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Let $A$ be an event divided into 4 events $A_i$ with the same probability. Why does the quantity of information of $A_i$ satisfy $$ I(A_i) = I(A) + \log (4)?$$

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  • $\begingroup$ What does $I(A)$ stand for? $\endgroup$ Nov 13, 2018 at 21:23
  • $\begingroup$ I(A) = -P(A) x log(P(A)) $\endgroup$
    – Sydney.Ka
    Nov 13, 2018 at 21:36
  • $\begingroup$ Your equation doesn't seem to hold for your formula. $\endgroup$ Nov 13, 2018 at 21:44

1 Answer 1

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I believe that the correct formula is $$ I(A) = \log \frac{1}{\Pr[A]}. $$

In this case, we have $$ I(A_i) = \log \frac{4}{\Pr[A]} = \log \frac{1}{\Pr[A]} + \log 4 = I(A) + \log 4. $$

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  • $\begingroup$ Yes, I'm sorry. But why for X a random variable with the probability distribution p1, ...pn, and Y with the probability distribution (p1)/4,....(pn)/4, we have the same relation H(X)=H(Y)+log(4) ? $\endgroup$
    – Sydney.Ka
    Nov 13, 2018 at 21:54
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    $\begingroup$ Not that you've seen what calculation you need to do, I encourage you to try again on your own. $\endgroup$ Nov 13, 2018 at 22:02
  • $\begingroup$ Also, you can try to use the chain rule in some way. $\endgroup$ Nov 13, 2018 at 22:03
  • $\begingroup$ I tried but I got a different result. H(Y) = -ΣP(Y=k) x log(P(Y=K)) = -Σpk/4x log(pk/4) = H(X)/4 + Σpk/4 x log(4)= H(X) /4+ log(4)/4 x Σpk = H(X) /4+ log(4)/4 $\endgroup$
    – Sydney.Ka
    Nov 13, 2018 at 22:26
  • $\begingroup$ Well then, perhaps a different relation holds. $\endgroup$ Nov 13, 2018 at 23:23

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