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I proved that

I(X,Y) <= 1 - H(p)

to the following way: enter image description here

How can I prove if I start in that way I(X,Y) = H(X) - H(X|Y), I know we can deduct I(X,Y) = H(X) - H(X|Y), H(X,Y) = H(X,Y) - H(Y) and H(X,Y) = H(Y|X) + H(X) => I(X,Y) = H(X) - H(Y|X) - H(X) + H(Y) but this is the first equality.

If it is impossible why, can i prove it a logic way?

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You would require the Bayes rule to compute the probabilities. Since this problem is small, it can be done. But in general, the identity $I(X;Y) = H(X)-H(X|Y) = H(Y)-H(Y|X)=I(Y;X)$ is very useful in one direction and painful in the other. Try other channels like the multiplicative channel and noisy typewriter. Refer "Elements of Information Theory" by Cover and Thomas for more details. –  Gautam Shenoy Nov 13 '12 at 12:32
    
I already download the book, but only I find the first in the book. Thanks for your answer. –  Tatar Elemér Nov 13 '12 at 12:50
    
Look at the later chapters. Not just the first two. Look at the index as well. It may be in the problem section. –  Gautam Shenoy Nov 14 '12 at 5:10

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