Poisson limit theorem

Let $B_n$ have the binomial distribution with trial count $m_n\to \infty$ and success Probability $p_n\to 0$. If $m_n{p}_n\to \lambda \in [0,\infty )$, thenThis is the Poisson limit theorem: many independent rare trials retain a finite total expected count. For a proof using characteristic functions,The logarithmic expansion error vanishes because $m_n{p}_n^2=(m_n{p}_n)p_n\to 0$. The limit is the Poisson distribution characteristic function, so the Lévy continuity theorem proves convergence. The case $\lambda =0$ is the degenerate count zero; it also follows directly from Markov's inequality.^[ref]

For example, $\mathrm{Bin}(100,0.01)$ is approximated by $\mathrm{Poisson}(1)$, and its zero-count Probability $0.99^{100}$ is about $0.366$, close to $e^{-1}\approx 0.368$. The limiting mean is the limit of the product, not either individual parameter. independent unequal-probability Bernoulli trials have a related Poisson approximation when their largest success Probability vanishes and their total mean converges; dependence requires additional control, as in the Stein-Chen method.