The Question
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Suppose that $X\_n$ are independent random variables, $|X\_n| \leq 1$, and $\mathbb{E}[X\_n] = 0$. Let $S\_n = \sum\_{i=1}^n X\_i$ and let $s\_n = \sqrt{\sum\_k \sigma^2(X\_k)}$ be the standard deviation of $S\_n$. Does $P(1 + |S\_n| <\epsilon s\_n)$ go to zero uniformly in $n$, as $\epsilon \rightarrow 0$? In other words, if you permit the abuse of notation, is $s\_n = O(1 + |S\_i|)$ in probability?


Some Background
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I'm interested in estimating the size of $S\_n$, in probability. Chebyshev's Inequality gives: $S\_N = \mathcal{O}(s\_n)$ in probability. In fact, if we take $T\_n = \max\_{1 \leq i \leq n} |S\_i|$, then we also have $T\_n = O(s\_n)$ in probability by Kolmogorov's inequality.


In the other direction, taking advantage of $|X\_i| \leq 1$, we have: $P(1 + T\_n < \epsilon s\_n)$ converges to zero uniformly in $n$, as $\epsilon \rightarrow 0$. A similar statement is used in a proof of Kolmogorov's 3 Series theorem, but I don't know what this theorem itself is called. Thus, by abuse of notation, $s\_n = O(1 + T\_n)$. Since $T\_n$ and $|S\_n|$ often have similar size (for example, with Ottaviani's inequality, you can prove that $P(T\_n > \lambda) \leq 4 \max\_{1 \leq i \leq n} P(|S\_i| > \lambda/4)$), I would expect that a similar statement is true for $|S\_n|$. However, I can't seem to bridge that gap.


\*For reference, when I write $X\_\alpha = O(Y\_\alpha)$ when $X\_\alpha, Y\_\alpha$ are random variables and $Y\_\alpha \geq 0$, I mean that $\sup\_\alpha P(|X\_\alpha| \leq K Y\_\alpha)$ converges to 0 as $K \rightarrow \infty$.



Is a sum of a bounded random variables the same order as its standard deviation?
$\newcommand\ep\epsilon$Let $Z$ denote a standard normal random variable. The condition $|X\_i|\le1$ implies that $A\_3:=\sum\_{i=1}^n E|X\_i|^3\le\sum\_{i=1}^n E|X\_i|^2 =s\_n^2$. Also note that $P(1+|S\_n|\le\epsilon s\_n)=0$ unless $1\le\ep s\_n$.


Therefore and in view of the [Berry--Esseen inequality with Shevtsova's constant factor $0.5600$](https://en.wikipedia.org/wiki/Berry%E2%80%93Esseen_theorem#Non-identically_distributed_summands), for real $\ep\ge0$,
$$\begin{align}
&P(1+|S\_n|\le\epsilon s\_n) \\
&\le1(1\le\ep s\_n)P(|S\_n|\le\epsilon s\_n) \\ 
&\le
1(1\le\ep s\_n)
\Big(P(|Z|\le\ep)+0.5600\,\frac{A\_3}{s\_n^3}\Big) \\ 
&\le 1(1\le\ep s\_n)\Big(\frac2{\sqrt{2\pi}}\,\ep+\frac{0.5600}{s\_n}\Big) \\ 
&\le\Big(\frac2{\sqrt{2\pi}}+0.5600\Big)\ep
\le1.4\ep. \end{align}$$
So, $P(1+|S\_n|\le\epsilon s\_n)\to0$ uniformly in $n$ as $\ep\downarrow0$. Thus, $P(1+|S\_n|>\epsilon s\_n)\to1$ uniformly in $n$ as $\ep\downarrow0$.


