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Rate in terms of noise entropy [Under Information theory > Shannon's paper]

Let us put down our findings from article on Information rate with noise and try to combine them.

1) R = H(x)Hy(x)

2) H(x,y)= H(x) + Hy(x)

3) H(x,n) = H(x) + H(n)

Here destination (y) symbols are produced by impact of noise (n) symbols on source (x) symbols. This mean each y symbol correspond to combination of certain x symbol and n symbol. So number of combinations of x and y are same as number of combinations of x and n. In other words, average number of bits per symbol required to represent combined x and n signals is same as average number of bits per symbols required to represent combined x and y signals.

That is: 4) H(x,y)= H(x,n)

Combining above equations, we get R = H(y)H(n). This theorem 16 (page 43) from Shannon's paper.

Can we directly try to come out with above equation ? Consider following diagram: Each Sx is combined with each Sn (based on probabilities) to give us Sy. As x and n are independent, whatever are the individual probabilities (uncertainities or entropies) of x and n, the diagram hold true. If n is not present or it has zero entropy (uncertainty), each y symbol will map to each x symbol. Due to presence of n's symbols (non-zero entropy), more number of bits are needed for y as shown in above diagram. If you examine the diagram and try b = 1,2,4,...., it will be evident that to get actual bits per symbol, we must deduct noise introduced n's bits per symbol from y's observed bits per symbol.

That is: R = H(y)H(n)