3
$\begingroup$

Let $X_1,\ldots X_n$ be i.i.d. random variables and denote by $S_n$ their sum. Assuming that $\mathbb{E}S_n=0$, $\mathbb{E}S^2_n=1$ and that $\mathbb{E}|X_i|^3=b$, the Berry-Esseen Theorem (in the i.i.d. case) gives us the estimate

$$|P(Z<x)-P(S_n<x)|\leq cb/\sqrt{n},$$

where $c$ is an absolute constant and $Z$ has the standard normal distribution.

I wanted to know what happens if we do not assume finiteness of the $3-$rd or $(2+\varepsilon)$-th moments? Could it still be true that

$$|P(Z<x)-P(S_n<x)|\leq c'/\sqrt{n},$$

where $c'$ depends only on the distribution of $X_1$?

It is easy to construct examples of $2-3$ point distributions that would make $c'$ arbitrary large (for those examples). But once we fix the distribution, is it in general reasonable to assume that that the is a constant so that the aforementioned inequality holds?

$\endgroup$
1
  • $\begingroup$ Is $S_n$ their sum or sum divided by $\sqrt{n}$? $\endgroup$ Commented Mar 14, 2016 at 19:50

1 Answer 1

7
$\begingroup$

Your question is unclear. E.g., if you allow $c$ to depend on the distribution of $X_1$, you can let $c=1/E|X_1|^3$. Then $cb=cnE|X_1|^3=n\ge1$, so that the inequality in question trivially holds.

Your question can be given a more nontrivial meaning by reinterpreting it, say as follows. Let $X,X_1,\ldots X_n$ be i.i.d. random variables and let $S_n:=X_1+\dots+X_n$. Assuming that $EX=0$, $EX^2=1$, and $E|X|^3=b$, the Berry--Esseen theorem yields $$\Delta_n:=\sup_{x\in\mathbb{R}}|P(S_n/\sqrt n<x)-P(Z<x)|\leq cb/\sqrt n,\tag{1}$$
where $c$ is an absolute constant and $Z$ has the standard normal distribution. Does $(1)$ hold if we do not assume finiteness of the $3$rd or $(2+\varepsilon)$th moments but allow the constant $c$ to depend on the distribution of $X$?

The answer to this question is negative, which can be seen in a variety of ways. For instance, a result by Ibragimov [1] implies that, if $\Delta_n=O(n^{-h/2})$ for some $h\in(0,1]$, then $EX^2I\{|X|\ge z\}=O(z^{-h})$ as $z\to\infty$ (whence $E|X|^{2+h-\delta}<\infty$ for all $\delta\in(0,h)$); here $I$ denotes the indicator function. In particular, if $\Delta_n=O(n^{-1/2})$, then $E|X|^{3-\delta}<\infty$ for all $\delta\in(0,1)$.

[1] Ibragimov, I. A. On the accuracy of approximation by the normal distribution of distribution functions of sums of independent random variables. Theor. Probability Appl. 11 (1966), 559--579.

$\endgroup$
2
  • $\begingroup$ This is indeed a useful comment and I will mend the formulation. Yet it is not what I exactly mean, although very close. This time I am sure that it is more meaningful. $\endgroup$
    – TOM
    Commented Mar 14, 2016 at 5:56
  • $\begingroup$ Alas, your reformulation has not improved the question. With $ES_n^2=1$ and $E|X_1|^3=b$, the Berry--Esseen bound is $cnb$ (not $cb/\sqrt n$), and (in view, say, of the Edgeworth expansion) the factor $n$ in $cnb$ cannot be improved under any moment finiteness condition. Also, as long as you assume $ES_n^2=1$, you cannot say "we fix the distribution [of $X_1$]" -- because then $EX_1^2=1/n$, which depends on $n$. I still don't see a meaningful reinterpretation of the question other than the one given in my answer. $\endgroup$ Commented Mar 14, 2016 at 14:34

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .