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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

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!

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    $\begingroup$ Edgeworth expansions give extra terms that can be used to estimate the rate of convergence. $\endgroup$
    – Brendan McKay
    Commented Feb 16, 2014 at 8:14
  • $\begingroup$ Oh, never heard about these. Indeed they look very relevant. Thanks! $\endgroup$
    – Daniel Soudry
    Commented Feb 17, 2014 at 1:59
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    $\begingroup$ 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. $\endgroup$
    – Brendan McKay
    Commented Feb 17, 2014 at 3:32
  • $\begingroup$ 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 (en.wikipedia.org/wiki/Edgeworth_series): 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. $\endgroup$
    – Daniel Soudry
    Commented Feb 17, 2014 at 18:20
  • $\begingroup$ 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. $\endgroup$
    – Brendan McKay
    Commented Feb 18, 2014 at 7:57

1 Answer 1

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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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    $\begingroup$ I believe you should expect faster convergence if you rule out the lattice distributions, though these are limits of non-lattice distributions. $\endgroup$
    – Douglas Zare
    Commented Feb 15, 2014 at 22:03
  • $\begingroup$ Hi Douglas - sounds interesting. Care to explain why this is so (or point to some reference)? $\endgroup$
    – Daniel Soudry
    Commented Feb 15, 2014 at 23:11
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    $\begingroup$ @Daniel Soudry: See stats.stackexchange.com/questions/30468/… $\endgroup$
    – Douglas Zare
    Commented Feb 17, 2014 at 7:05

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