Poisson parameter equal to product of binomial parameters

Question

I'm wondering why the parameter $\lambda$ of a Poisson distribution ($Poisson(\lambda)$) is equal to the product between the parameters of a binomial distribution ($Bin(n, \, p)$):

$$ \lambda = n \, p $$

Answer 1

Formally speaking, the parameter of a Poisson random variable is not equal to the product of parameters of the Binomial. It is do it this way because Poisson distribution can be considered as an approximation to Binomial for $n$ big and $p$ small enough so that $\lambda = np$ be of moderate size. Indeed, it can be proven that if $X$ has binomial distribution $Bin(n,p)$, and $\lambda=np$ then $$P(X=k)=\binom{n}{k}\bigg(\frac{\lambda}{n}\bigg)^k\bigg(1 - \frac{\lambda}{n}\bigg)^{n-k}\to e^{-\lambda}\frac{\lambda^k}{k!}$$ when $n\to\infty$.