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Math Help - Sampling with replacement

  1. #1
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    Sampling with replacement

    A = {1, 2, ..., N} is a set of N integers.

    I'm sampling from A, one element each time, with replacement.

    How many sampling may I expect to perform, in average, before obtaining one element that already came out?

    I would appreciate some guidance. Thanks!
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  2. #2
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    Quote Originally Posted by paolopiace View Post
    A = {1, 2, ..., N} is a set of N integers.

    I'm sampling from A, one element each time, with replacement.

    How many sampling may I expect to perform, in average, before obtaining one element that already came out?

    I would appreciate some guidance. Thanks!
    The probability of the ith trial failing given the failure of preceding trials is \frac{N+1-i}{N}. The probability of every trial failing up to and including the y-1th trial is therefore

    \sum_{i=1}^{y-1}\frac{N+1-i}{N}=\frac{N+1}{N}\sum_{i=1}^{y-1}-\frac{1}{N}\sum_{i=1}^{y-1}i

    =\frac{(N+1)(y-1)}{N}-\frac{(y-1)(y-2)}{2N}

    =\frac{(y-1)(2N+4-y)}{2N}.

    The probability of the ith trial succeeding given the failure of preceding trials is \frac{i-1}{N}. Let Y=y denote the event that every trial up to and including the y-1th trial fails and then the yth trial succeeds. Then

    p(y)=\frac{(y-1)(2N+4-y)}{2N}\left(\frac{y-1}{N}\right)=\frac{(y-1)^2(2N+4-y)}{2N^2}.

    Therefore \mathbb{E}(Y)=\sum_{y=1}^{N}yp(y)=\sum_{y=1}^{N}\f  rac{y(y-1)^2(2N+4-y)}{2N^2}.

    This can be simplified using common summation identities.
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