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As far as I understood independence, A and B should not be independent since if either of them happens then we can tell something about the other one. But if they are independent then B and C should also be. From mathematical proof it is explainable, but I didn't understand the intuition behind it.

The example is from “Elementary Bayesian Statistics” by Gordon Antelman, Chapter 2

The example is from “Elementary Bayesian Statistics” by Gordon Antelman, Chapter 2

Text:

For a fair die $U = (1,2,3,4,5,6)$. Let three events be defined as:

  • $A \equiv (1,2)$, so $P(A)=2/6$

  • $B \equiv (2,4,6)$, so $P(B)=3/6$, and

  • $C \equiv (4,5,6)$, so $P(A)=3/6$

Then

  • $A \cap B = (2)$ and $P(A \cap B) = 1/6 = P(A)P(B) = (1/3)(1/2)$, so A and B are independent.

  • $B \cap C = (4,6)$ and $P(B \cap C) = 2/6 \ne P(B)P(C) = (1/2)(1/2)$, so B and C are not independent.

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    $\begingroup$ Rename the faces of the die from $1,2,3,4,5,6$ to $(1,0),(1,1),(2,0),(2,1),(3,0),(3,1),$ thereby identifying them with the Cartesian product $\{1,2,3\}\times\{0,1\}$ and notice the distribution is the product of the uniform distributions on these two sets. Writing the coordinates as $(x,y),$ $A$ is the event $x=1$ while $B$ is the event $y=1,$ making their independence intuitively obvious. In some sense, all independent events occur in this way. $\endgroup$
    – whuber
    Jan 16 at 18:56
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Other explanation why $\color{red}{A = \{1, 2\}} $ and $\color{blue}{B = \{2, 4, 6\}}$ are independent:

Someone rolled a die:

  • The probability of the event $\color{red}A$ is $\color{red}{2\mkern-0.1ex/6}$, i.e. $\color{red}{1\mkern-0.1ex/3}$.
  • Someone tells you that the result is an $\color{blue}{\text{even}}$ number (the event $\color{blue}B$). In spite of this new info the probability of $\color{red}A$ is still $\color{red}{1\mkern-0.1ex/3}$
    ($\color{blue}2 \in \color{red}A$, but $\color{blue}{4} \notin \color{red}A$, $\color{blue}6 \notin \color{red}A$).

In the opposite way —

  • The probability of event $\color{blue}B$ is $\color{blue}{3\mkern-0.1ex/6}$, .i.e. $\color{blue}{1\mkern-0.1ex/2}$.
  • Someone tells you that the result is a number $\color{red}1$ or $\color{red}2$ (the event $\color{red}A$). In spite of this new info the probability of $\color{blue}B$ is still $\color{blue}{1\mkern-0.1ex/2}$
    ($\color{red}2 \in \color{blue}B$, but $\color{red}{1} \notin \color{blue}B$).

Note:

You may use the same scheme to see that the events $\color{blue}B$ and $\color{green}C$ are not independent.

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  • $\begingroup$ Thank you for this explanation. So what I understood is: if B happens then the probability of A will still be 1/3 because {2}/{2,4,6} and if A happens then also the probability of B will be same (1/2) because {2}/{1,2}. And in case of B and C, if B happens i.e. {2,4,6} the probability of C will be 2/3 because {4,6}/{2,4,6} and if C happens, the probability of B will also be 2/3 {4,6}/{4,5,6} in which case the probability of C (1/2) and B (1/2) has changed respectively and that's why they are not independent. Please correct me if I am wrong. $\endgroup$
    – someUser
    Jan 16 at 20:37
  • $\begingroup$ @someUser, exactly, nothing to correct. What should I do when someone answers my question?. $\endgroup$
    – MarianD
    Jan 16 at 20:40
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You probably confuse indepenent events for mutually exclusive events.

... A and B should not be independent since if either of them happens then we can tell something about the other one.

It is not true. What does mean “something”? The probability of “the other one” didn't change!

The correct formulation should be

  • “A and B are independent — if either of them happens, then it doesn't change the probability of the other”.

As I wrote, you are probably confused mutually exclusiveness (the empty intersection, distinctiveness) for the independence of two events.

But they are two independent things:

  • the events $\{1\}$ and $\{2, 4, 6\}$ are mutually exclusive, but they are not independent,

  • the events $\{2\}$ and $\{2, 4, 6\}$ are neither mutually exclusive, nor independent,

  • the events $\{1, 2\}$ and $\{2, 4, 6\}$ are not mutually exclusive, but they are independent,

  • the events $\emptyset$ and $\{2, 4, 6\}$ are mutually exclusive, and they are independent.


Independence can arise in two distinct ways:

  • we explicitly assume that events are independent (e.g. rolling a die again), or
  • we derive independence by verifying that it fulfills the formula $P(A \cap B) = P(A)P(B)$.

Generally, there is no way to “see” the independence (e.g. by looking in a Venn diagram).

By other words, if we may “see” the independence of two events, then they certainly fulfill the formula. But the opposite is not true — the formula is fulfilled, the independence is guaranteed, but there is not a visible reason, why.

The idea of independent events arises from the evident, noticable independence, but is not limited to it.

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  • $\begingroup$ So how can A $\cap$ B and B $\cap$ C be explained in current scenario? $\endgroup$
    – someUser
    Jan 16 at 19:16
  • $\begingroup$ @someUser, don't try to “see” the independence by interpreting intersections of 2 events. Most teachers don't teach their student what I wrote in my answer — independence is sometime visible (and understandable), sometime not. $\endgroup$
    – MarianD
    Jan 16 at 19:34

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