# A Learning Game-Based Approach to Task-Dependent Edge Resource Allocation
## 4 Incentives under Information Sharing Conditions
### Stackelberg Equilibrium Analysis under Information Sharing Conditions
For communication resource Equation (10), the derivative with respect to \(r_{m}\) can be obtained as follows:  
\[\frac{\partial u_{2,m}}{\partial r_{m}}=\frac{-R_{2}r_{m}}{(\sum_{m=1}^{M}r_{ m})^{2}}+\frac{R_{2}}{\sum_{m=1}^{M}r_{m}}-\beta_{m}, \tag{22}\]  
\[\frac{\partial^{2}u_{2,m}}{\partial r_{m}^{2}}=-\frac{2R_{2}\sum_{n=1,n\
eq m }^{M}r_{n}}{(\sum_{m=1}^{M}r_{m})^{3}}<0. \tag{23}\]  
Since Equation (23) is a constant and less than zero, utility function \(u_{2,m}\) is a concave function; thus, there is an NE in the non-cooperative game between the follower parties over communication resources. From reference [36], if the game of communication resources satisfies Theorem 1, the uniqueness of the non-cooperative game's NE can be demonstrated. Theorem 1: _Given strategy \(R_{2}\) and set \(\bar{\mathbf{S}}=\{m\in\mathcal{M}|\bar{r}_{m}>0\}\) of follower parties, we let \(\bar{r}=(\bar{r}_{1},\bar{r}_{2},...,\bar{r}_{n})\) be a Nash equilibrium (NE) strategy if the following four conditions are satisfied. If so, then the NE is proven to be a unique NE as follows:_  
_(1). \(|\bar{\mathbf{S}}|\geq 2\),__(2). \(\bar{r}_{m}= \begin{cases}0&if\quad m\
otin\bar{\mathbf{S}}\\ \frac{(|M|-1)\bar{R}_{2}}{\sum_{m=1}^{M}\bar{\beta}_{m}}(1-\frac{(|M|-1)\bar{ \beta}_{m}}{\sum_{m=1}^{M}\bar{\beta}_{m}})&\text{otherwise}\end{cases} ,\) _(3). If \(\beta_{m}\leq\max_{j\in\bar{\mathbf{S}}}\{\beta_{j}\}\), then \(m\in\bar{\mathbf{S}}\),_  
_(4). Suppose the unit rate transmission cost of the edge server satisfies \(\beta_{1}\leq\beta_{2}\leq...\leq\beta_{n}\), and let \(h\) be the largest integer in [2,n] such that \(\beta_{h}<\frac{\sum_{j=1}^{h}\beta_{j}}{h-1}\), then \(\bar{\mathbf{S}}=\{1,2,...,h\}\)._  
Clearly, the non-cooperative game between parties over communication resources satisfies these four conditions. Proof of Theorem 1.: When assuming \(|\mathbf{S}|=0\), no ES participates in the game and the game is not established. Therefore, it can be inferred that \(\mathbf{S}\geq 1\). Now, we suppose \(\mathbf{S}=1\); this implies that \(k\in\mathcal{M}\), \(\bar{r}_{k}>0\). According to Equation (10), the utility of ES k is \(R_{2}-\bar{r}_{k}\beta_{k}\), which means that ES \(k\) can unilaterally modify its strategy to increase its utility, thereby contradicting the NE. Thus, Condition 1 is proved. To next prove Condition 2 and to accumulate Equation (22), we obtain the following:  
\[\sum_{m=1}^{M}\bar{r}_{m}=\frac{(|M|-1)R_{2}}{\sum_{m=1}^{M}\bar{\beta}_{m}}. \tag{24}\]  
By substituting Equation (24) back into Equation (22), as well as by setting the result to zero while considering \(\bar{r}_{j}=0\) for any \(j\in\mathcal{M}\setminus\bar{\mathbf{S}}\), we obtain the following:  
\[\bar{r}_{m}= \begin{cases}0&if\quad m\
otin\bar{\mathbf{S}}\\ \frac{(|M|-1)R_{2}}{\sum_{m=1}^{M}\bar{\beta}_{m}}(1-\frac{(|M|-1)\bar{\beta}_ {m}}{\sum_{m=1}^{M}\bar{\beta}_{m}})&\text{otherwise}\end{cases} . \tag{25}\]  
The proof of Condition 2 is thus complete. For (3) given \(|\bar{\mathbf{S}}|\) for any \(i\in\bar{\mathbf{S}}\), we can deduce that \(\bar{r}_{i}>0\). According to Equation (25), \(\bar{r}_{i}>0\) implies \(\frac{(|M|-1)\bar{\beta}_{m}}{\sum_{m=1}^{M}\bar{\beta}_{m}}<1\), and hence we can obtain the following:  
\[\beta_{i}<\frac{\sum_{j\in\bar{\mathbf{S}}}\beta_{j}}{|\bar{\mathbf{S}}|-1}, \forall i\in\bar{\mathbf{S}}. \tag{26}\]  
This tells us  
\[\max_{i\in\bar{\mathbf{S}}}\beta_{i}<\frac{\sum_{j\in\bar{\mathbf{S}}}\beta_{ j}}{|\bar{\mathbf{S}}|-1}. \tag{27}\]  
Assuming that there is \(\beta_{q}<\max_{j\in\bar{\mathbf{S}}}\{\beta_{j}\}\), but \(q\
otin\bar{\mathbf{S}}\), then--according to Equation (25)--we have \(\bar{r}_{q}=0\). Hence, Equation (22) can be rewritten as follows:  
\[\frac{R_{2}}{\sum_{j\in\bar{\mathbf{S}}}\bar{r}_{j}}-\beta_{q}=\frac{\sum_{j \in\bar{\mathbf{S}}}\bar{r}_{j}}{|\bar{\mathbf{S}}|-1}-\beta_{q}>\max_{i\in \bar{\mathbf{S}}}\{\beta_{i}\}-\beta_{q}. \tag{28}\]  
This implies that ESs can unilaterally modify their strategy to increase their utility, which contradicts the NE definition.