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

## 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 $$K \rightarrow \infty$$.

$$\newcommand\ep\epsilon$$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 $$\ep\downarrow0$$.
• Thank you! I did mix up the inequality on $|S_n|$, as you noted. Jan 8, 2022 at 23:34