Re: P(A|B&C)=[/P(A|B)*P(A|C)]/P(A) ?

From: Duncan Smith <buzzard@xxxxxxxxxxxxxxxxxxxxx>
Date: Sat, 27 Jan 2007 15:30:00 +0000

nulli0.9@xxxxxx wrote:

I have a question.
if B and C are independent. Does the following relation hold?:

P(A|B&C)=[/P(A|B)*P(A|C)]/P(A)

If not could someone tell me why the following proof is not correct:

P(A|B&C)
= P(A&B&C) / P(B&C)
= P(A&B&C) / (P(B)*(P(C))
= P(B&C|A)*P(A) / (P(B)*P(C))
= P(B|A)*P(C|A)*P(A) / (P(B)*P(C))

P(B,C) = P(B)*P(C) does not imply P(B,C|A) = P(B|A)*P(C|A)

= P(B|A)*P(C|A)*P(A)*P(A) / (P(A)*P(B)*P(C))
= P(A&B)*P(A&C) / (P(A)*P(B)*P(C))
= (P(A|B)*P(A|C)) /P(A) q.e.d.

References:
- P(A|B&C)=[/P(A|B)*P(A|C)]/P(A) ?
  - From: nulli0.9@xxxxxx

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