2

How to prove that mutual independence of events implies pairwise independence?

That is, prove that if $P(A \cap B \cap C) = P(A)P(B)P(C)$ then A, B are independent. B, C are independent. C, A are independent.

I tried using the sum rule to prove but couldn't separate two events from the third.

Elf
  • 83
  • I tried that but couldn't get anywhere. $\mathbb P(A\cap C)=\mathbb P((A\cap C\cap B )\cup(A\cap C\cap B^c)) = \mathbb P(A\cap C\cap B ) + \mathbb P (A\cap C\cap B^c)$. Now consider $$\mathbb P (A\cap C\cap B^c) = \mathbb P (A\cap C)\mathbb P (B^c | A\cap C) = \mathbb P (A\cap C)\left(1 - \mathbb P (B | A\cap C) \right) = \mathbb P (A\cap C)\left(1 - \dfrac{\mathbb P (B \cap A\cap C)}{\mathbb P (A\cap C)}\right) = 1 - \mathbb P(A\cap C\cap B )$$.

    But this approach led to nowhere :(. How should I modify it?

     – Elf Oct 12 '21 at 13:43
  • 1
    See https://math.stackexchange.com/a/3129523/356933 . While it is true that mutual independence implies pairwise independence, the definition of mutual independence encompasses the equations for pairwise independence. Hence, the proof is trivial because mutual independence is a strong definition. However, $P(A\cap B \cap C)=P(A)P(B)P(C)$ is not enough to imply the other equations. – ProfOak Oct 12 '21 at 14:28
  • 1
    In fact, mutual independence doesn't means that $\mathbb P(A\cap B\cap C)=\mathbb P(A)\mathbb P(B)\mathbb P(C)$, but it mean that $\mathbb P(U\cap V)=\mathbb P(U)\mathbb P(V)$ for all $U,V\in \sigma ({A,B,C})$. – Surb Oct 12 '21 at 14:51

1 Answers1

0

Let prove that $A$ and $B$ are independent. I let you adapt for $A$ and $C$. \begin{align*} \mathbb P(A\cap B)&=\mathbb P\big((A\cap B\cap C)\cup (A\cap B\cap C^c)\big)\\ &=\mathbb P(A\cap B\cap C)+\mathbb P(A\cap B\cap C^c)\\ &=\mathbb P(A)\mathbb P(B)\mathbb P(C)+\mathbb P(A)\mathbb P(B)\mathbb P(C^c)\\ &=\mathbb P(A)\mathbb P(B)\big(\mathbb P(C)+\mathbb P(C^c)\big)\\ &=\mathbb P(A)\mathbb P(B). \end{align*} I let you justify each step.

Surb
  • 57,262
  • 11
  • 68
  • 119
  • 1
    Doesn't the following provide a counter example? https://math.stackexchange.com/a/3129523/356933 – ProfOak Oct 12 '21 at 14:11
  • I think I got it. Mutual independence implies $\mathbb P(C | A \cap B) = \mathbb P(C)$ hence $\mathbb P(C^c | A \cap B) = \mathbb P(C^c)$. From this, $\mathbb(A \cap B \cap C) = \mathbb P(A) \mathbb P(B) \mathbb P(C)$. – Elf Oct 12 '21 at 14:15
  • @zugzug : Indeed, thank you. I will erase my answer. (if Elf don't accept my answer anymore...) – Surb Oct 12 '21 at 14:49
  • @Surb please don't erase your answer. Your answer seems consistent. Why is it wrong? – Elf Oct 12 '21 at 14:59
  • @Elf: It's not true that $\mathbb P(A\cap B\cap C^c)=\mathbb P(A)\mathbb P(B)\mathbb P(C^c)$. – Surb Oct 12 '21 at 15:25