Rate splitting transfers Ozarow's sum point to lower-correlation corners

First temporarily lower the total powers to $P_1^{\circ }<P_1$, $P_2^{\circ }<P_2$ close to the budgets. The unique root and corner rates vary continuously with these powers: any convergent subsequence of roots as the powers converge must solve the limiting strictly decreasing root equation, whose root is unique. Hence still $0<\rho <{\rho }_{\star }(P_1^{\circ },P_2^{\circ },N)$ when the reductions are small. Write ${\rho }_{\ast }^{\circ }$ for this temporary sum-corner root and $C(t)=\frac{1}{2}\ln (1+t)$.

For fixed temporary powers define $\mathrm{\Phi }(r)=P_1^{\circ }{P}_2^{\circ }(1-r^2{)}^2-N(P_1^{\circ }+P_2^{\circ })r^2-2Nr\sqrt{P_1^{\circ }{P}_2^{\circ }}\mathrm{.}$ Expanding the root equation in [F3] shows $\mathrm{\Phi }({\rho }_{\ast }^{\circ })=0$. Its derivative is strictly negative on $[0,1)$, so $\mathrm{\Phi }(\rho )>0$. On $0\le \alpha \le 1-{\rho }^2$, set $G(\alpha )=1+\frac{P_2^{\circ }(1-{\rho }^2\mathrm{/}(1-\alpha ))}{N+\alpha P_1^{\circ }},L=\frac{N+P_1^{\circ }+P_2^{\circ }+2\rho \sqrt{P_1^{\circ }{P}_2^{\circ }}}{N+P_1^{\circ }(1-{\rho }^2)}\mathrm{.}$ The function $G$ is continuous and strictly decreasing, $G(1-{\rho }^2)=1<L$, and direct cross-multiplication gives $G(0)\ge L$ exactly when $\mathrm{\Phi }(\rho )\ge 0$. Thus the intermediate value theorem gives an $\alpha \in [0,1-{\rho }^2)$ satisfying $L=1+\frac{P_2^{\circ }(1-{\rho }^2\mathrm{/}(1-\alpha ))}{N+\alpha P_1^{\circ }}\mathrm{.}$ For $0<\rho <{\rho }_{\ast }^{\circ }$ the inequality at $\alpha =0$ is strict, so $\alpha >0$.

Put $P_0=\alpha P_1^{\circ }$, $P_1^{\mathrm{\prime }\mathrm{\prime }}=(1-\alpha )P_1^{\circ }$, $N^{\mathrm{\prime }}=N+P_0$, and ${\rho }_f=\rho \mathrm{/}\sqrt{1-\alpha }$. The root for the residual feedback channel is exactly ${\rho }_f$: indeed $N^{\mathrm{\prime }}+P_1^{\mathrm{\prime }\mathrm{\prime }}(1-{\rho }_f^2)=N+P_1^{\circ }(1-{\rho }^2),$ while the equation in step 2.1 gives $\frac{N^{\mathrm{\prime }}+P_2^{\circ }(1-{\rho }_f^2)}{N^{\mathrm{\prime }}}=L\mathrm{.}$ Multiplying these two normalized factors yields $(1+\frac{P_1^{\mathrm{\prime }\mathrm{\prime }}(1-{\rho }_f^2)}{N^{\mathrm{\prime }}})(1+\frac{P_2^{\circ }(1-{\rho }_f^2)}{N^{\mathrm{\prime }}})=1+\frac{P_1^{\mathrm{\prime }\mathrm{\prime }}+P_2^{\circ }+2{\rho }_f\sqrt{P_1^{\mathrm{\prime }\mathrm{\prime }}{P}_2^{\circ }}}{N^{\mathrm{\prime }}},$ so uniqueness in [F3] identifies ${\rho }_f$ as the residual sum-corner root.

Choose any $R_0$ with $0<R_0<C(P_0\mathrm{/}N)$ and set $L_0=\lceil e^{n{R}_0}\rceil$. Generate an independent private codebook $A^n(m_0)$, $m_0\in [L_0]$, with IID $N(0,P_0)$ coordinates. User 1 splits its message into $m_0$ and a feedback-layer message; it adds $A_i(m_0)$ to its feedback-layer input on each recursive use. User 2 uses the feedback layer with residual power $P_2^{\circ }$. On the three startup uses send only the feedback layer, using the message-point/noise construction of the sum-corner scheme. The startup noises have variance $N$, but this construction only needs the resulting initial Gaussian pair: choose the shared-noise coefficient and the two scales so its variances are $P_1^{\mathrm{\prime }\mathrm{\prime }},P_2^{\circ }$ and its correlation is ${\rho }_f$. On recursive uses, the codebook-ensemble private symbols are IID $N(0,P_0)$, independent of the feedback messages, past outputs and fresh channel noise before each current use. Thus they are an additional independent Gaussian noise of variance $P_0$, and the recursive feedback layer sees variance $N^{\mathrm{\prime }}$. The causal feedback-layer maps are affine in the observed past outputs and message points, and adding the fixed codeword is Borel-measurable, so they satisfy the feedback-code interface. The covariance update [F4] and the calculation in [F3] apply to this initial pair and recursive noise, giving feedback-layer rates arbitrarily close to $R_{1f}=C\text{\hspace{0.17em}⁣}\left(\frac{P_1^{\mathrm{\prime }\mathrm{\prime }}(1-{\rho }_f^2)}{N^{\mathrm{\prime }}}\right),R_2=C\text{\hspace{0.17em}⁣}\left(\frac{P_2^{\circ }(1-{\rho }_f^2)}{N^{\mathrm{\prime }}}\right)\mathrm{.}$

The feedback receiver first decodes both feedback-layer messages as in the sum-corner scheme, using nearest-grid decoding with ties assigned to the least-indexed grid point. For the strict feedback-layer rates in step 4.1, [F3] gives vanishing ensemble-average error. On the event that this stage is correct, the receiver reconstructs the transmitted feedback-layer inputs recursively and subtracts them. For the error bound, compare the private decoder on that event with a genie decoder given the true feedback-layer messages; their outputs agree whenever the first stage is correct. The genie residual is $A_i(m_0)+Z_i$. Choose $R_0<\tau <C(P_0\mathrm{/}N)$. Decode $m_0$ as the unique codeword whose log likelihood ratio $\mathrm{ı}(a^n;y^n):={\sum }_{i=1}^n\ln \frac{{\varphi }_N(y_i-a_i)}{{\varphi }_{N+P_0}(y_i)}$ exceeds $n\tau$; if there is not exactly one such codeword, output the fixed default $m_0=1$. For a true IID Gaussian symbol $A_i\sim N(0,P_0)$ and independent $Z_i\sim N(0,N)$, the summand equals $\frac{1}{2}\ln \frac{N+P_0}{N}-\frac{Z_i^2}{2N}+\frac{(A_i+Z_i{)}^2}{2(N+P_0)}\mathrm{.}$ Its mean is $C(P_0\mathrm{/}N)$ and its variance is finite because it is a quadratic polynomial in a finite-variance Gaussian pair. For a false codeword $A^{\mathrm{\prime }n}$ independent of the output $Y^n$, its likelihood ratio has expectation $\mathbb{E}\frac{p_{Y^n\mathrm{\mid }{A}^{\mathrm{\prime }n}}(Y^n\mathrm{\mid }{A}^{\mathrm{\prime }n})}{p_{Y^n}(Y^n)}=\int p_{A^{\mathrm{\prime }n}}(a^n)\int p_{Y^n}(y^n)\frac{p_{Y^n\mathrm{\mid }{A}^{\mathrm{\prime }n}}(y^n\mathrm{\mid }{a}^n)}{p_{Y^n}(y^n)}d{y}^{n}d{a}^n=1,$ since $p_{Y^n}$ is the $N(0,N+P_0{)}^n$ density and each conditional density integrates to one. Markov's inequality and a union bound make the false-pass probability at most $L_0{e}^{-n\tau }$, which tends to zero because $\tau >R_0$; Chebyshev's inequality makes the true-codeword miss probability tend to zero because $\tau <C(P_0\mathrm{/}N)$. Thus the genie private-decoder error tends to zero in ensemble mean, and the total error is at most the feedback-layer error plus this genie error.

The exact rate algebra is $C(P_0\mathrm{/}N)+R_{1f}=C\text{\hspace{0.17em}⁣}\left(\frac{P_1^{\circ }(1-{\rho }^2)}{N}\right),$ because $N^{\mathrm{\prime }}+P_1^{\mathrm{\prime }\mathrm{\prime }}(1-{\rho }_f^2)=N+P_1^{\circ }(1-{\rho }^2)$. Also step 2.1 gives $R_2=C\text{\hspace{0.17em}⁣}\left(\frac{P_1^{\circ }+P_2^{\circ }+2\rho \sqrt{P_1^{\circ }{P}_2^{\circ }}}{N}\right)-C\text{\hspace{0.17em}⁣}\left(\frac{P_1^{\circ }(1-{\rho }^2)}{N}\right)\mathrm{.}$ Taking $R_0$ and feedback-layer rates up to their strict limits gives the temporary-power corner.

It remains to fix a deterministic private codebook satisfying both expected block-power constraints. For a fixed private word $a^n$, the recursive causal feedback equations are linear in the past outputs and message points, so the full transmitted vector of user $j$ has the form $T_j{a}^n+B_j\xi$, where $\xi$ collects the channel noises and the independent centered message-grid points. Averaging over $\xi$ gives expected energy $\mathbb{E}\mathrm{\parallel }{T}_j{a}^n+B_j\xi {\mathrm{\parallel }}^2=(a^n{)}^T{K}_j{a}^n+c_j,K_j:=T_j^T{T}_j⪰0,c_j:=\mathbb{E}\mathrm{\parallel }{B}_j\xi {\mathrm{\parallel }}^2\mathrm{.}$ The startup contribution is $O(1)$ and the ensemble expected energy is $n{P}_j^{\circ }+O(1)$, so $P_0\mathrm{tr}{K}_j+c_j=n{P}_j^{\circ }+O(1)$ and $\mathrm{tr}{K}_j=O(n)$. For $L_0$ independent Gaussian private words, the variance of their message-averaged quadratic contribution divided by $n^2$ is at most $\frac{2P_0^2\mathrm{tr}(K_j^2)}{n^2{L}_0}\le \frac{2P_0^2(\mathrm{tr}{K}_j{)}^2}{n^2{L}_0}=O(L_0^{-1}),$ using the Gaussian fourth-moment identity after diagonalizing $K_j$. Since $R_0>0$, $L_0$ is exponential in $n$; the average powers therefore concentrate at $P_j^{\circ }<P_j$. The ensemble error tends to zero by steps 4.1 and 5.1, so Markov's inequality makes the event of nonvanishing error negligible. A deterministic codebook satisfying both power constraints and vanishing error exists.

Let $P_j^{\circ }\uparrow P_j$ and use continuity of the corner rates in step 5.2. Taking closure proves the stated point for the original powers. Exchanging user labels proves the other dominant corner.