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Thread: Independent events

  1. #1
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    Independent events

    I am given that $\displaystyle P(A) >0, P(B) >0, P(C) > 0$ (strictly greater) and

    $\displaystyle P(B)P(C) = \frac{P(B \cap C)P(A \cap B)}{P(A \cap (B \cap C))} + \frac{P(B \cap C^{C})P(A \cap B)P(C)}{P(A \cap (B \cap C^{C})P(C^{C})}$

    I am to deduce whether B and C could be independent.

    I am thinking no.

    In order for the right side to be equal to $\displaystyle P(B \cap C)$ this means at least

    $\displaystyle
    \frac{P(B \cap C^{C})P(A \cap B)P(C)}{P(A \cap (B \cap C^{C})P(C^{C})} = 0 $

    Now, $\displaystyle P(C) \neq 0$ since it was given that $\displaystyle P(C) > 0$
    $\displaystyle P ( A \cap B) \neq 0$, since this would make $\displaystyle \frac{P(B \cap C)P(A \cap B)}{P(A \cap (B \cap C))} = 0 $

    Therefore $\displaystyle P(B \cap C^{C}) = 0$

    However $\displaystyle P(B \cap C^{C}) = 0 $ implies $\displaystyle B$ and $\displaystyle C^{C}$ are not independent in this case because $\displaystyle P(B) > 0$ and $\displaystyle P(C^{C}) > 0 $( since $\displaystyle P(C) > 0$)

    $\displaystyle B$ and $\displaystyle C^{C}$ not independent implies $\displaystyle B $and $\displaystyle C$ aren't independent.

    Is this sound reasoning? I am skeptical because this based of the converse of a true statement.

    Thank you,

    -Jame
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  2. #2
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    Wait

    $\displaystyle P(C^{C})$ is not necessarily greater than zero. ( i.e. $\displaystyle if P(C) = 1$)

    Darn, well at least what I said still works if $\displaystyle P(C) \neq 1 $
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  3. #3
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    Edit: I forgot that we were assuming that $\displaystyle P(A | B \cap C) \neq P(A|B)$.

    So the $\displaystyle P(C)$ is not 1 and the $\displaystyle P(C^{C})$ is not zero.
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  4. #4
    MHF Contributor matheagle's Avatar
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    Did you try a few numbers to see whether there might be a counter example?
    maybe drawing a Venn diagram and throwing in some numbers where P(BC)=P(B)P(C) might shed light one way or another.
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