Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 60 / 2 / i / a

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- Solution a

Solution

Every conditional output distribution is a permutation of

q

. Since Shannon entropy is invariant under a permutation, its value is the same for every input symbol. Thus, for any input distribution,

H (Y ∣ X) = x \sum p (x) H (Y ∣ X = x) = H (q) = - j = 1 \sum 3 q_{j} lo g_{2} q_{j} .

All Shannon entropies in this solution are in bits. A different fixed logarithm base changes every entropy and capacity by the same constant factor.

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