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Give an example of a probability space (Ω,Pr) and pairwise independent events A, B, and C which are not mutually independent.

This is my understanding of what pairwise independent events are: Events $A_1, A_2,..., A_k$ are pairwise independent if for all i,j, $A_i $ and $A_j$ are independent: $Pr(A_i\cap A_j) = Pr(A_i)Pr(A_j)$

Events $A_1, A_2,..., A_k$ are mutually independent if for all $I\subset 1,2,...,k, Pr(\bigcap ._{i\subset I} A_i) = \prod\limits_{i\subset I} Pr(A_i) $ so $ Pr(A_1)\cap Pr(A_2) ... Pr(A_k) = Pr(A_1)Pr(A_2)...Pr(A_k)$

user1045890
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  • The question is a bit confusing, and some of the notion does not look right (e.g. $Pr(A_1)∩Pr(A_2)...$). – Jay Zha Mar 13 '17 at 03:37
  • \prod\limits_{i\subset I} produces $\prod\limits_{i\subset I}$. More formatting tips can be found here. As for the content of your question, if you search a bit harder on this site you will surely find some examples. How about the uniform probability measure on ${1,2,3,4}$. Let $A={1,2}$ and $B={1,3}$. Are they independent? Can you come up with a third that is also pairwise independent that is similar to $A$ and $B$ in some regard? What would $Pr(A\cap B\cap C)$ be if mutually indpndt – JMoravitz Mar 13 '17 at 03:43
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    If you prefer some flavor, instead of ${1,2,3,4}$ look instead at ${HH,HT,TH,TT}$, the sample space for flipping two fair coins in sequence, and $A$ is the event the first coin is a head and $B$ is the event the second coin is a head. $C$ could be something which has something to do with both flips... perhaps counting something... – JMoravitz Mar 13 '17 at 03:48
  • An example is here – Ross Millikan Mar 13 '17 at 03:48
  • @JMoravitz Ah, you already commented the same thing here , that's nice! (+1) – Jay Zha Mar 13 '17 at 04:03

1 Answers1

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I suggest @Ashley to refine your question, but meanwhile according to your title, I'll give an example of "pairwise but not mutually independent".

You toss a fair coin twice, let

$A_1$=Both tosses give the same outcome (HH or TT).

$A_2$=The first toss is a heads (HT HH).

$A_3$=The second toss is heads (TH HH).

We have:

$\mathbb P(A_1)=\mathbb P(A_2)=\mathbb P(A_3)=\frac{1}{2}$

$\mathbb P(A_1\cap A_2)=\mathbb P(A_1\cap A_3)=\mathbb P(A_2 \cap A_3)=\frac {1}{4}$

Thus, $\mathbb P(A_i \cap A_j)=\mathbb P(A_i)\mathbb P(A_j), \forall \,i,\,j \in \{1,2,3\},\, and \,i \ne j$

However, they are NOT mutually independent, by noticing that:

$\mathbb P(A_1\cap A_2\cap A_3)=\frac{1}{4}\ne \mathbb P(A_1)\mathbb P(A_2)\mathbb P(A_3)=\frac{1}{8}$

Jay Zha
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