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Poisson distibution

  1. Apr 14, 2006 #1
    Hi do u know if the poisson distribution has always the same value for EX(mean value) and variance?
  2. jcsd
  3. Apr 14, 2006 #2


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    Do you mean "how do you know that the mean and variance of a Poisson distribution are the same"? Do the math!

    For given parameter, [itex]\lambda[/itex], the Poisson Distribution is
    [tex]P_\lambda(n)= \lambda^n \frac{e^{-\lambda}}{n!}[/tex]
    where n can be any positive integer.
    The mean is given by
    [tex]\Sigma_{n=1}^\infty \lambda^n \frac{e^{-\lambda}}{(n-1)!}[/tex]
    [tex]= \lambda e^{-\lambda}\Sigma_{n=1}^\infty \frac{\lambda^{n-1}}{(n-1)!}[/tex]
    and taking j= n-1,
    [tex]= \lambda e^{-\lambda}\Sigma_{j= 0}^\infty \frac{\lambda^j}{j!}[/tex]
    It is easy to recognise that sum as Taylor's series for [itex]e^\lambda[/itex] so the sum is just [itex]\lambda[/itex].

    The variance is given by
    [tex]\Sigma_{n=1}^\infty e^{-\lambda}n^2 \frac{\lambda^n}{n!}- \lambda^2[\tex]
    [tex]= e^{-\lambda}\Sigma_{n=1}^\infty \frac{n^2\lambda^n}{n!}-\frac{n\lambda^n}{n!}+ \frac{n\lambda^n}{n!}-\lambda^2[/tex]
    [tex]= e^{-\lambda}\Sigma_{n=1}^\infty \frac{n(n-1)\lambda^n}{n!}+ \frac{n\lambda^n}{n!}[/tex]
    [tex]= \lambda^2 e^{-\lambda}\Sigma_{n=2}^\infty \frac{\lambda^{n-2}}{(n-2)!}+ \lamba e^{-\lambda}\Sigma_{n=1}^\infty \frac{\lambda^{n-1}}{(n-1)!}- \lambda^2[/tex]
    Now we can recognize both of those sums as Taylor's series for [itex]e^{\lamba}[/itex] and so the variance is [itex]\lambda^2+ \lambda- \lamba^2= \lambda[/itex].
    Last edited by a moderator: Apr 14, 2006
  4. Apr 18, 2006 #3
    Thx a lot really .. u seem to be really pro :)
    I have read somewhere that sometimes we can assume that a poisson distribution is the same as the gaussian one
  5. Apr 18, 2006 #4


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    Poisson dist. can be approximated by Gaussian when the mean is large (compared to 1). However, Poisson is discrete, while Gaussian is continuous.
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