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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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    Any numerical analysis book. I like Numerical Recipes, http://nr.com/, chapter 9. Obsolete versions are free on-line2011-09-28

2 Answers 2

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$F(k,\lambda) = Q(k+1,\lambda)$ (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...](http://dx.doi.org/10.1137/0907069) you can also start up with [Wilson-Hilferty](http://www.johndcook.com/wilson_hilferty.html) and polish off with Newton-Raphson.2011-09-28
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You can use Excel solver in the analysis toolpak add-in. Use the evolutionary feature. Minimize the absolute value of the difference between your desired F and the cell that calculates the CDF of your best guess at lambda. Give it about 30 seconds to run and you'll have your answer.