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CSE5313 Coding and information theory for data science (Lecture 17)
Shannon's coding Theorem
Shannon’s 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 outputYreveals all information about inputX.- rate of
R=I(X;Y)=H(Y)is possible. (same as information compression)
- rate of
- If
H(Y|X)=H(X), thenYreveals no information aboutX.- rate of
R=I(X;Y)=0no information is transferred.
- rate of
Note
Compression is transmission without noise.
Side notes for Cryptography
Goal: Quantify the amount of information that is leaked to the eavesdropper.
- Let:
Mbe the message distribution.- Let
Zbe the cyphertext distribution.
- How much information is leaked about
mto the eavesdropper (who seesoperatorname{Enc}(m))? - Idea: One-time pad.