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Let $A$, $B$, $C$ be three events such that $P(A\cup B\cup C)=1$.

Suppose that their event-intersection $I=A\cap B\cap C$ is such that it holds the constraint

$$ P(I)=\frac{P(I|A)P(I|B)}{P(I|A)+P(I|B)}. $$

From this assumption, I would conclude that the knowledge about the occurrence (success/failure) of the event $A$ and of the event $B$ is enough to determine the probability that the event $I$ takes place or not, independently on what is the occurrence of the event $C$.

Is this interpretation correct?

Otherwise, what kind of information gives us such imposition (e.g. about their mutual exclusivity)?

So far, I was able to prove that such statement corresponds to $P(A)+P(B)=1$, which however seems not to involve the same interpretation (at least, not at first sight).

I am not an expert of conditional probability, therefore this question may be obvious for an expert, and I apologize in that case.

However, many thanks for your suggestions!

A similar problem is discussed here A problem of conditional probability, where the focus was on whether or not $P(I)=0$, whereas here I am more interested on the correct interpretation of the constraint. Sorry for some overlap!

  • Could you perhaps say something about how you arrived at this interpretation of the constraint? – joriki Aug 01 '18 at 06:52
  • Sure: $P(A)+P(B)=1$, $1=\frac{1}{P(A)+P(B)}$, $P(I)=\frac{P^2(I)}{P(I)P(A)+P(I)P(B)}$, $P(I)=\frac{P^2(I)}{P(I|B)P(B)P(A)+P(I|A)P(A)P(B)}$, $P(I)=\frac{P(I|A)P(I|B)}{P(I|A)+P(I|B)}\frac{P(A)P(B)}{P(A)P(B)}$, $P(I)=\frac{P(I|A)P(I|B)}{P(I|A)+P(I|B)}$. –  Aug 01 '18 at 07:01
  • I extensively used the fact that $P(I|A)P(A)=P(A|I)P(I)$, and $P(A|I)=1$. Same for $B$. –  Aug 01 '18 at 07:12
  • That was a misunderstanding. I meant not how you showed that it's equivalent to $P(A)+P(B)=1$ (though that's also interesting) but how you arrived at the interpretation about which you're asking whether it's correct. – joriki Aug 01 '18 at 07:14
  • @joriki Ah, sorry. Well, my interpretation is based on my idea of conditional probability, which may be wrong. According to my understanding, $P(U|V)$ answers the question "if we know that $V$ takes place, what is the probability that $U$ takes places?". Therefore, the constraint above, which involves only $P(I|A)$ and $P(I|B)$, gives the exact probability to get a success of $I$, only on the basis of this kind of knowledge: "if we know that $A$ takes place", "if we know that $B$ takes place", no matter what happens about the event $C$. –  Aug 01 '18 at 07:20
  • That's for sure wrong, but I don't understand why!!! –  Aug 01 '18 at 07:27
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    I see. Then consider $P(I)=P(I\mid A)$. You could apply the same argument to that. But that's clearly not a statement that $I$ is independent of $C$ -- in fact, it's a statement that $I$ is independent of $A$. Whether an expression involves only conditioning on $A$ and $B$ has nothing to do with other events. In fact, even the trivial statement $P(I)=P(I)$ doesn't contain anything other than conditioning on $A$ and $B$, and here it's even more obvious that this tells us nothing about $C$. – joriki Aug 01 '18 at 07:33
  • I understand! Thanks for your observation! I would delete this post, but your answer may be useful to other users. –  Aug 01 '18 at 07:35

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With reference to your comments under the question:

Consider $P(I)=P(I\mid A)$. You could apply the same argument to that. But that's clearly not a statement that $I$ is independent of $C$ – in fact, it's a statement that $I$ is independent of $A$. Whether an expression involves only conditioning on $A$ and $B$ has nothing to do with other events. In fact, even the trivial statement $P(I)=P(I)$ doesn't contain anything other than conditioning on $A$ and $B$, and here it's even more obvious that this tells us nothing about $C$.

joriki
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