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For an integer $I$, the input-output relationship of a discrete memoryless channel is given by:

$Y = X + Z$ (mod $I$, i.e. sum indicates a modular addition)

where $I ≥ 2$, and

$X$ is an integer chosen from the alphabet $A_x = \{1,\dots,2I\}$,

$Z$ is noise which is a uniform Bernoulli random variable. This means that $A_z = \{0,1\}$, and

$$\Pr\{Z = 0\} = \Pr\{Z = 1\} = 0.5.$$

How can we calculate the capacity of this channel?

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This answer assumes that $Y = X + Z \bmod 2I$, which seems more reasonable than the current $Y = X + Z \bmod I$, which would mean that $X$ is effectively chosen from an alphabet of size $I$ rather than $2I$.

The capacity is $\log I$.

For the upper bound, $$ I(X;Y) = H(Y) - H(Y|X) = H(Y) - H(Z) = H(Y) - 1 \leq \log(2I) - 1 = \log I. $$

For the lower bound, if $X$ is chosen uniformly among the $I$ values $1,3,\ldots,2I-1$ then you can recover $X$ from $Y$ (without any error!), and so $I(X;Y) = H(X) = \log I$.

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  • $\begingroup$ One thing though, for mod $ I$ we will have $I$ possible "different " values for $Y$. Shouldn't we take this into consideration? $\endgroup$ – Niousha Feb 8 at 22:31
  • $\begingroup$ My answer actually assumes that the calculation is modulo $2I$. Otherwise the input alphabet has effective size $I$ rather than $2I$. $\endgroup$ – Yuval Filmus Feb 8 at 22:42

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