<p>Look at the <a href="https://en.wikipedia.org/wiki/Noisy-channel_coding_theorem#Strong_converse_for_discrete_memoryless_channels">strong converse</a> to Shannon's theorem:</p>

<blockquote>
  <p>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.  </p>
</blockquote>

<p>Also, look at <a href="https://en.wikipedia.org/wiki/Rate%E2%80%93distortion_theory">rate distortion theory.</a></p>

<p>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. </p>

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

CC BY-SA 3.0
2016-08-28T21:36:40.467Z
2016-08-29T10:35:14.507Z
2016-08-29T10:35:14.507Z
Peter Shor 
<p>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 &gt; C $ is used, the probability of error is bounded away from zero. </p>

<p>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 ? </p>

CC BY-SA 3.0
2016-08-23T13:53:32.430Z
Not all nonzero numbers are equal.
Boris Bukh
CC BY-SA 3.0
2016-08-23T13:55:39.620Z
But how do we know that the probability of error caused by transmitting at rates above C will be intolerable ?
Dina Abdelhadi
CC BY-SA 3.0
2016-08-23T14:36:32.933Z
This sounds like an engineering question, not a theoretical CS one.
Peter Taylor
CC BY-SA 3.0
2016-08-25T06:47:25.343Z
@PeterTaylor it can be tcs
Turbo
CC BY-SA 3.0
2016-08-25T06:49:19.347Z
@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
Turbo
CC BY-SA 3.0
2016-08-25T23:31:15.607Z
This paper talks about the achievable rates for any given error probability at finite blocklenghts. http://people.lids.mit.edu/yp/homepage/data/finite_block.pdf
Devil
<it.information-theory><shannon>
Why don't we transmit at rates higher than the Shannon capacity if we are going to get a nonzero probability of error anyways ?
Dina Abdelhadi
# Why don't we transmit at rates higher than the Shannon capacity if we are going to get a nonzero probability of error anyways ?

(Asked by: username_0 on 2016-08-23)

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 ?

**Tags:** it.information-theory, shannon

> **Comments:**
> **username_1**: Not all nonzero numbers are equal.  
> **username_0**: But how do we know that the probability of error caused by transmitting at rates above C will be intolerable ?  
> **username_2**: This sounds like an engineering question, not a theoretical CS one.  
> **username_3**: @PeterTaylor it can be tcs  
> **username_3**: @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  
> **username_4**: This paper talks about the achievable rates for any given error probability at finite blocklenghts. http://people.lids.mit.edu/yp/homepage/data/finite_block.pdf  

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## 1 answers

### Answer by username_5 on 2016-08-28. Score: 7 | Accepted answer

Look at the [strong converse](https://en.wikipedia.org/wiki/Noisy-channel_coding_theorem#Strong_converse_for_discrete_memoryless_channels) 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.](https://en.wikipedia.org/wiki/Rate%E2%80%93distortion_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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