Probability of at least 2 variable is greater than zero in a multivariate Poisson

Question

The problem I am facing...

I have an $N$-variate Poisson distribution, $p(k_1, k_2, ... k_N)$, and I know that the variables are independent, i.e. $p(k_1, k_2, ... k_N) = p(k_1)p(k_2)...p(k_N)$.

I am looking for: The probability that at least 2 of the variables are greater than zero (denote by $P_2$).

For example,

For $N=2$, $$P_2 = p(k_1>0)p(k_2>0)$$

For $N=3$, $$P_2 = p(k_1>0)p(k_2>0)p(k_3>0) + p(k_1>0)p(k_2>0)p(k_3=0) + p(k_1>0)p(k_2=0)p(k_3>0) + p(k_1=0)p(k_2>0)p(k_3>0)$$

I am trying to generalize this to any $N>2$. Note that I can also use the fact that $p(k=0)=e^{-\lambda}$ and $p(k>0) = 1 - e^{-\lambda}$ (although the $\lambda$ for each variable is not assumed to be the same).

I would like a single formula to compute $P_2$. Any suggestions on how to proceed?

Answer 1

Thew complementary event is that $0$ or $1$ variables are greater than $0$, so

\begin{eqnarray} P_2 &=& 1-\prod_{j=1}^Np(k_j=0)-\sum_{m=0}^N\frac{p(k_m\gt0)}{p(k_m=0)}\prod_{j=1}^Np(k_j=0) \\ &=& 1-\exp\left(-\sum_{j=0}^N\lambda_j\right)\left(1+\sum_{m=0}^N\left(\mathrm e^{\lambda_m}-1\right)\right)\;. \end{eqnarray}