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Thread: [SOLVED] Mutually Indpendent Question

  1. #1
    mtb
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    [SOLVED] Mutually Indpendent Question

    Hi,

    I just needed a bit of help with the following question.

    Show that if A, B and C are mutually independent, then A and (B U C) are independent.

    From what I understand if A, B and C are mutually independent then

    P(AnBnC) = P(A)P(B)P(C)
    P(AnB) = P(A)P(B)
    P(AnC) = P(A)P(C)
    P(BnC) = P(B)P(C)


    To prove A and (B U C) are independent do I have to prove P(A|(BUC) ) = P(A) because I don't know what to do after
    P((BUC)nA)/P(BUC) = P[(BnA)U(AnC)]/P(BUC)



    THANK YOU
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  2. #2
    MHF Contributor

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    Your definition of "mutual independence" is indeed correct. As you can see, it involves intersections. So you must reduce $\displaystyle P(A\cap(B\cup C))$ to an expression using intersections. The usual trick is to use the complement since $\displaystyle (B\cup C)^c=B^c\cap C^c$ (where $\displaystyle B^c=\Omega\setminus B$ if $\displaystyle \Omega$ is the whole probability space).

    So here are the first steps (that you should justify): $\displaystyle P(A\cap(B\cup C))=P(A)-P(A\cap(B\cup C)^c)=P(A)-P(A\cap B^c \cap C^c)$.

    You will need to know that if $\displaystyle A, B, C$ are mutually independent, then $\displaystyle A, B^c, C^c$ are mutually independent as well. Either you know it already or you'll have to prove it.

    Laurent.
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  3. #3
    mtb
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    Can I continue like this


    $\displaystyle

    P(A)-P(A\cap B^c \cap C^c) = P(A)-P(A)P( B^c \cap C^c) = P(A)[1 -P( B^c \cap C^c)]

    $

    $\displaystyle

    = P(A)(B\cup C)

    $

    Therefore they are independent
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  4. #4
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    I'd rather insert an intermediate equality $\displaystyle P(A)-P(A)P(B^c)P(C^c)=P(A)-P(A)P(B^c\cap C^c)$ in what you wrote, since you don't know directly that $\displaystyle P(A\cap B^c\cap C^c)=P(A)P(B^c\cap C^c)$ (this is almost what you were asked to prove!). Then this is OK.
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  5. #5
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    $\displaystyle \begin{array}{rcl}
    {P\left[ {A(B \cup C)} \right]} & = & {P\left( {AB \cup AC} \right)} \\
    {} & = & {P(AB) + P(AC) - P(ABC)} \\
    {} & = & {P(A)P(B) + P(A)P(C) - P(A)P(B)P(C)} \\
    {} & = & {P(A)\left[ {P(B) + P(C) - P(BC)} \right]} \\
    {} & = & {P(A)\left[ {P(B \cup C)} \right]} \\ \end{array} $
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  6. #6
    mtb
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    Thanks for all the help!!!
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