## 3 System Model and Performance Analysis
\begin{table}
\begin{tabular}{c c} \hline \hline
**Symbol** & **Definition** \\ \hline \(w_{i}\) & the size of the competition window of the \(i\)th backoff phase \\ \(w_{0}\) & the initial value of the competition window \\ \(m\) & maximum backoff phase \\ \(d_{i}\) & the value of \(w_{i}\) minus \(w_{i-1}\) \\ \(p_{1}\) & the probability of collision when patients compete fairly \\ \(p_{2}\) & the probability of collision on the remaining subchannel \\ \(p_{3}\) & the probability of collision for SU \\ \(\theta\) & the transmission probability \\ \(t_{succ}\) & the average time to successfully transfer data \\ \hline \hline \end{tabular}
\end{table} 
Table 2: The list of the main symbols in the model. The STA collides on the remaining subchannel and goes from state \((i^{*},0)\) to state \((i+1,r)\): \[P\big{\{}(i+1,r)\big{|}(i^{*},0)\big{\}}=\frac{p_{2}}{d_{i+1}};\ \ \ 0\leq i\leq m-1,\ \ w_{i}\leq r\leq w_{i+1}-1.\] (5)
5. The STA has reached the last stage of the backoff process and remains in that state after an unsuccessful transmission on the remaining subchannel: \[P\big{\{}(m,r)|(m^{*},0)\big{\}}=\frac{p_{2}}{d_{m}};\ \ \ 0\leq i\leq m,\ w_{m-1}\leq r\leq w_{m}-1.\] (6)
6. The STA has no collision on the remaining subchannel and moves from state \((i^{*},0)\) to state \((i-1,r)\): \[P\big{\{}(i-1,r)|(i^{*},0)\big{\}}=\frac{1-p_{2}}{d_{i-1}};\ \ \ 0\leq i\leq m,\ \ w_{i-2}\leq r\leq w_{i-1}-1.\] (7)  
Based on the above transition probabilities, we calculated the probability of STA in state \((i,j)\), where \(\pi(i,j)\) denotes the stationary probability of the Markov chain model. Figure 7: The discrete-time two-dimensional Markov chain for high-priority STAs. Establishing the stationary equation according to state \((i,0)\), we have \(\pi(i,w_{i-1})=\pi(i,0)\). From the state \((i^{*},0)\), the stationary equation is \(\pi(i^{*},0)(1-p_{2})+\pi(i^{*},0)p_{2}=\pi(i,0)p_{1}\), simplified to obtain:  
\[\pi(i^{*},0)=p_{1}\pi(i,0). \tag{8}\]  
Similarly, the stationary equation is established from the states \((i,w_{i-1}+k)\), \(0\leq k\leq d_{i}-1\). \[\pi(i,w_{i-1}+k)=\pi(i,w_{i-1}+k+1)+\pi(i+1,0)\frac{1-p_{1}}{d_{i}}+\pi(i+1^{* },0)\frac{1-p_{2}}{d_{i}}+\pi(i-1^{*},0)\frac{p_{2}}{d_{i}}. \tag{9}\]  
When \(i=0\), summing Equation (9) left and right, we can obtain \(\pi(1,0)=p_{1}p_{2}\pi(0,0)/(1-p_{1}p_{2})\). Similarly,  
\[\pi(i,0)=\bigg{(}\frac{p_{1}p_{2}}{1-p_{1}p_{2}}\bigg{)}^{i}\pi(0,0). \tag{10}\]  
By eliminating \(\pi(i,w_{i-1}+k+1)\) from Equation (9), we can obtain:  
\[ \begin{array}{ll}\pi(i,w_{i-1}+k)&=(d_{i}-k)\pi(i+1,0)\frac{1-p_{1}}{d_{i}} +(d_{i}-k)\pi(i+1^{*},0)\frac{1-p_{2}}{d_{i}}\\ &+(d_{i}-k)\pi(i-1^{*},0)\frac{p_{2}}{d_{i}},\ \ \ 0\leq i\leq m-1,\ 0\leq k \leq d_{i}-1.\end{array}  \tag{11}\]  
Substitute Equations (8) and (10) into Equation (11), we have the following:  
\[\pi(i,w_{i-1}+k)=\frac{d_{i}-k}{d_{i}}\pi(i,0),\ \ \ 0\leq i\leq m-1,\ 0\leq k \leq d_{i}-1. \tag{12}\]  
So, we can obtain:  
\[\sum\limits_{j=w_{i-1}}^{w_{j}-1}\pi(i,j)=\sum\limits_{k=0}^{d_{i}-1}\frac{d_ {i}-k}{d_{i}}\pi(i,0)=\frac{d_{i}+1}{2}\pi(i,0). \tag{13}\]  
Considering the fact that the sum of all stationary probabilities is equal to one, we can obtain:  
\[ \begin{array}{ll}1&=\sum\limits_{i=0}^{m}\pi(i,0)+\frac{d_{i}+1}{2}\pi(i,0 )+\pi(i^{*},0)=\sum\limits_{i=0}^{m}\pi(i,0)(\frac{d_{i}+3}{2}+p_{1})\\ &=\sum\limits_{i=0}^{m}(\frac{d_{i}+3}{2}+p_{1})\big{(}\frac{p_{1}p_{2}}{1-p_ {1}p_{2}}\big{)}^{i}\pi(0,0).\end{array}  \tag{14}\]  
Solving Equation (14), we have:  
\[\pi(0,0)=\left(\frac{w_{0}}{4}\frac{1-\bigg{(}\frac{2p_{1}p_{2}}{1-p_{1}p_{2} }\bigg{)}^{m+1}}{1-\frac{2p_{1}p_{2}}{1-p_{1}p_{2}}}+\bigg{(}\frac{3}{2}+p_{1} \bigg{)}\frac{1-\bigg{(}\frac{p_{1}p_{2}}{1-p_{1}p_{2}}\bigg{)}^{m+1}}{1-\frac {p_{1}p_{2}}{1-p_{1}p_{2}}}\right)^{-1}. \tag{15}\]  
Thus, we obtain the transmission probability.