Covert communication with noise and channel uncertainties

Abstract

Covert communication is critical to guarantee a strong security and secure user privacy. In this work, we consider adversary’s noise and channel uncertainties and analyze their impact on adversary’s optimum detection performance and the throughput of covert messages. We determine the throughput of covert messages and its gain and loss by having adversary’s channel uncertainty and fading channel, respectively. The results show that fading is essential to hide information, particularly for low noise uncertainty or at high SNR. The improvement of adversary’s optimum detection performance, hence the covert throughput, under channel uncertainty is more significant for larger noise uncertainty. The covert throughput is gained by having channel uncertainty roughly 12–19% when the noise uncertainty is about 1–2 dB.

Appendices

Appendix A

Equation (12) is derived in this “Appendix”:

$$\begin{aligned} \xi (\hat{g})&= 1 - \frac{\ln (\lambda )}{2 \ln (\alpha )} + \int _{\left( \frac{\lambda }{\hat{\sigma }_w^2} - \frac{1}{\alpha }\right) \frac{1}{\gamma }}^\infty \ln \left( \frac{\hat{\sigma }_w^2}{\alpha } \right) \frac{f_{X| \hat{g}}(x|\hat{g}) dx}{2 \ln (\alpha )} \\&\quad + \int _0^{\left( \frac{\lambda }{\hat{\sigma }_w^2} - \frac{1}{\alpha }\right) \frac{1}{\gamma }} \ln (\lambda - x P) \frac{f_{X| \hat{g}}(x|\hat{g}) dx}{2 \ln (\alpha )} \\&= 1 - \frac{\ln (\alpha \lambda /\hat{\sigma }_w^2)}{2\ln (\alpha )} \\&\quad + \int _0^{\left( \frac{\lambda }{\hat{\sigma }_w^2} - \frac{1}{\alpha }\right) \frac{1}{\gamma }} \ln \left( \frac{\lambda - x P}{\hat{\sigma }_w^2/\alpha }\right) \frac{f_{X| \hat{g}}(x|\hat{g}) dx}{2 \ln (\alpha )}. \end{aligned}$$

(48)

Fig. 9

$\eta _n/\eta _p$ varies with $\epsilon$ for some $\alpha$

Appendix B

The pseudo-convexity of $\xi (\hat{g})$ is proved in this “Appendix”. It follows from [28] that a function f(x) is strictly pseudo-convex if, for any value of $x=x_0$ such that $d f(x)/d x =0$, we have $d^2 f(x)/d x^2 > 0$.

We have

$$\begin{aligned} 2\ln (\alpha ) \frac{d \xi (\hat{g})}{d\lambda } = - \frac{1}{\lambda } + \int _0^{\left( \frac{\lambda }{\hat{\sigma }_w^2} - \frac{1}{\alpha }\right) \frac{1}{\gamma }} \frac{f_{X| \hat{g}}(x|\hat{g}) dx}{\lambda - xP}, \end{aligned}$$

(49)

and

$$\begin{aligned} 2\ln (\alpha ) \frac{d^2 \xi (\hat{g})}{d\lambda ^2}&= \frac{1}{\lambda ^2} + \frac{\alpha }{\hat{\sigma }_w^2} f_{X| \hat{g}}\left( {\left[ \frac{\lambda }{\hat{\sigma }_w^2} - \frac{1}{\alpha }\right] \frac{1}{\gamma }} \bigg | \hat{g}\right) \\&\quad - \int _0^{\left( \frac{\lambda }{\hat{\sigma }_w^2} - \frac{1}{\alpha }\right) \frac{1}{\gamma }} \frac{P f_{X| \hat{g}}(x|\hat{g}) dx }{(\lambda - x P)^2} \\&> \frac{\alpha }{\hat{\sigma }_w^2} f_{X| \hat{g}} \left( {\left[ \frac{\lambda }{\hat{\sigma }_w^2} - \frac{1}{\alpha }\right] \frac{1}{\gamma }} \bigg | \hat{g}\right) \\&\quad - \int _0^{\left( \frac{\lambda }{\hat{\sigma }_w^2} - \frac{1}{\alpha }\right) \frac{1}{\gamma }} f_{X| \hat{g}}(x|\hat{g}) d\left( \frac{1}{\lambda - x P} \right) \\&= \frac{f_{X|\hat{g}}(0|\hat{g})}{\lambda } - \int _0^{\left( \frac{\lambda }{\hat{\sigma }_w^2} - \frac{1}{\alpha }\right) \frac{1}{\gamma }} \frac{f'_{X|\hat{g}}(x|\hat{g}) dx }{\lambda - x P}, \end{aligned}$$

(50)

respectively. Then, for any $\lambda =\lambda _0$ such that

$$\begin{aligned} \frac{d \xi (\hat{g})}{d \lambda } \bigg |_{\lambda = \lambda _0}&= - \frac{1}{\lambda _0} + \int _0^{\left( \frac{\lambda _0}{\hat{\sigma }_w^2} - \frac{1}{\alpha }\right) \frac{1}{\gamma }} \frac{f_{X| \hat{g}}(x|\hat{g}) dx}{\lambda _0 - x P} \end{aligned}$$

(51)

$$\begin{aligned}&\quad \quad \quad = 0, \end{aligned}$$

(52)

it is inferred from (50) that

$$\begin{aligned} \frac{d^2 \xi (\hat{g})}{d \lambda ^2}\bigg |_{\lambda =\lambda _0}> & \int _0^{\left( \frac{\lambda _0}{\hat{\sigma }_w^2} - \frac{1}{\alpha }\right) \frac{1}{\gamma }} \frac{f_{X| \hat{g}}(x|\hat{g}) dx}{\lambda _0 - x P} \\&\times \left( f_{X|\hat{g}}(0|\hat{g}) -\frac{d(\ln f_{X| \hat{g}}(x|\hat{g}))}{dx} \right) . \end{aligned}$$

(53)

Since $I_0(z) = \sum ^\infty _{k=0} (z/2)^{2k}/(k!)^2$, we obtain from (10),

$$\begin{aligned}&\frac{d(\ln f_{X| \hat{g}}(x|\hat{g}))}{dx} \\&= \frac{d\left( -\frac{x+|\hat{g}|^2}{\kappa \sigma _g^2}+\ln \left( \frac{1}{\kappa \sigma _g^2}\right) +\ln \left( I_0 \left( \frac{2|\hat{g}|}{\kappa \sigma _g^2} \sqrt{x} \right) \right) \right) }{dx} \\&= -\frac{1}{\kappa \sigma _g^2} + \frac{ \left( \sum ^\infty _{k=0} \frac{1}{(k!)^2} \frac{|\hat{g}|^{2k} x^k}{(\kappa \sigma _g^2)^{2k}} \right) '}{I_0 \left( \frac{2|\hat{g}|}{\kappa \sigma _g^2} \sqrt{x} \right) }, \end{aligned}$$

(54)

Due to the log-concavity of the non-central Chi-square [29], i.e. $(\ln (f_{X| \hat{g}}(x|\hat{g})))'' < 0$, (54) decreases with increasing x and hence, we obtain

$$\begin{aligned} f_{X|\hat{g}}(0|\hat{g}) -\frac{f'_{X|\hat{g}}(x|\hat{g})}{f_{X| \hat{g}}(x|\hat{g})}\ge & f_{X|\hat{g}}(0|\hat{g}) -\frac{f'_{X|\hat{g}}(x|\hat{g})}{f_{X| \hat{g}}(x|\hat{g})}\bigg |_{x=0} \\&= \frac{1}{\kappa \sigma _g^2} e^{-\frac{|\hat{g}|^2}{\kappa \sigma _g^2}}-\frac{1}{\kappa \sigma _g^2} \left( 1 - \frac{|\hat{g}|^2}{\kappa \sigma _g^2}\right) , \end{aligned}$$

(55)

which, since $e^{-t} \ge 1-t$, yields $f_{X|\hat{g}}(0|\hat{g}) \ge 0$. Hence, we obtain from (53) that $d^2 \xi /d \lambda ^2|_{\lambda =\lambda _0} > 0$, yielding the pseudo-convexity of $\xi (\hat{g})$.

Appendix C

In this “Appendix”, we derive (25). We have

$$\begin{aligned} \xi (0)&= 1 - \frac{\ln (\alpha \lambda /\hat{\sigma }_w^2)}{2\ln (\alpha )} - \frac{1}{2 \ln (\alpha )} \\&\times \int _0^{\left( \frac{\lambda }{\hat{\sigma }_w^2} - \frac{1}{\alpha }\right) \frac{1}{\gamma }} \ln \left( \alpha \left[ \frac{\lambda }{\hat{\sigma }_w^2} - x \gamma \right] \right) d\left( e^{-\frac{x}{\sigma _g^2}} \right) , \end{aligned}$$

(56)

which applying integration by part into (56) yields

$$\begin{aligned} \xi (0)&= 1 - \frac{\gamma \hat{\sigma }_w^2}{2\ln (\alpha )} \int _0^{\left( \frac{\lambda }{\hat{\sigma }_w^2} - \frac{1}{\alpha }\right) \frac{1}{\gamma }} \frac{e^{-x/\sigma _g^2} }{\lambda -x \gamma \hat{\sigma }_w^2} dx \\&= 1 - \frac{1}{2\ln (\alpha )} \int _0^{\left( \frac{\lambda }{\hat{\sigma }_w^2} - \frac{1}{\alpha }\right) \frac{1}{\gamma }} \frac{e^{-x/\sigma _g^2}}{\frac{\lambda }{\gamma \hat{\sigma }_w^2}-x} dx \\&= 1 - \frac{e^{-\frac{\lambda }{\sigma _g^2 P}}}{2\ln (\alpha )} \int _{\frac{1}{\alpha \sigma _g^2\gamma }}^{\frac{\lambda }{\sigma _g^2 P}} \frac{e^{x}}{x } dx \\&= 1 - \frac{e^{-\frac{\lambda }{\sigma _g^2 P}}}{2\ln (\alpha )} \left[ \mathrm{Ei}\left( \frac{\lambda }{\sigma _g^2 P}\right) - \mathrm{Ei}\left( \frac{1}{\alpha \sigma _g^2\gamma } \right) \right] . \end{aligned}$$

(57)

Appendix D

We firstly prove $\xi _{\min } < 0.9$ for $1.0002 \le \alpha \le 3.16$ if $\lambda ^\perp \ge \alpha \hat{\sigma }_w^2$. Then, we can conclude that for $\xi _{\min } \ge 0.9$ and $1.0002 \le \alpha \le 3.16$, we obtain $\lambda ^\perp < \alpha \hat{\sigma }_w^2$.

Since the left hand side (LHS) of (26) increases with $\lambda$ and is 0 when $\lambda =\lambda ^\perp$, then

$$\begin{aligned} 0\ge & \int ^{\alpha /(\sigma _g^2 \gamma )}_{1/(\alpha \sigma _g^2 \gamma )} \frac{e^x}{x} dx - \frac{\sigma _g^2 \gamma }{\alpha } e^{\frac{\alpha }{\sigma _g^2 \gamma }} \\&= \frac{e^x}{x} \bigg |_{x=1/(\alpha \sigma _g^2 \gamma )}^{x=\alpha /(\sigma _g^2 \gamma )} + \int ^{\alpha /(\sigma _g^2 \gamma )}_{1/(\alpha \sigma _g^2 \gamma )} \frac{e^x}{x^2} dx - \frac{\sigma _g^2 \gamma }{\alpha } e^{\frac{\alpha }{\sigma _g^2 \gamma }} \\&= \int ^{\alpha /(\sigma _g^2 \gamma )}_{1/(\alpha \sigma _g^2 \gamma )} \frac{e^x}{x^2} dx - \alpha \sigma _g^2 \gamma e^{\frac{1}{\alpha \sigma _g^2 \gamma }} \\\ge & \left( \frac{\sigma _g^2 \gamma }{\alpha } \right) ^2 \left( e^{\frac{\alpha }{\sigma _g^2 \gamma }} - e^{\frac{1}{\alpha \sigma _g^2 \gamma }} \right) - \alpha \sigma _g^2 \gamma e^{\frac{1}{\alpha \sigma _g^2 \gamma }}. \end{aligned}$$

(58)

for $\lambda ^\perp \ge \alpha \hat{\sigma }_w^2$. In (58), we applied $\int ^b_a e^x dx /x^2 \ge (e^b-e^a)/b^2$. From (58), we obtain

$$\begin{aligned} - \left( \frac{\alpha ^2-1}{\alpha ^4} + \frac{\alpha -\alpha ^{-1}}{\sigma _g^2 \gamma } \right) e^{- \left( \frac{\alpha ^2-1}{\alpha ^4} + \frac{\alpha -\alpha ^{-1}}{\sigma _g^2 \gamma } \right) } \le - \frac{\alpha ^2-1}{\alpha ^4} e^{-\frac{\alpha ^2-1}{\alpha ^4}}, \end{aligned}$$

(59)

which, from [26], yields

$$\begin{aligned} \frac{\alpha -\alpha ^{-1}}{\sigma _g^2 \gamma } \le -W_{-1}\left( -\frac{\alpha ^2-1}{\alpha ^4} e^{-\frac{\alpha ^2-1}{\alpha ^4}} \right) -\frac{\alpha ^2-1}{\alpha ^4}, \end{aligned}$$

(60)

or equivalently,

$$\begin{aligned} \sigma _g^2 \gamma \ge \frac{\alpha -\alpha ^{-1}}{-W_{-1}\left( -\frac{\alpha ^2-1}{\alpha ^4} e^{-\frac{\alpha ^2-1}{\alpha ^4}} \right) -\frac{\alpha ^2-1}{\alpha ^4}}. \end{aligned}$$

(61)

Applying the inequality $\int ^b_a e^x dx /x \ge (e^b-e^a)/b$, we have

$$\begin{aligned} \mathrm{Ei}\left( \frac{\alpha }{\sigma _g^2 \gamma }\right) - \mathrm{Ei}\left( \frac{1}{\alpha \sigma _g^2 \gamma } \right)&= \int ^{\alpha /(\sigma _g^2 \gamma )}_{1/(\alpha \sigma _g^2 \gamma )} \frac{e^x}{x} dx \\&> \frac{\sigma _g^2 \gamma }{\alpha } \left( e^{\frac{\alpha }{\sigma _g^2 \gamma }} - e^{\frac{1}{\alpha \sigma _g^2 \gamma }} \right) , \end{aligned}$$

(62)

which, from (28), yielding for $\lambda ^\perp \ge \alpha \hat{\sigma }_w^2$,

$$\begin{aligned} \xi _{\min }< & 1 - \frac{1}{2 \ln (\alpha )} \frac{\sigma _g^2 \gamma }{\alpha } \left( 1 - e^{-\frac{\alpha -\alpha ^{-1}}{\sigma _g^2 \gamma }} \right) \\< & 1- \frac{\alpha -\alpha ^{-1}}{2 \alpha \ln (\alpha )} \frac{1 - e^{W_{-1}\left( -\frac{\alpha ^2-1}{\alpha ^4} e^{-\frac{\alpha ^2-1}{\alpha ^4}} \right) +\frac{\alpha ^2-1}{\alpha ^4}}}{-W_{-1}\left( -\frac{\alpha ^2-1}{\alpha ^4} e^{-\frac{\alpha ^2-1}{\alpha ^4}} \right) -\frac{\alpha ^2-1}{\alpha ^4}}, \end{aligned}$$

(63)

which is derived from (61) and the fact that $\sigma _g^2 \gamma \left( 1 - e^{-\frac{\alpha -\alpha ^{-1}}{\sigma _g^2 \gamma }} \right)$ is an increasing function of $\sigma _g^2 \gamma$. Moreover, we obtain

$$\begin{aligned} \frac{\alpha -\alpha ^{-1}}{2 \alpha \ln (\alpha )} \frac{1 - e^{W_{-1}\left( -\frac{\alpha ^2-1}{\alpha ^4} e^{-\frac{\alpha ^2-1}{\alpha ^4}} \right) +\frac{\alpha ^2-1}{\alpha ^4}}}{-W_{-1}\left( -\frac{\alpha ^2-1}{\alpha ^4} e^{-\frac{\alpha ^2-1}{\alpha ^4}} \right) -\frac{\alpha ^2-1}{\alpha ^4}} \ge 0.1, \end{aligned}$$

(64)

for $1.0002 \le \alpha \le 3.16$. Accordingly, $\xi _{\min } < 0.9$ for $1.0002 \le \alpha \le 3.16$.

Appendix E

Equation (40) is derived in this “Appendix”. Since the LHS of (26) increases with $\lambda$, it follows from (26) and (38) that

$$\begin{aligned}&\int _{\frac{1}{\alpha \sigma _g^2\gamma }}^{ \frac{1}{2 \epsilon \ln (\alpha )}} x^{-1} e^{x} dx - e^{ \frac{1}{2 \epsilon \ln (\alpha )}} 2 \epsilon \ln (\alpha ) \\&\quad \le \int _{\frac{1}{\alpha \sigma _g^2\gamma }}^{ \frac{\lambda ^\perp }{\sigma _g^2 P}} x^{-1} e^{x} dx - \frac{e^{\lambda ^\perp /(\sigma _g^2 P) }}{\lambda ^\perp /(\sigma _g^2 P)} \\&\quad = 0, \end{aligned}$$

(65)

for $\lambda ^\perp \ge \sigma _g^2 P/[2 \epsilon \ln (\alpha )]$. Since $\int e^z z^{-1} dt \simeq e^z ( z^{-1} + z^{-2})$ for $z \gg 1$ [25], we can approximate the LHS of (65) as

$$\begin{aligned}&\int _{\frac{1}{\alpha \sigma _g^2\gamma }}^{ \frac{1}{2 \epsilon \ln (\alpha )}} x^{-1} e^{x} dx - 2 \epsilon \ln (\alpha ) e^{ \frac{1}{2 \epsilon \ln (\alpha )}} \\&\quad \simeq [2 \epsilon \ln (\alpha )]^2 e^{ \frac{1}{2 \epsilon \ln (\alpha )}} - [\alpha \sigma _g^2\gamma + (\alpha \sigma _g^2\gamma )^2] e^{\frac{1}{\alpha \sigma _g^2\gamma }} \\&\quad \simeq [2 \epsilon \ln (\alpha )]^2 e^{ \frac{1}{2 \epsilon \ln (\alpha )}} - \alpha \sigma _g^2\gamma e^{\frac{1}{\alpha \sigma _g^2\gamma }}, \end{aligned}$$

(66)

for $\sigma _g^2 \gamma \ll 1$. Therefore, it follows from (65) and (66) that

$$\begin{aligned} \frac{e^{-\frac{1}{\alpha \sigma _g^2\gamma }}}{\alpha \sigma _g^2\gamma } \lesssim [2 \epsilon \ln (\alpha )]^{-2}e^{-\frac{1}{2 \epsilon \ln (\alpha )}}, \end{aligned}$$

(67)

yielding (40).