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I'm attempting to acquire an intuitive understanding of why the content in the question of the title is correct, however I am unable to do so. Is there way of thinking about the result that makes sense?

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  • $\begingroup$ How do you know it is not necessarily independent of B∩C ? $\endgroup$ – anonymous2 Dec 6 '16 at 16:11
  • $\begingroup$ It is mentioned in a first course in probability by Sheldon Ross. $\endgroup$ – David Dec 6 '16 at 16:15
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    $\begingroup$ Does he consider null sets to be independent? $\endgroup$ – anonymous2 Dec 6 '16 at 16:16
  • $\begingroup$ He only wrote "Suppose now that $E$ is independent of $F$ and is also independent of $G$. Is $E$ then necessarily independent of $FG$? The answer, somewhat surprisingly, is no, as the following example demonstrates.". Here $FG$ = $F\cap G$. $\endgroup$ – David Dec 6 '16 at 16:18
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Three events can be pairwise independent but not jointly independent. Think about two fair coin tosses $X_1$ and $X_2$ such that

$$X_i= \begin{cases} 1 & \text{if heads show up} \\ 0 & \text{otherwise} \end{cases} $$

And define $X_3=(X_1+X_2) \text{ mod } 2$

Then you will see that

$P(X_1=1,X_2=1)=P(X_1=1)P(X_2=1)=1/4$

$P(X_1=1,X_3=1)=P(X_1=1)P(X_3=1)=1/4$

$P(X_1=1,X_2=1,X_3=1)=0\ne P(X_1=1)P(X_2=1,X_3=1)=1/8$

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    $\begingroup$ In other words: Flip two coins, let B be the event that the first coin is heads, let C be the event that the second coin is heads, and let A be the event that both coins come up the same. $\endgroup$ – Nate Dec 6 '16 at 16:25
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    $\begingroup$ Yes, that's the proper English for what i said :) Though the "modulo" definition can be easily generalised to build $n$ events/random variables so that any $n-1$ are independent. $\endgroup$ – Momo Dec 6 '16 at 16:28
  • $\begingroup$ So A is the joint probability of B and C, and this answer shows how shifting the joint probabilities of $(B,C)$ to $(A,B)$ is how the intersection of $(B,C)$ is not necessarily independent of A? $\endgroup$ – user304051 Dec 6 '16 at 16:36
  • $\begingroup$ That is a very nice exempel Momo, thank you. Generally speaking however I still find the result unintuitive. But I suppose I will settle for this. $\endgroup$ – David Dec 6 '16 at 16:38
  • $\begingroup$ @David Maybe you can think of it this way...(maybe) The set $A$ is independent of sets $(B,C)$ and all elements of the sets are probabilities of an event with respect to that set. The elements of set $C$ and set $B$ are a fair coin toss. Set $A$ is the contingent probability that the event in $B$ is equal to the event in $C$. Set $A$ is a "conditional container" set, so if the event of $B$ and event of $C$ are equivalent, then the rule for membership in $A$ is satisfied. $\endgroup$ – user304051 Dec 6 '16 at 16:42
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Consider the following Venn diagram. $P(A)=P(A|B)=P(A|C)=\frac 58$ but $P(A|BC)=1$enter image description here

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  • $\begingroup$ Very nice example, thank you. $\endgroup$ – David Dec 6 '16 at 16:46
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i could give you a situation where it might happen. A might be disjoint with B intersection C. Thus A and "B intersection C" will definitely be dependent.

There might be many other cases where independence doesn't hold true.

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  • $\begingroup$ What? If A is disjoint with B∩C, they are independent. This doesn't answer the question. $\endgroup$ – anonymous2 Dec 6 '16 at 16:20
  • $\begingroup$ what i'm saying is if A is disjoint with BC then A and BC are dependent. Hence we have atleast one situation where A and BC are not independent $\endgroup$ – aman_cc Dec 6 '16 at 16:22
  • $\begingroup$ Why would they be dependent if they are disjoint? $\endgroup$ – anonymous2 Dec 6 '16 at 16:26
  • $\begingroup$ @anonymous2 because there intersection will be null and hence conditional probability will necessarily be O $\endgroup$ – aman_cc Dec 6 '16 at 16:35
  • $\begingroup$ Thanks to Momo. In his example if I define A as 2nd coin is head. B as 1st coin is head. C as exactly 1 tail in 2 tosses That would be a specific example of situation I was saying $\endgroup$ – aman_cc Dec 6 '16 at 16:43

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