I'm looking for a proof as to why the Poisson-Binomial distribution can be approximated by the Poisson distribution.
A Poisson-Binomial distribution is a sum of independent Bernoulli trials that are not necessarily identically distributed: https://en.wikipedia.org/wiki/Poisson_binomial_distribution
There's a Poisson limit theorem on Wikipedia (https://en.wikipedia.org/wiki/Poisson_limit_theorem) that gives me a solution for the binomial distribution.
The proof of the limit theorem starts with: $\lim_{n \to \infty} {n\choose k} p^k (1-p)^{n-k} = \lim_{n \to \infty} {n\choose k} (\frac{\lambda}{n})^k (1- \frac{\lambda}{n})^{n-k}$.
It uses the fact that $ np = \lambda$ (n trials with possibility p) and therefore $ p = \frac{\lambda}{n} $ and works from that, showing that $\lambda$ works out as the Poisson parameter. In the case of the Poisson-Binomial distribution, the expected value is: $\lambda = \sum_{i=1}^{n} p_i$, where $p_i$ are the probabilities for the different Bernoulli experiments.
I'm not sure how I can use $\lambda = \sum_{i=1}^{n} p_i$ in the proof from Wiki. Or is there an entirely different way?
Thanks in advance!