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In probability theory, one way to define independence is:

If $P(AB) = P(A)P(B)$, then $A$ and $B$ are independent events. (this is referred as "formal independent" in this question).

Let $A$ represent the event that the toss of a coin gives head, with $P(A) = \frac12$, and $B$ represent the event that today will rain, with $P(B) = \frac15$.

Question 1: without saying nothing more, is $P(AB)$ calculable?

Question 2: one reasonable answer to question 1 is that $P(AB) = \frac1{10}$, this answer is given based on our common sense that, the result of the toss has nothing to do with the weather, (this is referred as "real world independence" in this post), my question is: in real world applications of the probability theory, if two events are real-world-independent, then we treat them as formal-independent (often to utilize $P(AB) = P(A)P(B)$), how do you justify this methodology? (I understand that, in the case of real world application of a certain math theory, you need to make an approximation of the problem at hand, i.e., you make a math model of the problem, but in the case of the formal vs. real-world independence, I don't think it is an approximation/modeling issue, it seems to me that there is something more going on). You often see the argument like this "since $A$ and $B$ are (real-world) independent, (by the definition of formal-independent), $P(AB) = P(A)P(B) = \text{some number}$", you don't often see sentence like this "since $P(AB) = c$, and $P(A)P(B)=c$, $A$ and $B$ are independent events".

Question 3: can you provide an example, where, $0 \lt P(A), P(B) \lt 1$, and $P(AB) = P(A)P(B)$, but $A$ and $B$ are not real-world-independent?

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    $\begingroup$ Not following. That's the definition of independence. Why should there be any difference between "formal" and "real world" versions of the notion? $\endgroup$ – lulu Aug 15 '18 at 14:08
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    $\begingroup$ To be sure, it's easy to produce examples where, for simplicity, people assume independence when it really isn't there. For instance, it's routine to imagine that birth gender is independent of the birth gender of siblings, but I don't think this is strictly true in the real world. But I don't think this is what you are asking about. $\endgroup$ – lulu Aug 15 '18 at 14:11
  • $\begingroup$ Questions like what are asked here are why I do not teach $P(AB) = P(A)P(B)$ when introducing independence. Instead, I teach $P(A \mid B) = P(A)$ and $P(B \mid A) = P(B)$ when introducing independent events for the first time, and then teach the multiplication rule as a consequence of the definition of conditional probability. This latter definition is a lot more intuitive: given $B$, the probability of $A$ doesn't change; and similarly, given $A$, the probability of $B$ doesn't change. $\endgroup$ – Clarinetist Aug 15 '18 at 15:25
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  1. No. There is no way to say whether the events of rain and heads are correlated. Perhaps the increased humidity associated with rain affect the drag coefficient of air in such a way as to increase the probability of heads. That is, we cannot make an exact conclusion.

  2. The justification is that the forces underlying the two systems interact only weakly, so the randomness in one system is approximately unaffected by the randomness in the other. The conclusion is always approximate; we never conclude $P(AB)=P(A)P(B)$ exactly, except in trivial cases.

    I like to think of it like this: imagine a coupled differential equation $y_1'=F_1(y_1,y_2)$, $y_2'=F_2(y_1,y_2)$. To say that $y_1$ and $y_2$ are "real-world indpendent" would mean there are functions $\tilde F_1$ and $\tilde F_2$ so that $\tilde F_1(y_1)\approx F(y_1,y_2)$ uniformly in $y_2$, and the same for $\tilde F_2$, in such a way that the solutions to $\tilde y_1'=\tilde F_1(\tilde y_1)$ and $\tilde y_2'=\tilde F_2(\tilde y_2)$ are within $\epsilon$ of $y_1$ and $y_2$ on the interval $[0,T]$, for some $\epsilon$ too small to be detected by human measurement, for some $T$ longer than the lifetime of humanity. If $F_1$ and $F_2$ are random functions, this means that $P(AB)\approx P(A)P(B)$, where $A$ is an event for $y_1$ and $B$ for $y_2$.

  3. On a frictionless pool table, two elastically colliding pool balls are set in motion in random directions. They will almost surely land in a pocket. Let $A$ be the event that the first ball lands in a corner pocket, and $B$ that the second ball lands in a corner pocket. I would not call these events real world independent since the pool balls interact. But it should be the case that $P(AB)=P(A)P(B)$.

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