Flips versus erasures in a binary channel 2026-10-14
A binary erasure channel marks its failed transmissions with a special output symbol, whereas a binary symmetric channel changes a bit without announcing whether it changed. At the same nominal failure probability , the two communication channels therefore leave different uncertainty. With a fair input, an unerased output determines the input exactly, while the erasure flag leaves one bit of conditional entropy. Thus and for the binary erasure channel. This is the binary case of capacity of a finite-alphabet erasure channel.
For the binary symmetric channel at , neither possible received symbol rules out either input, and . Its channel capacity is only . A receiver benefits from knowing where information was lost, even when the fraction of affected transmissions is the same.
This comparison concerns the specified transition probabilities, rather than a general ordering of every error mechanism. In particular, at , an erasure flag conveys no input information, while deterministic inversion still conveys a full bit. Reporting only an error percentage loses the distinction between hidden corruption, announced loss, and a predictable reversible transformation.
Past exam of the mathematics course of the University of Cambridge 2015 iii Paper 66 4 iii Solution Created 2026-10-03 Updated 2026-10-06
Use the Shannon second coding theorem in its channel capacity form . The finite-alphabet erasure channel erases independently of the input. On an unerased output Bob knows exactly; on the erasure output his posterior distribution is still . Its classical conditional entropy is thereforeEquivalently, and , so the binary entropy terms cancel in the mutual information.
For an alphabet of size , Shannon entropy is at most , with equality for the uniform distribution. Maximizing the displayed mutual information proves the capacity of a finite-alphabet erasure channel:This includes , which transmits the whole symbol, and , which transmits no information.
