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October 5th, 2014, 03:49 AM   #1
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entropy of combined events

Hi there, suppose I toss a fair coin twice. Suppose that I can perfectly observe every outcome, so in the end I observe each of the 4 events {HH},{TH},{HT},{TT}. The informational content measured by entropy would be -1/2log(1/2)*4=2.

But now suppose that I know that I would be able to observe only 2 events:
E1:={H,H}
E2:={TH,HT,TT}

How do I redefine the measure of entropy to account for such cases?

Many thanks,
D
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October 5th, 2014, 04:27 AM   #2
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$\mathbb{P}\left(\{H,H\}\right) = \frac{1}{4}$ and $\mathbb{P}\left(\{TH,HT,TT\}\right) = \frac{1}{2^6}$,

$H = -\frac{1}{4}\log_{2}\left(\frac{1}{4}\right) -\frac{1}{2^6}\log_{2}\left(\frac{1}{2^6}\right) = \frac{1}{2}+\frac{6}{64}=\frac{38}{64}=\frac{19}{3 2}$.

For the first one,
all the events observed can occur with the same probability $\frac{1}{4}$, so

$H =-4\frac{1}{4}\log_{2}\left(\frac{1}{4}\right)=2$, which means that we get full information 2bits (two sides of the coin).

Last edited by ZardoZ; October 5th, 2014 at 04:48 AM.
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