Covert Communication in Continuous-Time Systems

Consider a scenario shown in Fig. 2 where transmitter Alice (“a”) wants to transmit a message to intended recipient Bob (“b”) reliably without being detected by a warden Willie (“w”). A jammer (“j”) assists the communication by actively sending jamming signals, but without any coordination with Alice.

For $t\in[0,T]$, we consider continuous-time channels where Alice and the jammer send symbols using pulse-shaped waveforms. Since Alice can send pulses at any time in the continuous time interval $[0,T]$, she and Bob share an infinite length key [4] encoding those locations unknown to Willie. If Alice decides to transmit, she maps her message to waveform $x_{a}(t)$ restricted to approximate bandwidth ¹¹1See Appendix B for a discussion of the bandwidth of the constructions $W$ under an average power constraint of $\sigma_{a}^{2}$. The jammer transmits regardless if Alice transmitted or not. It sends waveform $x_{j}(t)$ that is also restricted to approximate bandwidth $W$ under an average power constraint of $\sigma_{j}^{2}$. The channels between each transmitter and receiver pair are assumed to be AWGN, and thus the signal observed by Willie is given by:

$\displaystyle z(t)=$

$\displaystyle\left\{\begin{array}[]{lr}\frac{x_{a}(t-\tau_{a})}{d_{aw}^{r/2}}+\frac{x_{j}(t-\tau_{j})}{d_{jw}^{r/2}}+N^{(w)}(t),\,\text{Alice transmits}\\ \frac{x_{j}(t-\tau_{j})}{d_{jw}^{r/2}}+N^{(w)}(t),\,\text{Alice does not transmit}\end{array}\right.$

(3)

where $d_{xy}$ is the distance between a transmitter $x$ and a receiver $y$, $r$ is the path-loss exponent, $\tau_{a}$ and $\tau_{j}$ are time delays of Alice’s and the jammer’s signal, respectively, and $N^{(w)}(t)$ is the noise observed at Willie’s receiver, which is a zero-mean stationary Gaussian random process with power spectral density $N_{0}^{(w)}/2$. Bob observes the channel output $y(t)$ at time $t$, which is analogous to $z(t)$ but with the substitution of the noise $N^{(b)}(t)$ for $N^{(w)}(t)$, and $N^{(b)}(t)$ is a zero-mean stationary Gaussian random process with power spectral density of $N_{0}^{(b)}/2$.

We consider two scenarios: 1) the path-loss $d_{aw}^{r}$ between Alice and Willie is known; and 2) the path-loss $d_{aw}^{r}$ is unknown. In both scenarios, if not specified, we assume the path-loss between any transmitter and receiver pair is one, without loss of generality.

We employ random coding arguments and generate codewords by independently drawing symbols from a zero-mean complex Gaussian distribution with variance $\sigma_{a}^{2}$. If Alice decides to transmit, she selects the codeword corresponding to her message, sets $f_{i}$ to the $i^{th}$ symbol of that codeword, and transmits the symbol sequence $\mathbb{f}=\{f_{1},f_{2},\ldots\}$. The jammer transmits zero-mean complex Gaussian symbol sequence $\mathbb{v}=\{v_{1}.v_{2},\ldots\}$, with variance $\sigma_{j}^{2}$. Here we choose $\sigma_{j}^{2}=\sigma_{a}^{2}$, i.e., Alice and the jammer use this same average transmit power, so that Alice can possibly hide her signal in the jammer’s interference.

Let $n=\lfloor WT\rfloor$ be an integer. Over the time interval $[0,T]$, Alice sends $M_{a}$ symbol pulses, where $M_{a}$ follows a binomial distribution with mean $\alpha n$, i.e., $M_{a}\sim B(n,\alpha)$, with constant $0\leq\alpha<1$. Alice’s codeword length is chosen to be an integer close to $\alpha n-\epsilon_{c}n$ ($\epsilon_{c}$ is a small positive constant), such that as $n\to\infty$, there are enough pulses over $[0,T]$ for all of Alice’s codeword symbols. The jammer sends $M_{j}$ pulses, where $M_{j}$ follows a binomial distribution with mean $\beta n$, i.e., $M_{j}\sim B(n,\beta)$, with $\beta$ is uniformly distributed over $[\mu,\mu+\Delta]$, $0\leq\mu<\mu+\Delta\leq 1$ and $\Delta\geq\alpha$. Let $\tau_{k},k=1,2,\ldots,M_{a}$ and $\tau_{k}^{\prime},k=1,2,\ldots,M_{j}$ be be independent and identically distributed (i.i.d.) sequences of pulse delays of Alice and the jammer, respectively. The delays are drawn uniformly over $[0,T]$. Alice’s waveform within interval $[0,T]$ is then given by:

$\displaystyle x_{a}(t)=\sum_{k=1}^{M_{a}}f_{k}p(t-\tau_{k})$

(4)

where $p(t)$ is a unit-energy pulse shaping filter with bandwidth $W$. Obviously, a waveform restricted to $[0,T]$ cannot have a finite bandwidth, we provide a brief discussion on this issue in Appendix B. The jammer’s waveform within interval $[0,T]$ is given by:

$\displaystyle x_{j}(t)=\sum_{k=1}^{M_{j}}v_{k}p(t-\tau^{\prime}_{k}).$

(5)

For an AWGN channel, Willie observes the signal $z(t)$ given in (3).

To obtain an achievability result for covert communications, Willie should be assumed to employ an optimal detector. We will find an upper bound to the performance of that optimal detector by assuming a genie provides Willie additional information; in particular, we assume that Willie not only knows how the system is constructed (including $\alpha$, the distribution of $\beta$ and the transmission power $\sigma_{a}^{2}$ and $\sigma_{j}^{2}$ of the symbols), but also knows the number of pulses and the exact locations (timing) of each pulse on the channel in $[0,T]$. The only thing he does not know is from whom each pulse is sent. In the next section, we will prove that the optimal test for Willie is a threshold test on the number of pulses he observed.

Given the construction above, Willie’s test is between the two hypotheses $H_{0}$ and $H_{1}$ where he has complete statistical knowledge of his observations when either hypothesis is true. We denote: Alice’s decision on transmission as $D$ (which corresponds to hypothesis $H_{0}$ when Alice decides to transmit, or $H_{1}$ when she decides not to); the total number of pulses sent during time $T$ as $M$; the locations (over $[0,T])$ of the pulses as a vector $\mathbb{L}$; and the height (square root of the power) of the pulses as a vector $\mathbb{S}$, i.e., $\mathbb{S}$ is the vector of the original symbols sent. We want to first show that $M$ is a sufficient statistic for Willie’s detection. The random variables $D$, $M$, $\mathbb{L}$ and $\mathbb{S}$ form a Markov chain shown in Fig. 3, which illustrates the transition from Alice’s state $D$ to Willie’s received signal $z(t)$. The transitions of the Markov chain are:

• •

  $D\longrightarrow M$: The conditional distribution of $M$, given $D$, is binomial with mean $\beta n$ when Alice does not transmit, and binomial with mean $\beta n+\alpha n$ when Alice transmits.

• •

  $M\longrightarrow\mathbb{L},\mathbb{S}$: Given $M$, the distribution of $L_{m}$ for $m=1,2,\ldots,M$ is uniform over $[0,T]$. The distribution of $S_{m}$ for $m=1,2,\ldots,M$ is zero-mean Gaussian with variance $\sigma_{a}^{2}=\sigma_{j}^{2}$.

Given the pulse locations and the height of the pulses, the signal $z(t)$ observed at Willie’s receiver can be constructed from the pulse-shaping function $p(t)$ and the AWGN of Willie’s channel. From the Markov chain shown in Fig. 3, we see that $z(t)$ conditioned on $M$ is independent of $D$. Thus, $M$ is a sufficient statistic for Willie to make an optimal decision on Alice’s presence. Therefore, by applying the Neyman-Pearson criterion, the optimal test for Willie to minimize his probability of error is the likelihood ratio test (LRT) [11]:

$\displaystyle\Lambda(M=m)=\frac{P_{M|H_{1}}(m)}{P_{M|H_{0}}(m)}\overset{H_{1}}{\underset{H_{0}}{\gtrless}}\gamma$

(6)

where $\gamma=P(H_{0})/P(H_{1})$, and $P_{M|H_{1}(m)}$ and $P_{M|H_{0}(m)}$ are the probability mass functions (pmfs) of the number of pulses given that Alice transmitted or did not transmit, respectively. Given the LRT above, we want to show that this is equivalent to a threshold test on the number of pulses Willie observes at his receiver, which is true if the LRT exhibits monotonicity in $M$. We employ the concept of stochastic ordering [12] to derive the desired monotonicity result. We say that $X$ is smaller than $Y$ in the likelihood ratio order (written as $X\leq_{lr}Y$) when $\frac{f_{Y}(x)}{f_{X}(x)}$ is non-decreasing over the union of their supports, where $f_{Y}(x)$ and $f_{X}(x)$ are pmfs or probability density functions (pdfs) of $Y$ and $X$, respectively [8].

We let:

$\displaystyle b\overset{\Delta}{=}\left\{\begin{array}[]{lr}\beta n,\,\text{when Alice does not transmit}\\ \beta n+\alpha n,\,\text{when Alice transmits}\end{array}\right.$

(9)

and introduce two random variables $B_{0}$ and $B_{1}$ with pdfs given by:

$\displaystyle f_{B_{\rho}}(b)=$

$\displaystyle\left\{\begin{array}[]{lr}\frac{1}{\Delta},\,\mu n<b\leq\mu n+\Delta n,\,\rho=0\\ \frac{1}{\Delta},\,\mu n+\alpha n<b\leq\mu n+\alpha n+\Delta n,\,\rho=1\\ 0,\,\text{else}\end{array}\right.$

(13)

The LRT in (6) can be written as:

$\displaystyle\Lambda(M=m)=\frac{E_{B_{1}}\left[P_{M(b)}(m)\right]}{E_{B_{0}}\left[P_{M(b)}(m)\right]}$

where $M(b)$ follows a binomial distribution, i.e., $M(b)\sim B(n,\frac{b}{n})$.

Dividing both sides of (14) by $n$ yields the equivalent test:

$\displaystyle\frac{M}{n}\overset{H_{1}}{\underset{H_{0}}{\gtrless}}\gamma_{n}$

where $\gamma_{n}=\gamma^{\prime}/n$. For any finite $n$, there is an optimal threshold $\gamma_{n}$ such that it minimizes Willie’s probability of error in detecting Alice’s existence. However, we will show that for any $\gamma_{n}$ Willie chooses, he will not be able to detect Alice as $n\to\infty$; that is, for any $\epsilon>0$, there exists a construction such that $P_{FA}+P_{MD}>1-\epsilon$ for $n$ large enough.

Recall that $\beta$ is a uniform random variable on $[\mu,\mu+\Delta]$, with $0\leq\mu<\mu+\Delta\leq 1$ and constant $\Delta\geq\alpha$. Let $P_{FA}(u)$ and $P_{MD}(u)$ be Willie’s probability of false alarm and missed detection conditioned on $\beta=u$, respectively. Then:

$\displaystyle P_{FA}(u)$

$\displaystyle=P\left(\frac{M}{n}\geq\gamma_{n}\mid H_{0}\right)=P\left(\frac{1}{n}\sum_{i=1}^{n}R_{i}\geq\gamma_{n}\mid H_{0}\right)$

where, under $H_{0}$, $R_{i}$ is a Bernoulli random variable that takes value one with probability $u$. By the weak law of large numbers, $\frac{1}{n}\sum_{i=1}^{n}R_{i}$ converges in probability to $u$. Thus, for any $\eta>0$, there exists $N_{0}$ such that, for $n\geq N_{0}$, we have:

$\displaystyle P\left(\frac{1}{n}\sum_{i=1}^{n}R_{i}\in\left(u-\eta,u+\eta\right)\mid H_{0}\right)>1-\frac{\epsilon}{2}\,.$

Therefore, for $n>N_{0}$, $P_{FA}(u)>1-\frac{\epsilon}{2}$ for any $\gamma_{n}<u-\eta$. Analogously, we write:

$\displaystyle P_{MD}$

$\displaystyle=P\left(\frac{M}{n}\leq\gamma_{n}\mid H_{1}\right)=P\left(\frac{1}{n}\sum_{i=1}^{n}R_{i}\leq\gamma_{n}\mid H_{1}\right)$

Likewise, by the weak law of large numbers, for any $\eta>0$, there exists $N_{1}$ such that, for $n\geq N_{1}$, we have:

$\displaystyle P\left(\frac{1}{n}\sum_{i=1}^{n}R_{i}\in\left(u+\alpha-\eta,u+\alpha+\eta\right)\mid H_{1}\right)>1-\frac{\epsilon}{2}\,.$

Therefore, for $n>N_{1}$, $P_{MD}(u)>1-\frac{\epsilon}{2}$ for any $\gamma_{n}>\alpha+u+\eta$.

Define the set $\mathcal{A}=\{u:u-\eta<\gamma_{n}<\alpha+u+\eta\}$. We have established that, for any $u\in\mathcal{A}^{c}$ and $n>\max(N_{0},N_{1})$, $P_{FA}(u)+P_{MD}(u)>1-\frac{\epsilon}{2}$. The probability of $\mathcal{A}$ has the following upper bound:

$\displaystyle P(A)$

$\displaystyle=P(\gamma_{n}-\alpha-\eta<\beta<\gamma_{n}+\eta)\leq\frac{2\eta+\alpha}{\Delta}\,.$

By choosing $\eta=\frac{\epsilon\Delta}{4}$ and $\alpha=\frac{\epsilon\Delta}{2}$, we have $P(\mathcal{A})\leq\frac{\epsilon}{2}$, i.e., $P(\mathcal{A}^{c})>1-\frac{\epsilon}{2}$. Hence,

	$\displaystyle P_{FA}+P_{MD}$	$\displaystyle=E_{\beta}[P_{FA}(\beta)+P_{MD}(\beta)]$
		$\displaystyle\geq E_{\beta}[P_{FA}(\beta)+P_{MD}(\beta)\mid\mathcal{A}^{c}]P(\mathcal{A}^{c})$
		$\displaystyle>1-\frac{\epsilon}{2}\,.$

Thus, Alice can send an average of $\alpha n=\frac{\epsilon\Delta}{2}\lfloor WT\rfloor$ pulses with a constant power and remain covert from Willie. Note that since the maximum interference from the jammer at Bob can be upper bounded by a constant, reliability is also achieved under the same construction. Therefore, $\mathcal{O}(WT)$ bits can be transmitted covertly and reliably from Alice to Bob.

For achievability, we derive an upper bound to the performance of Willie’s optimal detector by assuming a genie provides Willie extra knowledge on the exact power range $\left[\frac{P_{a}}{d_{aw}^{2}},\frac{P_{a}+\Delta_{P_{a}}}{d_{aw}^{r}}\right]$ received from Alice, the distribution of the number of the jammer’s power levels (including all of the parameters), and the values of all power levels employed by the jammer and Alice (if she decided to transmit), but not which power level is employed by whom.

In this section, we show that the number of power levels in the range of $\left[\frac{P_{A}}{d_{aw}^{2}},\frac{P_{A}+\Delta_{P_{A}}}{d_{aw}^{r}}\right]$, which we term the detection region, is a sufficient statistic for Willie in deciding between hypothesis $H_{0}$ or $H_{1}$. Fig. 4 illustrates the power levels received at Willie.

Let $K_{1}$ be the number of power levels inside the detection region, $K_{2}$ be the number of power levels outside the detection region, i.e. $K=K_{1}+K_{2}$. By construction, all of the power levels sent by the jammer form a Poisson point process with $K\sim Pois\left(\lambda_{j}\right)$ on $[P_{j},P_{j}+\Delta_{P_{j}}]$. Note that for a Poisson point process, generating $K$ power levels with mean $\lambda_{j}$ and placing them uniformly over $[P_{j},P_{j}+\Delta_{P_{j}}]$ is equivalent to: generating $K_{1}$ power levels with mean $\frac{\Delta_{P_{a}}}{\Delta_{P_{j}}d_{aw}^{r}}\lambda_{j}$ and placing them uniformly inside the detection region, and generating $K_{2}$ power levels with mean $\left(1-\frac{\Delta_{P_{a}}}{\Delta_{P_{j}}d_{aw}^{2}}\right)\lambda_{j}$ and placing them uniformly outside the detection region. This is critical in the proof below.

Recall that $D$ denotes Alice’s decision on transmission, $L$ denotes the locations (in $[0,T]$) of all of the pulses sent, and $\mathbb{S}$ denotes the height of the pulses. We also denote the values of all power levels (within and outside the detection region) as a vector $\mathbb{V}$. The random variables $D$, $K_{1}$, $\mathbb{V}$, $\mathbb{L}$ and $\mathbb{S}$ form a Markov chain shown in Fig. 5, which illustrates the transition from Alice’s state $D$ to Willie’s received signal $z(t)$. The transitions of the Markov chain are:

• •

  $D\longrightarrow K_{1}$: $K_{1}$ and $K_{1}-1$ are characterized by a Poisson process with mean $\frac{\Delta_{P_{a}}\lambda_{j}}{\Delta_{P_{j}}d_{aw}^{r}}$ when Alice does not transmit, and a Poisson process with mean $\frac{\Delta_{P_{a}}\lambda_{j}}{\Delta_{P_{j}}d_{aw}^{r}}$ when she does transmit.

• •

  $K_{1}\longrightarrow\mathbb{V},\mathbb{L}$: Let $V_{k},k=1,2,\ldots,K_{1}$ be the values of power levels within the detection region, and $V_{k},k=K_{1}+1,K_{1}+2,\ldots,K$ the power values outside the detection region. Given $K_{1}$, the conditional distribution of $V_{k},k=1,2,\ldots,K_{1}$, is uniform within the detection region. Note that $K_{2}$ is independent of $D$ since the pulses sent with power levels outside the detection region can only come from the jammer, no matter if Alice transmits or not. Given $K_{2}$ (Poisson with mean $\left(1-\frac{\Delta_{P_{A}}}{\Delta_{P_{J}}d_{aw}^{2}}\right)\lambda_{j}$), the distribution of $V_{k},k=K_{1}+1,K_{1}+2,\ldots,K$, is uniform outside the detection region. Let $\{L_{k,m}:k=1,\ldots,K_{1},m=1,\ldots,M_{n}\}$ denote the locations (in $[0,T]$) of pulses sent with power within the detection region, and $\{L_{k,m}:k=K_{1}+1,\ldots,K,m=1,\ldots,M_{n}\}$ denote the locations of pulses sent with power outside the detection region. Given $K_{1}$, the distribution of $L_{k},m$ for $k=1,2,\ldots,K_{1}$ and all $m$ is uniform over $[0,T]$. Given $K_{2}$, the distribution of $L_{k}$ for $k=K_{1}+1,K_{1}+2,\ldots,K$ and all $m$ is also uniform over $[0,T]$, which is independent from $D$.

• •

  $\mathbb{V},\mathbb{L}\longrightarrow\mathbb{S},\mathbb{L}$: The conditional distribution of $S_{k,m}$, for $k=1,2,\ldots,K,m=1,2,\ldots,M_{n}$, given $V_{k}$, is a zero-mean Gaussian random variable with variance $V_{k}$.

Given the pulse locations and the height of the pulses, the signal $z(t)$ can be constructed from $p(t)$ and the AWGN of Willie’s channel. From the Markov chain shown in Fig. 5, we see that $z(t)$ conditioned on $K_{1}$ is independent of $D$. Therefore, $K_{1}$ is a sufficient statistic for Willie to decide between hypotheses $H_{0}$ and $H_{1}$.

In particular, hypotheses $H_{0}$ and $H_{1}$ can be characterized as:

• •

  $H_{0}$: the number of power levels within the detection region follows $Pois\left(\frac{\lambda_{j}\Delta_{P_{a}}}{\Delta_{P_{j}}d_{aw}^{r}}\right)$;

• •

  $H_{1}$: the number of power levels within the detection region follows $Pois\left(\frac{\lambda_{j}\Delta_{P_{a}}}{\Delta_{P_{j}}d_{aw}^{r}}\right)+1$.

Let $P_{0}$ and $P_{1}$ denote the distribution of the number of power levels observed by Willie (within Willie’s detection range) given $H_{0}$ and $H_{1}$, respectively:

$\displaystyle P_{0}(k)=\frac{\lambda^{k}e^{-\lambda}}{k!},\,\,\,k\geq 0$

(15)

and

$\displaystyle P_{1}(k)=\frac{\lambda^{k-1}e^{-\lambda}}{(k-1)!},\,\,\,k\geq 1$

(16)

where $\lambda=\frac{\lambda_{j}\Delta_{P_{a}}}{\Delta_{P_{j}}d_{aw}^{r}}$. Theorem 13.1.1 in [13] shows that for the optimal hypothesis test,

$\displaystyle P_{FA}+P_{MD}=1-\mathcal{V}_{T}(P_{0},P_{1})$

where

$\displaystyle\mathcal{V}_{T}(P_{0},P_{1})=\frac{1}{2}\sum_{k}|P_{0}(k)-P_{1}(k)|$

is the total variation distance between $P_{0}$ and $P_{1}$, where the sum is over all $k$ in the support of $P_{0}\cup P_{1}$. Therefore, by the definition of covertness, if

$\displaystyle\mathcal{V}_{T}(P_{0},P_{1})\leq\epsilon,$

(17)

Alice achieves covert communications.

Given (15) and (16), we derive:

	$\displaystyle\mathcal{V}_{T}(P_{0},P_{1})=\frac{1}{2}\sum_{k=1}^{\infty}|P_{0}(k)-P_{1}(k)|+\frac{1}{2}P_{0}(0)$
	$\displaystyle=\frac{1}{2}\sum_{k=1}^{\infty}\frac{\lambda^{k-1}e^{-\lambda}}{(k-1)!}\left|\frac{\lambda}{k}-1\right|+\frac{1}{2}e^{-\lambda}$
	$\displaystyle=\frac{1}{2}\Bigg{[}\sum_{k=1}^{\lambda}\frac{\lambda^{k-1}e^{-\lambda}}{(k-1)!}\left(\frac{\lambda}{k}-1\right)+\sum_{k=\lambda+1}^{\infty}\frac{\lambda^{k-1}e^{-\lambda}}{(k-1)!}\left(1-\frac{\lambda}{k}\right)$
	$\displaystyle\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\,\,\,\,\,+e^{-\lambda}\Bigg{]}$
	$\displaystyle=\frac{1}{2}\Bigg{[}\sum_{k=1}^{\lambda}\frac{\lambda^{k}e^{-\lambda}}{k!}-\sum_{k=1}^{\lambda}\frac{\lambda^{k-1}e^{-\lambda}}{(k-1)!}+\sum_{k=\lambda+1}^{\infty}\frac{\lambda^{k-1}e^{-\lambda}}{(k-1)!}$
	$\displaystyle\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad-\sum_{k=\lambda+1}^{\infty}\frac{\lambda^{k}e^{-\lambda}}{k!}+e^{-\lambda}\Bigg{]}$
	$\displaystyle=\frac{1}{2}\Bigg{[}\sum_{k=1}^{\lambda}\frac{\lambda^{k}e^{-\lambda}}{k!}+\sum_{k=\lambda}^{\infty}\frac{\lambda^{k}e^{-\lambda}}{k!}-\sum_{k=0}^{\lambda-1}\frac{\lambda^{k}e^{-\lambda}}{k!}$
	$\displaystyle\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad-\sum_{k=\lambda+1}^{\infty}\frac{\lambda^{k}e^{-\lambda}}{k!}+e^{-\lambda}\Bigg{]}$
	$\displaystyle=\frac{1}{2}\left(\sum_{k=1}^{\infty}\frac{\lambda^{k}e^{-\lambda}}{k!}+\frac{\lambda^{\lambda}e^{-\lambda}}{\lambda!}-\sum_{k=0}^{\infty}\frac{\lambda^{k}e^{-\lambda}}{k!}+\frac{\lambda^{\lambda}e^{-\lambda}}{\lambda!}+e^{-\lambda}\right)$
	$\displaystyle=\frac{\lambda^{\lambda}e^{-\lambda}}{\lambda!}$

Using Stirling’s approach, this can be upper bounded as:

$\displaystyle\frac{\lambda^{\lambda}e^{-\lambda}}{\lambda!}$

$\displaystyle\leq\frac{\lambda^{\lambda}e^{-\lambda}}{\sqrt{2\pi}\lambda^{\lambda+1/2}e^{-\lambda}}=\frac{1}{\sqrt{2\pi\lambda}}\,.$

Thus, if

$\displaystyle\lambda\geq\frac{1}{2\pi\epsilon^{2}},$

i.e.,

$\displaystyle\lambda_{j}\geq\frac{\Delta_{P_{j}}d_{aw}^{r}}{2\pi\Delta_{P_{a}}\epsilon^{2}}\,,$

(18)

covertness is achieved. This implies that one of the two strategies can be employed: 1) Alice chooses a $\Delta_{P_{a}}$ and the jammer can use an upper bound on $d_{aw}^{r}$ to choose $\lambda_{j}$; 2) the jammer chooses a $\lambda_{j}$ and Alice can use $d_{aw}^{r}$ to choose $\Delta_{P_{a}}$.

Since the maximum interference from the jammer at Bob can be upper bounded by a constant, reliability is achieved under the same construction. Thus, under this construction, Alice can achieve covert and reliable communications when the path-loss between her and Willie is unknown. Also, since under the above construction, Alice can send $M_{n}=\lfloor\alpha\lfloor WT\rfloor\rfloor$ pulses with a constant power (which does not decrease with $WT$), $\mathcal{O}(WT)$ bits can be transmitted covertly and reliably from Alice to Bob.