Physically degraded two-user DM broadcast capacity region

Fix a finite law $P_{UX}=P_U{P}_{X\mid U}$ and rates with $R_2<I(U;Y_2)$ and $R_1<I(X;Y_1\mid U)$. Generate independent IID cloud words $u^n(m_2)\sim P_U^n$ and, conditionally on each cloud, IID satellite words $x^n(m_1,m_2)\sim {\prod }_i{P}_{X\mid U}(\cdot \mid u_i(m_2))$. Send the satellite indexed by $(m_1,m_2)$.

For a converse code, let $U_i:=(M_2,Y_1^{i-1})$. Fano [F7], independence in [F2], and the chain rule give $n{R}_1\le I(M_1;Y_1^n\mid M_2)+\ln 2+P_{e}n{R}_1,n{R}_2\le I(M_2;Y_2^n)+\ln 2+P_{e}n{R}_2\mathrm{.}$ The information terms are at most $n\ln \mathrm{\mid }{\mathcal{Y}}_1\mathrm{\mid }$ and $n\ln \mathrm{\mid }{\mathcal{Y}}_2\mathrm{\mid }$, respectively, so the Fano inequalities bound each rate whenever $P_e<1\mathrm{/}2$ and their normalized remainders vanish along a reliable sequence. Expanding the first information and using the no-feedback encoder and memorylessness gives $I(M_1;Y_1^n\mid M_2)\le {\sum }_{i}I(X_i;Y_{1i}\mid U_i)$.

Receiver 2 outputs the unique cloud passing the test ${\sum }_i\ln [P_{Y_2\mid U}(y_i\mid u_i)\mathrm{/}{P}_{Y_2}(y_i)]>n{\tau }_2$, with $R_2<{\tau }_2<I(U;Y_2)$, and outputs the fixed default cloud $1$ if zero or multiple clouds pass. Receiver 1 outputs the own-message component of the unique pair passing both tests: the cloud density with $Y_1$ exceeds $n{\tau }_1$, and the conditional satellite density ${\sum }_i\ln [P_{Y_1\mid X,U}(y_i\mid x_i,u_i)\mathrm{/}{P}_{Y_1\mid U}(y_i\mid u_i)]$ exceeds $n{\tau }_X$. In each test, a summand with zero numerator is $-\mathrm{\infty }$; on the codebook support a positive numerator has positive denominator. If zero or multiple pairs pass, receiver 1 outputs the fixed default message $1$. Choose thresholds with $R_2<{\tau }_1<I(U;Y_1)$ and $R_1<{\tau }_X<I(X;Y_1\mid U)$. The chain rule gives $I(U;Y_1)=I(U;Y_2)+I(U;Y_1\mid Y_2)$ because $I(U;Y_2\mid Y_1)=0$ by [F1]; the last conditional information is nonnegative by [F5].

For the weak receiver, adding variables to the first argument of mutual information and then using physical degradation gives $I(M_2;Y_2^n)\le {\sum }_{i}I(M_2,Y_1^{i-1},Y_2^{i-1};Y_{2i})={\sum }_{i}I(U_i;Y_{2i})\mathrm{.}$ The equality holds because, conditional on $(M_2,Y_1^{i-1})$, the past weak outputs are generated from $Y_1^{i-1}$ independently of the current output.

The true cloud and satellite information densities are sums of IID finite-valued terms on their positive supports. Their means are the four mutual informations used in step 2.1; Chebyshev's inequality makes every true threshold pass with probability tending to one.

A false cloud word is independent of $Y_1^n$ and $Y_2^n$; under the corresponding product reference law each cloud likelihood ratio has expectation one, so a threshold $n{\tau }_1$ or $n{\tau }_2$ is passed with probability at most $e^{-n{\tau }_j}$. A false satellite in the true cloud is independent conditionally on that cloud and has conditional likelihood-ratio expectation one, so its threshold probability is at most $e^{-n{\tau }_X}$. There are at most $e^{n{R}_2+o(n)}$ wrong clouds and $e^{n{R}_1+o(n)}$ wrong satellites per true cloud. The strict threshold gaps and union bound make both receiver error probabilities tend to zero in ensemble mean.

The random codebook ensemble is finite at each blocklength; since its mean joint error tends to zero, at least one deterministic codebook has error no larger than that mean. If $R_2=0$, one cloud is used and its cloud threshold is omitted; if $R_1=0$, one satellite per cloud is used and its satellite threshold is omitted. The remaining packing bounds above still apply; taking closure gives every nonnegative rate pair in the asserted union.

Let $Q$ be uniform on $\{1,\dots ,n\}$ and independent, and put $U=(Q,U_Q)$, $X=X_Q$, $Y_j=Y_{jQ}$. The channel factorization [F1] gives $U\to X\to Y_1\to Y_2$. Step 1.2 gives $R_1\le I(X;Y_1\mid U)+{ϵ}_n$, and step 2.2 gives $R_2\le I(U_Q;Y_{2Q}\mid Q)+{ϵ}_n\le I(U;Y_2)+{ϵ}_n$ by nonnegative $I(Q;Y_2)$. The alphabets of $U$ are finite. Let $n\to \mathrm{\infty }$ and take closure.