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Shannon capacity $C$ is the upper limit on a rate $R$ defined as the number of information symbols $k$ divided by the number of transmitted symbols $n$, that can be transmitted over a channel such that as $n \rightarrow \infty$, the probability of error goes to zero. If a rate $R > C $ is used, the probability of error is bounded away from zero.

Since $n$ is finite in practical applications, there is some nonzero probability of error.Thus, why would it matter if we transmit at a rate higher than $C$, Shannon's theorem predicts that such a rate will have a nonzero probability of error, but we already have a nonzero probability of error due to the fact that $n$ is finite.In other words, why don't we transmit at rates higher than $C$ if we are going to get a nonzero probability of error anyways ?

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    $\begingroup$ Not all nonzero numbers are equal. $\endgroup$
    – Boris Bukh
    Commented Aug 23, 2016 at 13:53
  • $\begingroup$ But how do we know that the probability of error caused by transmitting at rates above C will be intolerable ? $\endgroup$ Commented Aug 23, 2016 at 13:55
  • $\begingroup$ This sounds like an engineering question, not a theoretical CS one. $\endgroup$ Commented Aug 23, 2016 at 14:36
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    $\begingroup$ @PeterTaylor it can be tcs $\endgroup$
    – Turbo
    Commented Aug 25, 2016 at 6:47
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    $\begingroup$ This paper talks about the achievable rates for any given error probability at finite blocklenghts. people.lids.mit.edu/yp/homepage/data/finite_block.pdf $\endgroup$
    – Devil
    Commented Aug 25, 2016 at 23:31

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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 some constant $c$ depending on the channel.

Also, look at rate distortion theory.

This gives a formula for the highest rate at which you can transmit if you want to make sure that at most $\epsilon$ fraction of the transmitted bits are received erroneously.

Both of these quantify the cost of errors incurred when you transmit at $R>C$, and in most cases it's not worth it.

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