For simplicity, let
$$ h_{i} = T_{1} \left(l^{2} + d_{i}^{2}\right)^{-T_{2}}, $$
(28)
where \(T_{1} = \frac {S_{r} (m+1)l^{m+1}}{2\pi }\), \(T_{2} = \frac {m+3}{2}\). For indoor VLC system, \(m = -\frac {\ln (2)}{\ln \left (\cos \left (\Phi _{\frac {1}{2}}\right)\right)}>0\). Thus
$$ T_{1} > 0,~ T_{2} > 0. $$
(29)
Note that in (28) and (33), the subscript of i=0 is for the legitimate receiver, while the subscripts of i≥1 are for the eavesdroppers.
Here, we divide the circle \(\mathcal {S}_{i}\), as shown in Fig. 8. For i∈[0,N], we have
$$ \begin{aligned} &~~~~ \mathcal{S}_{i} = \mathcal{S}_{il} + \mathcal{S}_{ir}, \\ & \int\!\!\!\!\int\limits_{\mathcal{S}_{il}} f(d_{i}) ~\mathrm{d} \tau_{i} =\int\!\!\!\!\int\limits_{\mathcal{S}_{ir}} f(d_{i}) ~\mathrm{d} \tau_{i} = \frac{1}{2}\int\!\!\!\!\int\limits_{\mathcal{S}_{i}} f(d_{i}) ~\mathrm{d} \tau_{i}, \end{aligned} $$
(30)
which means that the possibilities of the receiver di(xi,yi) locating at \(\mathcal {S}_{il}\), \(\mathcal {S}_{ir}\) are equal. When di(xi,yi) lies in the half-circle \(\mathcal {S}_{il}\), we have
$$ t_{i} - u_{i} \leq d_{i} \leq t_{i}. $$
(31)
When di(xi,yi) lies in the half-circle \(\mathcal {S}_{ir}\), we have
$$ t_{i} \leq d_{i} \leq t_{i} + u_{i}. $$
(32)
By using the polar coordinate system, let
$$ \begin{aligned} & x_{i} = t_{i1} + \rho_{i} \cos\theta_{i}, \\ & y_{i} = t_{i2} + \rho_{i} \sin\theta_{i}. \end{aligned} $$
(33)
Proof of upper bounds (20) and (23)
We derive the upper bounds based on (13) and (10). Then, we have
$$ \begin{aligned} & \bar{\mathcal{C}}_{s}(h_{B}, h_{E}) = \int\!\!\!\!\int\limits_{\mathcal{S}_{0}} \int\!\!\!\!\int\limits_{\mathcal{S}_{i}} \mathcal{C}_{s}(d_{0}, d_{i}) f(d_{0}) f(d_{i}) ~\mathrm{d} \tau_{0} \mathrm{d} \tau_{i} \\ & \leq \int\!\!\!\!\int\limits_{\mathcal{S}_{0}} \int\!\!\!\!\int\limits_{\mathcal{S}_{i}} \frac{1}{2} \log \frac{h_{B}^{2} A^{2} + \sigma^{2} }{h_{E}^{2} A^{2} + \sigma^{2}} f(d_{0}) f(d_{i}) ~\mathrm{d} \tau_{0} \mathrm{d} \tau_{i} \\ & = \frac{1}{2} \underbrace{ \int\!\!\!\!\int\limits_{\mathcal{S}_{0}} \log (h_{B}^{2} A^{2} + \sigma^{2}) f(d_{0}) ~\mathrm{d} \tau_{0} \int\!\!\!\!\int\limits_{\mathcal{S}_{i}} f(d_{i}) ~\mathrm{d} \tau_{i} }_{w_{1}} \\ & ~~ - \frac{1}{2} \underbrace{ \int\!\!\!\!\int\limits_{\mathcal{S}_{0}} f(d_{0}) ~\mathrm{d} \tau_{0} \int\!\!\!\!\int\limits_{\mathcal{S}_{i}} \log (h_{E}^{2} A^{2} + \sigma^{2}) f(d_{i}) ~\mathrm{d} \tau_{i} }_{w_{2}} \end{aligned} $$
(34)
Based on (28), the terms of w1,w2 are given by
$$ \begin{aligned} w_{1} & = \int\!\!\!\!\int\limits_{\mathcal{S}_{0}} \log \left[\sigma^{2} + A^{2} T_{1}^{2} \left(l^{2}+d_{0}^{2}\right)^{-2 T_{2}}\right] \\ & \hspace{5mm} \cdot f(d_{0}) ~\mathrm{d} \tau_{0} \int\!\!\!\!\int\limits_{\mathcal{S}_{i}} f(d_{i}) ~\mathrm{d} \tau_{i}, \end{aligned} $$
(35)
$$ \begin{aligned} w_{2} & = \int\!\!\!\!\int\limits_{\mathcal{S}_{i}} \log \left[\sigma^{2} + A^{2} T_{1}^{2} \left(l^{2}+d_{i}^{2}\right)^{-2 T_{2}}\right] \\ & \hspace{5mm} \cdot f(d_{i}) ~\mathrm{d} \tau_{i} \int\!\!\!\!\int\limits_{\mathcal{S}_{0}} f(d_{0}) ~\mathrm{d} \tau_{0}. \end{aligned} $$
(36)
Next, we derive the upper bounds under two typical distribution.
1) When Gaussian distribution.
For (35) and (36), let us denote
$$ w_{1} = \frac{1}{4\pi^{2} \zeta_{0}^{2} \zeta_{i}^{2}} ~ a_{1}(d_{0}) \cdot a_{3}, $$
(37)
$$ w_{2} = \frac{1}{4\pi^{2} \zeta_{0}^{2} \zeta_{i}^{2}} ~ a_{1}(d_{i})\cdot a_{2}, (i\geq 1) $$
(38)
where the terms of a1(d0),a1(d1),a2,a3 are given in the following,
$$ \begin{aligned} a_{1}(d_{i}) = & \int\!\!\!\!\int\limits_{\mathcal{S}_{i}} \left(\log \left[\sigma^{2} + A^{2} T_{1}^{2} \left(l^{2}+d_{i}^{2}\right)^{-2 T_{2}} \right] \right) \\ & ~~~~ \cdot e^{-\frac{(x_{i} - t_{i1})^{2} + (y_{i} - t_{i2})^{2} }{2\zeta_{0}^{2}}} ~\mathrm{d} \tau_{0}, (i \geq 0). \end{aligned} $$
(39)
$$ \begin{aligned} a_{2} & = \int\!\!\!\!\int\limits_{\mathcal{S}_{0}} e^{-\frac{(x_{0} - t_{01})^{2} + (y_{0} - t_{02})^{2} }{2\zeta_{0}^{2}}} ~\mathrm{d} \tau_{0} \\ & = \int_{0}^{2\pi}\!\!\!\! \int_{0}^{u_{0}} e^{-\frac{\rho_{0}^{2}}{2\zeta_{0}^{2}}} \rho_{0} ~\mathrm{d} \rho_{0} \mathrm{d} \theta_{0} = 2\pi \zeta_{0}^{2} \left(1- e^{-\frac{u_{0}^{2}}{2\zeta_{0}^{2}}}\right). \end{aligned} $$
(40)
Similarly,
$$ \begin{aligned} a_{3} = \int\!\!\!\!\int\limits_{\mathcal{S}_{i}} e^{-\frac{(x_{i} - t_{i1})^{2} + (y_{i} - t_{i2})^{2} }{2\zeta_{i}^{2}}} ~\mathrm{d} \tau_{i} = 2\pi \zeta_{i}^{2} \left(1- e^{-\frac{u_{i}^{2}}{2\zeta_{i}^{2}}}\right). \end{aligned} $$
(41)
With T2>0 in (29), a2(d0) can be upper-bounded by
$$ \begin{aligned} & a_{1}(d_{0}) \underset{=}{(c_{1})} \int\!\!\!\!\int\limits_{\mathcal{S}_{0l}} + \int\!\!\!\!\int\limits_{\mathcal{S}_{0r}} \\ & \underset{\leq}{(c_{2})} \log \left[\sigma^{2}+ A^{2} T_{1}^{2} \left[ l^{2} + (t_{0} - u_{0})^{2}\right]^{-2 T_{2}} \right] \\ & ~~~ \cdot \int\!\!\!\!\int\limits_{\mathcal{S}_{0l}} \!\! e^{-\frac{(x_{0} - t_{01})^{2} + (y_{0} - t_{02})^{2} }{2\zeta_{0}^{2}}} \!\! \mathrm{d} \tau_{0} \!\! + \!\! \int\!\!\!\!\int\limits_{\mathcal{S}_{0r}} \!\! e^{-\frac{(x_{0} - t_{01})^{2} + (y_{0} - t_{02})^{2} }{2\zeta_{0}^{2}}} \!\! \mathrm{d} \tau_{0} \\ & ~~~ \cdot \log \left[\sigma^{2}+ A^{2} T_{1}^{2} \left[ l^{2} +t_{0}^{2}\right]^{-2 T_{2}} \right] \\ & = \frac{a_{2}}{2} \left(\log \left[\sigma^{2}+A^{2} T_{1}^{2} [ l^{2} + (t_{0} - u_{0})^{2}]^{-2 T_{2}}\right] +\right.\\ & ~~~~\left.\log \left[\sigma^{2}+A^{2} T_{1}^{2} (l^{2} + t_{0}^{2})^{-2 T_{2}}\right] \right), \end{aligned} $$
(42)
where the transformation of (c1) is due to (30) and the approximation of (c2) is based on (31), (32). Note that a2 is calculated in (40).
The term of a1(di) can be bounded by
$$ \begin{aligned} & a_{1}(d_{i}) \underset{=}{(c_{3})} \int\!\!\!\!\int\limits_{\mathcal{S}_{il}} + \int\!\!\!\!\int\limits_{\mathcal{S}_{ir}} \\ &\underset{\geq}{(c_{4})}\log \left[\sigma^{2}+ A^{2} T_{1}^{2}\left(l^{2}+t_{i}^{2}\right)^{-2 T_{2}} \right] \\ & ~~ \cdot \int\!\!\!\!\int\limits_{\mathcal{S}_{i}} \!\! e^{-\frac{(x_{i} - t_{i1})^{2} + (y_{i} - t_{i2})^{2} }{2\zeta_{i}^{2}}} \!\! \mathrm{d} \tau_{i} \!\! + \!\! \int\!\!\!\!\int\limits_{\mathcal{S}_{i}}\!\! e^{-\frac{(x_{i} - t_{i1})^{2} + (y_{i} - t_{i2})^{2} }{2\zeta_{i}^{2}}} \!\! \mathrm{d} \tau_{i} \\ & ~~ \cdot \log \left[\sigma^{2}+ A^{2} T_{1}^{2} \left(l^{2}+(t_{i}+u_{i})^{2}\right)^{-2 T_{2}} \right] \\ & = \frac{a_{3}}{2}\left(\log \left[\sigma^{2} + A^{2} T_{1}^{2} (l^{2}+t_{i}^{2})^{-2 T_{2}} \right] +\right. \\ & ~~~~ \left.\log \left[\sigma^{2} + A^{2} T_{1}^{2} \left(l^{2}+(t_{i}+u_{i})^{2}\right)^{-2 T_{2}} \right] \right), \end{aligned} $$
(43)
where the transformation of (c3) is due to (30) and the approximation of (c4) is based on (31), (32). Note that a3 is calculated in (41).
Finally, combining (34) with (42), (43), we can obtain
$$ \begin{aligned} & \bar{\mathcal{C}}_{s}(h_{B}, h_{E}) \leq \frac{1}{2}(w_{1} - w_{2}) \\ & = \frac{1}{8\pi^{2} \zeta_{0}^{2} \zeta_{i}^{2}} ~ [ a_{1}(d_{0}) \cdot a_{3} - a_{1}(d_{i}) \cdot a_{2} ] \\ & \leq \frac{a_{2} * a_{3}}{16\pi^{2} \zeta_{0}^{2} \zeta_{i}^{2}} \cdot \left(\log \left[\sigma^{2}+A^{2} T_{1}^{2} [ l^{2} + (t_{0} - u_{0})^{2}]^{-2 T_{2}}\right] \right.\\ & \hspace{7mm} + \log \left[\sigma^{2}+A^{2} T_{1}^{2} \left(l^{2} + t_{0}^{2}\right)^{-2 T_{2}}\right] - \\ & \hspace{10mm} \log \left[\sigma^{2} + A^{2} T_{1}^{2} (l^{2}+t_{i}^{2})^{-2 T_{2}} \right] - \\ & \hspace{10mm} \left.\log \left[\sigma^{2} + A^{2} T_{1}^{2} (l^{2}+(t_{i}+u_{i})^{2})^{-2 T_{2}} \right] \right) \\ & = \frac{\left(1- e^{-\frac{u_{0}^{2}}{2\zeta_{0}^{2}}}\right)\left(1- e^{-\frac{u_{i}^{2}}{2\zeta_{i}^{2}}}\right)}{4} \mathcal{G}_{1}(u_{0},t_{0},u_{i},t_{i}). \end{aligned} $$
(44)
This completes the proof of (20).
2) When uniform distribution.
For the ith (i≥0) receiver,
$$ \int\!\!\!\!\int\limits_{\mathcal{S}_{i}} f(d_{i}) ~\mathrm{d} \tau_{i} = \frac{1}{\pi u_{i}^{2}} \int_{0}^{2\pi}\!\!\! \int_{0}^{u_{i}} \rho_{i}~ \mathrm{d} \rho_{i} \mathrm{d} \theta_{i} = 1. $$
(45)
Then, using the similar method of proofing (44), we have
$$ \begin{aligned} & \bar{\mathcal{C}}_{s}(h_{B}, h_{E}) \leq \frac{1}{2}(w_{1} - w_{2}) \\ & \leq \frac{ \int\!\!\!\!\int\limits_{\mathcal{S}_{0}} \!\! f(d_{0}) \mathrm{d} \tau_{0}}{4} \cdot \left(\log\left[\sigma^{2} \!+ \! A^{2} T_{1}^{2} \left[ l^{2} \! +\! (t_{0} \,-\, u_{0})^{2}\right]^{-2 T_{2}}\right] \right.\\ & \hspace{6mm} + \left.\log \left[\sigma^{2}+A^{2} T_{1}^{2} \left(l^{2} + t_{0}^{2}\right)^{-2 T_{2}}\right] \right) \\ & \hspace{3mm}- \frac{ \int\!\!\!\!\int\limits_{\mathcal{S}_{i}} \!\! f(d_{i}) \mathrm{d} \tau_{i}}{4} \left(\log \left[\sigma^{2} + A^{2} T_{1}^{2} \left(l^{2}+t_{i}^{2}\right)^{-2 T_{2}} \right] \right.\\ & \hspace{6mm} +\left. \log \left[\sigma^{2} + A^{2} T_{1}^{2} \left(l^{2}+(t_{i}+u_{i})^{2}\right)^{-2 T_{2}} \right] \right)\\ & = \frac{1}{4} \mathcal{G}_{1}(u_{0},t_{0},u_{i},t_{i}). \end{aligned} $$
(46)
This completes the proof of (23).
Proof of lower bound (21) and (24)
We derive the upper bounds based on (12) and (10). Then, we have
$$ \begin{aligned} & \bar{\mathcal{C}}_{s}(h_{B}, h_{E}) = \int\!\!\!\!\int\limits_{\mathcal{S}_{0}} \int\!\!\!\!\int\limits_{\mathcal{S}_{i}} \mathcal{C}_{s}(d_{0}, d_{i}) f(d_{0}) f(d_{i}) ~\mathrm{d} \tau_{0} \mathrm{d} \tau_{i} \\ & \geq \int\!\!\!\!\int\limits_{\mathcal{S}_{0}} \int\!\!\!\!\int\limits_{\mathcal{S}_{i}} \frac{1}{2} \log \frac{6 h_{B}^{2} A^{2} + 3 \pi e \sigma^{2}}{\pi e h_{E}^{2} A^{2} + 3 \pi e \sigma^{2}} f(d_{0}) f(d_{i}) ~\mathrm{d} \tau_{0} \mathrm{d} \tau_{i} \\ & = \underbrace{ \int\!\!\!\!\int\limits_{\mathcal{S}_{0}} \frac{ \log \left(6 h_{B}^{2} A^{2} + 3 \pi e \sigma^{2}\right)}{2} f(d_{0}) \mathrm{d} \tau_{0} }_{b_{1}(d_{0})} \underbrace{ \int\!\!\!\!\int\limits_{\mathcal{S}_{i}} \! f(d_{i}) \mathrm{d} \tau_{i}}_{b_{3}} \\ & ~ - \!\! \underbrace{ \int\!\!\!\!\int\limits_{\mathcal{S}_{0}} f(d_{0}) \mathrm{d} \tau_{0} }_{b_{2}} \underbrace{ \int\!\!\!\!\int\limits_{\mathcal{S}_{i}} \frac{ \log \left(\pi e h_{E}^{2} A^{2} + 3 \pi e \sigma^{2}\right)}{2} f(d_{i}) \mathrm{d} \tau_{i}}_{b_{1}(d_{i})} \\ & = b_{1}(d_{0}) \cdot b_{3} - b_{2} \cdot b_{1}(d_{i}). \end{aligned} $$
(47)
1) When Gaussian Distribution. It is easy to know that
$$ b_{2} = a_{2}/\left(2\pi \zeta_{0}^{2}\right), ~ b_{3} = a_{3}/\left(2\pi \zeta_{i}^{2}\right), $$
(48)
where a2,a3 are in (40) and (41). Applying to the similar method of proofing (44), we have
$$ {}\begin{aligned} b_{1}(d_{0}) \geq & \frac{b_{2}}{4}\left(\log \left[3\pi e \sigma^{2} +6 A^{2} T_{1}^{2} \left(l^{2}+t_{0}^{2}\right)^{-2 T_{2}}\right] + \right.\\ & \left.\log \left[3\pi e \sigma^{2} +6 A^{2} T_{1}^{2} \left(l^{2}+(t_{0}+u_{0})^{2}\right)^{-2 T_{2}} \right] \right), \end{aligned} $$
(49)
$$ {}\begin{aligned} b_{1}(d_{i}) \leq & \frac{b_{3}}{4}\left(\log \left[3\pi e \sigma^{2} + \pi e A^{2} T_{1}^{2} \left(l^{2}+t_{i}^{2}\right)^{-2 T_{2}}\right] + \right.\\ & \left.\log \left[3\pi e \sigma^{2} + \pi e A^{2} T_{1}^{2} \left(l^{2}+(t_{i}-u_{i})^{2}\right)^{-2 T_{2}} \right] \right), \end{aligned} $$
(50)
Then, the average secrecy capacity is lower-bounded by
$$ \begin{aligned} & \bar{\mathcal{C}}_{s}(h_{B}, h_{E}) = \int\!\!\!\!\int\limits_{\mathcal{S}_{0}} \int\!\!\!\!\int\limits_{\mathcal{S}_{i}} \mathcal{C}_{s}(d_{0}, d_{i}) f(d_{0}) f(d_{i}) ~\mathrm{d} \tau_{0} \mathrm{d} \tau_{i} \\ & \geq b_{1}(d_{0}) \cdot b_{3} - b_{2} \cdot b_{1}(d_{i}) \\ & \geq \frac{\left(1- e^{-\frac{u_{0}^{2}}{2\zeta_{0}^{2}}}\right)\left(1- e^{-\frac{u_{i}^{2}}{2\zeta_{i}^{2}}}\right)}{4} \\ & \hspace{4mm} \cdot \left(\log \left[3\pi e \sigma^{2} +6 A^{2} T_{1}^{2} \left(l^{2}+t_{0}^{2}\right)^{-2 T_{2}}\right]\right. \\ & \hspace{7mm} + \log \left[3\pi e \sigma^{2} +6 A^{2} T_{1}^{2} \left(l^{2}+(t_{0}+u_{0})^{2}\right)^{-2 T_{2}} \right] \\ & \hspace{7mm} - \log \left[3\pi e \sigma^{2} + \pi e A^{2} T_{1}^{2} \left(l^{2}+t_{i}^{2}\right)^{-2 T_{2}}\right] \\ & \hspace{7mm} -\left. \log \left[3\pi e \sigma^{2} + \pi e A^{2} T_{1}^{2} \left(l^{2}+(t_{i}-u_{i})^{2}\right)^{-2 T_{2}} \right] \right)\\ & \triangleq \frac{\left(1- e^{-\frac{u_{0}^{2}}{2\zeta_{0}^{2}}}\right)\left(1- e^{-\frac{u_{i}^{2}}{2\zeta_{i}^{2}}}\right)}{4} \cdot \mathcal{G}_{1}(u_{0},t_{0},u_{i},t_{i}). \end{aligned} $$
(51)
2) When Uniform Distribution. It is easy to get
$$ \begin{aligned} & \bar{\mathcal{C}}_{s}(h_{B}, h_{E}) \geq \frac{1}{4} \mathcal{G}_{1}(u_{0},t_{0},u_{i},t_{i}). \end{aligned} $$
(52)
This completes the proofs of (21) and (24
Proof of lower bounds (22) and (25)
For simplicity, let us denote the lower bound in (11) as
$$ \begin{aligned} \mathcal{C}_{L1}(h_{B}, h_{E}) \triangleq \frac{p_{1}(h_{B})}{2}- p_{2}(h_{E})- p_{3}, \end{aligned} $$
(53)
where the functions of p1(x),p2(x),p3 are
$$ p_{1}(x) = \log\left(1 + \frac{2 x^{2} A^{2}}{\pi e \sigma^{2}} \right), $$
(54)
$$ {}p_{2}(x) = \left(1- 2\mathcal{Q}\left(\frac{\delta + x A}{\sigma} \right) \right) \log \frac{2(xA + \delta)}{\sqrt{2\pi \sigma^{2}} \left(1-2\mathcal{Q}\left(\frac{\delta}{\sigma}\right) \right)}, $$
(55)
$$ p_{3} = \left(\mathcal{Q}\left(\frac{\delta}{\sigma}\right) + \frac{\delta}{\sqrt{2\pi \sigma^{2}}} e^{-\frac{\delta^{2}}{2\sigma^{2}}} - \frac{1}{2} \right). $$
(56)
We can observe that p1(hB) has no relation with hE, while p2(hE) has no relation with hB. Moreover, p3 has no relation with hB,hE. Then, the corresponding \(\bar {\mathcal {C}}_{s}(h_{B}, h_{E})\) can be lower-bounded as
$$ \begin{aligned} & \bar{\mathcal{C}}_{s}(h_{B}, h_{E}) \geq \int\!\!\!\!\int\limits_{\mathcal{S}_{0}} \int\!\!\!\!\int\limits_{\mathcal{S}_{i}} \mathcal{C}_{L1}(h_{B}, h_{E}) f(d_{0},d_{i}) ~\mathrm{d} \tau_{0} \mathrm{d} \tau_{i} \\ & = \int\!\!\!\!\int\limits_{\mathcal{S}_{0}} \int\!\!\!\!\int\limits_{\mathcal{S}_{i}} \left(\frac{p_{1}(h_{B})}{2} - p_{2}(h_{E}) - p_{3} \right) f(d_{0},d_{i}) ~\mathrm{d} \tau_{0} \mathrm{d} \tau_{i} \\ & = \int\!\!\!\!\int\limits_{\mathcal{S}_{0}} \frac{ p_{1}(h_{B})}{2} \cdot f(d_{0}) ~\mathrm{d} \tau_{0} \int\!\!\!\!\int\limits_{\mathcal{S}_{i}} f(d_{i}) ~\mathrm{d} \tau_{i} \\ & \hspace{5mm} - \int\!\!\!\!\int\limits_{\mathcal{S}_{i}} p_{2}(h_{E}) \cdot f(d_{i}) ~\mathrm{d} \tau_{i} \int\!\!\!\!\int\limits_{\mathcal{S}_{0}} f(d_{0}) ~\mathrm{d} \tau_{0} \\ & \hspace{5mm} - p_{3} \int\!\!\!\!\int\limits_{\mathcal{S}_{0}}f(d_{0})~\mathrm{d} \tau_{0} \int\!\!\!\!\int\limits_{\mathcal{S}_{i}} f(d_{i}) ~\mathrm{d} \tau_{i}. \end{aligned} $$
(57)
Applying to the similar method of proofing (44), (57) can be further approximated as
$$ {}\begin{aligned} & \bar{\mathcal{C}}_{s}(h_{B}, h_{E}) \geq \int\!\!\!\!\int\limits_{\mathcal{S}_{0}}f(d_{0})~\mathrm{d} \tau_{0} \int\!\!\!\!\int\limits_{\mathcal{S}_{i}} f(d_{i}) ~\mathrm{d} \tau_{i} \\ & \cdot \left[ \frac{1}{4} p_{1}(h_{0L}) + \frac{1}{4} p_{1}(h_{0R})- \frac{1}{2} p_{2}(h_{iL}) -\frac{1}{2} p_{2}(h_{iR}) - p_{3} \right], \end{aligned} $$
(58)
where
$$ h_{0L} = T_{1} \left(l^{2} + t_{0}^{2}\right)^{-T_{2}}, ~ h_{0R} = T_{1} \left(l^{2} + \left(t_{0}+u_{0}\right)^{2}\right)^{-T_{2}}, $$
(59)
$$ h_{iL} = T_{1} \left(l^{2} + (t_{i}-u_{i})^{2}\right)^{-T_{2}}, ~ h_{iR} = T_{1} \left(l^{2} + t_{i}^{2}\right)^{-T_{2}}. $$
(60)
When Uniform Distribution, \(\int \!\!\!\!\int \limits _{\mathcal {S}_{0}}f(d_{0})~\mathrm {d} \tau _{0} \int \!\!\!\!\int \limits _{\mathcal {S}_{i}} f(d_{i}) ~\mathrm {d} \tau _{i} = 1\). When Gaussian Distribution, \(\int \!\!\!\!\int \limits _{\mathcal {S}_{0}}f(d_{0})~\mathrm {d} \tau _{0} \int \!\!\!\!\int \limits _{\mathcal {S}_{i}} f(d_{i}) ~\mathrm {d} \tau _{i} = \left (1- e^{-\frac {u_{0}^{2}}{2\zeta _{0}^{2}}}\right)\left (1- e^{-\frac {u_{i}^{2}}{2\zeta _{i}^{2}}}\right) \).
This completes the proofs of (22) and (25).