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Thread: Conditioning on more than one event

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    Conditioning on more than one event




    Lately, I've been studying conditional probability. I'm working with the following theorems:

    1. P(A|B)=P(A,B)/P(B) (the comma represents intersection)
    2. P(A1,A2,...,An)=P(A1)P(A2|A1)...P(An|A1,A2,...An-1)
    3. P(A|B)=[P(B|A)P(A)]/P(B) (Bayes' Rule)

    I'm curious about conditioning with more than one event:

    P(B|A1,A2,...,An) for n events.

    In particular, I don't know the stipulations for the A1,A2,...,An
    As I understand conditional probability, I am updating prior probability with the evidence I collect. Can someone give me a formal definition for conditioning with more than one event and some working theorems?
    "I have discovered a truly remarkable proof of this theorem which this margin is too narrow to contain." Pierre de Fermat

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    Re: Conditioning on more than one event

    I think you need to be a bit more specific with your question. I'm not quite clear on what you're asking. There really are no stipulations on the events that you condition on.
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    Re: Conditioning on more than one event

    Quote Originally Posted by Dason View Post
    I'm not quite clear on what you're asking.
    Is there an explicit formula for P(B|A1,A2,...,An)? I can't find anything online. My book only discusses P(B|A). Maybe I'm describing Bayes' Rule with multiple variables? Something like this is what I'm looking for http://cs.wellesley.edu/~anderson/wr...aive-bayes.pdf
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    Re: Conditioning on more than one event

    Does it help if I tell you that typically A_1, A_2, \ldots, A_n is shorthand for A_1 \cap A_2 \cap \ldots \cap A_n? Just think of it like any other set. Or do something like define A = A_1 \cap \ldots \cap A_n and then apply what you know.
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    Re: Conditioning on more than one event


    I was reading more about Bayes' rule:

    P(C|A)=[P(A|C)P(C)]/P(A)

    Suppose I know that an event B has occurred:

    P(C|A,B)=[P(A|B,C)P(C|B)]/P(A|B).

    Of course, these events are arbitrary and I might be omitting important mathematical details. But this formula is the same as the first, just conditioned on event B.

    I'm hoping someone can confirm this? Bayes' Rule can be used while conditioning on multiple events? This is what I was getting at with my OP.
    "I have discovered a truly remarkable proof of this theorem which this margin is too narrow to contain." Pierre de Fermat

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