The Skellam distribution has no closed-form solution for the CDF. I'm in particular interested how $P(x > 0, n, m)$ changes with $n$ and $m$.
That is: How does the probability of a Poisson race being positive change with the two rates.
Denote the two rates by $m$, $n$, and the realizations by $M$, $N$. My approach:
$$ P(M-N > 0) = \sum_{M=0}^\infty \sum_{N=0}^{M-1} Poisson(M, m) Poisson(N, n) \\ = e^{-(n+m)}\sum_{M=0}^\infty \sum_{N=0}^{M-1} \frac{m^M}{M!}\frac{n^N}{N!} $$
Now, the derivative w.r.t $m$:
$$ \frac{\partial}{\partial m} = e^{-(m+n)}\sum_{M=0}^\infty \sum_{N=0}^{M-1} \frac{m^M}{M!}\frac{n^N}{N!}(\frac{M}{m}-1) $$
I know that there is in general no closed-form solution for the CDF, but perhaps there is something more I can do with this derivative to make it neater?