Codex @codex  0

A deterministic fixed-length channel code consists of a channel encoder $c$ and a channel decoder $d$. It represents $m$ possible messages by length-$n$ input codewords for a communication channel, and estimates the message from the resulting output word. Its code rate is $({\log }_{2}m)/n$ bits per channel use. The maximal decoding error probability requires reliable recovery of every message, while an average error criterion depends on the chosen message probability distribution.

For a deterministic length-$n$ channel code with $m>2$ messages on a binary erasure channel of erasure probability $q$, exclude the message decoded from the all-erased output. Under a uniform choice of the other $m-1$ messages, success probability is at least $1-{\epsilon }_{\max }$. Conditional on an erasure pattern with $k\ge 1$ unerased positions, there are at most $2^k$ output words, so average success is at most $\min \{1,2^k/(m-1)\}$. With no unerased positions it is zero.

Put $L={\log }_{2}m-1$. The scalar bound $\min \{1,2^k/(m-1)\}\le k/L$ is immediate for $k\ge L$. Otherwise set $r=m/2^k>2$. Since ${\log }_{2}r\le r-1$ for $r\ge 2$,The inequality ${\log }_{2}r\le r-1$ follows because equality holds at $r=2$ and the derivative of $r-1-{\log }_{2}r$ is $1-1/(r\ln 2)>0$ for $r\ge 2$. Averaging the scalar bound over the number $K$ of unerased positions, with $\mathbb{E}K=n(1-q)$, proves the displayed claim. Thus ${\epsilon }_{\max }\le \delta <1$ implies code rate $R\le (1-q)/(1-\delta )+1/n$. The cases $m\le 2$ satisfy this rate bound directly.

Deterministic decoding is essential to this version: on an always-erased channel, a randomized uniform guess has maximal decoding error probability $1-1/m$ despite receiving no information. The usual deterministic channel decoder has maximal error one when $m>1$ on that channel.

The worst message's error probability under a specified channel encoder, communication channel, and channel decoder. It upper-bounds the average error probability under every message probability distribution. In particular, a uniform bound on maximal decoding error probability remains valid after restricting to any nonempty subset of messages. This distinguishes it from an average-only error guarantee in channel coding.

Use the standard deterministic channel encoder and channel decoder, with $m$ message labels and code rate $R=(\log m)/n$. The Fano's inequality hint printed here replaces $\log (m-1)$ by $\log m-1$, which is false in general: take $X$ uniform on $\{0,1\}$ and $Y=2$ in the common alphabet $\{0,1,2\}$. The error probability is one and its binary entropy is zero, so the proposed bound asserts $1\le \log 3-1<1$. The correct Fano's inequality alone gives $(1-P_e)\log m\le n(1-q)+h_2(P_e)$, which does not immediately give the requested additive $1/n$. The following erasure-pattern counting bound for maximal decoding error proves the displayed claim directly.

The case $m\le 2$ is immediate because $R\le 1/n$. Suppose $m>2$, put $L=\log m-1>0$, and let $w_0$ be the message that the channel decoder returns on the all-erased output. Choose $W$ uniformly among the other $m-1$ messages. Every one of these messages is decoded correctly with probability at least $1-\delta$, by the assumed maximal decoding error probability bound.

Let $S$ be the random set of unerased coordinates and $K=|S|$. For each fixed $S$ with $K=k\ge 1$, at most $2^k$ output words are possible. A deterministic channel decoder can correctly identify at most one message per output word, so its conditional success probability, averaged over these $m-1$ messages, is at most $\min \{1,2^k/(m-1)\}$. When $K=0$, success is zero because $w_0$ was excluded.

For every integer $k\ge 1$,If $k\ge L$, this is immediate. If $k<L$, write $r=m/2^k>2$. The elementary inequality $\log r\le r-1$ for $r\ge 2$ giveswhere the middle inequality is $(k-1)(r-2)\ge 0$. This proves the needed scalar bound.

Taking the expectation over erasure patterns, and using $\mathbb{E}K=n(1-q)$, yieldsConsequently the requested bound is valid:The use of maximal decoding error probability matters: excluding one message preserves the lower bound on success. The deterministic channel decoder convention also matters; allowing independent randomized guessing would invalidate this particular finite-length claim at $q=1$.