= 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 $H(X\mid Y)=h_2(0.1)\approx0.46899559$. Its <channel capacity> is only $1-h_2(0.1)\approx0.53100441$. A receiver benefits from knowing \i[where] information was lost, even when the fraction of affected transmissions is the same.
\Image[/binary-channel-errors-versus-erasures.png]
{title=Announced erasures preserve more information than hidden bit flips at the same low failure probability}
{description=Solid arrows carry <probability> $0.9$ and dashed arrows carry <probability> $0.1$ in the two illustrated laws of <transition probabilities>. The erasure flag is distinct from both <bits>. The lower plot compares <channel capacity> for $0\leq p\leq1/2$, using $1-p$ for the <binary erasure channel> and $1-h_2(p)$ for the <binary symmetric channel>.}
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.
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