# Thread: Poisson distribution parameter estimation given a CDF ratio

1. ## 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

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

for a given $R$.

Some ideas to solve this?
Simo

2. Originally Posted by Simo
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

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

for a given $R$.

Some ideas to solve this?
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 $\sum_{x=0}^{R-1}\frac{e^{-\lambda}\lambda^{x}}{x!} = \frac{k}{k+1}$.

3. 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!

4. 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!