6
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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 $k=n$.

$\endgroup$
  • $\begingroup$ @user659895: Are you following? Is everything clear? Any feedback? $\endgroup$ – W-t-P May 23 at 8:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.