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So in my head I think that this is not possible. If A is independent of B and C and B is independent of A and C and C is independent of A and B then all 3 of them are independent. Is this right? I have a feeling I am wrong and I have been stick for a while.

If I am wrong could you give me a hint to point me in the right direction?

user120767
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  • You only have A is independent of $B \cap C$ and $B$ is independent of $A \cap C$ and $C$ is independent of $A \cap B$, and this does not imply all three are independent. – Henry Oct 29 '15 at 09:08
  • But I also fail to see why this is the case. – user120767 Oct 29 '15 at 09:10
  • There is an answer here: http://math.stackexchange.com/questions/266557/pairwise-independent-is-weaker-that-independent – uniquesolution Oct 29 '15 at 09:21
  • That example does not work as $P(A\cap B\cap C)={1\over4}$ as well so $A$ is not independent of $B\cap C$ – cr001 Oct 29 '15 at 09:29
  • Here a concrete example showing that conditional independence does not imply indepedence. Although it is for two events and one conditioning event, that result could be generalized to more events, as @cr001 showed you. The nice thing of the example in the link, though, is that the intuition (in terms of the concept of conditioning) about why that is the case is more evident. – Carlos H. Mendoza-Cardenas Oct 29 '15 at 10:10

1 Answers1

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$P(A\cap B\cap C)=P(A)P(B\cap C)=P(B)P(C\cap A)=P(C)P(A\cap B)\neq P(A)P(B)P(C)$ is what you want.

Let's say we have $6$ mutually exclusive event $x_1,x_2,x_3,x_4,x_5,x_6$ each has possibility $1\over 6$. You can consider dice rolling as a real world example.

Let

$A=\{x_1,x_2,x_3,x_4\}$ (either of the four happening)

$B=\{x_1,x_2,x_3,x_5\}$ (either of the four happening)

$C=\{x_1,x_2,x_4,x_5\}$ (either of the four happening)

Then $P(A)=P(B)=P(C)={2\over3}$

$P(A\cap B)=P(B\cap C)=P(A\cap C)={1\over2}$

$P(A\cap B \cap C)={1\over3}$

cr001
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