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Math Help - Proof of Binomial to Poisson using PGF

  1. #1
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    Proof of Binomial to Poisson using PGF

    Hi,

    Does anyone know how to show, by means of probability generating functions, that for large n and small p, (n →∞, p→0), such that np →λ, that a binomial distribution will converge to a poisson with parameter λ?

    Thanks in advance for your help
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  2. #2
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    I believe so. This is something I had to prove sometime back. It's a cool proof.

    Let np={\lambda}. Therefore, p=\frac{\lambda}{n}

    Start with the binomial b(x;n,p)=\binom{n}{x}\left(\frac{\lambda}{n}\right  )^{x}\left(1-\frac{\lambda}{n}\right)^{n-x}

    =\frac{n(n-1)(n-2)......(n-x+1)}{x!}\left(\frac{\lambda}{n}\right)^{x}\left(1-\frac{\lambda}{n}\right)^{n-x}

    Now, if we divide one of the x factors, n, in \left(\frac{\lambda}{n}\right)^{x} into each factor of the product n(n-1)(n-2)....(n-x+1), and then write

    \left(1-\frac{\lambda}{n}\right)^{n-x}=\left[\left(1-\frac{\lambda}{n}\right)^{\frac{-n}{\lambda}}\right]^{-\lambda}\left(1-\frac{\lambda}{n}\right)^{-x}

    we get:

    \frac{\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)....\left(1-\frac{x-1}{n}\right)}{x!}\cdot {\lambda}^{x}\left[\left(1-\frac{\lambda}{n}\right)^{\frac{-n}{\lambda}}\right]^{-\lambda}\left(1-\frac{\lambda}{n}\right)^{-x}

    Now, let n\to {\infty} while x and lambda stay fixed.

    we get:

    \left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)....\left(1-\frac{x-1}{n}\right)\to 1

    \left(1-\frac{\lambda}{n}\right)^{-x}\to 1

    Now, we get the famous 'e' limit:

    \left(1-\frac{\lambda}{n}\right)^{\frac{-n}{\lambda}}\to e

    Therefore, hence, and heretofore, we get:

    p(x;{\lambda})=\frac{{\lambda}^{x}e^{-\lambda}}{x!}

    and the limiting distribution becomes the Poisson distribution.
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  3. #3
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    Hi,


    First of all thanks so much for the proof.

    I do have a few qns:
    The problem did ask for proof via probability generating functions, and I can't help but think that there is a more simple proof to it than the method you have shown (i do very much appreciate that it is indeed, quite a cool proof). So do you have any idea how this might work?

    [ pgf of binomial(n,p) = (pz + 1-p)^n and pgf of poisson(λ) e^λ(z-1) ]

    Also about this proof:

    I'm guessing the bit about the famous 'e' limit can be shown via taylor series' expansion??

    I fully understand the rest of the proof though, its very thorough and helpful^^

    cheers
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  4. #4
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    The proof I gave is via a probability generating function. That is what a binomial is.

    Don't worry about deriving the 'e' limit. Though you can if you wish. It's derivation can be found in a calc book.


    One can also show, via repeated integration by parts, that \sum_{x=0}^{n}\frac{{\lambda}^{x}e^{-\lambda}}{x!}=\frac{1}{n!}\int_{\lambda}^{\infty}t  ^{n}e^{-t}dt

    This right side is called the incomplete gamma function

    It is so cool how all this stuff ties together.
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  5. #5
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    Hi

    Correct me if i'm wrong.. you started with a probability mass function which is given by

    <br /> <br />
b(x;n,p)=\binom{n}{x}\left(\frac{\lambda}{n}\right  )^{x}\left(1-\frac{\lambda}{n}\right)^{n-x}<br />


    the probability generating function on the other hand, is given by G(z) = E(z^X) which results in G(z) = (pz + 1- p)^n for the binomial with parameters n,p. G(z) = e^λ(z-1) for a poisson.

    I'm thinking the problem requires the equivalence of both functions as n tends to infinity and p tends to zero but I'm not confident of taking both limits (simultaneously?) without making major conceptual errors.

    Really hope I've got this right please let me know if I'm mistaken.
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