# Probability of being away from mean for independent random variables

Let $X_1,X_2,\ldots,X_n$ be independent random variables drawn uniformly from $[-1,1]$. The (weak) law of large numbers says that $$\lim_{n\rightarrow\infty}\text{Pr}\left(\left|\frac{X_1+\ldots+X_n}{n}\right|>0.1\right)=0.$$ But for what functions $f(n)$ can we say that $$\text{Pr}\left(\left|\frac{X_1+\ldots+X_n}{n}\right|>0.1\right)<f(n)?$$ For example, is it true for $f(n)=c/2^n$ for some constant $c$?

I thought this would be answered by the central limit theorem, but it doesn't seem to follow directly (or at least I don't understand it well enough.)

• One that doesn't allow for negative values for one... – Zach466920 Jul 30 '15 at 19:46
• This link might be helpful? en.m.wikipedia.org/wiki/Irwin–Hall_distribution – David Quinn Jul 30 '15 at 19:48

Using Hoeffding's inequality $$\text{Pr}\left(\left|\bar X\right|>\epsilon\right)\le 2\exp\left(-\frac{1}{2}\epsilon^2n\right)$$
• Isn't it $n^2$ instead of $n$? – nan Jul 30 '15 at 20:12
• @nan: nope. you have $\text{const}\times \frac{n^2}{n}$ – d.k.o. Jul 30 '15 at 20:15
We have $\Bbb E X_j^2=1/2$. Hence $$\Bbb E\frac{|X_1+\dots+X_n|^2}{n^2}=\frac1{2n}.$$So Chebyschev's inequality says the probabilty you ask about is no larger than...