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Math Help - Neg Binomial and Binomial algebra relationship

  1. #1
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    Neg Binomial and Binomial algebra relationship

    Hi could anyone give me a bit of help on this, i know it is probably simple but i am a bit unsure:

    Find an algebraic relationship between the negative binomial probability P(x) and the binomial probability for the probability of n successes in x trials.

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  2. #2
    MHF Contributor matheagle's Avatar
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    The real relationship is between the geometric and the Neg Bi.
    The rv in the geo is the trial in which the first success appears.
    The rv in the neg bi is the trial in which the r^{th} success appears.
    So, if r=1, the neg bi becomes the geo.
    And the geo is a sum of r i.i.d. geo's.
    Both of these rvs have infinite support.
    BUT with the Binomial there is a fixed number of trials.
    In the other two we can keep waiting for Godot, which may never happen.
    I work with the St petersburg game and thats an example where we wait a long time for a big pay out.
    It has infinite expectation, which confused the Bernoulli's 295 years ago.
    BUT my solution to that game does work.
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  3. #3
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    Hi thanks for your reply, i think the question however is what is the algebraic relationship between:

    Binomial

    n! multiplied by p^x (1-p)^n-x
    x!(n-x)!

    Negative Binomial

    (x-1)! multiplied by p^n (1-p)^x-n
    (n-1)!(x-n)!

    How would I explain this??
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  4. #4
    MHF Contributor matheagle's Avatar
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    Both of these distributions are based on repeat independent trials where there are two possible outcomes. The probability of 'success' is p and the probability of 'failure' is q=1-p. You can switch the success and failure when answering these questions.
    Now with the binomial, you have a fixed number of trials, n.
    The {n \choose x} is how many ways you can place x successes and (n-x) failures in those n trials.
    The negative binommial is the same in regards to success and failures.
    BUT here we are waiting for the n^{th} success
    USING your notation, since you have p^n
    You have
    P(X=x)={x-1\choose n-1}p^n(1-p)^{x-n}
    since that last success appears in the x^{th} trial
    So we have p-1 more successes to place within the previous x-1 trials.
    But there are still n successes, hence the p^n, which represents n successes all with probability p.
    Likewise we have x-n failures all with probability 1-p.
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