Simplifying Summation and Factorial

  • Thread starter rwinston
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Main Question or Discussion Point

I was looking at the web page containing a derivation for the Poisson distribution:

http://en.wikipedia.org/wiki/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

[tex]
\lim_{n\rightarrow\infty}\frac{n!}{k!(n-k)!}\left(\frac{\lambda}{n}\right)^k\left(1-\frac{\lambda}{n}\right)^{n-k}
[/tex]

and

[tex]
=\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}}
[/tex]

Does the main simplification come from:

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

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

?
 
Last edited:

Answers and Replies

154
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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.
 
mathman
Science Advisor
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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.
 
HallsofIvy
Science Advisor
Homework Helper
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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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