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Why don't we transmit at rates higher than the Shannon capacity if we are going to get a nonzero probability of error anyways ?
Look at the strong converse to Shannon's theorem:
for rates above the channel capacity, if $n$ bits are to be transmitted, the probability of error is exponentially close to 1, so $1-e^{c n}$ for ...
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7
votes
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Does Huffman coding always produce shorter codes than the Shannon code?
This is actually problem 5.12 in Cover and Thomas's information theory textbook; show that the probability distribution ${1/12,1/4,1/3,1/3}$ gives a counterexample.
And if you want a really nice ...
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