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Thread: Bayes Theorem Limits

  1. #1
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    Bayes Theorem Limits

    Let ($\displaystyle A_n$) and ($\displaystyle B_n$) be in A with $\displaystyle A_n$-->A and $\displaystyle B_n$-->B, with P(B)>0 and P($\displaystyle B_n$)>0. Show that:

    a) $\displaystyle lim_{n->\infty}P(A_n|B)$=P(A|B)

    b) $\displaystyle lim_{n->\infty}P(A|B_n)$=P(A|B)

    c) $\displaystyle lim_{n->\infty}P(A_n|B_n)$=P(A|B)

    So, I know that:

    $\displaystyle P(A_n|B)$=$\displaystyle \frac{P(B|A_n)P(A_n)}{P(B)}$=$\displaystyle \frac{P(B \cap A_n)}{P(B)}$

    $\displaystyle P(A|B_n)$=$\displaystyle \frac{P(B_n|A)P(A)}{P(B_n)}$=$\displaystyle \frac{P(B_n \cap A)}{P(B_n)}$

    $\displaystyle P(A_n|B_n)$=$\displaystyle \frac{P(B_n|A_n)P(A_n)}{P(B_n)}$=$\displaystyle \frac{P(B_n \cap A_n)}{P(B_n)}$

    $\displaystyle P(A|B)$=$\displaystyle \frac{P(B|A)P(A)}{P(B)}$=$\displaystyle \frac{P(B \cap A)}{P(B)}$

    My problem really is how to show the limit of $\displaystyle P(B|A_n)$, $\displaystyle P(B_n|A)$, and $\displaystyle P(B_n|A_n)$ is$\displaystyle P(B|A)$. They seem obvious, but I can't figure out how to show it.
    Last edited by azdang; Sep 16th 2009 at 12:33 PM.
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