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Simplifying Summation and Factorial

  1. Mar 19, 2008 #1
    I was looking at the web page containing a derivation for the Poisson distribution:


    which derives it as the limiting case of the binomial distribution. There is a simplification step which I am missing, which is the step(s) between



    =\lim_{n\to\infty} \underbrace{\left({n \over n}\right)\left({n-1 \over n}\right)\left({n-2 \over n}\right) \cdots \left({n-k+1 \over n}\right)}\ \underbrace{\left({\lambda^k \over k!}\right)}\ \underbrace{\left(1-{\lambda \over n}\right)^n}\ \underbrace{\left(1-{\lambda \over n}\right)^{-k}}

    Does the main simplification come from:

    \frac{n!}{k!(n-k)!} \left( \frac{\lambda}{n} \right)^k


    Last edited: Mar 19, 2008
  2. jcsd
  3. Mar 19, 2008 #2
    No, too complicated. The "simplifications" they have done are
    1) split the last factor in two with powers n and -k
    2) move 1/k! into the 2nd to last factor
    3) write n!/(n-k)! explicitly as k terms while stealing the (1/n)^k from the 2nd to last term.
  4. Mar 19, 2008 #3


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    If k is fixed, (n/n)....((n-k+1)/n) ->1, (1-L/n)n) -> e-L, and (1-L/n)-k ->1.

    This results in the Poisson term for k.
  5. Mar 20, 2008 #4


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    [tex]\frac{n!}{k!(n-k)!}\left(\frac{\lambda}{n}\right)^n= \frac{\lambda^k}{k!}\frac{n!}{(n-k)!n^k}= \frac{\lambda^k}{k!}\frac{n(n-1)\cdot\cdot\cdot(n-k+1)}{n^k}= \frac{\lambda^n}{k!}\frac{n}{n}\frac{n-1}{n}\cdot\cdot\cdot\frac{n-k+1}{n}[/tex]
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