One consequence of the Berry-Esseen approach to the Central Limit theorem, states the following. Let $X_1,\ldots ,X_n$ be real valued random variables, all independent, with ${\mathbb E}(X_i) = 0$, ${\mathbb E}(X_i^2) = \sigma _i^2$ which satisfies $|X_i| \leq c$ for all $i\in [n]$ and $\sum _{i\in [n]} \sigma _i^2 = 1$. Set $S_n = \sum _{i\in [n]} X_i$. Then for all $t \in {\mathbb R}$ we have \begin{equation*} {\mathbb P}(S_n \leq t) = {\mathbb P}({\cal Z} \leq t) + O(c), \end{equation*} where ${\cal Z}$ is a standard normal random variable. My question is simply whether this bound can be improved if we additionally know that ${\mathbb E}(X_i^3) = 0$ for all $i\in [n]$? Intuitively, it seems that this may be improved to \begin{equation*} {\mathbb P}(S_n \leq t) = {\mathbb P}({\cal Z} \leq t) + O(c^2). \end{equation*} Any help / references are appreciated. Thanks!
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$\begingroup$ This is not a central limit theorem type result that allows control by moments(via, say Edgeworth expansion), the nature of its proof will not allow a tighter bound. So could you explain how it is "intuitive"? $\endgroup$– Henry.LCommented Apr 25, 2017 at 21:46
2 Answers
Consider $X_i$ which are equally likely to be $n^{-1/2}$ or $-n^{-1/2}$.
These variables are essentially as nice as you could possibly want in terms of moments: The odd moments are all equal to $0$, and $E(|X_i|^k)$ is as small as possible given the variance.
However, the usual Berry-Esseen inequality is sharp up to constants for them ($S_n$ has jumps corresponding to single values taken on with probability proportional to $n^{-1/2}$, while $\mathcal{Z}$ is continuous).
This suggests that, in general, small moments are not in and of themselves enough to improve Berry Esseen.
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2$\begingroup$ On the other hand, if one looks at smoothed versions ${\bf E} F(S_n) = {\bf E} F({\mathcal Z}) + O(c)$ of Berry-Esseen, then more matching moments definitely help; this is easiest to see using the Lindeberg exchange proof of this smoothed version, see e.g. terrytao.wordpress.com/2010/01/05/… $\endgroup$ Commented Apr 25, 2017 at 23:48
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$\begingroup$ Thanks Kevin and Terry. I feel slightly stupid having missed this, but I was indeed thinking of matching moments in the Lindeberg exchange proof. The difference between discrete and smooth seem to be key here. $\endgroup$ Commented Apr 26, 2017 at 10:43
If the random variables have densities, and the third moment is zero, then (provided the fourth moment is finite) the Berry-Esseen theorem holds with rate $1/N$, see for instance
Edit: Results of this form (often given in terms of Edgeworth-style expansions) are well known in the literature. See for instance Osipov's theorem, Theorem 5.18 of
Petrov, Valentin V., Limit theorems of probability theory. Sequences of independent random variables, Oxford Studies in Probability. 4. Oxford: Clarendon Press. ix, 292 p. (1995). ZBL0826.60001.
Theorem 5.18 involves a bound which is only effective when the characteristic function of the random variables has modulus bounded away from $1$ on $\mathbb{R} - [-\delta,\delta]$. This bound is effective whenever the random variables have densities.
