Secrecy Capacity Analysis for Indoor Visible Light Communications with Input-Dependent Gaussian Noise

Abstract

This paper mainly focus on the performance of secrecy capacity in the physical layer security (PLS) for the eavesdropping channel in visible light communication (VLC) system. In this system, due to the effects of thermal and shoot noises, the main interference of the channel is not only from additive white Gaussian noise (AWGN), but also dependent on the input signal. Considering a practical scenery, based on the input-dependent Gaussian noise, the closed-form expression of the upper and lower bounds of secrecy capacity are derived under the constraints of non-negative and average optical intensity. Specifically, since the entropy of the output signal is always greater than the input signal, on this basis, the derivation of lower bound is using the variational method to obtain a better input distribution. The upper bound is derived by the dual expression of channel capacity. We verified the performance of secrecy capacity through numerical results. The results show that the upper and lower bounds are relatively tight when optical intensity is high, which proves validity of the expression. In the low signal-to-noise ratio (SNR) scheme, the result of bounds with more input-dependent noise is better than less noise. And in the high SNR scheme, the result of bounds with less input-dependent noise outperforms that noise is more.

6 Appendix

1.1 6.1 Appendix A

For expression (7), according to [10], we have

$$\begin{aligned} \mathcal {H}\left( {{Y_\mathrm{{B}}}} \right)&\ge \mathcal {H}\left( {{H_\mathrm{{B}}}X} \right) + {f_{\mathrm{{low}}}}\left( {\xi P} \right) \\&= \mathcal {H}\left( X \right) + \ln \left( {{H_\mathrm{{B}}}} \right) + {f_{\mathrm{{low}}}}\left( {\xi P} \right) \nonumber \end{aligned}$$

(17)

According to Theorem 17.2.3 in [12], an upper bound of $\mathcal {H}\left( {{Y_\mathrm{{E}}}} \right)$ is given by

$$\begin{aligned} \mathcal {H}\left( {{Y_\mathrm{{E}}}} \right) \le \frac{1}{2}\ln \left[ {2\pi e{\mathop {\mathrm {var}}} \left( {{Y_{\mathrm {E}}}} \right) } \right] \end{aligned}$$

(18)

Substituting (17) and (18) into (7), ${C_s}$ can be written as

$$\begin{aligned} {C_s} \!\ge \! \mathcal {H}\left( X \right) \!+\! \ln \left( {{H_\mathrm{{B}}}} \right) \!+\! {f_{{\mathrm {low}}}}\left( {\xi P} \right) \!-\! \mathcal {H}\left( {{Y_\mathrm{{B}}}\left| X \right. } \right) \!-\! \frac{1}{2}\ln \left[ {2\pi e{\mathop {\mathrm {var}}} \left( {{Y_{\mathrm {E}}}} \right) } \right] \mathrm{{ \!+\! }}\mathcal {H}\left( {{Y_{\mathrm {E}}}\left| X \right. } \right) \end{aligned}$$

(19)

where ${f_{{\mathrm {low}}}}\left( {\xi P} \right)$, $\mathcal {H}\left( {{Y_{\mathrm {B}}}\left| X \right. } \right)$ and $\mathcal {H}\left( {{Y_{\mathrm {E}}}\left| X \right. } \right)$ are given by

$$\begin{aligned} {f_{{\mathrm {low}}}}\left( {\xi P} \right) = \frac{1}{2}\ln \left( {{H_{\mathrm {B}}} + \frac{{2\varsigma _1^2\sigma _{\mathrm {B}}^2}}{{\xi P}}} \right) - \frac{{\xi P + \varsigma _1^2\sigma _{\mathrm {B}}^2}}{{\varsigma _1^2\sigma _{\mathrm {B}}^2}} + \frac{{\sqrt{\xi P\left( {{H_{\mathrm {B}}}\xi P + 2\varsigma _1^2\sigma _{\mathrm {B}}^2} \right) } }}{{\varsigma _1^2\sigma _{\mathrm {B}}^2}} \end{aligned}$$

(20)

$$\begin{aligned} \mathcal {H}\left( {{Y_{\mathrm {B}}}\left| X \right. } \right)&= \frac{1}{2}{E_{{f_X}}}\left\{ {\ln \left[ {2\pi e\sigma _{\mathrm {B}}^2\left( {1 + \varsigma _1^2X} \right) } \right] } \right\} \nonumber \\&= \frac{1}{2}\ln \left( {2\pi e\sigma _{\mathrm {B}}^2} \right) {\mathrm { + }}\frac{1}{2}{E_{{f_X}}}\left[ {\ln \left( {1 + \varsigma _1^2X} \right) } \right] \end{aligned}$$

(21)

$$\begin{aligned} \mathcal {H}\left( {{Y_{\mathrm {E}}}\left| X \right. } \right) = \frac{1}{2}\ln \left( {2\pi e\sigma _{\mathrm {E}}^2} \right) {\mathrm { + }}\frac{1}{2}{E_{{f_X}}}\left[ {\ln \left( {1 + \varsigma _2^2X} \right) } \right] \end{aligned}$$

(22)

Then ${C_s}$ in (19) can be written as

$$\begin{aligned} {C_s}&\ge \mathcal {H}\left( X \right) + \frac{1}{2}\ln \left( {{H_{\mathrm {B}}} + \frac{{2\varsigma _1^2\sigma _{\mathrm {B}}^2}}{{\xi P}}} \right) - \frac{{\xi P + \varsigma _1^2\sigma _{\mathrm {B}}^2}}{{\varsigma _1^2\sigma _{\mathrm {B}}^2}}\nonumber \\&+ \frac{{\sqrt{\xi P\left( {{H_{\mathrm {B}}}\xi P + 2\varsigma _1^2\sigma _{\mathrm {B}}^2} \right) } }}{{\varsigma _1^2\sigma _{\mathrm {B}}^2}} - \frac{1}{2}\ln \left[ {2\pi e{\mathop {\mathrm {var}}} \left( {{Y_{\mathrm {E}}}} \right) } \right] \; + \frac{1}{2}\ln \left( {\frac{{\sigma _{\mathrm {E}}^2}}{{\sigma _{\mathrm {B}}^2}}} \right) \\&+ \frac{1}{2}{E_{{f_X}}}\left\{ {\ln \left( X \right) } \right\} + \frac{1}{2}{E_{{f_X}}}\left\{ {\ln \left( {\frac{{1 + \varsigma _2^2X}}{{X + \varsigma _1^2{X^2}}}} \right) } \right\} \nonumber \end{aligned}$$

(23)

where the ${E_{{f_X}}}\left\{ {\ln \left[ {{{\left( {1 + \varsigma _2^2X} \right) } / {\left( {X + \varsigma _1^2{X^2}} \right) }}} \right] } \right\}$ is tends to zero when X is infinite.

We select an input distribution ${f_X}\left( x \right)$ to maximizes the under the input constrains (3) and (4). Such an optimization problem as following can be solved to find the better input PDF.

$$\begin{aligned} \begin{array}{l} \mathop {\max }\limits _{{f_X}\left( x \right) } \left\{ {\mathcal {H}\left( X \right) {\mathrm { + }}\frac{1}{2}{E_{{f_X}}}\left[ {\ln \left( X \right) } \right] } \right\} \\ {\mathrm {s}}{\mathrm {.t}}{.}\;\;\;\int _0^\infty {{f_X}\left( x \right) {\mathrm {d}}x} = 1 \\ \;\;\;\;\;\;E\left( X \right) = \int _0^\infty {x{f_X}\left( x \right) {\mathrm {d}}x} = \xi P \end{array} \end{aligned}$$

(24)

Then an optimal distribution problem can be transformed as

$$\begin{aligned} \begin{array}{l} \mathop {\max }\limits _{{f_X}\left( x \right) } F\left[ {{f_X}\left( x \right) } \right] \buildrel \varDelta \over = \int _0^\infty {\left\{ {\frac{1}{2}\ln \left( x \right) - {\mathrm {ln}}\left[ {{f_X}\left( x \right) } \right] } \right\} {f_X}\left( x \right) {\mathrm {d}}x} \\ {\mathrm {s}}{\mathrm {.t}}{.}\;\;\;\int _0^\infty {{f_X}\left( x \right) {\mathrm {d}}x} = 1 \\ \;\;\;\;\;\;E\left( X \right) = \int _0^\infty {x{f_X}\left( x \right) {\mathrm {d}}x} = \xi P \end{array} \end{aligned}$$

(25)

This problem can be solved by variational method. Assuming that optimal result in (24) is ${f_X}\left( x \right)$, then define a perturbation function as

$$\begin{aligned} {\tilde{f}_X}\left( x \right) = {f_X}\left( x \right) {\mathrm { + }}\varepsilon \eta \left( x \right) \end{aligned}$$

(26)

where $\varepsilon$ is a variable, and $\eta \left( x \right)$ is a function, where x is the independent variable of the function. And the perturbation function in (26) should also satisfy the constraints in (24). So we have

$$\begin{aligned} \left\{ {\begin{array}{c} {\int _0^\infty {\eta \left( x \right) {\mathrm {d}}x = 0} } \\ {\int _0^\infty {x\eta \left( x \right) {\mathrm {d}}x} = 0} \end{array}} \right. \end{aligned}$$

(27)

Then define a function $\rho \left( \varepsilon \right)$, where $\varepsilon$ is the independent variable of the function.

$$\begin{aligned} \rho \left( \varepsilon \right) = F\left[ {{{\tilde{f}}_X}\left( x \right) } \right] = F\left[ {{f_X}\left( x \right) \mathrm{{ + }}\varepsilon \eta \left( x \right) } \right] \end{aligned}$$

(28)

The extremum value is obtained when $\varepsilon = 0$, the first variation can be expressed as

$$\begin{aligned} {\left. {\frac{{{\mathrm {d}}\rho \left( \varepsilon \right) }}{{{\mathrm {d}}\varepsilon }}} \right| _{\varepsilon = 0}} = \int _0^\infty {\left\{ {\frac{1}{2}\ln \left( x \right) - {\mathrm {ln}}\left[ {{f_X}\left( x \right) } \right] - 1} \right\} \eta \left( x \right) {\mathrm {d}}x} = 0 \end{aligned}$$

(29)

So we have

$$\begin{aligned} {f_X}\left( x \right) = \sqrt{x} {e^{ - cx - b}} \end{aligned}$$

(30)

where b and c are the free parameters. Submitting (30) into the constrains in (25), we have

$$\begin{aligned} \left\{ {\begin{array}{c} {b = \ln \left[ {\frac{{\sqrt{2\pi } {{\left( {\xi P} \right) }^{\frac{3}{2}}}}}{{3\sqrt{3} }}} \right] } \\ {c = \frac{3}{{2\xi P}}}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \end{array}} \right. \end{aligned}$$

(31)

And ${f_X}\left( x \right)$ in (30) can be written as

$$\begin{aligned} {f_X}\left( x \right) = \frac{3}{{\xi P}}\sqrt{\frac{3}{{2\pi \xi P}}} \sqrt{x} {e^{ - \frac{3}{{2\xi P}}x}} \end{aligned}$$

(32)

The unknowns $\mathcal {H}\left( X \right)$, $\frac{1}{2}{E_{{f_X}}}\left\{ {\ln \left( X \right) } \right\}$ and $\frac{1}{2}{E_{{f_X}}}\big \{ {\ln }$ ${\left[ {{{\left( {1 + \varsigma _2^2X} \right) } / {\left( {X + \varsigma _1^2{X^2}} \right) }}} \right] } \big \}$ in (23) can be solved as following

$$\begin{aligned} \mathcal {H}\left( X \right)&= - \int _0^\infty {{f_X}\left( x \right) \ln \left[ {{f_X}\left( x \right) } \right] } {\mathrm {d}}x\nonumber \\&= - \frac{{3\sqrt{3} }}{{\sqrt{2\pi } {{\left( {\xi P} \right) }^{\frac{3}{2}}}}}\left\{ {\ln \left[ {\frac{{3\sqrt{3} }}{{\sqrt{2\pi } {{\left( {\xi P} \right) }^{\frac{3}{2}}}}}} \right] \int _0^\infty {\sqrt{x} {e^{ - \frac{3}{{2\xi P}}x}}} {\mathrm {d}}x} \right. \nonumber \\&\quad + \left. {\frac{1}{2}\int _0^\infty {\sqrt{x} {e^{ - \frac{3}{{2\xi P}}x}}\ln \left( x \right) } {\mathrm {d}}x - \frac{3}{{2\xi P}}\int _0^\infty {\sqrt{x} {e^{ - \frac{3}{{2\xi P}}x}}x} {\mathrm {d}}x} \right\} \\&\le \frac{1}{2} + \sqrt{\frac{6}{{\pi \xi P}}} - \ln \left[ {\frac{{3\sqrt{3} }}{{\sqrt{2\pi } {{\left( {\xi P} \right) }^{\frac{3}{2}}}}}} \right] \nonumber \end{aligned}$$

(33)

$$\begin{aligned}&\frac{1}{2}{E_{{f_X}}}\left\{ {\ln \left( X \right) } \right\} + \frac{1}{2}{E_{{f_X}}}\left\{ {\ln \left( {\frac{{1 + \varsigma _2^2X}}{{X + \varsigma _1^2{X^2}}}} \right) } \right\} \nonumber \\&= \frac{1}{2}{E_{{f_X}}}\left\{ {\ln \left( {\frac{{1 + \varsigma _2^2X}}{{1 + \varsigma _1^2X}}} \right) } \right\} \nonumber \\&\le \frac{1}{2}\int _0^\infty {\frac{{3\sqrt{3} }}{{\sqrt{2\pi } {{\left( {\xi P} \right) }^{\frac{3}{2}}}}}\sqrt{x} {e^{ - \frac{3}{{2\xi P}}x}}\left( {\frac{{1 + \varsigma _2^2x}}{{1 + \varsigma _1^2x}} - 1} \right) {\mathrm {d}}x}\\&= \frac{1}{2}\left( {\frac{{\varsigma _2^2}}{{\varsigma _1^2}} - 1} \right) erfc\left( 0 \right) = \frac{{\varsigma _2^2}}{{2\varsigma _1^2}} - \frac{1}{2}\nonumber \end{aligned}$$

(34)

As for ${\mathop {\mathrm {var}}} \left( {{Y_{\mathrm {E}}}} \right)$, we have

$$\begin{aligned} {\mathop {\mathrm {var}}} \left( {{Y_{\mathrm {E}}}} \right)&{\mathrm { = var}}\left( {{H_{\mathrm {E}}}X} \right) + {\mathop {\mathrm {var}}} \left( {\sqrt{{H_{\mathrm {E}}}X} {Z_{2}}} \right) {\mathrm { + }}{\mathop {\mathrm {var}}} \left( {{Z_{\mathrm {E}}}} \right) \nonumber \\&= H_{\mathrm {E}}^2{\mathrm {var}}\left( X \right) + {H_{\mathrm {E}}}{\mathop {\mathrm {var}}} \left( {\sqrt{X} {Z_{2}}} \right) {\mathrm { + }}{\mathop {\mathrm {var}}} \left( {{Z_{\mathrm {E}}}} \right) \end{aligned}$$

(35)

The ${\mathrm {var}}\left( X \right)$ and ${\mathop {\mathrm {var}}} \left( {\sqrt{X} {Z_{2}}} \right)$ can be written as

$$\begin{aligned}&\begin{array}{c} {\mathrm {var}}\left( X \right) = E\left( {{X^2}} \right) - {\left[ {E\left( X \right) } \right] ^2}= \frac{2}{3}{\xi ^2}{P^2} \end{array} \end{aligned}$$

(36)

$$\begin{aligned}&\begin{array}{c} {\mathop {\mathrm {var}}} \left( {\sqrt{X} {Z_{2}}} \right) = E\left[ {{{\left( {\sqrt{X} } \right) }^2}Z_2^2} \right] - {\left[ {E\left( {\sqrt{X} {Z_{2}}} \right) } \right] ^2}=\frac{{16}}{{\sqrt{2\pi } }}\xi P\varsigma _2^5\sigma _{\mathrm {E}}^5 \end{array} \end{aligned}$$

(37)

Submitting (36) and (37) into (35), we have

$$\begin{aligned} {\mathop {\mathrm {var}}} \left( {{Y_{\mathrm {E}}}} \right) = \frac{2}{3}H_{\mathrm {E}}^2{\xi ^2}{P^2}\mathrm{{ + }}\frac{{16}}{{\sqrt{2\pi } }}{H_{\mathrm {E}}}\xi P\varsigma _2^5\sigma _{\mathrm {E}}^5\mathrm{{ + }}\sigma _{\mathrm {E}}^2 \end{aligned}$$

(38)

Then substituting (33), (34) and (38) into (23), (8) can be derived.

1.2 6.2 Appendix B

Expression (14) can be written as

$$\begin{aligned} \begin{array}{l} \!\!\!\!\!\!{C_s} \!\le \! \underbrace{{E_{{X^*}}}\!\!\left\{ {\int _{ - \infty }^\infty {\int _{ - \infty }^\infty {{f_{\left. {{Y_{\mathrm {B}}}{Y_{\mathrm {E}}}} \right| X}}\left( {\left. {{y_{\mathrm {B}}},{y_{\mathrm {E}}}} \right| X} \right) \ln \left[ {{f_{\left. {{Y_{\mathrm {B}}}} \right| X{Y_{\mathrm {E}}}}}\left( {\left. {{y_{\mathrm {B}}}} \right| X,{y_{\mathrm {E}}}} \right) } \right] {\mathrm {d}}{y_{\mathrm {B}}}{\mathrm {d}}{y_{\mathrm {E}}}} } } \right\} }_{{I_1}}\\ \;\;\;\;\;\;\underbrace{ - {E_{{X^*}}}\!\!\left\{ {\int _{ - \infty }^\infty {\int _{ - \infty }^\infty {{f_{\left. {{Y_{\mathrm {B}}}{Y_{\mathrm {E}}}} \right| X}}\left( {\left. {{y_\mathrm{{B}}},{y_{\mathrm {E}}}} \right| X} \right) \ln \left[ {{g_{\left. {{Y_{\mathrm {B}}}} \right| {Y_{\mathrm {E}}}}}\left( {\left. {{y_{\mathrm {B}}}} \right| {y_\mathrm{{E}}}} \right) } \right] \mathrm{{d}}{y_\mathrm{{B}}}\mathrm{{d}}{y_\mathrm{{E}}}} } } \right\} }_{{I_2}} \end{array} \end{aligned}$$

(39)

${I_1}$ can be written as

$$\begin{aligned} {I_1}&= {E_{{X^*}}}\left\{ {\int _{ - \infty }^\infty {\int _{ - \infty }^\infty {{f_{\left. {{Y_\mathrm{{B}}}} \right| X{Y_\mathrm{{E}}}}}\left( {\left. {{y_\mathrm{{B}}}} \right| X,{y_\mathrm{{E}}}} \right) \ln \left[ {{f_{\left. {{Y_\mathrm{{B}}}} \right| X{Y_\mathrm{{E}}}}}\left( {\left. {{y_\mathrm{{B}}}} \right| X,{y_\mathrm{{E}}}} \right) } \right] \mathrm{{d}}{y_\mathrm{{B}}}\mathrm{{d}}{y_\mathrm{{E}}}} } } \right\} \nonumber \\&= - \mathcal {H}\left( {\left. {{Y_\mathrm{{B}}}} \right| {X^*},{Y_\mathrm{{E}}}} \right) \\&= - \left[ {\mathcal {H}\left( {\left. {{Y_\mathrm{{B}}}} \right| {X^*}} \right) + \mathcal {H}\left( {\left. {{Y_\mathrm{{E}}}} \right| {X^*},{Y_\mathrm{{B}}}} \right) - \mathcal {H}\left( {\left. {{Y_\mathrm{{E}}}} \right| {X^*}} \right) } \right] \nonumber \end{aligned}$$

(40)

Obtained by (21) and (22), $\mathcal {H}\left( {\left. {{Y_\mathrm{{B}}}} \right| {X^*}} \right)$ and $\mathcal {H}\left( {\left. {{Y_\mathrm{{E}}}} \right| {X^*}} \right)$ can be written as

$$\begin{aligned} \left\{ {\begin{array}{c} {\mathcal {H}\left( {\left. {{Y_\mathrm{{B}}}} \right| {X^*}} \right) = H\left( {\left. {{Y_\mathrm{{B}}}} \right| X} \right) = \frac{1}{2}\ln \left( {2\pi e\sigma _\mathrm{{B}}^2} \right) \mathrm{{ + }}\frac{1}{2}{E_{{f_X}}}\left[ {\ln \left( {1 + \varsigma _1^2X} \right) } \right] }\\ {\mathcal {H}\left( {\left. {{Y_\mathrm{{E}}}} \right| {X^*}} \right) = H\left( {\left. {{Y_\mathrm{{E}}}} \right| X} \right) = \frac{1}{2}\ln \left( {2\pi e\sigma _\mathrm{{E}}^2} \right) \mathrm{{ + }}\frac{1}{2}{E_{{f_X}}}\left[ {\ln \left( {1 + \varsigma _2^2X} \right) } \right] } \end{array}} \right. \end{aligned}$$

(41)

As for $H\left( {\left. {{Y_\mathrm{{E}}}} \right| {X^*},{Y_\mathrm{{B}}}} \right)$, the conditional PDF ${f_{\left. {{Y_\mathrm{{E}}}} \right| {Y_\mathrm{{B}}}}}\left( {\left. {{y_\mathrm{{E}}}} \right| {y_\mathrm{{B}}}} \right)$ can be expressed as

(42)

$\mathcal {H}\left( {\left. {{Y_\mathrm{{E}}}} \right| {X^*},{Y_\mathrm{{B}}}} \right)$ can be expressed as

(43)

Substituting (43) and (41) into (40), we have

$$\begin{aligned} {I_1}&\le \frac{1}{2}\mathrm{{ + }}\frac{1}{{\sqrt{\pi }}} - \frac{{\sqrt{2} }}{2} + \ln \left( {\frac{{\sigma _\mathrm{{E}}^{}}}{{\sigma _\mathrm{{B}}^{}}}} \right) - \frac{{\varsigma _2^2}}{{2\varsigma _1^2}}\nonumber \\&\quad - \frac{{\ln \left( {2\pi } \right) }}{{\sqrt{\pi }}} - \left( {\frac{1}{{\sqrt{\pi }}} + \frac{{\sqrt{2} }}{2}} \right) \left( {\frac{{H_\mathrm{{E}}^2}}{{H_\mathrm{{B}}^2}}\sigma _\mathrm{{B}}^2 + \sigma _\mathrm{{E}}^2} \right) \\&\quad - \xi P\left( {\frac{1}{2} + \frac{{\sqrt{2\pi } }}{4}} \right) \left( {\frac{{H_\mathrm{{E}}^2}}{{H_\mathrm{{B}}^2}}{H_\mathrm{{B}}}\varsigma _1^2\sigma _\mathrm{{B}}^2 + {H_\mathrm{{E}}}\varsigma _2^2\sigma _\mathrm{{E}}^2} \right) \nonumber \end{aligned}$$

(44)

One of the difficulties in solving ${I_2}$ is that the input signal X has no peak intensity constraints so it is difficult to find the bounds of upper bound because the signal range can be arbitrarily large. In order to get ${I_2}$, ${g_{\left. {{Y_\mathrm{{B}}}} \right| {Y_\mathrm{{E}}}}}\left( {\left. {{y_\mathrm{{B}}}} \right| {y_\mathrm{{E}}}} \right)$ can be chosen as [10]

$$\begin{aligned} {g_{\left. {{Y_\mathrm{{B}}}} \right| {Y_\mathrm{{E}}}}}\left( {\left. {{y_\mathrm{{B}}}} \right| {y_\mathrm{{E}}}} \right) = \frac{1}{{2{s^2}}}{e^{ - \frac{{\left| {{y_\mathrm{{B}}} - \mu {y_\mathrm{{E}}}} \right| }}{{{s^2}}}}} \end{aligned}$$

(45)

where s and $\mu$ are free parameters.

Let $p \!=\! \left[ {{{H_\mathrm{{E}}^2}/{H_\mathrm{{B}}^2}} + \left( {{{H_\mathrm{{E}}^2}/ {{H_\mathrm{{B}}}}}} \right) x\varsigma _1^2} \right] \sigma _\mathrm{{B}}^2\mathrm{{ + }} \left( {1 + {H_\mathrm{{E}}}x\varsigma _2^2} \right) \sigma _\mathrm{{E}}^2$ and $q = \left( {1 + {H_\mathrm{{B}}}x\varsigma _1^2} \right) \sigma _\mathrm{{B}}^2$. Therefore, ${f_{\left. {{Y_\mathrm{{B}}}{Y_\mathrm{{E}}}} \right| X}}\left( {\left. {{y_\mathrm{{B}}},{y_\mathrm{{E}}}} \right| X} \right)$ can be written as

$$\begin{aligned} {f_{\left. {{Y_\mathrm{{B}}}{Y_\mathrm{{E}}}} \right| X}}\left( {\left. {{y_\mathrm{{B}}},{y_\mathrm{{E}}}} \right| X} \right) = \frac{1}{{\sqrt{2\pi q} }}{e^{ - \frac{{{{\left( {{y_\mathrm{{B}}} - {H_\mathrm{{B}}}x} \right) }^2}}}{{2q}}}} \times \frac{1}{{\sqrt{2\pi p} }}{e^{ - \frac{{{{\left( {{y_\mathrm{{E}}} - \frac{{{H_\mathrm{{E}}}}}{{{H_\mathrm{{B}}}}}{y_\mathrm{{B}}}} \right) }^2}}}{{2p}}}} \end{aligned}$$

(46)

Substituting (46) into (39), ${I_2}$ can be written as

(47)

Then ${I_2}$ can be upper-bounded by

(48)

Case1: when $\mu < 0$, ${I_3}$ is given by

(49)

Case2: when , ${I_3}$ is given by

(50)

So ${I_3}$ can be lower-bounded by

(51)

Case3: when , ${I_3}$ is given by

(52)

From three cases, ${I_3}$ can be expressed as

(53)

Substituting (53) into (48) ${I_2}$ can be written as

(54)

Then substituting (44) and (54) into (39), the secrecy capacity (15) can be derived.