If you have two events A and B that are independent, then it is said that $p(A)p(B)=p(A\cap B)$, and illustrated in a venn diagram as two areas that do not overlap.

The opposite goes for dependent events A and B, where if A is dependent on B, then $p(A|B)p(B)=p(A\cap B)$ and in the venn diagram, the areas representing these events do overlap.

My questions are:

  1. Are there any instances where event A does overlap B but is not dependent on B?
  2. Are there instances where A does not overlap with B, but is dependent on B?

My guess here is that there are no such cases, where if one instance is satisfied then it by definition calls for the other. For instance, regarding question 1, if A overlaps B, then a dependence between the two exists. For question 2, if A does not overlap B, then it cannot be dependent on it.

Are these ideas correct, or are there any situations where this might not work?

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    $\begingroup$ "If you have two events A and B that are independent" ..."then illustrated in a venn diagram as two areas that do not overlap." This isn't true. In fact, this is only true if one of the two probabilities $P(A), P(B)$ is 0. Independent events are not the same as mutually exclusive events. $\endgroup$ – Ilham Apr 16 '15 at 22:31
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    $\begingroup$ Your understanding is incorrect. For independent events, $A$ and $B$ must_ overlap (except in trivial cases when $P(A)$ or $P(B)$ are zero). For non-overlappiing sets $A$ and $B$, $A\cap B = \emptyset$ and so $P(A\cap B) = 0$, implying that if we insist that $P(A\cap B) = P(A)P(B)$, then at least one of $P(A)$ and $P(B)$ must be zero. What you are thinking about is the case when $A$ and $B$ are mutually exclusive or disjoint and so on the Venn diagram, the sets are shown as nonoverlapping. $\endgroup$ – Dilip Sarwate Apr 16 '15 at 22:32
  • $\begingroup$ Thanks for the responses. I believe I was confusing disjoint/mutual exclusivity with independence, so there are some partial truths to my train of thought, but not in all situations. I have to work the ideas of disjointedness in to how I'm thinking. $\endgroup$ – Topher Apr 16 '15 at 22:42

I think you are confusing independent events with disjoint events.

Independent events are defined by $P(A\cap B) = P(A)P(B)$. $A$ and $B$ will overlap in the Venn diagram, except in the case of $P(A) = 0$ or $P(B) = 0$.

Disjoint events are given by $P(A\cap B) = 0$, meaning it is impossible for the events to occur together. Here the Venn diagram areas have no overlap.

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  • $\begingroup$ Ok thats what I've been wrestling with. I was seeing independence as the same as disjointedness. In a quick look at them I can see the difference, but will have to read up on it a bit more and work it over in my mind. Thanks! Bummer...I thought I was onto something there. :) $\endgroup$ – Topher Apr 16 '15 at 22:40
  • $\begingroup$ @Topher If it helps to say things in more common language: $A$ and $B$ are independent if knowing one does not help you find anything new about the other. $A$ and $B$ are disjoint if they can never happen at the same time. $\endgroup$ – eigenchris Apr 16 '15 at 22:48

"Are there any instances where event $A$ does overlap $B$ but is not dependent on $B$?"

Yes, consider choosing a ball out of a set of two balls, one red and one blue, with replacement, twice. Let the event that the first ball chosen is red be $A$, that the second ball is red be $B$. (We can define the events more rigorously using a probability space, but I think you want the intuition).

Then $A$ is independent of $B$ since $P(A) = \frac{1}{2} = P(A|B)$. But the events do "overlap", i.e. the intersection is non-empty ($A\cap B \neq \emptyset$). Also, $P(A \cap B) = \frac{1}{4} > 0$.

Any two events that are mutually exclusive and which both have non-zero probability of occuring are dependent on each other e.g. choosing a blue red ball (let this event be $C$) and a red ball on the first choice. $P(A|C) = \frac{P(A\cap C)}{P(C)} = 0 \neq \frac{1}{2} = P(A)$

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  • $\begingroup$ Thanks! That's a clean example. I'm still a bit hazy on the differences by definition, but this helps clarify it. $\endgroup$ – Topher Apr 16 '15 at 22:44
  • $\begingroup$ @Topher I just edited in an example for your second question too. $\endgroup$ – Ilham Apr 16 '15 at 22:47

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