Conditional type packing bounds for source-indexed DM-MAC codebooks

For a fixed conditioning block, group candidate tuples by their joint empirical type with that block. If the wrong set is $S=\{1\}$, the varying letters are $(S_1,X_1)$ and the fixed block is $(Q,S_2,X_2,Y)$; for $S=\{2\}$ interchange indices. The number of joint types is at most $(n+1{)}^K$ by [F2]. For any one type, splitting its positions according to the fixed conditioning letters gives a product of conditional type classes, each of size at most the exponential of its empirical conditional entropy by [F2]. Thus the candidate count is at most $(n+1{)}^K\exp [n{H}_{\widehat{P}}(S_S,X_S\mid Q,S_{S^c},X_{S^c},Y)]$.

For each such candidate, its codeword probability is the product of the conditional kernels $P_{X_j\mathrm{\mid }{S}_{j}Q}$. On a support-restricted $\delta$-typical type, its negative log probability divided by $n$ differs from ${\sum }_{j\in S}H(X_j\mid S_j,Q)$ by at most ${\omega }_1(\delta )$: write the logarithm as the finite sum of empirical frequencies times $-\ln P_{X_j\mathrm{\mid }{S}_{j}Q}(x_j\mathrm{\mid }{s}_j,q)$; the summands are bounded on the finite positive support, and [F1] controls every frequency. Also $H_{\widehat{P}}(S_S,X_S\mid Q,S_{S^c},X_{S^c},Y)$ differs from the target conditional entropy by at most ${\omega }_2(\delta )$, because finite-alphabet entropy is continuous, including at zero via $-t\ln t\to 0$. Multiplying the count from step 1.1 by the largest candidate probability gives exponent at most $H_P(S_S,X_S\mid Q,S_{S^c},X_{S^c},Y)-{\sum }_{j\in S}H(X_j\mid S_j,Q)+\omega (\delta )\mathrm{.}$

For $S=\{1\}$, expand the conditional entropy in step 2.1 by [F4]. Since $X_2$ is conditionally independent of $S_1$ given $(S_2,Q)$, and $X_1$ is conditionally independent of $(S_2,X_2)$ given $(S_1,Q)$, the factorization in the Given law gives $H(S_1,X_1\mid Q,S_2,X_2,Y)-H(X_1\mid S_1,Q)=H(S_1\mid S_2)-I(S_1,X_1;Y\mid X_2,S_2,Q)\mathrm{.}$ The channel factorization [F6] implies $I(S_1;Y\mid X_1,X_2,S_2,Q)=0$, so the last mutual information is $I(X_1;Y\mid X_2,S_2,Q)$. Combining with step 2.1 proves the $N_1$ bound. Exchanging indices proves the $N_2$ bound.

For $S=\{1,2\}$, the same conditional entropy expansion gives $H(S_1,S_2,X_1,X_2\mid Q,Y)-H(X_1\mid S_1,Q)-H(X_2\mid S_2,Q)=H(S_1,S_2)-I(X_1,X_2;Y\mid Q),$ because $Q$ is independent of the source pair, the two codebook kernels factor given $(S_1,S_2,Q)$, and the channel depends on the source only through $(X_1,X_2)$. Substitution in step 2.1 proves the $N_{12}$ bound.