From indicator random variable into probability using fundamental bridge?

639 Views Asked by At

I'm a newbie studying probability. Let I be an indicator random variable. A and B be an event. Then:

$$I_{A \cup B} = I_A + I_B - I_A I_B$$

I want to go from this equation into this equation:

$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$

From the fundamental bridge which says $P(A) = E(I_A)$ we can use Expectation and Linearlity property to derive the second equation from the first equation. Then:

$$I_{A \cup B} = I_A + I_B - I_A I_B$$ Becomes: $$E(I_{A \cup B}) = E(I_A + I_B - I_A I_B) $$ $$P(A \cup B) = E(I_A + I_B - I_A I_B)$$ how can I solve the right hand side??? Now my question is you can see on the right 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, why? can someone please explain to me in very detail?

if it is valid then we will get: $$P(A \cup B) = E(I_A) + E(I_B) - E(I_A) E (I_B)$$ Then: $$P(A \cup B) = P(A) + P(B) - P(A)P(B)$$

My second question is, since $P(A)$ is independent from $P (B)$ (because indicator random variable is distributed as Bernoulli distribution), so the above equation can be thought of as:

$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$

which coherence with the union of event theory.

Am I understand correctly? or I have make some mistake here? Something I miss understand? please give me step by step consistent and precise explanation.