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The answer is probably well-known, but I cannot find anything definite in the literature.

Suppose we have the usual ingredients of a CLT, i.e. the series

$$X_N = \sum_{n=1}^N x_n $$

where $x_n$ are i.i.d.'s. The CLT says that $X_N/ \sqrt{N}$ approaches a normal distribution.

Some of the literature states $X_N = O(\sqrt{N})$. That's the interpretation I would like for my purposes! But it seems more like the CLT implies it is $O(\sqrt{N})$ with probability equal to 1. In mathematical physics, we would be un-inclined to make such a distinction. In pure math, what is the rigorous way to state these things? Are there any delicate issues involved?

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    $\begingroup$ As a mathematical physicist, I must protest: we would indeed be careful with such distinctions. Maybe not a theoretical physicist. $\endgroup$
    – Robert Israel
    Commented Feb 13, 2017 at 20:57
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    $\begingroup$ It's not even $O(\sqrt{N})$ with probability 1. Do you know about the law of the iterated logarithm? $\endgroup$
    – Nate Eldredge
    Commented Feb 13, 2017 at 20:58
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    $\begingroup$ Yes, it does; law of the iterated logarithm again. $\endgroup$
    – Christian Remling
    Commented Feb 13, 2017 at 21:04
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    $\begingroup$ Watch the quantifiers. It's clear that CLT implies $O(N^{1/2+\epsilon})$ with high probability. That is, for every $\epsilon$ and $C$ we have $P(|X_N| \le C N^{1/2+\epsilon}) \to 1$ as $N \to \infty$. When you say "$O(N^{1/2+\epsilon"})$ with probability 1" I read that as "for $P$-almost every $\omega$, there exists $C = C(\omega)$ such that $|X_N(\omega)| \le C N^{1/2+\epsilon}$ for all $N$". The latter statement is a lot stronger. You could perhaps get it from a version of CLT with good bounds on the convergence rate, via Borel-Cantelli, but I don't think it's trivial. $\endgroup$
    – Nate Eldredge
    Commented Feb 13, 2017 at 22:18
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    $\begingroup$ @AndréLeClair: The party seems over here by now, but one more comment on your question above: even for a deterministic sequence $a_n=O(n^{1/2+\epsilon})$ for all $\epsilon>0$ will not imply that $a_n=O(n^{1/2})$ because the implied constant may depend on $\epsilon$ (trivial example: $a_n=n^{1/2}\log n$). $\endgroup$
    – Christian Remling
    Commented Feb 13, 2017 at 23:38

3 Answers 3

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The sharp general result in this direction is the classical law of the iterated logarithm (LIL). Suppose, after renormalizing if necessary, that the $x_n$ are iid with zero mean and unit variance. Then the LIL states that $$\limsup_{n \to \infty} \frac{X_N}{\sqrt{N \log \log N}} = \sqrt{2}, \quad \text{a.s.}$$ In your language, that says that with probability 1, $X_N$ is $O(\sqrt{N \log \log N})$, and that this cannot be improved to $O(\sqrt{N})$.

To be more careful, it says that for $P$-almost every $\omega$, there is a finite number $C(\omega)$ such that $|X_N(\omega)| \le C(\omega) \sqrt{N \log \log N}$ for all $N$.

LIL isn't a direct corollary of CLT, and I believe there are settings where either may hold while the other fails. So I don't think it's true that the CLT "implies" the result you desire, but in any case they are both true in your setting.

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There is a notion "big O in probability" that may be what you are bumping up against in the literature. The notation (for a sequence $(Y_n)$ of random variables and a sequence $(b_n)$ of positive constants) $Y_n=O_P(b_n)$ means that the sequence $(Y_n/b_n)$ is stochastically bounded, in that for each $\epsilon>0$ there is a cutoff $C>0$ such that $\Bbb P[|Y_n|/b_n>C]<\epsilon$ for all $n$. The CLT indeed implies that your sequence satisfies $X_N=O_P(\sqrt{N})$.

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As a complement, may be, the following statement citing from "R. I. Serfling, Approximation Theorems of Mathematical Statistics, John Wiley & Sons, 1980." is helpful for your question. enter image description here

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