Context of Poisson distribution: $n \rightarrow \infty$ and $\lambda := np$ where $p \in [0,1]$ can be thought of as probability of success, $n$ is the number of chances, and $\lambda$ as the average or expected value of success.
You can see the entire proof of turning a binomial distribution's probability mass function into Poisson's here in page 3, though I don't think you need to read it for this question:
https://mbernste.github.io/files/notes/Poisson.pdf
A key step within this proof is this relation:
$$e^{-\lambda} = \lim_{n\rightarrow \infty} \left(1-\frac{\lambda}{n}\right)^n$$
I understand this equality and its proof when $n$ and $\lambda$ are independent to each other, but have trouble accepting the equality when there is a relation, namely $\lambda = np$. Assuming $p$ is fixed and non-zero, $z:= -\frac{n}{\lambda}$, and using the limit definition of $e$, the proof for this equality goes like this:
$$\lim_{n\rightarrow \infty} \left(1 – \frac{\lambda}{n}\right)^n = \lim_{n\rightarrow \infty} \left(1 + \frac{1}{\left(-\frac{n}{\lambda}\right)}\right)^{\left(-\frac{n}{\lambda}\right)\left(-\lambda\right)}=\lim_{z\rightarrow \infty}\left(1+\frac{1}{z}\right)^{z\left(-\lambda\right)} = e^{-\lambda}$$
I feel this proof is incomplete or has some errors, specifically:
- Since $z := -\frac{n}{\lambda}$, shouldn't the limit be $z \rightarrow -\infty$ instead? (The equality remains true)
- Since $n \rightarrow \infty$ and $\lambda := np$, shouldn't $\lambda \rightarrow \infty$ as well such that $e^{-\lambda} = 0$?
Best Answer
What you have is a special case of the Poisson limit theorem.
The parameter $\lambda$ here is a fixed number, and $np=\lambda$ should be understood as $np_n=\lambda$. Note in particular that $p$ is not fixed.