Deriving condition to get correct asymptotic bound

Question

Suppose that $X\sim \text{Bin}(n,\theta)$. Note that $X$ is the sum of $n$ $iid$ Bernoulli($\theta$) random variables. By the local limit theorem (Theorem 7 here) for the sum of discrete random variables, $$ P(X=t)=\frac{1}{\sqrt{2\pi n\theta(1-\theta)}}\exp\left(-\frac{(t-n\theta)^2}{2n\theta(1-\theta)} \right)+o(n^{-1/2}) $$ for all $n\geq 1$ and uniformly in the integers $t$.

Suppose $t=n\theta+\sqrt{2\theta(1-\theta)n\log m+O(1)}$. I am interested in the relationship between $n$ and $m$, where we can assume, $m>>n>>1$. For example, in my application, $m\approx 20000$ and $n\approx 200$ seems to work well. Intuitively, $m$ should grow much faster than $n$.

I'm interested in finding a theoretical relationship between $n$ and $m$ such that quantity $mP(X=t)=O(1)$. Now if the $o(n^{-1/2})$ remainder were not there, then I can reason the following,

\begin{align*} mP(X=t)&=O(mn^{-1/2}\exp(-O(\log m)))\\ &=O\left(\exp\left(\log m-\log n^{1/2}-O(\log m)\right)\right)\\ &= O\left(\exp\left(\log m^{r}-\log n^{1/2}\right)\right)\text{ for some constant $r>0$}\\ &= O\left(\frac{m^{r}}{n^{1/2}}\right) \end{align*}

This suggests that $m\leq n^\gamma$, where $\gamma = \frac{1}{2r}>0$ for $r>0$ can be a reasonable relationship.

How do I handle that remainder $o(n^{-1/2})$?

Answer 1

To get what you want, you can use the refinement of the local central limit theorem due to Esseen, Theorem 5, page 63, which in your case yields \begin{align} P(X=t)&=\frac1{\sqrt{npq}}\phi(x)\Big(1+\frac1{\sqrt n}\,Q_k(x,1/\sqrt n)\Big)+o(1/n^{(k-1)/2}) %\\ \tag{1} \end{align} as $n\to\infty$ uniformly in integer $t$, where $p:=\theta\in(0,1)$, $q:=1-p$, $\phi(x):=\frac1{\sqrt{2\pi}}\,e^{-x^2/2}$, \begin{equation} x:=\frac{t-np}{\sqrt{npq}}, \end{equation} $k$ is any integer $\ge3$, and $Q_k(x,1/\sqrt n)$ is a polynomial in $x,1/\sqrt n$.

Suppose now that $m=n^c$, where $c=c(n)$ varies with $n$ so that $c\to c_0$ for some real $c_0>0$. Then for your $t=np+\sqrt{2pqn\ln m+O(1)}$ and $x$ as above we have \begin{equation} x=\sqrt{2\ln m+O(1/n)}=\sqrt{2c\ln n+o(1)}=o(n^a) \tag{2} \end{equation} for any real $a>0$, whence $\frac1{\sqrt n}\,Q_k(x,1/\sqrt n)=o(1)$ in (1).

Using the first equality in (2), we also have
\begin{equation} \phi(x)\le e^{-x^2/2}\le\frac{1+O(1/n)}m=\frac{1+o(1)}m. \end{equation} Using now (1) with $k>2c_0+2$ (so that $n^{(k-1)/2}\ge n^{c+1/2}=m\sqrt n$ for all large enough $n$), we have
\begin{equation} mP(X=t)=\frac m{\sqrt{npq}}\phi(x)(1+o(1))+o(m/n^{(k-1)/2}) =O\Big(\frac1{\sqrt{n}}\Big)+o\Big(\frac1{\sqrt{n}}\Big)=O\Big(\frac1{\sqrt{n}}\Big), \end{equation} which is better than the $O(1)$ that you desired.