Since I got into grad school to study computer sicence, I have been a T.A. and mid-term procter.

Plus, I should check students answering sheets.

To make answers, I would like to check the equation true which is

$P(A\cap B|C) = P(A|C)P(B|C)$ while A, B are independent.

I think that is true.

But I am still little bit not sure.

If that is true, could you prove why?

Thank you for helping in advance.

  • What does the "comma " mean? – SchrodingersCat Nov 3 '15 at 13:48
  • A intersection B. – Woonghee Lee Nov 3 '15 at 13:57
  • Try it out on $C=A\triangle B$. Then the LHS takes value $0$ but the RHS can easily take a positive value. – drhab Nov 3 '15 at 14:10
up vote 2 down vote accepted

Let $A$ and $B$ be the events that two independently tossed coins come up heads. Let $C$ be the event "exactly one coin comes up heads". Then the LHS is $0$, while the RHS is $1/4$.

  • Oh I am sorry that I missed somethings. A, B and C are random variables. Then isn't it changed your counterexample? – Woonghee Lee Nov 3 '15 at 15:07
  • I'm afriad that doesn't save it - events and random variables are two sides of the same coin (excuse the pun) and you can easily translate this into a statement about Bernoulli variables. – preferred_anon Nov 3 '15 at 15:44
  • I see. Maybe it is caused by my lack of English skill. Thank you very much. – Woonghee Lee Nov 3 '15 at 16:24
  • I search about it on wikipedia, en.wikipedia.org/wiki/Conditional_independence. It says, if A, B are conditionally independent on C, P(A,B|C) = P(A|C)P(B|C). Is it true that P(A,B) = P(A)P(B) then P(A,B|C) = P(A|C)P(B|C) and vice versa? – Woonghee Lee Nov 12 '15 at 17:56
  • No, that is not true. Independence of $A$ and $B$ doesn't imply that $A$ and $B$ are conditionally independent on $C$ - $C$ could in principle tell you everything about $A$ if you know $B$ as well. – preferred_anon Nov 12 '15 at 18:17

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