## 3 Outage Probability Performance Analysis
### Outage Probability of D2D User 2  
The outage events of \(D_{2}\) can be explained as below. The first is that \(D_{1}\) cannot detect \(x_{2}\). The second is that \(D_{2}\) cannot detect its own message \(x_{2}\) on the conditions that \(D_{1}\) can detect \(x_{2}\) successfully. Based on these, the outage probability of \(D_{2}\) is expressed as  
\[ \begin{split}& OP_{D2-bi}=\underbrace{\Pr\left(\left(\gamma_{SD2,k* }^{PDMA}<\varepsilon_{2}\cup\gamma_{SD1\gets 2,k*}^{PDMA}< \varepsilon_{1}\right),\;\;\right)}_{\gamma_{SD1,k*}^{PDMA}<\varepsilon_{2}} \\ &\hskip 14.226378 pt+\underbrace{\Pr\left(\left.\gamma_{SD2,k*}^{ PDMA}<\varepsilon_{2}\cup\max\left\{\gamma_{SD1\gets 2,k*}^{PDMA},\chi_{D2}^{PDMA} \right\}<\varepsilon_{1},\;\;\right)}_{\Psi_{2}},\end{split}  \tag{13}\]  
where \(\varepsilon_{2}=2^{2R_{2}}\), \(R_{2}\) is denoted as target rate for signal \(x_{2}\), and with the help of (5) and (7), terms \(\Psi_{1}\) and \(\Psi_{2}\) can be calculated, the first being  
\[ \begin{split}&\Psi_{1}=\Pr\left(\left(\gamma_{SD2,k*}^{PDMA}< \varepsilon_{2}\cup\gamma_{SD1\gets 2,k*}^{PDMA}<\varepsilon_{1} \right),\gamma_{SD1,k*}^{PDMA}<\varepsilon_{2}\right)\\ &=\underbrace{\left(1-\Pr\left(\gamma_{SD2,k*}^{PDMA}\geq \varepsilon_{2},\gamma_{SD1\gets 2,k*}^{PDMA}\geq\varepsilon_{1}\right) \right)}_{D_{11}},\underbrace{\left(1-\Pr\left(\gamma_{SD1,k*}^{PDMA}\geq \varepsilon_{2}\right)\right)}_{D_{12}}.\end{split}  \tag{14}\]Therefore, \(D_{11}\) can be expressed as  
\[ \begin{array}{l}D_{11}=\Pr\left(\left|g_{2,k}\right|^{2}\geq\frac{\varepsilon_{2 }\alpha\rho|f_{2}|^{2}+\varepsilon_{2}}{a_{2}\rho},\left|g_{2,k}\right|^{2} \geq\frac{\varepsilon_{1}\alpha\rho|f_{2}|^{2}+\varepsilon_{1}}{a_{1}\rho- \varepsilon_{1}a_{2}\rho}\right)\\ \\ =\Pr\left(\left|g_{2,k}\right|^{2}\geq\frac{\varepsilon_{2}\left(\alpha\rho|f_{ 2}|^{2}+1\right)}{a_{2}\rho},\left|g_{2,k}\right|^{2}\geq\frac{\varepsilon_{1} \left(\alpha\rho|f_{2}|^{2}+1\right)}{a_{1}\rho-\varepsilon_{1}a_{2}\rho}\right) \\ \\ =\Pr\left(\left|g_{2,k}\right|^{2}\geq\left(\alpha\rho|f_{2}|^{2}+1\right)\max \left(\frac{\varepsilon_{2}}{a_{2}\rho},\,\frac{\varepsilon_{1}}{a_{1}\rho- \varepsilon_{1}a_{2}\rho}\right)\right),\end{array}  \tag{15}\]  
where \(\theta=\max\left(\frac{\varepsilon_{2}}{a_{2}\rho},\,\frac{\varepsilon_{1}}{a _{1}\rho-\varepsilon_{1}a_{2}\rho}\right)\). It worth noting that, we can achieve important computations as below:  
\[ \begin{array}{l}D_{11}=\Pr\left(\left|g_{2,k}\right|^{2}\geq\left(\alpha\rho |f_{2}|^{2}+1\right)\theta\right)\\ \\ =\frac{1}{\lambda_{f_{2}}}\int_{0}^{\infty}\sum\limits_{k=1}^{K}\left(\begin{array} []{c}K\\ k\end{array}\right)\left(-1\right)^{k-1}\exp\left(-\frac{(\alpha\rho\
u+1) \theta k}{\lambda_{2}}-\frac{y}{\lambda_{f_{2}}}\right)dy\\ \\ =\sum\limits_{k=1}^{K}\left(\begin{array}{c}K\\ k\end{array}\right)\left(-1\right)^{k-1}\frac{\lambda_{2}}{\theta k\alpha\rho A _{f_{2}}+\lambda_{2}}\exp\left(-\frac{\theta k}{\lambda_{2}}\right).\end{array}  \tag{16}\]  
Similarly, \(D_{12}\) can be expressed as  
\[ \begin{array}{l}D_{12}=1-\Pr\left(\gamma_{SD1,k*}^{PDMA}\geq\varepsilon_{2 }\right)\\ \\ =1-\Pr\left(\left|g_{1,k}\right|^{2}\geq\frac{\varepsilon_{2}\alpha\rho|f_{1}| ^{2}+\varepsilon_{2}}{a_{1}\rho-\varepsilon_{2}a_{2}\rho}\right)\\ \\ =1-\frac{1}{\lambda_{f_{1}}}\int_{0}^{\infty}\sum\limits_{k=1}^{K}\left( \begin{array}{c}K\\ k\end{array}\right)\left(-1\right)^{k-1}\exp\left(-\frac{(\varepsilon_{2} \alpha\rho x+\varepsilon_{2})k}{(a_{1}\rho-\varepsilon_{2}a_{2}\rho)\lambda_{ 1}}-\frac{x}{\lambda_{f_{1}}}\right)dx\\ \\ =1-\sum\limits_{k=1}^{K}\left(\begin{array}{c}K\\ k\end{array}\right)(-1)^{k-1}v_{1},\end{array}  \tag{17}\]  
where \(v_{1}=\frac{(a_{1}\rho-\varepsilon_{2}a_{2}\rho)\lambda_{1}}{(a_{1}\rho- \varepsilon_{2}a_{2}\rho)\lambda_{1}+\varepsilon_{2}\alpha\rho k\lambda_{f_{1 }}}\exp\left(-\frac{\varepsilon_{2}k}{(a_{1}\rho-\varepsilon_{2}a_{2}\rho) \lambda_{1}}\right)\). From (16) and (17), we find the expression \(\Psi_{1}\)  
\[