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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

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

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!

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Edgeworth expansions give extra terms that can be used to estimate the rate of convergence. – Brendan McKay Feb 16 '14 at 8:14
Oh, never heard about these. Indeed they look very relevant. Thanks! – Daniel Soudry Feb 17 '14 at 1:59
ALso, if you can hold of P. Hall, Rates of convergence in the central limit theorem, there seems to be quite a lot of theory that is relevant. – Brendan McKay Feb 17 '14 at 3:32
Thanks again. For completeness, I will summarize my conclusions from the answer and all the helpful comments: 1) As a result of the Edgeworth expansion ( The convergence rate is $n^{-1/2}$, unless the third cumulant of the distribution is zero (as for the uniform distribution). In that case the convergence rate is $n^{-1}$, unless other cumulants are zero... and so on. 2) The reason that the approximation is bad at the tails is due to its relative precision - since the tails are usually are very small. – Daniel Soudry Feb 17 '14 at 18:20
Careful, Davide's example has zero third moment and has convergence rate $n^{-1/2}$. Probably you are looking at an Edgeworth expansion that excludes lattice distributions. There is a different expansion for those. – Brendan McKay Feb 18 '14 at 7:57
up vote 10 down vote accepted

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 $n^{-1/2}$.

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Nice answer - thank you! – Daniel Soudry Feb 15 '14 at 21:11
I believe you should expect faster convergence if you rule out the lattice distributions, though these are limits of non-lattice distributions. – Douglas Zare Feb 15 '14 at 22:03
Hi Douglas - sounds interesting. Care to explain why this is so (or point to some reference)? – Daniel Soudry Feb 15 '14 at 23:11
@Daniel Soudry: See… – Douglas Zare Feb 17 '14 at 7:05
Thanks Douglas, very relevant post. – Daniel Soudry Feb 17 '14 at 19:28

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