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Suppose we have binary channel from which we are able to receive zeroes and ones. We also know apriory probability $p$ of receiving "1". Then we can calculate information amount of each digit $q$ we receive:

q=0: $I=-\log_2(1-p)$ bits;

q=1: $I=-\log_2(p)$ bits.

Now imagine that the channel is "fussy": instead of receiving exact digits we receiving probability $q$ that transferred digit is "1". Previous example of "unfussy" channel is when $q$ can take only two values: $q\in\{0,1\}$.

What will be the amount of information of receiving probability $q$ given probability $p$?

Thanks!

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closed as off-topic by Stefan Kohl, Venkataramana, Andrey Rekalo, Ricardo Andrade, Suvrit Jan 18 at 5:08

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Not a research Q: read the FAQ –  Anthony Quas Jun 22 '12 at 20:28
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I have not found this question in the FAQ –  Anton Sukhinov Jun 23 '12 at 18:41