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Context: Let $\tilde B_n$ be standardized binomial distributed with $p\in(0,1)$ be the probability of success in the $n$ Binomial trials. So $P(\tilde B_n = x_k)=\binom nk p^k q^{n-k}$ for $x_k = \frac{k-np}{\sqrt{npq}}$. The mode of $\tilde B_n$ is $x_m$ with $m = \lfloor (n+1)p \rfloor$ or $m = \lfloor (n+1)p \rfloor-1$ [1].

Question: With Stirling's formula we can show $P(\tilde B_n = x_m)\sim \frac1{\sqrt n}$ [2]. Is there an easy proof two show $P(\tilde B_n=x_m) = O\left(\frac 1{\sqrt n}\right)$ without using Stirling's formula.

Reason for my question: I want to prove De Moivre-Laplace in an alternative way with using tools from the theory about ODEs. Thereby I need to show $P(\tilde B_n=x_m) = O\left(\frac 1{\sqrt n}\right)$, but I do not want to use Stirling's formula (because then I could also do the standard proof).

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Stephan Kulla
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1 Answers1

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Let me write $P_k=P(X=x_k)$.

The plan is to take the trivial estimate $1\ge \sum\limits_{k=m}^{m+d} P_k$ with some $d\approx \sqrt{n}$, and replace $P_k$ by $c\cdot P_m$. So, we need a suitable upper bound on $\frac{P_m}{P_k}$.

For $1\le j\le d\le\sqrt{n}$, $$ \frac{P_m}{P_{m+j}} = \frac{\binom{n}{m}p^{m}q^{n-m}}{\binom{n}{m+j}p^{m+j}q^{n-m-j}} = \frac{(m+1)(m+2)\cdots(m+j)}{(n-m)(n-m-1)\cdots(n-m-j+1)} \cdot \frac{q^j}{p^j} = \\ = \frac{\big(1+\frac1m\big)\big(1+\frac2m\big)\cdots\big(1+\frac{j}m\big) }{\big(1-\frac1{n-m}\big)\big(1-\frac2{n-m}\big)\cdots\big(1-\frac{j-1}{n-m}\big)} \cdot \left(\frac{mq}{(n-m)p}\right)^j < \\ < e^{O\big(\frac1m+\frac2m+\cdots+\frac{j-1}{n-m}+\frac1{n-m}+\frac2{n-m}+\cdots+\frac{j-1}{n-m}\big)} \cdot e^{j\cdot O\big(\frac1m+\frac1{n-m}\big)} \le e^{O(d^2/n)} \le O(1). $$ Hence, choosing $d=\lfloor\sqrt{n}\rfloor$, we get $P_{m+j} > c(p) P_m$ and $P_m \le O\left(\frac1{\sqrt{n}}\right)$.

Stephan Kulla
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