Ozarow capacity region of the two-user real AWGN MAC with feedback

For the converse normalize $V_{ji}=X_{ji}\mathrm{/}\sqrt{N}$ and write $Y_i$ for the invertibly rescaled output $Y_i^{\mathrm{p}\mathrm{h}\mathrm{y}\mathrm{s}}\mathrm{/}\sqrt{N}$; the normalized noise has variance one and $n^{-1}{\sum }_i\mathbb{E}{V}_{ji}^2\le s_j:=P_j\mathrm{/}N$. This output scaling preserves the feedback histories and the decoder's measurability. For fixed messages the Borel causal encoders and Gaussian transitions give the positive density $p(y^n\mid m_1,m_2)={\prod }_{i=1}^n{\varphi }_1(y_i-v_{1i}(m_1,y^{i-1})-v_{2i}(m_2,y^{i-1})),$ normalized by successive integration; averaging over the finite message sets gives positive conditional densities $p(y^n\mid m_j)$ and $p(y^n)$. Define each continuous mutual information below as the expectation of its log density ratio. Bayes' rule identifies $I(M_j;Y^n\mid M_{3-j})$ with $H(M_j\mid M_{3-j})-H(M_j\mid Y^n,M_{3-j})$, and similarly $I(M_1,M_2;Y^n)=H(M_1,M_2)-H(M_1,M_2\mid Y^n)$. Factoring the conditional densities over successive outputs and summing their log ratios gives the chain rules used in steps 2.1 and 4.1. These finite sums of log ratios are integrable: the conditional density of each output given both messages and its history is Gaussian, while after averaging over the finite messages each one-use conditional density is a finite Gaussian mixture. Its conditional entropy is finite by [F2], and its average is finite under the power constraints. Project the Borel joint decoder to estimate each $M_j$; these estimators take values in the respective message sets and have error at most $P_e$. Fano [F3] therefore gives $n{R}_j\le I(M_j;Y^n\mid M_{3-j})+\ln 2+P_{e}n{R}_j$ and, for the pair, $n(R_1+R_2)\le I(M_1,M_2;Y^n)+\ln 2+P_{e}n(R_1+R_2)$. Put $v_{ji}=\mathrm{Var}(V_{ji})$, $c_i=\mathrm{Cov}(V_{1i},V_{2i})$, and bars for time averages. By [F1], given $(M_2,Y^{i-1})$ the value $V_{2i}$ is known.

If $v_{2i}>0$, the conditional mean minimizes mean-square error, so $\mathbb{E}\mathrm{Var}(V_{1i}\mid M_2,Y^{i-1})\le \mathbb{E}{[V_{1i}-\mathbb{E}{V}_{1i}-\frac{c_i}{v_{2i}}(V_{2i}-\mathbb{E}{V}_{2i})]}^2=v_{1i}-\frac{c_i^2}{v_{2i}}\mathrm{.}$ The predictor is measurable from $(M_2,Y^{i-1})$ because $V_{2i}$ is. If $v_{2i}=0$, then $c_i=0$ and the bound is $v_{1i}$. Given $(M_1,M_2,Y^{i-1})$, the output density is a translate of ${\varphi }_1$, so its conditional differential entropy is $h(Z_i)$. Given $(M_2,Y^{i-1})$, it is a Gaussian convolution; integrating the conditional log density ratio gives $I(M_1;Y_i\mid M_2,Y^{i-1})=h(Y_i\mid M_2,Y^{i-1})-h(Z_i)$. Apply the scalar case of [F2] to this conditional mixture. When $v_{2i}>0$ it yields $I(M_1;Y_i\mid M_2,Y^{i-1})\le C\text{\hspace{0.17em}⁣}(v_{1i}-\frac{c_i^2}{v_{2i}}),C(t):=\frac{1}{2}\ln (1+t),$ and when $v_{2i}=0$ it yields $C(v_{1i})$. Interchanging the users gives the corresponding second-user bound.

Define $c_i^2\mathrm{/}{v}_{2i}:=0$ when $v_{2i}=0$; then also $c_i=0$ there. Jensen's inequality for the concave function $C$ and Cauchy–Schwarz yield $\frac{1}{n}{\sum }_{i}C\text{\hspace{0.17em}⁣}(v_{1i}-\frac{c_i^2}{v_{2i}})\le C\text{\hspace{0.17em}⁣}({\bar{v}}_1-\frac{{\bar{c}}^2}{{\bar{v}}_2})$ when ${\bar{v}}_2>0$, because $({\sum }_i{c}_i{)}^2\le ({\sum }_{i:v_{2i}>0}{c}_i^2\mathrm{/}{v}_{2i})({\sum }_i{v}_{2i})$. The zero-variance case uses $\bar{c}=0$ and the bound $C({\bar{v}}_1)$. Similarly, defining the zero quotient when $v_{1i}=0$, the other user is bounded by $C({\bar{v}}_2-{\bar{c}}^2\mathrm{/}{\bar{v}}_1)$ when ${\bar{v}}_1>0$, with $C({\bar{v}}_2)$ otherwise.

For the sum, condition on $Y^{i-1}$. Given both messages and that history the current output is a unit Gaussian translate, so integrating its conditional log density ratio gives $I(M_1,M_2;Y_i\mid Y^{i-1})=h(Y_i\mid Y^{i-1})-h(Z_i)\mathrm{.}$ The conditional output is the Gaussian convolution of $V_{1i}+V_{2i}$ with fresh unit noise. By [F2], its entropy is at most $\frac{1}{2}\ln (2\pi e[1+\mathbb{E}\mathrm{Var}(V_{1i}+V_{2i}\mid Y^{i-1})])$, and the law of total variance bounds the averaged conditional variance by $\mathrm{Var}(V_{1i}+V_{2i})=v_{1i}+v_{2i}+2c_i$. Therefore $I(M_1,M_2;Y_i\mid Y^{i-1})\le C(v_{1i}+v_{2i}+2c_i)\mathrm{.}$ Jensen across $i$ gives the bound $C({\bar{v}}_1+{\bar{v}}_2+2\bar{c})$.

If ${\bar{v}}_1{\bar{v}}_2>0$, set ${\rho }_n=\mathrm{\mid }\bar{c}\mathrm{\mid }\mathrm{/}\sqrt{{\bar{v}}_1{\bar{v}}_2}\in [0,1]$; otherwise set ${\rho }_n=0$. The individual right sides in step 3.1 are at most $C(s_j(1-{\rho }_n^2))$. Substituting these bounds in the exact Fano inequalities of step 1.1 gives $(1-P_e)R_j\le C(s_j)+\ln 2\mathrm{/}n$, so both rates are uniformly bounded for $P_e<1\mathrm{/}2$ and all Fano remainders divided by $n$ vanish. The sum right side in step 4.1 is at most $C(s_1+s_2+2{\rho }_n\sqrt{s_1{s}_2})$: replacing negative $\bar{c}$ by its absolute value only enlarges it, and ${\bar{v}}_j\le s_j$. Along a subsequence ${\rho }_n$ converges in $[0,1]$; continuity gives the stated outer region.

If either power is zero, the displayed union reduces to the single-user interval for the positive-power user, or to the origin if both vanish; [F6] achieves it and step 1 proves the converse. Assume henceforth $P_1,P_2>0$. Write $a_j(\rho )=C(s_j(1-{\rho }^2))$ and $b(\rho )=C(s_1+s_2+2\rho \sqrt{s_1{s}_2})$. The sum-corner equality in [F4] has a unique root ${\rho }_{\star }$; $a_j$ decrease with $\rho$ and $b$ increases. At $\rho =0$, [F6] achieves the displayed no-feedback region. At $\rho ={\rho }_{\star }$, [F4] achieves the corner $(a_1,a_2)$, while [F6] achieves the axis vertices $(a_1,0)$ and $(0,a_2)$ because $a_j({\rho }_{\star })\le C(s_j)$; the origin is achieved with singleton messages.

Fix $0<\rho <{\rho }_{\star }$. Then $b(\rho )<a_1(\rho )+a_2(\rho )$, and $b(\rho )>a_j(\rho )$ because at $\rho =0$ the sum bound is strictly greater than either individual bound and it increases while each individual bound decreases. Thus both endpoints of the active sum-rate face are nonnegative: $(a_1(\rho ),b(\rho )-a_1(\rho ))$ and $(b(\rho )-a_2(\rho ),a_2(\rho ))$. By [F5] the first is achievable; interchanging user labels gives the second. The axis vertices $(a_1(\rho ),0)$ and $(0,a_2(\rho ))$ are achievable by single-user Gaussian codes, since each $a_j(\rho )\le C(s_j)$ and [F6] supplies those rates.

For each $0<\rho <{\rho }_{\star }$, a deterministic public schedule that allocates long subblocks to the four vertices from step 7.1 and the origin achieves their convex hull. At $\rho ={\rho }_{\star }$, time-sharing the corner and the two axis vertices from step 6.1 with the origin achieves the full rectangle. Split each private message into independent submessages, use the corresponding code on each subblock, and let both senders ignore outputs from earlier subblocks when a new code starts. The block error is at most the sum of subblock errors and each average power is the weighted average of powers at most $P_j$. The convex hull for each parameter is exactly its polygon cut out by the two individual bounds and the active sum bound.

For $\rho >{\rho }_{\star }$, $a_j(\rho )\le a_j({\rho }_{\star })$ and $b(\rho )\ge b({\rho }_{\star })=a_1({\rho }_{\star })+a_2({\rho }_{\star })$. Thus the region at $\rho$ is a rectangle contained in the rectangle at ${\rho }_{\star }$ and adds no rates. Steps 6.1, 7.1 and 8.1 therefore achieve the full union in the Statement. Combining with the converse proves equality.