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Equipping a well-known axiom of probability, for any events $X,Y$ such that $P(Y)>0$ then $P(XY)=P(Y)P(X|Y)$; and statistical independence, two events having positive probabilities $X,Y$ are independent if and only if $P(X|Y)=P(X) \iff P(Y|X)=P(Y).$

Prove the following two statements are equivalent

$1.$ Three events $A, B$ and $C$ are independent if and only if occurrence of any of these events does not affect the probability of any event formed from other events.

$2.$ Three events $A, B$ and $C$ are independent if and only if the following four equations hold: $P(AB)=P(A)P(B), P(BC)=P(B)P(C), P(CA)=P(C)P(A)$ and $P(ABC)=P(A)P(B)P(C)$.

My attempt solution:

$1 \implies 2$: We first verify the equation $P(AB)=P(A)P(B)$. Observe that event $B$ can be obtained from $B$ and $C$ by writing $B=(B \cap C)\cup ( B \setminus C)$, so that $P(A|B)=P(A)$ (occurrence of $A$ does not affect the probability of any event formed from $B$ and $C$). Therefore, $P(AB)=P(B)P(B|A)=P(A)P(B)$. Similarly, $ P(BC)=P(B)P(C), P(CA)=P(C)P(A)$. In addition, $P(A)=P(A| B \cap C)=\dfrac{P(A\cap B \cap C)}{P(B \cap C)}=\dfrac{P(A\cap B \cap C)}{P(B)P(C)}$. It follows that, $P(ABC)=P(A)P(B)P(C)$. The proof of this part is completed.

I think my solution for "$1 \implies 2$" is fine. My question is how to prove "$2 \implies 1$"? I thought that the crucial step to prove "$2 \implies 1$" is $P(A)=P(A| B \cup C)$, is that true?

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is there anybody here? –  user53541 Feb 25 at 15:13

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