Alphabeta Math
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Capacity of the binary symmetric channel

Example

The binary symmetric channel has input and output alphabets {0,1}; independently at each use, the output is the input bit flipped with probability p∈[0,1]. Write h(p):=−pln⁡p−(1−p)ln⁡(1−p),0ln⁡0:=0. The channel capacity is ln⁡2−h(p) nats per use, attained by a uniform input.

Facts & Assumptions

Given: The binary symmetric channel with crossover probability p∈[0,1].

[F1]

Finite discrete memoryless channels: the row W(⋅∣x) assigns probability 1−p to x and p to 1−x.

[F2]

Operational capacity formula for a finite DMC: capacity is the maximum of finite mutual information over input laws.

[F3]

Finite entropy and mutual-information chain rules: I(X;Y)=H(Y)−H(Y∣X).

[F4]

Finite conditional entropy and Finite distributions, entropy, divergence, and stochastic matrices: conditional entropy is the finite weighted average of conditional pmf entropies, with 0ln⁡0=0.

[F5]

Finite KL divergence is nonnegative with equality exactly at equal laws: for a binary law Q=Bern⁡(q), D(Q∥Bern⁡(1/2))=ln⁡2−h(q)≥0, with equality exactly when q=1/2.

Verification

Given: An input bit X with P(X=1)=r∈[0,1] and the specified channel.

1.1F1F3F4givenalgebra

The output law is Bernoulli with parameter q=p+r(1−2p). Each row has entropy h(p), so [F3] and [F4] give I(X;Y)=h(q)−h(p).

2.1F2F5step 1.1algebracases∎

By [F5], 0≤D(Bern⁡(q)∥Bern⁡(1/2))=ln⁡2−h(q), so I(X;Y)≤ln⁡2−h(p). Taking r=1/2 gives q=1/2 and equality. Therefore [F2] gives capacity ln⁡2−h(p). If p=0 or p=1, h(p)=0 and the channel is a deterministic bijection; if p=1/2, every row is the same and the capacity is zero.

Depends on

Used by

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Sources