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Math Help - Proof of total probability theorem

  1. #1
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    Proof of total probability theorem

    Hi

    My question is

    State and prove the theorem of total probability ?

    I tried looking for this proof in my textbook but i can't seem to find it anywhere

    any help is appreichated

    thanks
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  2. #2
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    Quote Originally Posted by rpatel View Post
    Hi

    My question is

    State and prove the theorem of total probability ?

    I tried looking for this proof in my textbook but i can't seem to find it anywhere

    any help is appreichated

    thanks
    Are you looking for the proof of this theorem? i wasnt sure



    I am not sure if i can post external links on this website...

    Check out
    Baye's Theorem explained with proof | TutorVista

    it has a proof of this theorem
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  3. #3
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    are bayes theorem and theorem of total probability the same thing ?

    I though total probability theorem had something to do with 'paritions'.
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  4. #4
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    Quote Originally Posted by rpatel View Post
    are bayes theorem and theorem of total probability the same thing ?

    I though total probability theorem had something to do with 'paritions'.
    They are different things

    in the website scroll down to Law of total probability
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  5. #5
    Moo
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    Hello,

    Law of total probability :
    For a partition (A_n)_{n\geq 1} of \Omega, the probability space, for any event B (and for higher levels, considering the measured space (\Omega,\mathcal{A},\mathbb{P}), for any B\in\mathcal{A}...), we have :

    \mathbb{P}(B)=\sum_{n=1}^\infty \mathbb{P}(B|A_n)\mathbb{P}(A_n)


    and the proof.... :

    \bigcup_{n=1}^\infty A_n=\Omega

    So B=B\cap \Omega=B\cap \bigcup_{n=1}^\infty A_n=\bigcup_{n=1}^\infty \{B\cap A_n\}

    But the last union is a disjoint union : \forall i\neq j~,~ \{B\cap A_i\}\cap \{B\cap A_j\}=\emptyset (very easy to prove)

    Thus the probability of the union is just the sum of the probabilities :

    \mathbb{P}(B)=\sum_{n=1}^\infty \mathbb{P}(B\cap A_n)

    Then just use the formula \mathbb{P}(A\cap B)=\mathbb{P}(B|A)\mathbb{P}(A)
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