# What is the relation between the quantity of information of $A_i$ and $A$, where $A=\bigcup_iA_i$?

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 $$I(A_i) = I(A) + \log (4)?$$

• What does $I(A)$ stand for? – Yuval Filmus Nov 13 '18 at 21:23
• I(A) = -P(A) x log(P(A)) – Sydney.Ka Nov 13 '18 at 21:36
• Your equation doesn't seem to hold for your formula. – Yuval Filmus Nov 13 '18 at 21:44

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

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

• 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) ? – Sydney.Ka Nov 13 '18 at 21:54
• Not that you've seen what calculation you need to do, I encourage you to try again on your own. – Yuval Filmus Nov 13 '18 at 22:02
• Also, you can try to use the chain rule in some way. – Yuval Filmus Nov 13 '18 at 22:03
• 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 – Sydney.Ka Nov 13 '18 at 22:26
• Well then, perhaps a different relation holds. – Yuval Filmus Nov 13 '18 at 23:23