 Flips versus erasures in a binary channel

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 $p=0.1$, 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 $H(X\mid Y)=p$ and $I(X;Y)=1-p=0.9$ for the binary erasure channel. This is the binary case of capacity of a finite-alphabet erasure channel.

For the binary symmetric channel at $p=0.1$, neither possible received symbol rules out either input, and $H(X\mid Y)=h_2(0.1)\approx 0.46899559$. Its channel capacity is only $1-h_2(0.1)\approx 0.53100441$. 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 $p=1$, 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.