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I've found the following exercise:

Suppose $A$,$B$ and $C$ are independent events and $P(A\cap B) \ne 0$. Show $P(C |A\cap B) = P(C)$

I've tried several times with no success and I would really appreciate some help.

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2 Answers 2

up vote 0 down vote accepted

By the definition of conditional probability, $$\Pr(C|A\cap B)=\frac{\Pr(A\cap B\cap C)}{\Pr(A\cap B)}.$$ By independence $\Pr(A\cap B\cap C)=\Pr(A)\Pr(B)\Pr(C)$ and $\Pr(A\cap B)=\Pr(A)\Pr(B)$.

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Another solution is almost the same as the one by Andre, but perhaps the following fact would be useful for you.

If $A,B,C$ are mutually independent then $C\perp A\cap B$. Indeed: $$ P(C\cap (A\cap B)) =P(C\cap A\cap B) = P(A)P(B)P(C) = P(C)P(A\cap B). $$

Now, if $F\perp G$ then $P(F|G) = P(F)$, so that choose $F=C$ and $G = A\cap B$ which gives you $$ P(C|A\cap B) = P(C),. $$

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I've not seen $\perp$ used before in this context, but I like it. Am assuming it denotes independence, is that right? Do you know of any probability references that use it? –  Assad Ebrahim May 19 at 8:09
1  
@AssadEbrahim: Yes, this means independence. I do remember using it back in Russia, however I am not sure I can recall precisely whether I saw such notation in English books on probability. I've seen in Kallenberg's Foundations they use of a similar symbol for conditional independence though. –  Ilya Jun 4 at 7:20

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