Cut-dependent Gaussian multicast gap to the cut-set bound

Fix a destination $d\in A^c$. Give the messages of nodes in $A^c$ to that destination, and also give it all outputs $Y_{A^c}^n$. The revealed messages are independent of the source messages in $A$, and the original decoder is still available. Since rates are bits and Fano's bound is in nats, [F4] gives the exact finite-code bound $(1-P_e)n\ln 2{\sum }_{k\in A\cap \mathcal{S}}{R}_k^{(n)}\le I(M_{A\cap \mathcal{S}};Y_{A^c}^n\mid M_{A^c})+\ln 2.$ By the conditional chain rule [F6], expand the mutual information across time. For time $i$, condition on $H_i=(M_{A^c},Y_{A^c}^{i-1})$. Each input in $A^c$ is then fixed by causality in [F2]. The current outputs depend on messages in $A$ only through the current inputs $X_{A,i}$, so conditional data processing [F6] gives $I(M_{A\cap \mathcal{S}};Y_{A^c,i}\mid H_i)\le I(X_{A,i};Y_{A^c,i}\mid H_i)\mathrm{.}$ The right side is the mutual information of $G(A)X_{A,i}+Z_{A^c,i}$ under the conditional input law given $H_i$.

For almost every $h$, the conditional input second moment given $H_i=h$ is finite, since its average is bounded by the node power constraints. The conditional output is a translate of a Gaussian convolution, so its density is bounded by the standard Gaussian noise peak; [F7] makes its differential entropy finite. Let $K_{i,h}$ be the conditional covariance of $X_{A,i}$ given $H_i=h$. The channel noise is independent with covariance $I$, so [F3] bounds each conditional mutual information in step 1.1, measured in nats, by $\frac{1}{2}\ln \det (I_{\mathrm{\mid }{A}^c\mathrm{\mid }}+G(A)K_{i,h}G(A{)}^T)\mathrm{.}$ Set $\bar{K}=n^{-1}{\sum }_i\mathbb{E}[K_{i,H_i}]$. Since covariance trace is at most the corresponding second moment and each node's average power is at most $P$, $\bar{K}⪰0$ and $\mathrm{tr}\bar{K}\le \mathrm{\mid }A\mathrm{\mid }P$.

For positive-definite matrices $B,B_0$, let $C=B_0^{-1\mathrm{/}2}B{B}_0^{-1\mathrm{/}2}$. Its eigenvalues ${\lambda }_j$ are positive by [F5], and $\ln {\lambda }_j\le {\lambda }_j-1$ because $u-1-\ln u$ has derivative $1-1\mathrm{/}u$, decreases on $(0,1]$, increases on $[1,\mathrm{\infty })$, and vanishes at $1$. Thus $\ln \det B\le \ln \det B_0+\mathrm{tr}(B_0^{-1}(B-B_0))\mathrm{.}$ Apply this with $B=I+G{K}_{i,h}{G}^T$ and $B_0=I+G\bar{K}{G}^T$, then average over $i,h$; the trace term averages to zero. Hence the average natural log determinant in step 2.1 is at most $\ln \det (I+G\bar{K}{G}^T)$. The spectral theorem gives $\bar{K}⪯\mathrm{tr}(\bar{K})I$, since every nonnegative eigenvalue is at most their sum; hence $G\bar{K}{G}^T⪯\mathrm{\mid }A\mathrm{\mid }PG{G}^T$. If $0\prec C⪯D$, then $C^{-1\mathrm{/}2}D{C}^{-1\mathrm{/}2}=I+E$ for some $E⪰0$, so its eigenvalues are at least one and $\det C\le \det D$. Applying this order property and dividing the Fano bound in step 1.1 by $n\ln 2$ yields $(1-P_e){\sum }_{k\in A\cap \mathcal{S}}{R}_k^{(n)}\le \frac{1}{2\ln 2}\ln \det (I_{\mathrm{\mid }{A}^c\mathrm{\mid }}+\mathrm{\mid }A\mathrm{\mid }PG{G}^T)+\frac{1}{n}\mathrm{.}$ Along a sequence of reliable codes $P_e\to 0$, this finite bound first bounds the cut rate sums, and then passage to their achievable lower limits gives ${\sum }_{k\in A\cap \mathcal{S}}{R}_k\le \frac{1}{2\ln 2}\ln \det (I_{\mathrm{\mid }{A}^c\mathrm{\mid }}+\mathrm{\mid }A\mathrm{\mid }PG{G}^T)\mathrm{.}$ No a priori bounded-rate hypothesis is needed.

Put $a=\mathrm{\mid }A\mathrm{\mid }$ and $b=\mathrm{\mid }{A}^c\mathrm{\mid }$; both are positive. For any positive-semidefinite $H$ of size $m$, $I_m+aPH⪯2a(I_m+(P\mathrm{/}2)H)$, so determinant monotonicity gives $\frac{1}{2\ln 2}\ln \det (I_m+aPH)\le \frac{1}{2}{\log }_2\det (I_m+(P\mathrm{/}2)H)+\frac{m}{2}{\log }_2(2a)\mathrm{.}$ The block determinant of $\left(\begin{array}{cc}{I}_b& \sqrt{aP}G\\ -\sqrt{aP}{G}^T& I_a\end{array}\right)$, computed by taking either identity block as the Schur complement pivot, gives $\det (I_b+aPG{G}^T)=\det (I_a+aP{G}^{T}G)$. Apply the previous inequality once in dimension $b$ and once in dimension $a$, then use this identity to take the smaller additive term. The cut-set outer bound is at most $\frac{1}{2}{\log }_2\det \text{\hspace{0.17em}⁣}(I+\frac{P}{2}G{G}^T)+\frac{1}{2}\min (a,b){\log }_2(2a)\mathrm{.}$

Write $D(A)=\frac{1}{2}{\log }_2\det (I+(P\mathrm{/}2)G(A)G(A{)}^T)$, so the sufficient lower bound in [F1] is $L(A)=D(A)-a\mathrm{/}2$. Step 4.1 gives the cut-set outer value $O(A)\le D(A)+\frac{1}{2}\min (a,b){\log }_2(2a)=L(A)+\mathrm{\Delta }(A)$. If $L(A)>0$, the exact NNC expression satisfies $N_d(A)\ge L(A)$, so $O(A)-N_d(A)\le \mathrm{\Delta }(A)$. If $L(A)\le 0$, use the separately achievable zero-rate tuple from [F1]; then $O(A)\le L(A)+\mathrm{\Delta }(A)\le \mathrm{\Delta }(A)$, so the outer cut is within $\mathrm{\Delta }(A)$ of that achievable rate. This proves the cutwise comparison without treating the determinant expression as the exact NNC rate and covers negative sufficient-cut values.