Let $D(\epsilon,C)$ be the collection of all random variables $X$ on $\mathbb{R}$ such that $E[X]=0$, $E[X^2]=1$, and $E[|X|^{2+\epsilon}]\leq C$. Define a function $L_{\epsilon,C}(n)$ by $$L_{\epsilon,C}(n) = \sup_{X\in D(\epsilon,C)} \sup_{x\in\mathbb{R}} \left\vert P\left(\frac{X_1+X_2+\ldots+X_n}{\sqrt{n}}<x\right)-P\left(Z < x\right)\right\vert$$ where $X_i$ are independent copies of $X$, and $Z$ is a standard normal.

Having made these definitions, the Berry-Esseen theorem tells us that $$L_{1,C}(n) = O(C n^{-1/2})$$ It is also not hard to use the Lindeberg-Feller theorem to show that for any fixed $\epsilon>0$ and $C\in \mathbb{R}$ $$\lim_{n\to\infty} L_{\epsilon,C}(n) = 0$$ Which leads to my question -- can we get any more quantitative result than that this goes to zero, without assuming $\epsilon=1$? Or, in the other direction, is it known that we can't?


1 Answer 1



To answer your question, one can use e.g. a result by Bikelis, which states the following: Let $X_1,\dots,X_n$ be independent zero-mean random variables such that $E|X_i|^{2+\ep}<\infty$ for some $\epsilon\in[0,1]$ and all $i$. Then for all real $x$ \begin{equation} \Big|P\Big(\sum_1^n X_i<x\sqrt B_n\Big)-P(Z<x)\Big| \le \frac A{B_n^{1+\ep/2}(1+|x|)^{2+\ep}}\,\sum_1^n E|X_i|^{2+\ep}, \end{equation} where $A$ is a finite universal constant, $B_n:=\sum_1^n EX_i^2$, and $Z$ is a standard normal random variable.

In you specific case, it follows that $L_{\epsilon,C}\le AC/n^{\epsilon/2}$.


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