# Differential Privacy in Federated Learning: An Evolutionary Game Analysis
## 4 Game Analysis
: _The population state \(\mathbf{x}^{*}=[x_{1}^{*},\ldots,x_{N}^{*}]\) is called a Nash equilibrium of the game if for all feasible states \(\mathbf{x}=[x_{1},\ldots,x_{N}]\) in the population, the inequality \((\mathbf{x}-\mathbf{x}^{*})^{T}\mathbf{\phi}(\mathbf{x}^{*})\leq 0\) holds, where \(\mathbf{\phi}(\mathbf{x}^{*})=[\phi_{1}(\mathbf{x}^{*}),\ldots,\phi_{N}(\mathbf{x}^{*})]\)[38]._  
In simpler terms, a Nash equilibrium is a fixed point of the ordinary differential equation, i.e.,  
\[\forall n\in\mathcal{N},\quad\dot{x}_{n}(t)=0.\]  
Suppose there exists another population state, \(\mathbf{x}^{\prime}\), that attempts to invade the state \(\mathbf{x}^{*}\) by attracting a fraction of players \(\lambda\in(0,1)\). If the following condition holds for all \(\lambda\in(0,1)\), then \(\mathbf{x}^{*}\) is an evolutionarily stable strategy (ESS). The definition is given as follows. **Definition 2**. : _If there exists a neighborhood \(\mathcal{B}\in\mathcal{X}\) such that_  
\[(\mathbf{x}-\mathbf{x}^{*})^{T}\mathbf{\phi}(\mathbf{x}^{*})=0\quad,\quad(\mathbf{x}^{*}-\mathbf{x})^{ T}\mathbf{\phi}(\mathbf{x})\geq 0,\forall x\in\mathcal{B}\]  
_then the population state \(\mathbf{x}^{*}\) is an ESS of the game._In practice, when \(N\) is very large, it is difficult to find a closed-form solution for \(x^{*}\). Thus, we study the case of two strategies (\(N=2\)) to illustrate stability in the game. Other cases can also be analyzed using a similar method, but due to the complexity of the calculations, they are not listed in this paper. In the numerical simulation section, cases with more than two strategies are still considered. Taking two strategies as an example, the share of choosing strategy 1 is \(x_{1}\), and the share of choosing strategy 2 is \(x_{2}\). Based on the replicator dynamics and the constraint conditions, the following equations can be listed:  
\[ \begin{cases}\Delta x_{1}(U_{1}-\bar{U})=0,\\ \Delta x_{2}(U_{2}-\bar{U})=0,\\ x_{1}+x_{2}=1.\end{cases}  \tag{14}\]  
The utility for selecting strategy \(S_{1}\), \(U_{1}\) can be expressed as  
\[U_{1}=\frac{r}{B+C\cdot\big{(}\sigma_{1}^{2}Ix_{1}+\sigma_{2}^{2}Ix_{2}\big{)} }\cdot\frac{1}{\sigma_{1}^{2}I\Big{(}\frac{x_{1}}{\sigma_{1}}+\frac{x_{2}}{ \sigma_{2}}\Big{)}}-\frac{\eta^{2}E_{1}^{2}}{2\sigma_{1}^{2}}+A \tag{15}\]  
When \(I\) is sufficiently large, the influence of \(B\) on \(U_{1}\) can be ignored, and we let  
\[a=\frac{r}{Cl^{2}},\quad b_{1}=\frac{\eta^{2}E_{1}^{2}}{2\sigma_{1}^{2}}, \quad b_{2}=\frac{\eta^{2}E_{2}^{2}}{2\sigma_{2}^{2}}. \tag{16}\]  
By substituting the above into Equation (15), we obtain the simplified expression of \(U_{1}\) as follows:  
\[U_{1}=\frac{a}{\big{(}\sigma_{1}^{2}x_{1}+\sigma_{2}^{2}x_{2}\big{)}\cdot \Big{(}\frac{x_{1}}{\sigma_{1}}+\frac{x_{2}}{\sigma_{2}}\Big{)}\sigma_{1}}-b _{1}+A. \tag{17}\]  
By substituting Equation (17) into Equation (14), we obtain that  
\[x_{1}^{*}\in\Bigg{\{}\frac{\sigma_{1}^{2}-\sigma_{2}^{2}-\sigma_{1}\sigma_{2} -\beta}{2\big{(}\sigma_{1}^{2}-\sigma_{2}^{2}\big{)}},0,1,\frac{\sigma_{1}^{2 }-\sigma_{2}^{2}-\sigma_{1}\sigma_{2}+\beta}{2\big{(}\sigma_{1}^{2}-\sigma_{2 }^{2}\big{)}}\Bigg{\}}, \tag{18}\]  
where  
\[\beta = \big{(}(4a\sigma_{1}^{2}-4a\sigma_{2}^{2}+b_{1}\sigma_{1}^{4}+b_ {1}\sigma_{2}^{4}-b_{2}\sigma_{1}^{4}-b_{2}\sigma_{2}^{4}+2b_{1}\sigma_{1} \sigma_{2}^{3}+2b_{1}\sigma_{1}^{3}\sigma_{2}-2b_{2}\sigma_{1}\sigma_{2}^{3}- 2b_{2}\sigma_{1}^{3}\sigma_{2} \tag{19}\] \[+ 3b_{1}\sigma_{1}^{2}\sigma_{2}^{2}-3b_{2}\sigma_{1}^{2}\sigma_{ 2}^{2}\big{)}/(b_{1}-b_{2})\big{)}^{0.5}.\]  
Thus, \(x_{2}^{*}=1-x_{1}^{*}\), and in all solutions, we should have \(x_{1}^{*}\in[0,1]\). Further, we will study the evolutionary stability of four fixed points. In the cases of \(x_{1}^{*}=0\) and \(x_{1}^{*}=1\), the shares of the two strategies are \((0,1)\) and \((1,0)\), respectively. In these two cases, the forms are similar. Therefore, we only need to check the case of \(x_{1}^{*}=0\).