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Math Help - Probability using Stirling's Approximation

  1. #1
    Member Maccaman's Avatar
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    Probability using Stirling's Approximation

    An atom travels across the integer web, commencing at 0 and ascending or descending depending on the outcome of a coin throw (ascending if heads, and descending if tails).
    (1) If after  2n steps ( n \leq 1 ) the atom is in the spot it commenced, what is the number of different paths it could have taken? And do these paths have the same probability?
    (2) If we assume that the probability of heads is  p , find  p_n , the probability that the atom is back where in its starting position after  2n steps.
    (3) Sterling’s approximation:

     n! \approx \sqrt{2 \pi n} (\frac{n}{e} )^{n}

    Prove that for large values of  n,  p_n ~  \frac{(4p(1-p))^{n}}{\sqrt{n \pi}} using Stirling’s approximation.
    Last edited by Maccaman; August 11th 2009 at 01:49 AM. Reason: incorrect latex
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Maccaman View Post
    An atom travels across the integer web, commencing at 0 and ascending or descending depending on the outcome of a coin throw (ascending if heads, and descending if tails).
    (1) If after  2n steps ( n \leq 1 ) the atom is in the spot it commenced, what is the number of different paths it could have taken? And do these paths have the same probability?
    Given that the atom has returned to its starting point after 2n steps there must be exactly n ups and n downs. So this is asking how many permutations are there of 2n objects where n are of one type and n of another (all the permutations are equally likely as the coint tosses are independendant).

    CB
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    Grand Panjandrum
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    Quote Originally Posted by Maccaman View Post
    An atom travels across the integer web, commencing at 0 and ascending or descending depending on the outcome of a coin throw (ascending if heads, and descending if tails).

    (2) If we assume that the probability of heads is  p , find  p_n , the probability that the atom is back where in its starting position after  2n steps.
    The probability that the atom is back in its starting position after 2n steps is the probability that in 2n trials with probability of success p that we have exactly n success, the number of success r has a binomial distribution B(2n,p) , so the required probability is:

    prob(2n)=b(n;2n,p)

    CB
    Last edited by CaptainBlack; August 11th 2009 at 04:13 AM.
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