I have $k$ Poisson RV, each with mean $\lambda_1, \dots, \lambda_i, \dots, \lambda_k$.
I am drawing without replacement $h$ many out of them.
I then add up the realization of all their outcomes, and wonder about that distribution.
Here's what I have so far. The probability of drawing the combination of any 3 RV in particular order is $(k-h)!/k!$. Denote $I = \{1, \dots k\}$
$$ Prob(i_1 + \dots + i_h = m) = \frac{(k-h)!}{k!} \sum_{i_1 \in I} \sum_{i_2 \in I \setminus i_1} \dots \sum_{i_h \in I \setminus \{i_1, \dots i_{h-1}\}} e^{-(\lambda_{i_1} + \dots \lambda_{i_h})}(\lambda_{i_1} + \dots \lambda_{i_h})^m \frac{1}{m!} $$
While this expression is good at explaining what I'm after, it's not useful for computing the sum and hopefully finding a simpler expression.
Can anyone drop some pointers on an alternative and hopefully better approach?