I came across this problem in the book "Introduction to Probability" by Dr. Joseph K. Blitzstein and Dr. Jessica Hwang. This problem deals with the concept of independent events in the context of the Simpson's Paradox. I seem to be a little stuck in the third part of the problem. Please let me know if my approach to the rest of the question is correct. Thank You!
The problem goes as follows:
Simpson’s paradox says that it is possible to have events $A, B, C$ such that $P(A | B,C) < P(A | B^c, C)$ and $P(A | B,C^c) < P(A | B^c, C^c)$, yet $P(A | B) > P(A | B^c)$.
(a) Can Simpson’s paradox occur if $A$ and $B$ are independent? If so, give a concrete example (with both numbers and an interpretation); if not, prove that it is impossible.
(b) Can Simpson’s paradox occur if $A$ and $C$ are independent? If so, give a concrete example (with both numbers and an interpretation); if not, prove that it is impossible.
(c) Can Simpson’s paradox occur if $B$ and $C$ are independent? If so, give a concrete example (with both numbers and an interpretation); if not, prove that it is impossible.
Answers:
(a) $P(A \cap B) = P(A) P(B)$
By this equation, $P(A|B,C) = P(A|C)$ and $P(A|B^c,C) = P(A|C)$. Hence, these two expressions are equal. So Simpson's Paradox does not apply here.
(b) $P(A \cap C) = P(A) P(C)$
By the above equation, $P(A|B,C) = P(A|B)$ and $P(A|B^c,C) = P(A|B^c)$. By the inequality $P(A|B,C^c) < P(A|B^c, C^c)$, $P(A|B)$ should be less than $P(A|B^c)$. This then contradicts the final result that $P(A|B) > P(A|B^c)$.
(c) Here I am not sure how to proceed. I am not sure how $B$ and $C$ being independent would affect the occurrence of $A$.
It cannot occur. If B, C are independent, then $B, B^C, C, C^C$ are pairwise independent. Then, $P(C|B)=P(C|B^C)=P(C), P(C^C|B)P(C^C|B^C)=P(C^C)$, by the law of total probability, $P(A|B) - P(A|B^C) = (P(A|C,B)P(C|B)+P(A|C^C,B)P(C^C|B)) - (P(A|C,B^C)P(C|B^C)+P(A|C^C,B^C)P(C^C|B^C) )= (P(A|C,B)- P(A|C,B^C))P(C) + P(A|C^C,B)P(C^C)<0$