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?

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

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



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

  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

Consider the following Venn diagram. $P(A)=P(A|B)=P(A|C)=\frac 58$ but $P(A|BC)=1$enter image description here

  • $\begingroup$ Very nice example, thank you. $\endgroup$ – David Dec 6 '16 at 16:46

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.