I'm studying probability. There is a Boole-Bonferrroni inequality in the book. The book says that we an use the fundamental bridge which is: $$P(A) = E(I_A)$$
Where A is an event and $I_A$ is an indicator random variables of A distributed as bernouli (Bern(p)) and E is an expectation.
The book "Introduction to Probability" by Joseph K. Blitzstein and Jessica Hwang in chapter 4, example 4.4.3, page 152 says that we can derive the second inequality from the first one:
$$I(A_1 \cup...\cup A_n) \le I(A_1) +...+ I(A_n)$$
$$P(A_1 \cup...\cup A_n) \le P(A_1) +...+ P(A_n)$$
I want to derive the second inequality from the first one, for example lets: $$I(A_1 \cup A_2) \le I(A_1) + I(A_2) $$
then we take expectation of both side, it becomes:
$$E(I(A_1 \cup A_2)) \le E(I_{A1} + I_{A2}) $$ $$E(I_{A1} + I_{A2} - I_{A1} I_{A2}) \le E(I_{A1}) + E(I_{A2})$$
Now my question is you can see on the left hand side there is a problem. From the Linearity of Expectation, I can break $E(A+B)$ into $E(A) + E(B)$ and I can also break E(cA) into cE(A) where c is a constant but can I also break $E(I_{A1} + I_{A2} - I_{A1} I_{A2})$ into $E(I_{A1}) + E(I_{A2}) - E( I_{A1}) E (I_{A2})$ ? is this valid? breaking $E(I_{A1} I_{A2})$ into $E(I_{A1}) E( I_{A2})$ is ok? is $I_{A1}$ and $I_{A2}$ considered as a constant??? if it is then we will get:
$$E(I_{A1}) + E(I_{A2}) - E( I_{A1}) E (I_{A2}) \le E(I_{A1}) + E(I_{A2})$$
which is equivalent to:
$$P(I_{A1}) + P(I_{A2}) - P( I_{A1}) P (I_{A2}) \le P(I_{A1}) + P(I_{A2})$$
then from the union formula which state that: $$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$
My second question is, since $P( I_{A1})$ is independent from $P (I_{A2})$. so this inequality can be think as: is it correct? $$P(I_{A1}) + P(I_{A2}) - P( I_{A1}) P (I_{A2}) \le P(I_{A1}) + P(I_{A2})$$ $$P(I_{A1}) + P(I_{A2}) - P( I_{A1} \cap I_{A2}) \le P(I_{A1}) + P(I_{A2})$$ $$P(I_{A1} \cup I_{A2}) \le P(I_{A1}) + P(I_{A2})$$
in conclusion: $$I(A_1 \cup A_2) \le I(A_1) + I(A_2) $$ can be used to derive this inequality:
$$P(I_{A1} \cup I_{A2}) \le P(I_{A1}) + P(I_{A2})$$
Please answer my two questions above and if any of this proof is wrong please tell me.
please explain to me in very basic notation and simple language, step by step. I'm newbie in this subject.
To get from $$I(A_1 \cup A_2) \leq I(A_1) + I(A_2)$$ to $$P(A_1 \cup A_2) \leq P(A_1) + P(A_2),$$ simply take the expectation of both sides and use linearity and the fundamental bridge (the expected value of the indicator of an event is the probability of the event): $$E(I(A_1 \cup A_2)) \leq E(I(A_1)) + E(I(A_2)),$$ which (by the fundamental bridge) is the desired inequality. The same idea works for general $n$.
The identity $I(A_1 \cup A_2) = I(A_1) + I(A_2) - I(A_1)I(A_2)$ that you mentioned is overkill for proving Boole's inequality. But it's very useful for proving inclusion-exclusion. Most books I've seen skip the proof of inclusion-exclusion for probability, or give tedious proofs by induction. I find it much more elegant to use indicator random variables, linearity, and the fundamental bridge.