## 3 Transmission Scenarios
### UR-MCJST Scenario
At the same time, non-altruistic CJs transmit the jamming signal \(Y_{\{CJ_{i_{1},\frac{6}{2}\}\}}\sim\mathcal{N}(0,1)\) to ensure that a secure communication between the S and D has been carried out over the wireless channel as shown in \[\beta=\frac{\mathbb{E}[Y^{2}_{\{CJ_{i_{1}},\frac{\alpha}{2}\}}]}{\mathbb{E}[Y^ {2}_{\{CJ_{i_{3}},1-\alpha\}}]}, \tag{1}\]3 During the first phase, the signals that will be received at \(R\) and \(D\) can be represented, respectively, as  
\[\beta=\frac{\mathbb{E}[Y^{2}_{\{CJ_{i_{1}},\frac{\alpha}{2}\}}]}{\mathbb{E}[Y^ {2}_{\{CJ_{i_{3}},1-\alpha\}}]}, \tag{1}\]4 \[\beta=\frac{\mathbb{E}[Y^{2}_{\{CJ_{i_{1}},\frac{\alpha}{2}\}}]}{\mathbb{E}[Y^ {2}_{\{CJ_{i_{3}},1-\alpha\}}]}, \tag{1}\]5  
and  
\[\beta=\frac{\mathbb{E}[Y^{2}_{\{CJ_{i_{1}},\frac{\alpha}{2}\}}]}{\mathbb{E}[Y^ {2}_{\{CJ_{i_{3}},1-\alpha\}}]}, \tag{1}\]6 \[\beta=\frac{\mathbb{E}[Y^{2}_{\{CJ_{i_{1}},\frac{\alpha}{2}\}}]}{\mathbb{E}[Y^ {2}_{\{CJ_{i_{3}},1-\alpha\}}]}, \tag{1}\]7  
where \(w_{r}\) and \(w^{(1)}_{d}\) are the AWGN noise at \(R\) and \(D\), respectively. Afterward, \(R\) will relay the received signal \(y_{r}\) to \(D\) using an amplify and forward strategy during the second phase; consequently, the signal received at the \(D\) can be represented as:  
\[\beta=\frac{\mathbb{E}[Y^{2}_{\{CJ_{i_{1}},\frac{\alpha}{2}\}}]}{\mathbb{E}[Y^ {2}_{\{CJ_{i_{3}},1-\alpha\}}]}, \tag{1}\]8  
where \(w^{(2)}_{d}\) denotes the AWGN noise at the \(D\). The amplification factor \(\eta_{r}\) can be computed as:  
\[\beta=\frac{\mathbb{E}[Y^{2}_{\{CJ_{i_{1}},\frac{\alpha}{2}\}}]}{\mathbb{E}[Y^ {2}_{\{CJ_{i_{3}},1-\alpha\}}]}, \tag{1}\]9By substituting (7) and (10) into (9), we obtain:  
\[P_{\text{s}} = \frac{1}{K}\sum_{t_{1}=1}^{\left\lfloor\frac{\alpha}{2}K\right \rfloor}\mathbb{E}[X^{2}_{\{t_{1},\frac{\alpha}{2}\}}]+\frac{1}{K}\sum_{t_{2} =\left\lfloor\frac{\alpha}{2}K+1\right\rfloor}^{\left\lfloor\frac{\alpha}{2} K+1\right\rfloor}\mathbb{E}[X^{2}_{\{t_{2},\frac{\alpha}{2}\}}] \tag{2}\]0 \[P_{\text{s}} = \frac{1}{K}\sum_{t_{1}=1}^{\left\lfloor\frac{\alpha}{2}K\right \rfloor}\mathbb{E}[X^{2}_{\{t_{1},\frac{\alpha}{2}\}}]+\frac{1}{K}\sum_{t_{2} =\left\lfloor\frac{\alpha}{2}K+1\right\rfloor}^{\left\lfloor\frac{\alpha}{2} K+1\right\rfloor}\mathbb{E}[X^{2}_{\{t_{2},\frac{\alpha}{2}\}}] \tag{2}\]1 \[P_{\text{s}} = \frac{1}{K}\sum_{t_{1}=1}^{\left\lfloor\frac{\alpha}{2}K\right \rfloor}\mathbb{E}[X^{2}_{\{t_{1},\frac{\alpha}{2}\}}]+\frac{1}{K}\sum_{t_{2} =\left\lfloor\frac{\alpha}{2}K+1\right\rfloor}^{\left\lfloor\frac{\alpha}{2} K+1\right\rfloor}\mathbb{E}[X^{2}_{\{t_{2},\frac{\alpha}{2}\}}] \tag{2}\]2  
By summing the two received signals, specifically \(y_{d}^{(1)}\) from Equation (8) and \(y_{d}^{(2)}\) from Equation (11), we obtain the following expression:  
\[P_{\text{s}} = \frac{1}{K}\sum_{t_{1}=1}^{\left\lfloor\frac{\alpha}{2}K\right \rfloor}\mathbb{E}[X^{2}_{\{t_{1},\frac{\alpha}{2}\}}]+\frac{1}{K}\sum_{t_{2} =\left\lfloor\frac{\alpha}{2}K+1\right\rfloor}^{\left\lfloor\frac{\alpha}{2} K+1\right\rfloor}\mathbb{E}[X^{2}_{\{t_{2},\frac{\alpha}{2}\}}] \tag{2}\]3 \[P_{\text{s}} = \frac{1}{K}\sum_{t_{1}=1}^{\left\lfloor\frac{\alpha}{2}K\right \rfloor}\mathbb{E}[X^{2}_{\{t_{1},\frac{\alpha}{2}\}}]+\frac{1}{K}\sum_{t_{2} =\left\lfloor\frac{\alpha}{2}K+1\right\rfloor}^{\left\lfloor\frac{\alpha}{2} K+1\right\rfloor}\mathbb{E}[X^{2}_{\{t_{2},\frac{\alpha}{2}\}}] \tag{2}\]4 \[P_{\text{s}} = \frac{1}{K}\sum_{t_{1}=1}^{\left\lfloor\frac{\alpha}{2}K\right \rfloor}\mathbb{E}[X^{2}_{\{t_{1},\frac{\alpha}{2}\}}]+\frac{1}{K}\sum_{t_{2} =\left\lfloor\frac{\alpha}{2}K+1\right\rfloor}^{\left\lfloor\frac{\alpha}{2} K+1\right\rfloor}\mathbb{E}[X^{2}_{\{t_{2},\frac{\alpha}{2}\}}] \tag{2}\]5  
Here, we define \(w_{d}\) as the sum of \(w_{d}^{(1)}\) and \(w_{d}^{(2)}\), and \(\eta_{r}\) is provided in Equation (10).