I would like to know the computational complexity of the following formula for the variance of thermal photon statistics.
$$P(n)=\sum_{d=1}^D\prod_{m=1}^M\frac{1}{(1+\langle n_m \rangle)(1+\langle n_m \rangle^{-1})^{n_{m,d}}}$$$$\langle n\rangle=\sum_{n=0}^\infty nP(n)$$
$$\left\langle (\Delta n)^2\right\rangle=\sum_{n=0}^\infty \left(n-\langle n\rangle\right)^2P(n)$$
This will take some explanation so here goes, $n$ is the number of photons and is the discrete random variable, $m$ is the mode index, $M$ is the total number of modes and can be anything from $1\to\infty$, $d$ is the distribution index, $D=\frac{(n+M-1)!}{n!(M-1)!}$ and is the total number of possible distributions, $\langle n_m\rangle$ is the mode distribution - for example if $\langle n_0\rangle=3$ that means there are 3 photons on average in mode $m=0$, $n_{m,d}$ is the number of photons in mode $m$ and distribution $d$.
The problem is given $M\ge1$ and the set of average photon populations $\langle n_m\rangle$ what is the variance?
As $M$ and increases in size the problem becomes more and more complex; $n_{m,d}$ is essentially a $M\times D$ array for every $n$.
Is finding the variance (the third equation) a P, NP, EXPTIME or some other type of problem?
EDIT: It appears to me that this problem is similar to Chess in that finding a solution takes a lot of time, as well as checking a given solution. Also looking into the computational complexity it seems that the third equation gets more complex as $\sim O(k^N)$ where $N$ replaces infinity in the second and third equation, and $M$ is set to $N$. $k\approx 4$ for small $N$ but slowly gets smaller for larger $N$.