Broadcast Channels with Confidential Messages: Channel Uncertainty, Robustness, and Continuity

Abstract

The broadcast channel with confidential messages (BCC) models the communication scenario in which a transmitter sends simultaneously common and confidential information to two receivers. The common information must be received by both receivers while the confidential information is designated for one receiver only and must be secured against the other one. The performance of this system is usually characterized by its secrecy capacity region determining the maximum transmission rates. In this chapter, the issue of whether this secrecy capacity region depends continuously on the system parameters or not is examined. In particular, this is done for compound channels, in which the users know only that the true channel realization is constant for the whole duration of transmission and this comes from a pre-specified uncertainty set. The secrecy capacity region of the compound BCC is shown to be robust in the sense that it is a continuous function of the uncertainty set. This means that small variations in the uncertainty set result in small variations in secrecy capacity.

Appendix

The following proofs of Lemmas 5.1 and 5.2 are adaptations of [2] and [25] where similar results were proved in the context of quantum information theory. However, we obtain bounds with better constants by restricting the analysis to classical probability distributions only.

5.1.1 Proof of Lemma 5.1

$\square$The proof of this lemma can also be found in [10, 11] and is given here for completeness. It follows [2] where a similar result is presented in the context of quantum information. However, we are able to get a better constant by using the fact that $H(Y|X)\ge 0$ for all $P_{XY}\in \mathscr {P}(\mathscr {X}\times \mathscr {Y})$. This is in contrast to the quantum version in [2].

Let $P_{XY},P_{\tilde{X}\tilde{Y}}\in \mathscr {P}(\mathscr {X}\times \mathscr {Y})$ be joint probability distributions with $\Vert P_{XY}-P_{\tilde{X}\tilde{Y}}\Vert \le \varepsilon$. We assume that

$$\begin{aligned} \sum _{x\in \mathscr {X}}\sum _{y\in \mathscr {Y}}\big |P_{XY}(x,y)-P_{\tilde{X}\tilde{Y}}(x,y)\big |=\varepsilon \end{aligned}$$

(5.19)

is satisfied with equality since otherwise $\varepsilon$ in (5.19) could be replaced with a smaller $\tilde{\varepsilon }<\varepsilon$ accordingly.

We define the function

$$\begin{aligned} f(x,y) =:\big |P_{XY}(x,y)-P_{\tilde{X}\tilde{Y}}(x,y)\big | \end{aligned}$$

(5.20)

and set

$$\begin{aligned} p^*(x,y) = (1-\varepsilon )P_{XY}(x,y)+f(x,y) \end{aligned}$$

for all $(x,y)\in \mathscr {X}\times \mathscr {Y}$ so that $p^*\in \mathscr {P}(\mathscr {X},\mathscr {Y})$ is a joint probability distribution on $\mathscr {X}\times \mathscr {Y}$.

Further, we set

$$\begin{aligned} \hat{p}(x,y) = \frac{1}{\varepsilon }f(x,y), \end{aligned}$$

(5.21a)

and

$$\begin{aligned} \hat{q}(x,y) = \frac{1}{\varepsilon }\big ((1-\varepsilon )\big [P_{XY}(x,y)-P_{\tilde{X}\tilde{Y}}(x,y)\big ]+f(x,y)\big ). \end{aligned}$$

(5.21b)

Next we check that $\hat{p}$ and $\hat{q}$ are well defined such that they are indeed probability distributions. $\hat{p}(x,y)\ge 0$ for all $(x,y)\in \mathscr {X}\times \mathscr {Y}$ is obviously true. It remains to verify that $\hat{q}(x,y)\ge 0$ for all $(x,y)\in \mathscr {X}\times \mathscr {Y}$ is also satisfied.

If $P_{XY}(x,y)\le P_{\tilde{X}\tilde{Y}}(x,y)$, then

$$\begin{aligned} -f(x,y)&\le P_{XY}(x,y)-P_{\tilde{X}\tilde{Y}}(x,y) \\&\le (1-\varepsilon )\big (P_{XY}(x,y)-P_{\tilde{X}\tilde{Y}}(x,y)\big ) \\&\le 0 \end{aligned}$$

so that $\hat{q}(x,y)\ge 0$. On the other hand, if $P_{XY}(x,y)>P_{\tilde{X}\tilde{Y}}(x,y)$, then

$$\begin{aligned} 0&<(1-\varepsilon )\big (P_{XY}(x,y)-P_{\tilde{X}\tilde{Y}}(x,y)\big ) \\&\le P_{XY}(x,y)-P_{\tilde{X}\tilde{Y}}(x,y) \\&\le f(x,y) \end{aligned}$$

so that $\hat{q}(x,y)\ge 0$ also in this case. From the definition of $\hat{p}$ and $\hat{q}$ in (5.21) and (5.19)–(5.20) it can further easily be verified that

$$\begin{aligned} \sum _{x\in \mathscr {X}}\sum _{y\in \mathscr {Y}}\hat{p}(x,y)= \sum _{x\in \mathscr {X}}\sum _{y\in \mathscr {Y}}\hat{q}(x,y)=1 \end{aligned}$$

which shows that $\hat{p}\in \mathscr {P}(\mathscr {X}\times \mathscr {Y})$ and $\hat{q}\in \mathscr {P}(\mathscr {X}\times \mathscr {Y})$ are joint probability distributions.

With this we can rewrite $p^*$ as

$$\begin{aligned} p^*(x,y)&= (1-\varepsilon )P_{XY}(x,y) + \varepsilon \hat{p}(x,y) \end{aligned}$$

(5.22a)

$$\begin{aligned}&= (1-\varepsilon )P_{\tilde{X}\tilde{Y}}(x,y) + \varepsilon \hat{q}(x,y) \end{aligned}$$

(5.22b)

for all $(x,y)\in \mathscr {X}\times \mathscr {Y}$. Next, we show that (5.22a) implies

$$\begin{aligned} \big |H(Y|X)-H(Y^*|X^*)\big | \le \varepsilon \log |\mathscr {Y}|+H_2(\varepsilon ). \end{aligned}$$

(5.23)

To do so, we use the fact that the conditional entropy is concave, i.e.,

$$\begin{aligned} H(Y^*|X^*) \ge (1-\varepsilon )H(Y|X) + \varepsilon H(\hat{Y}|\hat{X}). \end{aligned}$$

With this, we have

$$\begin{aligned} H(Y|X) - H(Y^*|X^*)&\le H(Y|X) - (1-\varepsilon )H(Y|X) - \varepsilon H(\hat{Y}|\hat{X}) \nonumber \\&= \varepsilon \big (H(Y|X)-H(\hat{Y}|\hat{X})\big ) \nonumber \\&\le \varepsilon H(Y|X) \nonumber \\&\le \varepsilon \log |\mathscr {Y}|. \end{aligned}$$

(5.24)

Using the concavity of the entropy

$$\begin{aligned} H(X^*) \ge (1-\varepsilon )H(X) + \varepsilon H(\hat{X}) \end{aligned}$$

and the upper bound on the joint entropy

$$\begin{aligned} H(X^*,Y^*) \le (1-\varepsilon )H(X,Y)+\varepsilon H(\hat{X},\hat{Y})+ H_2(\varepsilon ), \end{aligned}$$

we get

$$\begin{aligned} H(Y^*|X^*)&= H(X^*,Y^*) - H(X^*) \\&\le (1-\varepsilon )H(Y|X) + \varepsilon H(Y^*|X^*) + H_2(\varepsilon ) \end{aligned}$$

and further

$$\begin{aligned} H(Y|X) - H(Y^*|X^*)&\ge -\varepsilon \big (H(Y^*|X^*) - H(Y|X)\big ) - H_2(\varepsilon ) \nonumber \\&\ge -\varepsilon H(Y^*|X^*) - H_2(\varepsilon ) \nonumber \\&\ge -\varepsilon \log |\mathscr {Y}| - H_2(\varepsilon ). \end{aligned}$$

(5.25)

Now, (5.24) and (5.25) yield

$$\begin{aligned} \big |H(Y|X) - H(Y^*|X^*)\big | \le \varepsilon \log |\mathscr {Y}| + H_2(\varepsilon ) \end{aligned}$$

which shows (5.23). (By the same arguments, one can show that (5.22b) implies $|H(\tilde{Y}|\tilde{X})-H(Y^*|X^*)| \le \varepsilon \log |\mathscr {Y}|+H_2(\varepsilon )$.)

Finally, this yields

$$\begin{aligned}&\big |H(Y|X)-H(\tilde{Y}|\tilde{X})\big | \\&\qquad \qquad = \big |H(Y|X)-H(Y^*|X^*) + \big (H(Y^*|X^*)-H(\tilde{Y}|\tilde{X})\big )\big | \\&\qquad \qquad \le \big |H(Y|X)-H(Y^*|X^*)\big | + \big |H(\tilde{Y}|\tilde{X})-H(Y^*|X^*)\big | \\&\qquad \qquad \le 2\varepsilon \log |\mathscr {Y}| + 2H_2(\varepsilon ) \end{aligned}$$

which is (5.6), proving the lemma. $\square$

5.1.2 Proof of Lemma 5.2

$\square$The proof presented in the following is based on [10, Lemma 2]. Let $0\le k\le n$ be arbitrary. We define

$$\begin{aligned} P_{UVY_1^k\tilde{Y}_{k+1}^n}(u,v,y^k_1,{y}^n_{k+1})=:\sum _{x^n\in \mathscr {X}^n}\prod _{l=1}^k W(y_l|x_l)\prod _{l=k+1}^n \widetilde{W}(y_l|x_l)E(x^n|v)P_{V|U}(v|u)P_{U}(u). \end{aligned}$$

So we have

$$\begin{aligned} I(V;Y^n|U)-I(V;\tilde{Y}^n|U)=\sum _{k=0}^{n-1}\Big (I(V;Y_1^{k+1}\tilde{Y}_{k+2}^n|U)-I(V;Y^k_1\tilde{Y}_{k+1}^n|U)\Big ). \end{aligned}$$

For all $0\le k\le n-1$ it holds that

$$\begin{aligned}&I(V;Y^{k+1}_1\tilde{Y}_{k+2}^n|U)-I(V;Y^k_1\tilde{Y}_{k+1}^n|U) \nonumber \\&\qquad = I(V;Y^k_1|U)+I(V;Y_{k+1}\tilde{Y}_{k+2}^n|Y^k_1U)-I(V;Y^k_1|U)-I(V;\tilde{Y}_{k+1}^n|Y^k_1U)\nonumber \\&\qquad = I(V;Y_{k+1}\tilde{Y}_{k+2}^n|Y^k_1U)-I(V;\tilde{Y}_{k+1}^n|Y^k_1U)\nonumber \\&\qquad = I(V;\tilde{Y}_{k+2}^n|Y^k_1U)+I(V;Y_{k+1}|\tilde{Y}_{k+2}^nY^k_1U)\nonumber \\&\qquad \qquad \qquad -I(V;\tilde{Y}_{k+2}^n|Y^k_1U)-I(V;\tilde{Y}_{k+1}|\tilde{Y}_{k+2}^nY^k_1U)\nonumber \\&\qquad = I(V;Y_{k+1}|\tilde{Y}_{k+2}^nY^k_1U)-I(V;\tilde{Y}_{k+1}|\tilde{Y}_{k+2}^nY^k_1U)\nonumber \\&\qquad =H(Y_{k+1}|\tilde{Y}_{k+2}^nY^k_1U)-H(\tilde{Y}_{k+1}|\tilde{Y}_{k+2}^nY^k_1U)\nonumber \\&\qquad \qquad \qquad -H(VY_{k+1}|\tilde{Y}_{k+2}^nY^k_1U)+H(V\tilde{Y}_{k+1}|\tilde{Y}_{k+2}^nY^k_1U). \end{aligned}$$

(5.26)

We want to analyze the right-hand side of (5.26). For $0\le k\le n-1$, it holds that

$$\begin{aligned}&\Vert P_{UVY^{k+1}_1\tilde{Y}^{n}_{k+2}}-P_{UVY^{k}_1\tilde{Y}^{n}_{k+1}}\Vert \\&\qquad =\sum _{v\in \mathscr {V}}\sum _{u\in \mathscr {U}}\sum _{y^n\in \mathscr {Y}^n}\Big |P_{UVY^{k+1}_1\tilde{Y}^{n}_{k+2}}(u,v,y^{k+1}_1y^{n}_{k+2})-P_{UVY^{k}_1\tilde{Y}^{n}_{k+1}}(u,v,y^{k}_1y^{n}_{k+1})\Big |\\&\qquad =\sum _{v\in \mathscr {V}}\sum _{u\in \mathscr {U}}\sum _{y^n\in \mathscr {Y}^n}\Big |\sum _{x^n\in \mathscr {X}^n}\Big (\prod _{l=1}^{k+1} W(y_l|x_l)\prod _{l=k+2}^n \widetilde{W}(y_l|x_l)\\&\qquad \qquad \qquad -\prod _{l=1}^{k+1} W(y_l|x_l)\prod _{l=k+2}^n \widetilde{W}(y_l|x_l)\Big )E(x^n|v)P_{V|U}(v|u)P_{U}(u)\Big |\\&\qquad =\sum _{v\in \mathscr {V}}\sum _{u\in \mathscr {U}}\sum _{y^n\in \mathscr {Y}^n}\Big |\sum _{x^n\in \mathscr {X}^n}\prod _{l=1}^{k} W(y_l|x_l)\prod _{l=k+2}^n \widetilde{W}(y_l|x_l)\Big (W(y_{k+1}|x_{k+1})\\&\qquad \qquad \qquad -\widetilde{W}(y_{k+1}|x_{k+1})\Big )E(x^n|v)P_{V|U}(v|u)P_{U}(u)\Big |\\&\qquad \le \sum _{v\in \mathscr {V}}\sum _{u\in \mathscr {U}}\sum _{y^n\in \mathscr {Y}^n}\sum _{x^n\in \mathscr {X}^n}\prod _{l=1}^{k} W(y_l|x_l)\prod _{l=k+2}^n \widetilde{W}(y_l|x_l)\Big |W(y_{k+1}|x_{k+1})\\&\qquad \qquad \qquad -\widetilde{W}(y_{k+1}|x_{k+1})\Big |E(x^n|v)P_{V|U}(v|u)P_{U}(u)\\&\qquad =\sum _{v\in \mathscr {V}}\sum _{u\in \mathscr {U}}\sum _{x^n\in \mathscr {X}^n}\Big (\sum _{y^n\in \mathscr {Y}^n}\prod _{l=1}^{k} W(y_l|x_l)\prod _{l=k+2}^n \widetilde{W}(y_l|x_l)\Big |W(y_{k+1}|x_{k+1})\\&\qquad \qquad \qquad -\widetilde{W}(y_{k+1}|x_{k+1})\Big |\Big )E(x^n|v)P_{V|U}(v|u)P_{U}(u)\\&\qquad =\sum _{u\in \mathscr {U}}\sum _{x^n\in \mathscr {X}^n}\sum _{y_{k+1}\in \mathscr {Y}}\Big |W(y_{k+1}|x_{k+1})\\&\qquad \qquad \qquad -\widetilde{W}(y_{k+1}|x_{k+1})\Big |E(x^n|v)P_{V|U}(v|u)P_{U}(u)\\&\qquad <\varepsilon \sum _{v\in \mathscr {V}}\sum _{u\in \mathscr {U}}\sum _{x^n\in \mathscr {X}^n}E(x^n|v)P_{V|U}(v|u)P_{U}(u)=\varepsilon . \end{aligned}$$

This shows that the total variation between the joint probability distribution $P_{\textit{UVY}^k\tilde{Y}_{k+1}^n}$ and $P_{UVY^{k+1}\tilde{Y}_{k+2}^n}$ is smaller than $\varepsilon$. Then by Lemma 5.1 it holds that

$$\begin{aligned} \Big |H(Y_{k+1}|\tilde{Y}_{k+2}^nY^k_1U)-H(\tilde{Y}_{k+1}|\tilde{Y}_{k+2}^nY^k_1U)\Big |<2\varepsilon \log |\mathscr {Y}|+2H_2(\varepsilon ) \end{aligned}$$

(5.27)

and

$$\begin{aligned}&\Big |H(VY_{k+1}|\tilde{Y}_{k+2}^nY^k_1U)-H(V\tilde{Y}_{k+1}|\tilde{Y}_{k+2}^nY^k_1U)\Big |\nonumber \\&\qquad =\Big |H(V|\tilde{Y}_{k+2}^nY^k_1U)+H(Y_{k+1}|V\tilde{Y}_{k+2}^nY^k_1U) \nonumber \\&\qquad \qquad \qquad -H(V|\tilde{Y}_{k+2}^nY^k_1U)-H(\tilde{Y}_{k+1}|V\tilde{Y}_{k+2}^nY^k_1U)\Big | \nonumber \\&\qquad =\Big |H(Y_{k+1}|V\tilde{Y}_{k+2}^nY^k_1U)-H(\tilde{Y}_{k+1}|V\tilde{Y}_{k+2}^nY^k_1U)\Big |\nonumber \\&\qquad <2\varepsilon \log |\mathscr {Y}|+2H_2(\varepsilon ). \end{aligned}$$

(5.28)

Inserting (5.27) and (5.28) into (5.26) we obtain

$$\begin{aligned} \Big |I(V;Y^{k+1}_1\tilde{Y}_{k+2}^n|U)-I(V;Y^k_1\tilde{Y}_{k+1}^n|U)\Big |\le 4\varepsilon \log |\mathscr {Y}|+4H_2(\varepsilon ):=\delta _2(\varepsilon ,|\mathscr {Y}|). \end{aligned}$$

(5.29)

This gives in particular the following upper bound for the difference between $I(V;Y^n|U)$ and $I(V;\tilde{Y}^n|U)$:

$$\begin{aligned} \Big |I(V;Y^n|U)-I(V;\tilde{Y}^n|U)\Big |&\le \sum _{k=0}^{n-1}\Big |I(V;Y_1^{k+1}\tilde{Y}_{k+2}^n|U)-I(V;Y^k_1\tilde{Y}_{k+1}^n|U)\Big |\\&\le n\delta _2(\varepsilon ,|\mathscr {Y}|) \end{aligned}$$

proving the lemma. $\square$