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I undersrand that a probability space is composed of a sample space $\Omega$ containing all possible outcomes of an experiment, an event space $\mathcal F$ containing all events of interest (e.g. the power set of $\Omega$) and a probability measure $\mathbb P$ that assigns a probability $p$ to each event in $\mathcal F$.

  • We say two events $A$ and $B$ in $\mathcal F$ are disjoint iff $A\cap B = \emptyset$

  • On the other hand, two events are said to be independent iff $\mathbb P(A\cap B) = \mathbb P(A) \mathbb P(B)$. Meaning that additional knowledge of one event does not change the probability of occurrence of the other event. But I can't seem to fully wrap my mind around this. What does independent mean here? How can two events have a nonzero intersection and yet have nothing to do with one another?

  • For example: given $\Omega =\{1,2,3,4\}$ how can one intuitibwly realise that the events $A=\{1,2\}$, $B=\{1,3\}$ and $C=\{1,4\}$ are only pairwise independent and not independent?

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  • $\begingroup$ I think your sample space must have a random experiment associated with it to understand which events are independent or not. $\endgroup$ – StubbornAtom Sep 20 '16 at 18:11
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The usual way for two events to be independent is to have the space being a Cartesian product of two sets $A \times B$, and then have your events being $\{ (x,y) : x \in A_0 \}$ and $\{ (x,y) : y \in B_0 \}$ where $A_0 \subset A,B_0 \subset B$. These typically have nonempty intersection, namely $A_0 \times B_0$. More generally the space could be a Cartesian product of $n$ sets and your events could still be the projection onto one component being in some set.

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  • $\begingroup$ Sorry for the late response to your answer, thanks a lot. This is very much along the conceptual lines I was hoping for, though I m still slightly puzzled: which space are we exactly talking about? (Namely what is this separable space comprised of in case of indepdendent events of $\mathcal F$) Are we saying that events are referred to as indepdenent only if their intersection space (in the example i had used it would be $\{1\}$) can be factored out as a cartesian product? $\endgroup$ – user929304 Sep 21 '16 at 15:44
  • $\begingroup$ Independence has nothing to do with the set-theoretic structure of the probability space, it is just relative to the probability measure. But you can use Cartesian products as a convenient way to construct independent events/independent random variables. $\endgroup$ – Ian Sep 21 '16 at 15:47
  • $\begingroup$ You can make simple examples with finite sets where the probability of a single pair is the product of two functions, one applied to each component. Then you will have the phenomenon I mentioned before. $\endgroup$ – Ian Sep 21 '16 at 16:06
  • $\begingroup$ Oh :( I m still terribly confused still, where could i read up more along the lines of explanations you provided? :( $\endgroup$ – user929304 Sep 21 '16 at 20:06
  • $\begingroup$ I'm not sure, usually this makes sense to people after just playing with a couple of examples. Maybe a continuous example would help. Suppose you give the uniform length measure to two copies of $(0,1)$ and the uniform area measure to $(0,1)^2$, and the latter is your main probability space. Then $P((x,y) \in A \times B)=P(x \in A)P(y \in B)$: the area of a "rectangle" in $(0,1)^2$ is precisely the product of its "lengths". $\endgroup$ – Ian Sep 21 '16 at 20:11
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Two events, $A$ and $B$, are independent if the fact that $A$ occurs does not affect the probability of $B$ occurring. In addition to $A$ , $B$ and $C$ are independent if and only if \begin{cases} \mathbb P(A\cap B) = \mathbb P(A) \mathbb P(B)\\ \mathbb P(A\cap C) = \mathbb P(A) \mathbb P(C)\\ \mathbb P(B\cap C) = \mathbb P(B) \mathbb P(C)\\ \mathbb P(A \cap B \cap C) =\mathbb P(A) \mathbb P(B) \mathbb P(C)\\ \end{cases}

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  • $\begingroup$ Hi, thanks for your answer. This is sort of the textbook definition, which is correct. My question mainly aims at understanding the independence of events conceptually, e.g. along the lines provided by Ian (although I haven't fully understood it yet) $\endgroup$ – user929304 Sep 22 '16 at 13:18

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