## 3 Analytical Model
### Average Ergodic Rate  
Assume that the network adopts full buffer traffic model and all the users in the coverage of a BS share the entire frequency resource equally. Thus, the mean achievable downlink data rate of a typical UE \(l\) can be represented as follows:  
\[R_{l}=\frac{W_{l}}{E\left[N_{l}\right]}\,E\left[\log_{2}\left(1+\gamma_{l} \right)\right], \tag{6}\]where \(N_{l}\) is the mean load in a Voronoi cell and the expectation of \(N_{l}\) is \(E\left[N_{l}\right]=\left(A_{l}/\rho_{k}\right)+1\). When \(l=m_{u}\), then \(\rho_{k}=\rho_{1}=\lambda_{m}/\lambda_{u}\). When \(l\in\{p_{c},\rho_{e}\}\), then \(\rho_{k}=\rho_{2}=\lambda_{p}/\lambda_{u}\). \(W_{l}\) is the time-frequency resource that is allocated to user \(l\) and its value depends on the user type of user \(l\). Specifically, when \(l=p_{e}\), then \(W_{l}=\theta W\) and when \(l\in\{m_{u},p_{c}\}\), then \(W_{l}=\left(1-\theta\right)W\).  
Based on the analysis above, we get Lemma 3 in the following:  
**Lemma 3**.: _The average achievable downlink rate of a typical UE \(l\) can be further deduced as:_  
\[R_{l}=\frac{2\pi\lambda_{l}W_{l}}{A_{l}N_{l}}\int_{0}^{\infty}\int_{0}^{\infty }\exp\left(-\varphi_{l}-\pi r_{l}{}^{2}C_{l}\right)f_{l}\left(r_{l}\right)dr_{ l}dt, \tag{7}\]  
_where \(\tau=2^{t}-1\), \(\varphi_{l}=-\tau\sigma^{2}r_{l}{}^{2}P_{k}^{-1}\), \(C_{l}=\left\{\begin{array}{l}\lambda_{m}Z\left(\tau,\alpha,1\right)+\lambda _{p}Z\left(\tau,\alpha,B_{p}\right)\text{, when }l=m_{u}\\ \lambda_{m}\tilde{P}_{m}Z\left(\tau,\alpha,1\right)+\lambda_{p}Z\left(\tau, \alpha,1\right)\text{, when }l=p_{c}\\ \lambda_{p}Z\left(\tau,\alpha,1\right)\text{, when }l=p_{e}\end{array}\right.\), where \(Z\left(\tau,\alpha,\beta\right)=\tau^{2/\alpha}\int_{\left(\beta/\tau\right)^ {2/\alpha}}^{\infty}\frac{1}{1+\pi^{4/2}}dx\)._  
Proof: See Appendix C.  
**Corollary 1.** With noise ignored, and setting the large scale path loss exponent \(\alpha=4\), the average achievable downlink rate of a typical UE \(l\) can be simplified, respectively, according to its user type as follows:  
\[\begin{array}{l}R_{m_{u}}=\frac{(1-\theta)W}{\lambda_{m_{u}}N_{m_{u}}}\int _{0}^{\infty}\frac{1}{Q\left(\tau,4,1\right)+\frac{\lambda_{p}}{\lambda_{m}} \left(\beta\left(\tilde{P}_{p}\tau A,B_{p}\tilde{P}_{p}\right)\right)}dt,\\ R_{p_{c}}=\frac{(1-\theta)W}{A_{p}N_{p_{c}}}\int_{0}^{\infty}\frac{1}{Q\left( \tilde{P}_{m}\tau A,\tilde{P}_{m}\right)+\frac{\lambda_{m}}{\lambda_{p}} \left(\gamma\left(\tau,4,1\right)\right)}dt,\\ R_{p_{e}}=\frac{\theta W}{A_{p}N_{p_{c}}}\int_{0}^{\infty}\frac{1}{Q\left( \tau,4,1\right)+\left(B_{p}^{-1}\tilde{P}_{m}\right)^{1/2}\frac{\lambda_{m}} {\lambda_{p}}}\\ -\frac{1}{Q\left(\tau,4,1\right)+\left(\tilde{P}_{m}\right)^{1/2}\frac{ \lambda_{m}}{\lambda_{p}}}dt,\end{array} \tag{8}\]  
where \(Q\left(\tau,4,x\right)=\sqrt{x}+\sqrt{\tau}\text{tan}^{-1}\left(\sqrt{\tau/x}\right)\).  
Proof: when \(\alpha=4\), \(\sigma^{2}=0\), and let \(\varphi_{l}=0\) in Equation (7), then we can get \(Z\left(\tau,\alpha,\beta\right)=\sqrt{\tau}\int_{\sqrt{\theta}/\tau}^{\infty }\frac{1}{1+x}dx=\sqrt{\tau}\arctan\left(\sqrt{\tau/\beta}\right)\). Combining with Equation (7), we will obtain the desired results.