Is P(A,B|C) = P(A|C) P(B|C), if P(A,B) = P(A)P(B)?

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As stated in my subject line, I know that P(A|B) = P(A) and P(B|A) = P(B), i.e. A and B are separable as P(A,B) = P(A) P(B). I strongly suspect that this holds with a conditional added, but I can't find a way to formally prove it... can anyone prove this in a couple of lines via Bayes' rules? This is not a homework question, but part of my research and I can't find the answer anywhere.

Thanks to anyone who can help in advanced!
natski
 

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  • #2
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No, this isn't true. Consider two fair coins flipped independently, let A be the event that the first coin comes up heads, B the event that the second coin comes up heads, and C be the event that at least one of the coins comes up heads. Then P(A) = P(B) = 1/2, P(A,B) = P(A)P(B) = 1/4, but P(A|C) = P(B|C) = 2/3 and P(A,B|C) = 1/3 [itex]\neq[/itex] P(A|C)P(B|C) = 4/9
 
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No, this isn't true. Consider two fair coins flipped independently, let A be the event that the first coin comes up heads, B the event that the second coin comes up heads, and C be the event that at least one of the coins comes up heads. Then P(A) = P(B) = 1/2, P(A,B) = P(A)P(B) = 1/4, but P(A|C) = P(B|C) = 2/3 and P(A,B|C) = 1/3 [itex]\neq[/itex] P(A|C)P(B|C) = 4/9
Even more obvious is C= exactly one coin is a head. Then the condition C forces a complete dependence between A and B.
 

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