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I have a general doubt on independent events. After studying some basics on probability from A first course in probability, I don't see exactly why the relation for independent events comes from, specially if we derive it from the conditional probability equation and we think on terms of Venn diagrams.

So, if we have 2 events E and F, we say that the conditional probability of E given that F has occurred is $P(E|F)=\frac{P(EF)}{P(F)}$. So, the events E and F are independent if $P(E|F)=P(E)$, or $P(EF) = P(F)P(E)$. Actually, the first way of expressing it is pretty clear: doesn't matter if F occurred or not, the probability of E is the same. But numerically speaking, I fail to see it.

Moving on to Venn diagrams, and trying to portrait this result, the space that E must occupy inside of F needs to be the same as it occupies in proportion to the total sample space; but it means at the same time that they share some space. I can think of some cases where events share space but are independent (easily, taking a ace or a spade from a deck), but I fail to see how to integrate the idea of independence into the diagram.

I fail to understand the meaning of independent events, and how these can be differenced from the cases in which the conditional probability is the same as the event that is conditioned.

pdaranda661
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  • Not sure I see the question here. Of course Independent events must "share space". Mutually exclusive events aren't independent, since the probability that both occur is $0$ (I'm ignoring cases in which the conditioning event itself has probability $0$...those cases should be thought through separately). – lulu Jan 19 '24 at 22:03
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    Intuitively, "Independent" means that knowledge about one event tells you nothing about the other. Quantitatively, I like $P(E_1\cap E_2)=P(E_1)\times P(E_2)$ better than the other definition (since it handles the probability $0$ case better), though the conditional definition does mesh with the intuitive one better. – lulu Jan 19 '24 at 22:05
  • @lulu i see the intuition, but for me, numerically it seems to be that 2 events are independent when my argument on Venn diagrams (proportion of the part of E inside F respect to the total F is the same as the entire E respect to the sample space), which seems like a coincidence. So, isn't there cases where 2 events are dependant, but that proportion is also respected? Or is it that events are independent make that proportions be the same, but the opposite isn't true necessarily? – pdaranda661 Jan 19 '24 at 22:18
  • If I understand you correctly, the proportion computation is precisely $\frac {P(E_1\cap E_2)}{P(E_2)}=P(E_1)$. Again, you hit a problem if $P(E_2)=0$ but ignoring that situation, saying the proportions are the same is the same as saying they are independent. – lulu Jan 19 '24 at 22:22
  • Now, correlation is a more serious problem, intuition-wise. You can have two variables with $0$ correlation which determine each other exactly. Here is a discussion. In brief, the issue is that correlation detects specifically linear dependence, and might miss higher order relations. – lulu Jan 19 '24 at 22:24
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    Try this on. Suppose you “knew” E influenced F, yet the probability of F given E was indistinguishable from the probability of F alone. In what sense could E possibly be influencing F? That’s one way to see why the definition of independence is what it is. – RobinSparrow Jan 19 '24 at 22:25
  • One more possible, intuitive, explanation of $P(A\cup B)=P(A)\cdot P(B)$: probabilities of $A$ and $B$ do not affect each other. – zkutch Jan 19 '24 at 22:46
  • Perhaps the doubts arise because $P(EF)/P(E)$ and $P(EF)/P(F)$ are independent, but if $P(EF)/P(F)$ happens to be equal to $P(E)$, then $P(EF)/P(E)$ is equal to $P(F)$. This isn't apparent from a Venn diagram. – Rob Arthan Jan 19 '24 at 23:41

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