TheoremStatement: Literature-sourcedProof: AI-generatedPipeline-generated

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The error exponent is zero above finite-DMC capacity

Statement

For every finite unconstrained DMC and every $R > C (W)$ , $E_{a v}^{-} (R) = E_{a v}^{+} (R) = E_{m a x}^{-} (R) = E_{m a x}^{+} (R) = 0.$ These are error-probability exponents; no success-probability exponent is asserted.

Facts & Assumptions

Given: A finite DMC $W$ , $R > C (W)$ , and $M_{n} = ⌈ e^{n R} ⌉$ .

[F1]

Strong converse for finite-alphabet DMCs and Average and maximal error of a fixed-length code: for any code sequence with liminf rate above $C (W)$ , average success tends to zero, so both average and maximal errors tend to one; the two criteria are the uniform-message average and largest message error.

[F2]

Average and maximal error reliability exponents: $E_{k}^{\pm} (R)$ are the liminf and limsup of $- n^{- 1} \ln P_{k}^{*} (n, M_{n})$ , with $- \ln 0 = + \infty$ .

[F3]

Average and maximal error of a fixed-length code: maximal error is at least average error for every code.

Proof

1.1F1F2given

For each $n$ , choose a code attaining $P_{a v}^{*} (n, M_{n})$ ; the finite code space makes such a minimizer exist. Its rate is $n^{- 1} \ln M_{n} \to R > C (W)$ , so [F1] implies $P_{a v}^{*} (n, M_{n}) \to 1$ .

1.2F1F2F3givenalgebra

Choose likewise a code attaining $P_{m a x}^{*} (n, M_{n})$ . Its average error is no larger than its maximal error, and [F1] forces its average error to tend to one. Therefore its maximal error, bounded above by one and below by that average error, also tends to one; hence $P_{m a x}^{*} (n, M_{n}) \to 1$ .

2.1F2step 1.1step 1.2∎

If $p_{n} \to 1$ with $0 < p_{n} \leq 1$ eventually, then $- \ln p_{n} \to 0$ and $n^{- 1} (- \ln p_{n}) \to 0$ . Applying this to both optimal error sequences from steps 1.1 and 1.2 makes their liminf and limsup exponents all zero.

Depends on

Used by

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Dependency tree · two levels

14 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Yury Polyanskiy and Yihong Wu, Information Theory: From Coding to Learning, Chapter 22 §22.1 (standard reference, not scraped)
MIT 6.441 Information Theory, Lecture 11, Spring 2010 (standard reference, not scraped)