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Usually its stated that two events are independent if and only if their joint probability equals the product of their probabilities. i.e:

$$P(A \cap B) = P(A)P(B)$$

However, I was not sure if that was just a definition or if it had a proof. Usually the way I have seen it made sense is relating it to conditional independence:

$$P(A|B) = \frac{P(A \cap B)}{P(B)}$$

If independent then the distribution doesn't change if we are given that B has occurred:

$$P(A) = \frac{P(A \cap B)}{P(B)}$$

$$P(A)P(B) = P(A \cap B)$$

And then there is a proof for the statement but my concern is, if one of the two events has zero probability of occurring, then I was not sure what happened. For example, is the definition of independence only valid when P(A) and P(B) are non-zero? (since conditional probabilities don't really exists if the denominator is zero) Or Maybe $P(A \cap B) = P(A)P(B)$ is always true? Basically when does $$P(A \cap B) = P(A)P(B)$$ hold? Always?

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You may put it this way: two disjoint events are independent iff the probabilty of one event equals zero.

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As you noticed, if we use the definition "$A$ and $B$ are independent iff $P\left(A|B\right)=P\left(A\right)$", we might encounter some difficulties in case $P(B)=0$.
(here is a discussion about $P(A|B)$ in case $P(B)=0$)

On the other hand, the definition "$A$ and $B$ are independent iff $P(A\cap B)=P(A)P(B)$" doesn't introduce such problems, and thus it is the preferred definition.
Another perk of this definition is that it clearly shows that independence is symmetric.

So the latter is the (preferred) definition, and from it you can deduce the former.


(My answer is based on wikipedia. Also, here is a question that asked explicitly about the preferred definition.)

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  • $\begingroup$ Would you please say why you down-voted the answer (so that I could fix it)? $\endgroup$ – Oren Milman Sep 19 '18 at 7:04
  • $\begingroup$ I have given you an upvote. That at least compensates for the downvote! I this defining independence using conditional probabilities is a bad idea. $\endgroup$ – Kabo Murphy Sep 19 '18 at 7:25

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