## 4 Computation of \(\epsilon\) Proper Equilibria
### Mixed 0-1 Quadratic Formulations  
We now propose a sequence of parametrized 0-1 mixed quadratic programs with optimal solutions defining a sequence of \(\epsilon\)-proper equilibria while \(\sigma\) converges to 0. **Proposition 3**. : _A Nash equilibrium \(\hat{X}=(\hat{X}_{1},\ldots,\hat{X}_{n})\) is proper if and only if the following 0-1 mixed quadratic program \((\Omega_{>})\) is feasible for all \(\tilde{\sigma}>0\), and if \(\lim\limits_{\sigma\to 0^{+}}\epsilon(\sigma)=0\)._  
\[ \begin{array}{l}\epsilon(\sigma)=\min\ \epsilon\\ \text{subject to}\\ (\Omega_{>})\ \hat{x}_{ik}-\epsilon\leq x_{ik}\leq\hat{x}_{ik}+\epsilon,\ \forall\ i\in N,\ k\in\{1,2,\ldots,m_{i}\},\\ (X_{1},\ldots,X_{n})\ \in\ \Omega_{\epsilon}^{\sigma},\\ 0\leq\epsilon\leq 1.\end{array} \]  
Proof. : Assume that \((X_{1}(\sigma),\ldots,X_{n}(\sigma),\epsilon(\sigma))\) is the optimal solution of \((\Omega_{>})\) for a known Nash equilibrium \(\hat{X}=(\hat{X}_{1},\ldots,\hat{X}_{n})\). Proposition 1 asserts that \((X_{1}(\sigma),\ldots,X_{n}(\sigma))\) is an \(\epsilon(\sigma)\)-proper equilibrium. In addition, the convergence conditions are here reformulated through the minimization of \(\epsilon\) such that \(\hat{x}_{ik}-\epsilon\leq x_{ik}\leq\hat{x}_{ik}+\epsilon,\text{\&}k\in\{1,2, \ldots,m_{i}\},\text{\&}i\in N\), to make sure that the \(\epsilon\)-proper equilibrium converges towards \((\hat{X}_{1},\ldots,\hat{X}_{n})\). Hence, if the 0-1 mixed quadratic program \((\Omega_{>})\) is feasible for all \(\sigma>0\) and converging to 0, we may conclude from proposition (2) that there always exists an \(\epsilon\)-proper equilibrium \((X_{1},\ldots,X_{n})\in\Omega_{\epsilon}^{\sigma}\). In addition, if the Nash equilibrium \(\hat{X}=(\hat{X}_{1},\ldots,\hat{X}_{n})\) is proper, the optimal objective value \(\epsilon(\sigma)\) converges necessarily towards 0, when \(\sigma>0\) converges to 0, to force the solution \((X_{1}(\sigma),\ldots,X_{n}(\sigma))\) to converge towards \((\hat{X}_{1},\ldots,\hat{X}_{n})\) simultaneously. We can also add that \(\epsilon(0)=0\) in this case. Otherwise, if \(\epsilon(\sigma)\) does not converge towards 0, when \(\sigma>0\) converges to 0, such \(\epsilon\)-proper sequence \((X_{1}(\sigma),\ldots,X_{n}(\sigma))\) does not even exist, when \(\epsilon\) converges to 0. In this case, it is obvious that the given Nash equilibrium \(\hat{X}\) cannot be proper. In conclusion, if \(\epsilon(\sigma)\) converges to 0, while \(\sigma>0\) converges to 0, it is possible to obtain a sequence \((X_{1}(\sigma),\ldots,X_{n}(\sigma))\)\(\epsilon\)-proper converging to \((\hat{X}_{1},\ldots,\hat{X}_{n})\), when \(\epsilon(\sigma)\) converges to 0. Therefore, we have to compute the values of \(\epsilon(\sigma)\) for some small values of \(\sigma\). **Corollary 1**. : _If \((X_{1}(\sigma),\ldots,X_{n}(\sigma),\epsilon(\sigma))\) is the optimal solution to the 0-1 mixed quadratic program \((\Omega_{>})\), for some \(\sigma>0\), then \((X_{1}(\sigma),\ldots,X_{n}(\sigma))\) is an \(\epsilon_{\sigma}\)-proper equilibrium, and if \(0<\sigma^{\prime}<\sigma^{\prime\prime}\), then \(0\leq\epsilon(\sigma^{\prime})\leq\epsilon(\sigma^{\prime\prime})\)._  
Proof. : If \(0<\sigma^{\prime}<\sigma^{\prime\prime}\), then the 0-1 mixed quadratic program \((\Omega_{>^{\prime}})\), for \(\sigma^{\prime}>0\), is a relaxation of the same 0-1 mixed quadratic program \((\Omega_{>^{\prime\prime}})\), but for \(\sigma^{\prime\prime}>0\). The main difference between the two programs lies in fact in the conditions of \(\Omega_{\epsilon}^{\sigma^{\prime}}\) and \(\Omega_{\epsilon}^{\sigma^{\prime\prime}}\):  
\(\sigma^{\prime}\leq x_{ik}\), \(k\in\{1,2,\ldots,m_{i}\}\), \(i\in N\), and \(\sigma^{\prime\prime}\leq x_{ik}\), \(k\in\{1,2,\ldots,m_{i}\}\), \(i\in N\),  
\[\Rightarrow\ \sigma^{\prime}<\sigma^{\prime\prime}\leq x_{ik},\ \ \ k\in\{1,2,\ldots,m_{i}\},\ \ \ i\in N.\]  
Thus, \(\Omega_{\epsilon}^{\sigma^{\prime\prime}}\subseteq\Omega_{\epsilon}^{\sigma^{ \prime}}\) and \(0\leq\epsilon(\sigma^{\prime})\leq\epsilon(\sigma^{\prime\prime})\).