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Math Help - Binomial Approximation to Poisson Distribution

  1. #1
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    Post Binomial Approximation to Poisson Distribution

    Poisson distribution as a limiting case of binomial probability distribution under the following conditions
    1)n= number of trials indefinitely large. 2)p= constant probability of success for each trial is indefinitely small 3)np=m is finite

    In binomial distribution, the probability of x successes p(x)=ncx(p^x) q^(n-x),
    Put now p=m/n and q=1-p=1-(m/n) we get
    p(x)=n(n-1) (n-2)…(n-x+1)/x! (m/n)^x (1-(m/n)^(n-x)

    =1 [1-(1/n)][1-(2/n)]…[1-{(x-1)/n}] m^x. [(1-(m/n))^n]
    —————————————-—-—— ——————
    x! [(1-(m/n))^x]
    Now,for fixed x,as n -->∞
    [1-(1/n)][1-(2/n)]………[1-(x-1)/n] all tend to 1 and [1-(m/n)]^(-n) to. e^(-m)

    Can anyone tell me how [1-(m/n)]^(-x) is = e^(-m)
    Last edited by Vinod; September 23rd 2011 at 10:29 PM.
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  2. #2
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    Re: Binomial Approximation to Poisson Distribution

    Quote Originally Posted by Vinod View Post
    Poisson distribution as a limiting case of binomial probability distribution under the following conditions
    1)n= number of trials indefinitely large. 2)p= constant probability of success for each trial is indefinitely small 3)np=m is finite

    In binomial distribution, the probability of x successes p(x)=n_C_x (p^x) q^(n-x),
    Put now p=m/n and q=1-p=1-(m/n) we get
    p(x)=n(n-1) (n-2)…(n-x+1)/x! (m/n)^x (1-(m/n)^(n-x)

    =1 [1-(1/n)][1-(2/n)]…[1-{(x-1)/n}] m^x. [(1-(m/n))^n]
    —————————————-—-—— ——————
    x! [(1-(m/n))^x]
    Now,for fixed x,as n -->∞
    [1-(1/n)][1-(2/n)]………[1-(x-1)/n] all tend to 1 and [1-(m/n)]^(-n) to. e^(-m)

    Can anyone tell me how [1-(m/n)]^(-n) is = e^(-m)[/tex]
    See the overview >>here<<, it is one of the definitions of the exponential function (which you should have net in Calculus I(?)).

    CB
    Thanks from Vinod
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