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Thread: Poisson Distribution

  1. #1
    Super Member Aryth's Avatar
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    Poisson Distribution

    This problem considers the Poisson distribution, a probability distribution for a discrete random variable which was first used by Simeon-Denis Poisson to describe seemingly random criminal events in 1837 in Paris. If independent events have a constant tendency to occur and if the average rate of occurrence is a, then the probability that n events actually occur is given by

    $\displaystyle p_n = \frac{e^{-a}a^n}{n!}$ with $\displaystyle n = 0,1,2,...,\infty$

    (a) By noting that

    $\displaystyle e^{+a} = 1 + \frac{a}{1!} + \frac{a^2}{2!} + \frac{a^3}{3!} + ...$

    Show that

    $\displaystyle \sum_{n=0}^{\infty} p_n = 1$

    thereby verifying that the poisson distribution is normalized.

    I've done this already

    (b) By using $\displaystyle \frac{n}{n!} = \frac{1}{(n-1)!}$ and $\displaystyle a^n = aa^{n-1}$, show that

    $\displaystyle \sum_{n=0}^{\infty} np_n = a$

    thereby verifying that the average rate of occurrence, or the expectation value $\displaystyle <n>$, is equal to a.

    I've done this one as well

    (c) By using similar techniques, find $\displaystyle <n^2>$ and show, using $\displaystyle Std.Dev. = \Delta n$ and

    $\displaystyle (\Delta n)^2 = <n^2> - <n>^2$

    that the standard deviation of the poisson distribution is given by

    $\displaystyle \Delta n = \sqrt{a}$

    This is the one I need help with
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  2. #2
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    mr fantastic's Avatar
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    Quote Originally Posted by Aryth View Post
    This problem considers the Poisson distribution, a probability distribution for a discrete random variable which was first used by Simeon-Denis Poisson to describe seemingly random criminal events in 1837 in Paris. If independent events have a constant tendency to occur and if the average rate of occurrence is a, then the probability that n events actually occur is given by

    $\displaystyle p_n = \frac{e^{-a}a^n}{n!}$ with $\displaystyle n = 0,1,2,...,\infty$

    (a) By noting that

    $\displaystyle e^{+a} = 1 + \frac{a}{1!} + \frac{a^2}{2!} + \frac{a^3}{3!} + ...$

    Show that

    $\displaystyle \sum_{n=0}^{\infty} p_n = 1$

    thereby verifying that the poisson distribution is normalized.

    I've done this already

    (b) By using $\displaystyle \frac{n}{n!} = \frac{1}{(n-1)!}$ and $\displaystyle a^n = aa^{n-1}$, show that

    $\displaystyle \sum_{n=0}^{\infty} np_n = a$

    thereby verifying that the average rate of occurrence, or the expectation value $\displaystyle <n>$, is equal to a.

    I've done this one as well

    (c) By using similar techniques, find $\displaystyle <n^2>$ and show, using $\displaystyle Std.Dev. = \Delta n$ and

    $\displaystyle (\Delta n)^2 = <n^2> - <n>^2$

    that the standard deviation of the poisson distribution is given by

    $\displaystyle \Delta n = \sqrt{a}$

    This is the one I need help with
    Read this thread: http://www.mathhelpforum.com/math-he...sson-dist.html
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  3. #3
    Super Member Aryth's Avatar
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    Ah, thanks.
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  4. #4
    MHF Contributor matheagle's Avatar
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    There's a nice trick here.

    Instead of obtaining directly $\displaystyle E(X^2)$, find $\displaystyle E(X(X-1))$ instead.

    $\displaystyle E(X(X-1)) =\sum_{n=0}^{\infty}{n(n-1)e^{-a}a^n\over n!} =e^{-a}\sum_{n=2}^{\infty}{a^n\over (n-2)!} $

    $\displaystyle =e^{-a}a^2\sum_{n=2}^{\infty}{a^{n-2}\over (n-2)!}=e^{-a}a^2 e^a=a^2 $.

    Thus $\displaystyle E(X^2)-E(X) =a^2 $, so $\displaystyle E(X^2)=E(X)+a^2=a+a^2 $.

    Hence $\displaystyle V(X)=E(X^2)-(EX)^2 =a+a^2-a^2=a $.
    Last edited by matheagle; Jun 3rd 2009 at 10:38 PM.
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  5. #5
    Super Member Aryth's Avatar
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    Cool, I knew that... But it isn't statistics class (as you can see from the notation) so most of it was worked with $\displaystyle E(X^2)$. That is useful though, thanks.
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