# On Performance Analysis of Protective Jamming Schemes in Wireless Sensor Networks
## 4 Analysis on Eavesdropping Probability
### Analysis of Non-Jam Scheme  
According to the definition of the eavesdropping probability \(P(E)\), we need to derive the probability \(P_{e}\) that one transmission has been eavesdropped first. In particular, we have \(P_{e}\) of Non-Jam scheme as follows.  
**Theorem 1**.: _In Non-Jam scheme, the eavesdropping probability \(P_{e}\) that one transmission has been eavesdropped is_  
\[P_{e}=\int\limits_{r>0}e^{-\mu T_{p}r^{\alpha}\sigma^{2}-\pi r^{2}\lambda(\rho (T_{\mathcal{A}})+1)}2\pi\lambda rdr, \tag{3}\]  
_where \(\rho(T_{p},\alpha)=T_{p}^{-2/\alpha}\int_{T_{p}^{-2/\alpha}}^{\infty}\frac{1} {1+\mu^{\alpha/2}}d\mu\) and \(T_{p}=\frac{T}{Pl}\) for simplicity._  
Proof.: We denote the distance between the eavesdropper and its nearest transmitter by \(r\). Since the transmitters are distributed according to PPP, the probability density function (PDF) of \(r\) can be derived as the following steps.  
Firstly, we have the probability that no transmitter is closer than \(R\) given by  
\[P[r>R]=P[\text{No transmitter closer than }R]=e^{-\lambda\pi R^{2}}.\]  
Then, the cumulative distribution function (CDF) of \(r\) is \(P[r\leq R]=F_{R}(R)=1-e^{-\lambda\pi R^{2}}\). We next have the PDF of \(r\) as follows,  
\[f_{r}(r)=\frac{dF_{r}(r)}{dr}=e^{-\lambda\pi r^{2}}2\pi\lambda r.\]  
Since the channel gain is \(h\), the SINR at eavesdropper is  
\[\text{SINR}=\frac{P_{t}hr^{-\alpha}}{\sigma^{2}+I_{t}}, \tag{4}\]  
where \(I_{t}=\sum\limits_{i\in\Phi/t_{0}}P_{t}h_{i}R_{i}^{-\alpha}\).  
Then, the eavesdropping probability \(P_{e}\) that one transmission has been eavesdropped is  
\[\begin{split} P_{e}&=E_{r}[P(\text{SINR}>T|r)]\\ &=\int\limits_{r>0}P\left[\frac{P_{t}hr^{-\alpha}}{\sigma^{2}+I_{t }}>T|r\right]e^{-\lambda\pi r^{2}}2\pi\lambda rdr\\ &=\int\limits_{r>0}P[h>T_{p}r^{\alpha}(\sigma^{2}+I_{t})|r]e^{- \lambda\pi r^{2}}2\pi\lambda rdr.\end{split} \tag{5}\]  
Since \(h\) is a random variable following an exponential distribution with mean \(\frac{1}{\mu}\), the probability becomes  
\[\begin{split} P[h>T_{p}r^{\alpha}(\sigma^{2}+I_{t})|r]& =E_{I_{t}}[P[h>T_{p}r^{\alpha}(\sigma^{2}+I_{t})|r]]\\ &=E_{I_{t}}[\exp[-\mu T_{p}r^{\alpha}(\sigma^{2}+I_{t})]|r]\\ &=e^{-\mu T_{p}r^{\alpha}(\sigma^{2}+I_{t})}|r]\\ &=e^{-\mu T_{p}r^{\alpha}(\sigma^{2}+I_{t})}|r]\\ \end{split} \tag{6}\]  
where \(L(\cdot)\) denotes the Laplace transform.  
More specifically, we have  
\[\begin{split} L_{I_{t}}(s)&=E_{I_{t}}[e^{-sI_{t}}]\\ &=E_{\Phi,\{h_{i}\}}\Big{[}\exp(-s\sum\limits_{i\in\Phi/b_{0}}h_{ i}R_{i}^{-\alpha})\Big{]}\\ &=E_{\Phi}\Bigg{[}\prod\limits_{i\in\Phi/b_{0}}\frac{\mu}{\mu+sR _{i}^{-\alpha}}\Bigg{]}\\ &=\exp\left(-2\pi\lambda\int_{r}^{\infty}(1-\frac{\mu}{\mu+sv^{- \alpha}})vdv\right).\end{split}\]  
Replacing variable \(\mu\) with \((\frac{v}{rT_{p}^{1/\alpha}})^{2}\), we then have  
\[L[\mu T_{p}r^{\alpha}]=\exp(-\pi r^{2}\lambda\rho(T_{p},\alpha)), \tag{7}\]  
where \(\rho(T_{p},\alpha)=T_{p}^{-2/\alpha}\int_{T_{p}^{2/\alpha}}^{\infty}\frac{1}{1 +\mu^{\alpha/2}}d\mu\).  
It is shown in Theorem 1 that the eavesdropping probability \(P_{e}\) heavily depends on the channel conditions (such as the path loss and Rayleigh fading).