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Pass the output of a discrete memoryless channel of channel capacity $C$ through an independent finite-alphabet erasure channel with erasure probability $p$. Its capacity becomes $(1-p)C$.

To prove it, let $X$ be the original input, $Y$ the original output, $E$ the erasure indicator, and $Z$ the final output. Since the erasure marker is distinguishable, $E$ is determined by $Z$. It is independent of $X,Y$, so the chain rule for mutual information gives $I(X;Z)=I(X;E,Z)=I(X;Z\mid E)=(1-p)I(X;Y)$. Maximization over the same input distributions proves the formula. For $p<1$ the maximizing distributions are exactly the original channel's maximizers. For $p=1$, every input distribution attains the zero capacity. Input-dependent erasure need not satisfy this scaling identity.

For a finite-alphabet erasure channel with $k$ input symbols, $I(X;Y)=(1-f)H(X)$. Maximizing Shannon entropy at the uniform input gives channel capacity $C=(1-f){\log }_{2}k$ bits per use, by the Shannon second coding theorem.

Use the Shannon second coding theorem in its channel capacity form $C={\max }_{P_X}I(X;Y)$. The finite-alphabet erasure channel erases independently of the input. On an unerased output Bob knows $X$ exactly; on the erasure output his posterior distribution is still $P_X$. Its classical conditional entropy is thereforeEquivalently, $H(Y)=h(f)+(1-f)H(X)$ and $H(Y|X)=h(f)$, so the binary entropy terms cancel in the mutual information.

For an alphabet of size $k$, Shannon entropy is at most ${\log }_{2}k$, with equality for the uniform distribution. Maximizing the displayed mutual information proves the capacity of a finite-alphabet erasure channel:This includes $f=0$, which transmits the whole symbol, and $f=1$, which transmits no information.