1
$\begingroup$

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?

$\endgroup$

1 Answer 1

2
$\begingroup$

$\newcommand{\ep}{\epsilon}$

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 $L_{\epsilon,C}\le AC/n^{\epsilon/2}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.