Ozarow linear feedback scheme achieves the two-user Gaussian sum point

Put $a(r)=1-r^2$ and $F(r)=(1+s_{1}a(r))(1+s_{2}a(r))-(1+s_1+s_2+2r\sqrt{s_1{s}_2})$. Then $F(0)=s_1{s}_2>0$, $F(1)<0$, and $F^{\mathrm{\prime }}(r)=-2r[s_1(1+s_{2}a(r))+s_2(1+s_{1}a(r))]-2\sqrt{s_1{s}_2}<0.$ Continuity and strict decrease give a unique root ${\rho }_{\star }\in (0,1)$. Taking logarithms of its equation proves the rate-sum identity.

Choose any positive rates $0<R_j<\frac{1}{2}\ln (1+s_{j}a({\rho }_{\star }))$. By continuity of the unique root and rate expressions in $(P_1,P_2)$, choose temporary powers $0<P_j^{\mathrm{\prime }}<P_j$ close enough to $P_j$ that these strict inequalities still hold for $s_j^{\mathrm{\prime }}=P_j^{\mathrm{\prime }}\mathrm{/}N$ and their root ${\rho }^{\mathrm{\prime }}$. Continuity follows because every convergent subsequence of roots solves the limiting strictly decreasing equation, whose root is unique. Write ${\tilde{Z}}_i=Z_i\mathrm{/}\sqrt{N}$ and ${\tilde{Y}}_i=Y_i\mathrm{/}\sqrt{N}$, so ${\tilde{Y}}_i=V_{1i}+V_{2i}+{\tilde{Z}}_i$ and $\mathrm{Var}{\tilde{Z}}_i=1$. Use midpoint grids ${\theta }_j(m)=-1+(2m-1)\mathrm{/}{M}_j$ in $[-1,1]$, $M_j=\lceil e^{n{R}_j}\rceil$. The three startup transmissions in normalized units are $(V_{1,-2},V_{2,-2})=(0,{\theta }_2)$, $(V_{1,-1},V_{2,-1})=({\theta }_1,0)$ and $(V_{1,0},V_{2,0})=(0,0)$. User 1 learns ${\tilde{Z}}_{-1}$ by subtracting its own point from ${\tilde{Y}}_{-1}$; user 2 similarly learns ${\tilde{Z}}_{-2}$, and both learn ${\tilde{Z}}_0={\tilde{Y}}_0$ by feedback.

Set ${\lambda }^2={\rho }^{\mathrm{\prime }}\mathrm{/}(1-{\rho }^{\mathrm{\prime }})$, $U_1={\tilde{Z}}_{-1}+\lambda {\tilde{Z}}_0$, $U_2={\tilde{Z}}_{-2}+\lambda {\tilde{Z}}_0$, and $V_{j1}=\sqrt{s_j^{\mathrm{\prime }}\mathrm{/}(1+{\lambda }^2)}{U}_j$. Then both $U_j$ are known to their respective senders, have variance $1+{\lambda }^2$, and correlation ${\rho }^{\mathrm{\prime }}$, so $V_{j1}$ has variance $s_j^{\mathrm{\prime }}$ and the pair has correlation ${\rho }^{\mathrm{\prime }}$. For $i\ge 2$ define the causal innovations $E_{j,i-1}=V_{j,i-1}-\mathbb{E}[V_{j,i-1}\mid {\tilde{Y}}_1^{i-1}]$ using only the recursive outputs, and transmit $V_{1i}=\sqrt{s_1^{\mathrm{\prime }}\mathrm{/}\mathbb{E}{E}_{1,i-1}^2}{E}_{1,i-1}$ and $V_{2i}=-\sqrt{s_2^{\mathrm{\prime }}\mathrm{/}\mathbb{E}{E}_{2,i-1}^2}{E}_{2,i-1}$. Gaussian conditional means are affine functions of the observed histories, so these finite-message encoder maps are Borel-measurable; they are causal because the normalized history is an invertible rescaling of the physical output history.

The initial pair is jointly Gaussian and independent of the empty recursive history. Inductively, if the current pair is centered Gaussian with variances $s_1^{\mathrm{\prime }},s_2^{\mathrm{\prime }}$, correlation $r$, and independent of the earlier recursive outputs, its observation is the pair's sum plus fresh unit Gaussian noise. The Gaussian residual after conditioning on this observation is independent of the observation and earlier history. Formula [F3] therefore gives the next normalized opposite-sign correlation $r_{\mathrm{n}\mathrm{e}\mathrm{w}}=\frac{\sqrt{s_1^{\mathrm{\prime }}{s}_2^{\mathrm{\prime }}}(1-r^2)-r}{\sqrt{(1+s_1^{\mathrm{\prime }}(1-r^2))(1+s_2^{\mathrm{\prime }}(1-r^2))}},$ with both marginal variances restored to $s_j^{\mathrm{\prime }}$ by the chosen scales.

At $r={\rho }^{\mathrm{\prime }}$, the denominator in step 3.1 is $\sqrt{D}$, where $D=1+s_1^{\mathrm{\prime }}+s_2^{\mathrm{\prime }}+2{\rho }^{\mathrm{\prime }}\sqrt{s_1^{\mathrm{\prime }}{s}_2^{\mathrm{\prime }}}$. Expanding the root equation gives $(\sqrt{s_1^{\mathrm{\prime }}{s}_2^{\mathrm{\prime }}}(1-{\rho }^{\mathrm{\prime }2})-{\rho }^{\mathrm{\prime }}{)}^2={\rho }^{\mathrm{\prime }2}D\mathrm{.}$ The unsquared numerator is positive: let $r_0$ be the unique root of $r=\sqrt{s_1^{\mathrm{\prime }}{s}_2^{\mathrm{\prime }}}(1-r^2)$. Substitution gives $F(r_0)=r_0^2(1-s_1^{\mathrm{\prime }}-s_2^{\mathrm{\prime }})-2r_0\sqrt{s_1^{\mathrm{\prime }}{s}_2^{\mathrm{\prime }}}<0$; if $s_1^{\mathrm{\prime }}+s_2^{\mathrm{\prime }}<1$, use $r_0<\sqrt{s_1^{\mathrm{\prime }}{s}_2^{\mathrm{\prime }}}$, and otherwise both terms are nonpositive with the second strictly negative. Since $F$ is strictly decreasing, ${\rho }^{\mathrm{\prime }}<r_0$, so $\sqrt{s_1^{\mathrm{\prime }}{s}_2^{\mathrm{\prime }}}(1-{\rho }^{\mathrm{\prime }2})-{\rho }^{\mathrm{\prime }}>0$. Thus $r_{\mathrm{n}\mathrm{e}\mathrm{w}}={\rho }^{\mathrm{\prime }}$, and induction proves the same correlation and powers at every recursive use.

Conditional on the recursive past, the two current inputs are centered jointly Gaussian with variances $s_j^{\mathrm{\prime }}$ and correlation ${\rho }^{\mathrm{\prime }}$. Thus ${\tilde{Y}}_i\mid {\tilde{Y}}_1^{i-1}$ is Gaussian with variance $D$, while ${\tilde{Y}}_i\mid (U_j,{\tilde{Y}}_1^{i-1})$ is Gaussian with variance $v_j:=1+s_{3-j}^{\mathrm{\prime }}(1-{\rho }^{\mathrm{\prime }2})$ because $V_{ji}$ is determined by $(U_j,{\tilde{Y}}_1^{i-1})$. Define the conditional information here as the expectation of the log ratio of these two conditional Gaussian densities. Integrating their log densities, whose centered quadratic terms have expectation one, gives $I(U_j;{\tilde{Y}}_i\mid {\tilde{Y}}_1^{i-1})=\frac{1}{2}\ln (D\mathrm{/}{v}_j)$. The root equation then gives $I(U_j;{\tilde{Y}}_i\mid {\tilde{Y}}_1^{i-1})=\frac{1}{2}\ln \frac{D}{1+s_{3-j}^{\mathrm{\prime }}(1-{\rho }^{\mathrm{\prime }2})}=\frac{1}{2}\ln (1+s_j^{\mathrm{\prime }}(1-{\rho }^{\mathrm{\prime }2}))\mathrm{.}$ Factoring the positive conditional Gaussian densities over time and summing their log ratios gives the chain rule directly, so $I(U_j;{\tilde{Y}}_1^n)=\frac{n}{2}\ln (1+s_j^{\mathrm{\prime }}(1-{\rho }^{\mathrm{\prime }2}))$. Since $(U_j,{\tilde{Y}}_1^n)$ is jointly Gaussian, the same density-ratio calculation gives $I(U_j;{\tilde{Y}}_1^n)=\frac{1}{2}\ln \text{\hspace{0.17em}⁣}(\mathrm{Var}(U_j)\mathrm{/}\mathrm{Var}(U_j\mid {\tilde{Y}}_1^n))$: the conditional Gaussian variance is constant in the observed value, and the log-density ratio expectation is the difference of the two Gaussian entropies. Therefore $\mathrm{Var}(U_j\mid {\tilde{Y}}_1^n)=\mathrm{Var}(U_j)e^{-2I(U_j;{\tilde{Y}}_1^n)}$.

The receiver observes ${\tilde{Y}}_{-1}+\lambda {\tilde{Y}}_0={\theta }_1+U_1$ and ${\tilde{Y}}_{-2}+\lambda {\tilde{Y}}_0={\theta }_2+U_2$. It estimates ${\theta }_j$ by subtracting $\mathbb{E}[U_j\mid {\tilde{Y}}_1^n]$. The estimation error variance is the conditional variance from step 5.1. Since the adjacent grid spacing is $2\mathrm{/}{M}_j$, decode to a nearest grid point and resolve any tie by choosing the least-indexed point. Chebyshev's inequality makes the decoding error tend to zero whenever $R_j<\frac{1}{2}\ln (1+s_j^{\mathrm{\prime }}(1-{\rho }^{\mathrm{\prime }2}))$.

The $n$ recursive symbols use expected physical power $P_j^{\mathrm{\prime }}$ each, and the one nonzero startup symbol for user $j$ has physical energy at most $N$. Thus total expected energy is at most $n{P}_j^{\mathrm{\prime }}+N$, which is at most $(n+3)P_j$ for all sufficiently large $n$ because $P_j^{\mathrm{\prime }}<P_j$. The block has $n+3$ uses and rates $n{R}_j\mathrm{/}(n+3)\to R_j$. Letting $P_j^{\mathrm{\prime }}\uparrow P_j$ and then closing the strict rates proves achievability of the stated corner.