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The Poisson cdf can be written as $F(k, \lambda) = e^{-\lambda} \sum\limits_{i = 0}^{k}\frac{\lambda^{i}}{i!}$. I was wondering if it was possible to solve this equation for $\lambda$? That is, I know $k$ and the value of $F$, but I want to know what value of $\lambda$ makes the equation equal $F$.

Thanks.

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Shouldn't the lower limit of the sum be $i=0$? You can solve it numerically, but not (unless k=0) algebraically. –  Ross Millikan Sep 28 '11 at 17:05
    
Do you know of any resource that can give me more information on how to go about solve it numerically? –  Erich Peterson Sep 28 '11 at 17:08
    
Any numerical analysis book. I like Numerical Recipes, nr.com, chapter 9. Obsolete versions are free on-line –  Ross Millikan Sep 28 '11 at 17:12

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up vote 4 down vote accepted

$F(k,\lambda) = Q(k+1,\lambda)$ (see, for example, here), where $Q(s,x) = \Gamma(s,x)/\Gamma(s)$ is the regularized incomplete gamma function. This is a well-known enough special function that some common software packages have an implementation for its inverse in $x$. For example, Mathematica uses the command InverseGammaRegularized. Wolfram|Alpha uses the same command and is freely available, so that may be preferred.

As an illustration, suppose $F = 0.5$ and $k = 10$. Wolfram|Alpha says that InverseGammaRegularized[11,0.5] = 10.668522..., and we can verify this by calculating $$e^{-10.6685} \sum_{i=0}^{10} \frac{10.6685^i}{i!},$$ which Wolfram|Alpha says is $0.5$.

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Should you need to do things yourself... you can also start up with Wilson-Hilferty and polish off with Newton-Raphson. –  J. M. Sep 28 '11 at 17:40

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