along the unit vectors $\hat{a}$ and $\hat{c}$, respectively. Similarly, observer 2’s directions $D_{1}^{\prime }$ and $D_{2}^{\prime }$ are along the unit vectors $\hat{b}$ and $\hat{d}$, respectively. Without the loss of generality, we also assume that the unit vectors $\hat{a}=[a_{x},a_{y}]$, $\hat{b}=[b_{x},b_{y}]$, $\hat{c} =[c_{x},c_{y}]$, and $\hat{d}=[d_{x},d_{y}]$ are all located in the x–y plane. Observer 1’s measurement operators are then $\hat{\sigma }\cdot \hat{ a}$ and $\hat{\sigma }\cdot \hat{c}$, respectively. Similarly, observer 2’s measurement operators are $\hat{\sigma }\cdot \hat{b}$ and $\hat{\sigma }\cdot \hat{d}$, respectively. Here, $\hat{\sigma }=[\sigma _{x},\sigma _{y},\sigma _{z}]$ and $\sigma _{x},\sigma _{y},\sigma _{z}$ are Pauli spin matrices.

Consider the probability $\varepsilon _{1}$ in the (17) that corresponds to observers 1 and 2 measuring along the directions $\hat{a}$ and $\hat{b}$, respectively, and both obtaining the outcome $+1$. For this situation, we require the eigenstates of the operators $(\hat{\sigma }\cdot \hat{a})$ and $(\hat{\sigma }\cdot \hat{b})$, with the eigenvalue of $+1$ for both. These are found to be $\frac{\left| 0\right\rangle +(a_{x}+ia_{y})\left| 1\right\rangle }{\sqrt{2}}$ and $\frac{\left| 0\right\rangle +(b_{x}+ib_{y})\left| 1\right\rangle }{\sqrt{2}}$, respectively. From these, the eigenstate of the measurement operator $(\hat{ \sigma }\cdot \hat{a})\otimes (\hat{\sigma }\cdot \hat{b})$, with the eigenvalue $+1$, is obtained as

$$\begin{aligned} \left| \psi _{1}\right\rangle =\frac{1}{2}(\left| 00\right\rangle +(b_{x}+ib_{y})\left| 01\right\rangle +(a_{x}+ia_{y})\left| 10\right\rangle +(a_{x}+ia_{y})(b_{x}+ib_{y})\left| 11\right\rangle ), \end{aligned}$$

(34)

and the probability $\varepsilon _{1}$ is then obtained from $\left| \left\langle \psi _{1}\right| \left. \psi _{0}\right\rangle \right| ^{2}$. For the pure state (33), this becomes

$$\begin{aligned} \varepsilon _{1}=\frac{1}{4}\left| \alpha +\beta (b_{x}-ib_{y})+\gamma (a_{x}-ia_{y})+\delta (a_{x}-ia_{y})(b_{x}-ib_{y})\right| ^{2}. \end{aligned}$$

(35)

Similarly, for the probability $\varepsilon _{2}$, along with the eigenstate of $(\hat{\sigma }\cdot \hat{a})$ with eigenvalue $+1$ obtained above, we require the eigenstate of the operator $(\hat{\sigma }\cdot \hat{b})$ with the eigenvalues $-1$, which is $\frac{\left| 0\right\rangle -(b_{x}+ib_{y})\left| 1\right\rangle }{\sqrt{2}}$. From these, the eigenstate of the measurement operator $(\hat{\sigma }\cdot \hat{a})\otimes ( \hat{\sigma }\cdot \hat{b})$, with the eigenvalue $-1$, is obtained as

$$\begin{aligned} \left| \psi _{2}\right\rangle =\frac{1}{2}(\left| 00\right\rangle -(b_{x}+ib_{y})\left| 01\right\rangle +(a_{x}+ia_{y})\left| 10\right\rangle -(a_{x}+ia_{y})(b_{x}+ib_{y})\left| 11\right\rangle ), \end{aligned}$$

(36)

and, as before, the probability $\varepsilon _{2}$ is then obtained from $\left| \left\langle \psi _{2}\right| \left. \psi _{0}\right\rangle \right| ^{2}$. For the pure state (33), this becomes

$$\begin{aligned} \varepsilon _{2}=\frac{1}{4}\left| \alpha -\beta (b_{x}-ib_{y})+\gamma (a_{x}-ia_{y})-\delta (a_{x}-ia_{y})(b_{x}-ib_{y})\right| ^{2}, \end{aligned}$$

(37)

and the remaining probabilities $\varepsilon _{3},\varepsilon _{4} \cdots \varepsilon _{16}$ can be obtained similarly.