Context for my question: For one part of my thesis I try to find an upper bound for the error in the normal approximation of the binomial distribution following the standard proof of the De Moivre–Laplace theorem with Stirling's formula. To make it concrete: Let $B_n$ be binomially distributed and let $N$ have the standardized normal distribution. I want to find an upper bound for $$\epsilon_n = \sup_{a<b} \left|\mathcal P\left(a\le \frac{B_n-np}{\sqrt{np(1-p)}} \le b\right)-\mathcal P(a \le N \le b)\right|$$
I want to compare this error with the best known error estimation of the Berry-Essee theorem for the binomial distribution. So far I have only found a proof which shows that $\epsilon_n \in O\left(\frac 1{\sqrt n}\right)$. See this proof by Márton Balázs and Bálint Tóth (which also just considered $\left|\mathcal P\left(a\le \frac{B_n-np}{\sqrt{np(1-p)}} \le b\right)-\mathcal P(a \le N \le b)\right|$ without the supremum). Other proofs do not investigate the error at all (see for example this proof on Wikipedia).
My Question: Do you know any proof in a textbook / paper / article where the theorem by De Moivre and Laplace is proved with Stirling's formula and the total error is estimated? The value of any occurring constants in the error estimate shall also be calculated. Can you point me to this proof?
Update: I reasked the question on MO, see https://mathoverflow.net/questions/219253/finding-an-error-estimation-for-the-de-moivre-laplace-theorem-with-stirlings-fo
Best Answer
In fact, it follows form Balazs and Toth's paper that $$ \sup_{a_n\le c<d\le b_n}\left|\mathcal P\left(c\le \frac{B_n-np}{\sqrt{np(1-p)}} \le d\right)-\mathcal P(c \le N \le d)\right| \le C \frac{|a_n|^3+|b_n|^3+1}{\sqrt{n}}. $$
On the other hand, by Chernoff's (or Berstein's, or Hoeffding's) inequality, for $x\ge b_n>p$ we have $$ \mathcal P\left(\frac{B_n-np}{\sqrt{np(1-p)}} \ge x\right) \le Ke^{-Kx^2}\le Ke^{-K b_n^2}, $$ and a similar inequality may be written for $x\le a_n<-p$. Moreover, $\mathcal P\left(|N| \ge x\right)\le Me^{-Mx^2}$. Therefore, taking $a_n = - c\sqrt{\log n}$, $b_n = c\sqrt{\log n}$ with $c>0$ large enough, we get $$ \sup_{a<b} \left|\mathcal P\left(a\le \frac{B_n-np}{\sqrt{np(1-p)}} \le b\right)-\mathcal P(a \le N \le b)\right| = \mathcal O\left(\frac{\log^{3/2} n}{\sqrt{n}}\right), $$ which is not sharp, of course, but at least uniform.