# Shannon capacity with distribution different from Gaussian

If I understand correctly, in AWGN channel, for a given SNR, there is a code for $M$ input codeword $\mathbf{X}_{i, 1 \leq i \leq M}$ that $$\lim_{n \to \infty} R=\lim_{n \to \infty}\frac{\log(M)}{n} =\log(1+SNR)$$

(where $n$ is the length of $\mathbf{X}_i$) if $\mathbf{X}_i$ is drawn from multivariate Normal distribution.

However, I don't know if $\mathbf{X}_i$ is not Normal, does the Shannon channel coding theorem hold ?

• AFAIK, the derivation of the formula $C=\log_2(1+\mathrm{SNR})$ uses the fact that the input is normally distributed. I don't know if there is a more generic proof, though. Is that what you are asking for? Commented Jan 23, 2018 at 14:16
• You could think about it this way: every channel+encoder+decoder system has a capacity. When the noise is Gaussian and the input distribution is also Gaussian, then the capacity is the formula you provided. When these things change, the formula for the capacity will also change.
– MBaz
Commented Jan 23, 2018 at 18:58
• @Tendero yes the formula is given by taking Normal inputs. I want to ask the inverse that if the input is not Normal, can we prove that the rate is always lower the $C$ ? Because $C$ is taken by maximizing over all input distribution. Commented Jan 27, 2018 at 11:49
• @MBaz thanks, I wanted to talk about the case that the rate is maximized over all possible input distributions and all possible encoder-decoder pairs. Commented Jan 27, 2018 at 11:50
• @SabrinaCantu If the distribution is not normal, then the maximum rate is less than $C$.
– MBaz
Commented Jan 27, 2018 at 17:13

The more general definition of the capacity $C$ is $$C = \sup_{p_X(x)} \mathbf{I}(\mathbf{X};\mathbf{Y})$$ where $\mathbf{I}$ denotes the Mutual Information, $\mathbf{X}$ is the input to the channel and $\mathbf{Y}$ is the output of the channel, and the supremum is taken over all possible input distributions $p_X(x)$.