# Basic probability theory concepts

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?

• I think your sample space must have a random experiment associated with it to understand which events are independent or not. – StubbornAtom Sep 20 '16 at 18:11

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.
• 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? – user929304 Sep 21 '16 at 15:44
• 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". – Ian Sep 21 '16 at 20:11
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}