Is a sum of a bounded random variables the same order as its standard deviation?

Question

The Question

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

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 $K \rightarrow \infty$.

Answer 1

$\newcommand\ep\epsilon$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 $0.5600$, 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 $\ep\downarrow0$.