Prove that $A, B, C$ are independent (under certain conditions)

Question

Suppose:

1. $A$ is independent from both $B\cap C$ and $B \cup C$.

2. $B$ is independent from $A \cap C$.

3. $C$ is independent from $A \cap B$.

4. $P(A), P(B), P(C)$ are all greater than zero.

Prove that $A, B, C$ are independent.

I know I have to show that the 3 events are all pairwise independent and that $P(A\cap B \cap C) = P(A) P(B) P(C)$.

Working with 1., I got to the equality $P(A \cap B) + P(A \cap C) = P(A) [P(B\cap C) + P(B \cup C)]$, but I don't find it very useful. I tried several other approaches and they didn't work.

Could someone give me any hints? I'd also appreciate any intuition into how to solve these type of exercises. I always solve them by trial and error. Thanks!

Answer 1

I think you are on the right track. Let us start from the equation you got from hypothesis 1. \begin{eqnarray} P(A\cap B) &=& P(A) \cdot P(B\cup C) + P(A)\cdot P(B\cap C)-P(A\cap C)=\\ &=&P(A) [P(B) + P(C) - P(B\cap C)]+ P(A)\cdot P(B\cap C)-P(A\cap C)=\\ &=& P(A)\cdot P(B) + P(A)\cdot P(C) -P(A\cap C). \end{eqnarray} So we have \begin{equation} P(A\cap B) -P(A)\cdot P(B) = P(A)\cdot P(C) - P(A\cap C).\tag{1}\label{1} \end{equation} Now condition 4. makes me think that I could multiply both sides of \eqref{1} by $P(B)\cdot P(C)$ and get \begin{equation} P(B)\cdot P(C)\cdot P(A\cap B) -P(A)\cdot P^2(B)\cdot P(C) =\\ =P(A)\cdot P(B)\cdot P^2(C) - P(B)\cdot P(C)\cdot P(A\cap C).\tag{2}\label{2} \end{equation} By hypothesis 2. and 3. \begin{equation} P(A\cap B\cap C) = P(B) \cdot P(A\cap C) = P(C) \cdot P(A\cap B),\tag{3}\label{3} \end{equation} so that \eqref{2} yields \begin{equation} P(B) \cdot P(A\cap B\cap C) -P(A)\cdot P^2(B)\cdot P(C) =\\ =P(A)\cdot P(B)\cdot P^2(C)-P(C) \cdot P(A\cap B\cap C), \end{equation} that is \begin{equation} [P(B) +P(C)]\cdot [P(A\cap B\cap C) -P(A)\cdot P(B)\cdot P(C)] = 0. \end{equation} From the last equation you get \begin{equation} P(A\cap B\cap C) =P(A)\cdot P(B)\cdot P(C)\tag{4}\label{4}. \end{equation} Use now \eqref{4} and \eqref{3} to get pairwise independence.

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As for the general intuition, my suggestion is to always ask yourself: "did I use all the hypotheses?", "How can I translate into mathematical terms the hypotheses I have not yet used?" Note also here, the hidden use of the expression relating probability of union and probability of intersection.