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

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.

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 $$\lambda=pn$$, $$A=\frac{ (n p)^k}{k!}e^{-n p}+O\left(\frac{1}{n}\right)$$
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 $$k=n$$.