If $S_n$ is Binomial $(n,p)$ then $\mathbb P(S_n=k)\approx \frac{(np)^k}{k!}e^{-np}$.

Question

I was reading this post, and I have to admit that I was quite confused.

The question was : If $S_n$ is a Binomial r.v. with parameter $(n,p)$ s.t. $n$ large, $p$ very small and $np$ not to big (for instance $np\leq 10$), then $$\mathbb P(S_n=k)\approx \frac{(np)^k}{k!}e^{-np}.$$

---

What I completely agree is (using notation of the link I put) if $(B_m)$ is a sequence of $Binomial(m,p_m)$ where $\lim_{m\to \infty }mp_m=\lambda $, then $$\lim_{m\to \infty }\mathbb P(B_m=k)=\frac{\lambda ^k}{k!}e^{-\lambda }.$$ I can prove it without any problem. Now, if $np\leq 10$, $n$ big and $p$ small, I'm indeed confuse with $\mathbb P(S_n=k)\approx \frac{(np)^k}{k!}e^{-(np)}$.

Atempts

Let $n\in\mathbb N$ large and $p$ small s.t. $np\leq 10$. I set $\lambda =np$. Then, define the sequence $p_m=\frac{\lambda }{m}$, i.e. $mp_m=\lambda $ for all $m$. So now, $\mathbb E[S_n]=\mathbb E[B_m]$ for all $m$ and if $p_m$ is very small, then $p_m\approx p$ and thus $$\text{Var}(S_n)=np(1-p)=mp_m(1-p)\underset{(*)}{\approx} mp_m(1-p_m)=\text{Var}(B_m).$$

Therefore, if $m$ is big enough, then $B_m$ and $S_n$ are Binomial distributed with same expectation and very close variance.

Q1) Does this implies that $$\mathbb P(S_n=k)\approx \mathbb P(B_m=k) \ \ ?$$ i.e. that a Binomial is uniquely determined by its variance and expectation ?

Q2) In what the fact that $np\leq 10$ is relevant ?

I hope my question is clear, and if not, please let me know.

Answer 1

Using, as in the linked post, $p=\frac \lambda n$ $$A=p^k \binom{n}{k}p^k (1-p)^{n-k}=\binom{n}{k} \left(\frac{\lambda }{n}\right)^k \left(1-\frac{\lambda}{n}\right)^{n-k}$$ Taking logarithms and expanding as a Taylor series for large values of $n$ to get $$\log(A)=\left(k \log (\lambda )+\log \left(\frac{e^{-\lambda }}{k!}\right)\right)+\frac{-k^2-\lambda ^2+2 \lambda k+k}{2 n}+O\left(\frac{1}{n^2}\right)$$ Continuing with Taylor $$A=e^{\log(A)}=\frac{e^{-\lambda } \lambda ^k}{k!}\left(1+\frac{-k^2-\lambda ^2+2 \lambda k+k}{2 n} \right)+O\left(\frac{1}{n^2}\right)$$ that is to say $$A=\frac{e^{-\lambda } \lambda ^k}{k!}+O\left(\frac{1}{n}\right)$$ Back to $\lambda=pn$, $$A=\frac{ (n p)^k}{k!}e^{-n p}+O\left(\frac{1}{n}\right)$$

Answer 2

I think you need an extra assumption here, like $k=o(\sqrt n)$. Under this assumption, you have $kp=o(1)$, and then $$ \binom nk=\frac{n(n-1)\dotsb(n-(k-1)}{k!} = (1+o(1))\frac{n^k}{k!}, $$ $$ (1-p)^n = (1+o(1)) e^{-np}, $$ and $$ (1-p)^k = 1+o(1). $$ All this is not completely trivial, but not difficult either; say, the first estimate follows from \begin{multline*} n^k \ge n(n-1)\dotsb(n-(k-1)> \left(1-\frac kn\right)^k n^k \\ = e^{k\log(1-k/n)} n^k = e^{o(1)}n^k = (1+o(1))n^k \end{multline*} in view of $\log(1-k/n)=(1+o(1))(-k/n)=o(1/k)$.

As a result, $$ \mathbb P(S_n=k) = \binom nk p^k(1-p)^{n-k} = (1+o(1))\frac{n^k}{k!}\cdot p^k\cdot e^{-np} = (1+o(1))\frac{(np)^k}{k!}\,e^{-np}. $$

The assumption $k=o(\sqrt n)$ is essential and cannot be dropped: say, the approximation $\mathbb P(S_n=k)\approx \frac{(np)^k}{k!}e^{-np}$ is wrong if $k=n$.