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Thread: Poisson distribution parameter estimation given a CDF ratio

  1. #1
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    Poisson distribution parameter estimation given a CDF ratio

    Dear all,
    a particular estimation problem.
    I'm considering a Poisson distribution and I need to estimate it's parameter knowing the ratio $\displaystyle k$ between the sum of its values at the left and on the right of a given point. Just to simplify my question, I need to solve for $\displaystyle \lambda$ the following

    $\displaystyle \sum_{x=0}^{R-1}\frac{e^{-\lambda}\lambda^{x}}{x!}=k\sum_{x=R}^{\infty}%
    \frac{e^{-\lambda}\lambda^{x}}{x!}$

    for a given $\displaystyle R$.

    Some ideas to solve this?
    Thanks in advance!
    Simo
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  2. #2
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    Quote Originally Posted by Simo View Post
    Dear all,
    a particular estimation problem.
    I'm considering a Poisson distribution and I need to estimate it's parameter knowing the ratio $\displaystyle k$ between the sum of its values at the left and on the right of a given point. Just to simplify my question, I need to solve for $\displaystyle \lambda$ the following

    $\displaystyle \sum_{x=0}^{R-1}\frac{e^{-\lambda}\lambda^{x}}{x!}=k\sum_{x=R}^{\infty}%
    \frac{e^{-\lambda}\lambda^{x}}{x!}$

    for a given $\displaystyle R$.

    Some ideas to solve this?
    Thanks in advance!
    Simo
    Well here's the start of an idea anyway:

    Let $\displaystyle S = \sum_{x=0}^{R-1}\frac{e^{-\lambda}\lambda^{x}}{x!}$.

    Then you have:

    $\displaystyle S = k(1 - S) \Rightarrow S = \frac{k}{k+1}$.

    Therefore $\displaystyle \sum_{x=0}^{R-1}\frac{e^{-\lambda}\lambda^{x}}{x!} = \frac{k}{k+1}$.
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  3. #3
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    This start is certainly more clever than the work I've done until now...
    anyway, from this point, the only thing I can do is solve the equation in

    $\displaystyle \frac{\Gamma(R,\lambda)}{(R-1)!}=\frac{k}{k+1}$

    (where $\displaystyle \Gamma$ is the "upper" incomplete gamma function).
    At this point, for me, is quite impossible to solve the previous for $\displaystyle \lambda$

    In your opinion there are other way to simplify the series of the first post???

    Thanks for tour attention!
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  4. #4
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    solved...
    to manage the last equation is possible to see that
    $\displaystyle \sum_{x=0}^{R-1}\frac{e^{-\lambda}\lambda^{x}}{x!}=\frac{k}{k+1}$
    is, actually, the same as
    $\displaystyle Q(R,\lambda)=\frac{k}{k+1}$
    where $\displaystyle Q(R,\lambda)$ is the gamma regularized function. Even if it is not a "clean" approach, it can be numerically inverted thanks to theBoost C++ Libraries through the gamma_q_inv(.) function giving
    $\displaystyle \lambda=Q^{-1}\left( R,\frac{k}{k+1}\right)$

    I know, it is not a great solution...but is more than nothing

    Thank you all!
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