TITLE: Convergence rate of the central limit theorem near the center of the distribution
QUESTION [9 upvotes]: I'm looking for fast convergence rates for the central limit theorem - when we are not near the tails of the distribution.
Specifically, from the general convergence rates stated in the Berry–Esseen theorem 
http://en.wikipedia.org/wiki/Berry%E2%80%93Esseen_theorem
we know that, under certain conditions, the cumulative probability distribution of the scaled mean of a random sample $F_n(x)$ converges to the cumulative normal distribution $\Phi(x)$ with a convergence rate of $n^{-1/2}$, where n is the sample size.
However, as stated in
http://en.wikipedia.org/wiki/Central_limit_theorem
it is well known that "As an approximation for a finite number of observations, it [the central limit theorem] provides a reasonable approximation only when close to the peak of the normal distribution; it requires a very large number of observations to stretch into the tails."
Therefore, my question is:
If we are given additional assumption that $|x|<C$, where $C$ is some positive constant, can we improve the $n^{-1/2}$ convergence rate of the the Berry–Esseen theorem?
Thanks in advance!

REPLY [14 votes]: No, even in the most favorable case $(X_i)_{i\geqslant 0}$ iid with $\mathbb P(X_i=1)=\mathbb P(X_i=-1)=1/2$. Denoting $F_n$ the cumulative distribution function of $n^{-1/2}S_n$, we have by symmetry 
$$F_{2n}(0)=\frac 12(1+\mathbb P(S_{2n}=0)).$$
Since $\mathbb P(S_{2n}=0)=\binom{2n}n2^{-2n}$, denoting $\Phi$ the cdf of the standard normal distribution, 
$$\sup_{x\in\mathbb R}|F_{2n}(x)-\Phi(x)|\geqslant |F_{2n}(0)-\Phi(0)|\geqslant \frac 12\cdot \binom{2n}n2^{-2n}.$$
Using Stirling's formula, we obtain that the RHS behave asymptotically like $n^{-1/2}$.