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Can the conditional be 'distributed' over the the events?

PGupta
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2 Answers2

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

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

If we suppose that $A,B,C$ are all independent events relative to each other, it follows that $P(A\cap B\cap C) = P(A)P(B)P(C), P(A\cap C) = P(A)P(C), P(B\cap C) = P(B)P(C)$ such that:

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

So in short: no, if only $A$ and $B$ are independent relative to each other, we will not have equality

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Let us throw fair $6$-sided die twice. Let $A$ = first roll was $2$, $B$ = second roll was $3$, $C$ the sum of the rolls was $5$.

Because $C = \{(1,4),(2,3),(3,2),(4,1)\}$, we have

$$P(A\cap B\mid C) = \frac 14,\ P(A\mid C) = \frac 14,\ P(B\mid C) = \frac 14.$$

Ennar
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  • Thanks for the illustration; I see the point. However, a question here https://math.stackexchange.com/questions/198677/how-to-calculate-prdiseased-2-positive-tests is answered by considering P(TS/D)=P(T/D)*P(S/D), where T and S are independent. This is confusing me. – PGupta Aug 22 '17 at 13:49
  • @Teacher123, this is called conditional independence. It might be that this was meant in that question. – Ennar Aug 22 '17 at 14:08
  • @Teacher123, you might be interested in this question. – Ennar Aug 22 '17 at 14:18
  • Thanks, someone posted the same link above. My question is, are there any conditions under which conditional independence can be assumed? – PGupta Aug 22 '17 at 15:22
  • @Teacher123, well I'm not sure what exactly are you looking for. I don't think there is much to say about this when you are just concerned with events, and not random variables, or families of such. It kind of works just like usual independence; look at my example, I just arbitrarily assumed that two rolls are independent, but this is just a model I picked. Usually you suppose some kind of independence of some events/random variables and then examine independence of some other. – Ennar Aug 22 '17 at 15:41