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Thread: New to stats. Need help.

  1. #16
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    Ok that makes sense, I'm thinking then..

    (A1∩B)U(A2∩B)U...U(Aj∩B)=1. I feel like their is a more formal way of proving this however.

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    it will be
    (A1∩B)U(A2∩B)U...U(Aj∩B)=B
    In the long run, we're all dead.

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    Alright so then to relate it back the original equation

    ∑j-m P(Aj|B)=1.


    We have ∑j-m (A∩B)=B? So then how do we relate the two.

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    Quote Originally Posted by canadia View Post
    Vinux, I have another question for you if you dont mind.

    Suppose {Aj}j-m is a collection of mutally exclusive, collectively exhaustive events in the sample space, and let B an event with P(B) >0. Show that ∑j-m P(Aj|B)=1.

    Alright so I'm thinking P(Aj|B)= P(Aj∩ B)/P(B)

    And since they are mutually exclusive A∩ B= 0?

    Also, since they are collectively exhaustive AUB=S=1?
    From the question,
    Ai∩Aj = ∅ for i ≠ j
    ∪Aj = Ω (union over all Aj)
    Therefore, Ai∩B and Aj∩B are mutually exclusive also, for i ≠ j
    For any mutually exclusive events Ai∩B and Aj∩B, for i ≠ j,
    Pr{Ai∩B} + Pr{Aj∩B} = Pr{(Ai∩B)∪(Aj∩B)}
    By De Morgan's Law,
    (Ai∩B)∪(Aj∩B) = (Ai∪Aj)∩B

    Therefore,
    ∑Pr{Aj|B} = ∑Pr{Aj∩B}/Pr{B} = Pr{(∪Aj)∩B}/Pr{B} = Pr{Ω∩B}/Pr{B}
    = Pr{B}/Pr{B} = 1

  5. #20
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    Quote Originally Posted by canadia View Post
    Alright so then to relate it back the original equation

    ∑j-m P(Aj|B)=1.


    We have ∑j-m (A∩B)=B? So then how do we relate the two.
    P(Aj|B) and P(Aj∩B) are not same..

    In your case,
    ∑j-m (A∩B)=B => ∑j-m P(Aj ∩ B)=P(B) .

    Follow BGM's steps. you will reach answer.
    In the long run, we're all dead.

  6. #21
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    Thanks BGM and Vinux. BGM , is {(∪Aj)∩B the same as ∑j-m (A∩B)? By that i mean is{(∪Aj)∩B .. the union of all intersecions of A and B?

    And if you dont mind could you explain Pr{Ω∩B} = P(B)
    Last edited by canadia; 09-25-2009 at 10:56 AM.

  7. #22
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    The addition follows from the third probability axioms, assumption of σ-additivity

    For any events B, it must be a subset of the sample space Ω
    And B⊂Ω ⇒ Ω∩B = B

  8. #23
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    Sorry BGM, I had never seen the notation Ω, So Ω represents the sample space. and then Pr{(∪Aj)∩B=Pr(Ω∩B) because (UAj)∩B represents all the events in the sample space?

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    UAj = Ω because the collection of the events Aj is exhaustive
    {Aj} formed a partition of Ω as they are disjoint too.

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