Binomial to Poisson Approximation : Why does p have to be small

Question

I understandd that as n tends towards to infinity for a Binomial distrobution, it becomes a Poisson distobution and i have completed the proof for this.

However, I am not sure why when approximating, p has to be a relativley small value. Again, i understand why n must be large, but whats the purpose/ proof that the smaller p is, the better the approximation is.

Essentially waht I'm asking is, why does p have to be small when approximamting a Poisson from a binomial.

Could someone please help explain this

Thanks

Answer 1

In the proof, you need $np\to \lambda$. But if $n$ grows large and $np\to\lambda$, then we must have $p\to 0$, or else $np\to\infty$.

Answer 2

Intuitively, for approximating $\operatorname{Binomial}(n,p)\approx\operatorname{Poisson}(\lambda=np)$, the smaller $p$ is, the closer the variance $npq=\lambda(1-p)$ to $\lambda$, so you expect better approximation.

In the proof, you use $$ \frac{n(n-1)\dots(n-k+1)}{n^k}\left(1-\frac{\lambda}{n}\right)^{n-k}\approx e^{-\lambda} $$ to show $$ \binom{n}{k}p^k(1-p)^{n-k}\approx\frac{e^{-np}(np)^k}{k!} $$ If you analyse the error terms more carefully, you get some explicit bounds such as $$ \sum_{k=0}^\infty \left\lvert \binom{n}{k}p^k(1-p)^{n-k}-\frac{e^{-np}(np)^k}{k!} \right\rvert\leq Cp $$ where $C\leq 4$. So this justifies the motto "smaller $p$ gives better approximation".