# Why don't we transmit at rates higher than the Shannon capacity if we are going to get a nonzero probability of error anyways ?

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 ?

• Not all nonzero numbers are equal. – Boris Bukh Aug 23 '16 at 13:53
• But how do we know that the probability of error caused by transmitting at rates above C will be intolerable ? – Dina Aug 23 '16 at 13:55
• This sounds like an engineering question, not a theoretical CS one. – Peter Taylor Aug 23 '16 at 14:36
• @PeterTaylor it can be tcs – T.... Aug 25 '16 at 6:47
• @user2692162 look at gallager's cut off rate work or shannon's proof itself I think you will get explicit exponents of failure probability with length of code – T.... Aug 25 '16 at 6:49

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.
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.