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CSE5313 Coding and information theory for data science (Lecture 17)

Shannon's coding Theorem

Shannons coding theorem: For a discrete memoryless channel with capacity C, every rate R < C = \max_{x\in \mathcal{X}} I(X; Y) is achievable.

Computing Channel Capacity

X: channel input (per 1 channel use), Y: channel output (per 1 channel use).

Let the rate of the code be \frac{\log_F |C|}{n} (or \frac{k}{n} if it is linear).

The Binary Erasure Channel (BEC): analog of BSC, but the bits are lost (not corrupted).

Let \alpha be the fraction of erased bits.

Corollary: The capacity of the BEC is C = 1 - \alpha.

Proof

\begin{aligned}
C&=\max_{x\in \mathcal{X}} I(X;Y)\\
&=\max_{x\in \mathcal{X}} (H(Y)-H(Y|X))\\
&=H(Y)-H(\alpha)
\end{aligned}

Suppose we denote Pr(X=1)\coloneqq p.

Pr(Y=0)=Pr(X=0)Pr(no erasure)=(1-p)(1-\alpha)

Pr(Y=1)=Pr(X=1)Pr(no erasure)=p(1-\alpha)

Pr(Y=*)=\alpha

So,


\begin{aligned}
H(Y)&=H((1-p)(1-\alpha),p(1-\alpha),\alpha)\\
&=(1-p)(1-\alpha)\log_2 ((1-p)(1-\alpha))+p(1-\alpha)\log_2 (p(1-\alpha))+\alpha\log_2 (\alpha)\\
&=H(\alpha)+(1-\alpha)H(p)
\end{aligned}

So I(X;Y)=H(Y)-H(Y|X)=H(\alpha)+(1-\alpha)H(p)-H(\alpha)=(1-\alpha)H(p)

So C=\max_{x\in \mathcal{X}} I(X;Y)=\max_{p\in [0,1]} (1-\alpha)H(p)=(1-\alpha)

So the capacity of the BEC is C = 1 - \alpha.

General interpretation of capacity

Recall I(X;Y)=H(Y)-H(Y|X).

Edge case:

  • If H(X|Y)=0, then output Y reveals all information about input X.
    • rate of R=I(X;Y)=H(Y) is possible. (same as information compression)
  • If H(Y|X)=H(X), then Y reveals no information about X.
    • rate of R=I(X;Y)=0 no information is transferred.

Note

Compression is transmission without noise.

Side notes for Cryptography

Goal: Quantify the amount of information that is leaked to the eavesdropper.

  • Let:
    • M be the message distribution.
    • Let Z be the cyphertext distribution.
  • How much information is leaked about m to the eavesdropper (who sees operatorname{Enc}(m))?
  • Idea: One-time pad.

One-time pad