How to compute means μ1 and μ2 knowing sum of Skellam distributions f(k;μ1,μ2) and sum μ1+μ2, where k is from 2 to n?

Question

The probability mass function for the Skellam distribution for a count difference $ k=n_1-n_2 $ from two Poisson-distributed variables with means $\mu_1$ and $\mu_1$ is given by:

$$ f(k;\mu_1,\mu_2)= e^{-(\mu_1+\mu_2)} \left({\mu_1\over\mu_2}\right)^{k/2}I_{|k|}(2\sqrt{\mu_1\mu_2}) $$ where $I_k(z)$ is the modified Bessel function of the first kind.

I know the sum of these functions for k from 2 to n $$ \sum_{k=2}^n f(k;\mu_1,\mu_2) = a$$

I also know the sum $\mu_1 + \mu_2 = \mu$. So i can substitute $\mu_1$ for $\mu - \mu_2. $ So i have $$ a = \sum_{k=2}^n e^{-(\mu_1+\mu_2)} \left({\mu_1\over\mu_2}\right)^{k/2}I_{|k|}(2\sqrt{\mu_1\mu_2}) = \sum_{k=2}^n e^{-(\mu)} \left({\mu - \mu_2\over\mu_2}\right)^{k/2}I_{|k|}(2\sqrt{(\mu - \mu_2)\mu_2})$$

My question is how can i compute $\mu_2$ from this equation if it's possible.

Answer 1

Here is its numerical solution for required values of parameters done with Maple:

a := .22: n := 5: mu := 3.18:
 fsolve(a = add(exp(-mu)*((mu-mu2)/mu2)^(k/2)*BesselI(k, 2*sqrt((mu-mu2)*mu2)), k = 1 .. n));

$2.002275629 $