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A finite-alphabet erasure channel transmits the input symbol unchanged with probability $1-f$ and replaces it by a distinct erasure flag with probability $f$, independently of the symbol. Given the flag, the posterior remains the input distribution; otherwise the input is known. Thus its mutual information is $(1-f)H(X)$.

If a uniform index $J$ has $N$ possibilities and every proposed success neighbourhood contains at most $B<N$ possibilities, then any estimator has failure probability at least the displayed expression. Condition on its success indicator: conditional entropy is at most $\log 2+(1-p_e)\log B+p_e\log N$. Subtract this from $\log N$ to bound the mutual information and rearrange. A metric Hamming ball gives approximate recovery rather than exact index decoding.

Because $Z$ is determined by $X,Y$, its conditional entropy satisfies $H(Z\mid X,Y)=0$. Equate the two forms of the chain rule for conditional entropy to obtainConditioning cannot increase classical Shannon entropy: $H(Z)-H(Z\mid Y)=I(Z:Y)\ge 0$ by nonnegativity of mutual information. Thus $H(Z\mid Y)\le H(Z)=h(p_e)$. When $Z=0$, the value of $X$ is exactly $f(Y)$ and $H(X\mid Y,Z=0)=0$. When $Z=1$ and $Y=y$, the value $f(y)$ is excluded, leaving at most $|J_X|-1$ possibilities. By maximum entropy on a finite alphabet,Averaging the two conditional cases now yieldsand consequentlyThis is Fano's inequality via an error indicator. Events of zero probability contribute zero to the average and need no conditional distribution. For $|J_X|=1$, the inference is automatically correct and the entropy is zero; the displayed logarithmic form is intended for $|J_X|\ge 2$. Optimality of the guess is not needed for the inequality: it holds for every deterministic $f$.

Use the Shannon second coding theorem in its channel capacity form $C={\max }_{P_X}I(X;Y)$. The finite-alphabet erasure channel erases independently of the input. On an unerased output Bob knows $X$ exactly; on the erasure output his posterior distribution is still $P_X$. Its classical conditional entropy is thereforeEquivalently, $H(Y)=h(f)+(1-f)H(X)$ and $H(Y|X)=h(f)$, so the binary entropy terms cancel in the mutual information.

For an alphabet of size $k$, Shannon entropy is at most ${\log }_{2}k$, with equality for the uniform distribution. Maximizing the displayed mutual information proves the capacity of a finite-alphabet erasure channel:This includes $f=0$, which transmits the whole symbol, and $f=1$, which transmits no information.

In natural logarithms, Fano's inequality for a uniform index $J$ among $N$ alternatives states $\mathbb{P}(\hat{J}\ne J)\ge 1-(I(J;G)+\log 2)/\log N$. The mutual information is at most the average Kullback-Leibler divergence to any fixed reference law. With the supplied packing, nearest-neighbour decoding turns partition error below half the separation into correct decoding. Thus the packing alone proves a bound at radius $n(1/4-\epsilon /2)$, with the unavoidable $\log 2$ term. It does not automatically prove the much larger printed radius.

A list-decoding Fano inequality supplies the intended near-half error threshold. Assume $0<\epsilon <1/4$, since otherwise that threshold is nonpositive and its event is automatic. Take a uniform prior on all $N=\left(\genfrac{}{}{0ex}{}{n}{n/2}\right)/2$ unlabelled balanced partitions, and let $r=n(1/2-2\epsilon )$. A Hamming ball of radius less than $r$ around any estimated partition contains at most $B={\sum }_{j<r}\left(\genfrac{}{}{0ex}{}{n}{j}\right)$ such partitions. This follows by selecting the representative closer to the estimated group; each unlabelled partition contributes at most one such representative because $r<n/2$. The entropy bound for a Hamming ball gives $\log B\le nh(r/n)$, where $h$ is the binary entropy function measured in natural logarithms. Its curvature gives $\log 2-h(1/2-2\epsilon )\ge 8{\epsilon }^2$. Also $\left(\genfrac{}{}{0ex}{}{n}{n/2}\right)\ge 2^n/(n+1)$, soUse the all-$1/2$ independent-edge law $P_{\ast }$ as a reference. Only the $n^2/4$ crossing pairs differ; the preceding Bernoulli distribution divergence estimate gives $D_{\mathrm{KL}}(P_S\Vert P_{\ast })\le n{t}^2$. Hence $I(J;G)\le n{t}^2$. Whenever the denominator is positive, the list-decoding Fano inequality proves the finite-sample answerFor $4{\epsilon }^{2}n\ge \log (2(n+1))$, this is at least $1-t^2/(4{\epsilon }^2)-\log 2/(4{\epsilon }^{2}n)$. If in addition $n{t}^2\ge \log 2$, it implies the intended form with $c^{\prime }=1/2$. This proves the claimed information-theoretic scale with its needed finite-sample qualifications.

As a uniform finite-sample assertion for every positive $t$, the printed form is false. At $t=0$ every partition has the same data law, and a uniformly random balanced guess has a positive chance of being correct, giving error strictly below one. Such randomness can also be extracted from the finite graph sample: assign distinct graph outcomes to the finitely many balanced partitions; under the common law every selected outcome has positive probability. The worst error remains strictly below one for sufficiently small positive $t$, by continuity. The printed lower bound tends to one as $t\downarrow 0$ for any fixed universal $c^{\prime }$, which contradicts this. The entropy correction cannot be dropped without an asymptotic or signal-size condition.

For the distribution and POVM from part 2, apply the data-processing inequality for quantum relative entropy to the measurement channel. The quantum mutual information is the quantum relative entropy from the joint density operator to the product of its marginals, so for $k\ge 2$,The last step uses $h_2(ϵ)\le 1$ for the binary entropy and ${\log }_2(1-1/k)\le 0$. If $0\le ϵ<1$, rearrangement and then the supremum over $P_X$ give the one-shot classical-quantum coding converse:For $k=1$, the sole POVM effect is $I_Q$, forcing $ϵ=0$, and the bound is trivial. For $ϵ=1$, division by $1-ϵ$ is undefined: the undivided inequality remains valid and gives no size restriction. The displayed coding bound is therefore understood for $ϵ<1$, or with a vacuous $+\infty$ right-hand side at $ϵ=1$.

An alternative uses the original decoder directly. Let $M$ be the uniform message and $\hat{M}$ its measured estimate. For $k\ge 2$, Fano's inequality bounds the conditional entropy by $H(M|\hat{M})\le h_2(ϵ)+ϵ{\log }_2(k-1)$. The Holevo bound givesThus the same coding bound follows by bounding the classical mutual information obtainable from the POVM.

For a discrete random variable with probabilities $p_i$, its Shannon entropy in bits is $H(X)=-{\sum }_i{p}_i{\log }_2{p}_i$, with $0{\log }_{2}0=0$. Gibbs inequality states thatwith equality exactly when $p=q$. If $q_i=0<p_i$, the left side is infinite. Otherwise apply $-\ln u\ge 1-u$ to $u=q_i/p_i$ for $p_i>0$:Equality requires $q_i=p_i$ on the support and no remaining mass outside it. In particular, comparison with the uniform probability distribution on two outcomes gives binary Shannon entropy at most $1$.

Write $q=1-p_1$. For $q>0$, grouping the last two outcomes givesEquality holds if $p_2=p_3$; if $q=0$ it also holds, and then both are zero. Thus equality holds exactly when $p_2=p_3$.

For the discrete memoryless channel, assume $\alpha ,\beta \ge 0$ and $\alpha +\beta \le 1$, and set $w=1-\alpha -\beta$. Both conditional output probability distributions have Shannon entropy $H(w,\alpha ,\beta )$. The third output has probability $\beta$ regardless of the input; the first two have total probability $1-\beta$. Their Shannon entropy is maximized by making them equal, which a uniform input achieves. Hence the channel capacity isand thereforeThe usual zero-times-log convention includes $\beta =1$, when $C=0$. If $w=\alpha$, the rows coincide and every input achieves zero mutual information; otherwise the uniform input is the unique maximizing input.

At $p=1/2$, the output of the binary symmetric channel is independent of its input. It conveys no mutual information, and therefore

For a discrete memoryless channel with input $X$ and output $Y$, its information capacity iswhere $I(X;Y)$ is the mutual information and the maximum is over all input distributions. Capacity is measured in bits when logarithms have base two.