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 *d*_{i}(*x*_{i},*y*_{i}) locating at \(\mathcal {S}_{il}\), \(\mathcal {S}_{ir}\) are equal. When *d*_{i}(*x*_{i},*y*_{i}) lies in the half-circle \(\mathcal {S}_{il}\), we have

$$ t_{i} - u_{i} \leq d_{i} \leq t_{i}. $$

(31)

When *d*_{i}(*x*_{i},*y*_{i}) 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 *w*_{1},*w*_{2} 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 *a*_{1}(*d*_{0}),*a*_{1}(*d*_{1}),*a*_{2},*a*_{3} 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 *T*_{2}>0 in (29), *a*_{2}(*d*_{0}) 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 (*c*_{1}) is due to (30) and the approximation of (*c*_{2}) is based on (31), (32). Note that *a*_{2} is calculated in (40).

The term of *a*_{1}(*d*_{i}) 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 (*c*_{3}) is due to (30) and the approximation of (*c*_{4}) is based on (31), (32). Note that *a*_{3} 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 *i*th (*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 *a*_{2},*a*_{3} 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 *p*_{1}(*x*),*p*_{2}(*x*),*p*_{3} 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 *p*_{1}(*h*_{B}) has no relation with *h*_{E}, while *p*_{2}(*h*_{E}) has no relation with *h*_{B}. Moreover, *p*_{3} has no relation with *h*_{B},*h*_{E}. 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).