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 $$I(A_i) = \log \frac{4}{\Pr[A]} = \log \frac{1}{\Pr[A]} + \log 4 = I(A) + \log 4.$$