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Math Help - 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 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 \lambda the following

    \sum_{x=0}^{R-1}\frac{e^{-\lambda}\lambda^{x}}{x!}=k\sum_{x=R}^{\infty}%<br />
\frac{e^{-\lambda}\lambda^{x}}{x!}

    for a given 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 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 \lambda the following

    \sum_{x=0}^{R-1}\frac{e^{-\lambda}\lambda^{x}}{x!}=k\sum_{x=R}^{\infty}%<br />
\frac{e^{-\lambda}\lambda^{x}}{x!}

    for a given R.

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

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

    Then you have:

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

    Therefore \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

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

    (where \Gamma is the "upper" incomplete gamma function).
    At this point, for me, is quite impossible to solve the previous for \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
    \sum_{x=0}^{R-1}\frac{e^{-\lambda}\lambda^{x}}{x!}=\frac{k}{k+1}
    is, actually, the same as
    Q(R,\lambda)=\frac{k}{k+1}
    where 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
    \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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