Bounding the size of all consecutive sums of i.i.d. random variables Let $X_1, X_2, \ldots$ by an i.i.d sequence of bounded (for simplicity), mean zero random variables.  For any $a<b$, call $S_{a,b} = X_{a+1} + \cdots + X_b$.
I would like to show that for any $\epsilon >0$, there is a $\delta > 0$ such that for all $N \in \mathbb{N}$, 
$$
\mathbb{P} \left\{ |S_{a,b}| < \delta \; (b-a)^{1/2+\epsilon} \text{ for all } 0 \leq a < b \leq N \right\} > 1 - \epsilon.
$$
Is this a standard fact?  I haven't seen anything about asymptotics of all consecutive sums of random variables before.  It might follow easily from standard facts, but I don't see the trick...
Edit:  This event is the same as the event that the maximum of $|S_{a,b}|$ over all $a<b$ is bounded by $\delta (b-a)^{1/2+\epsilon}$.  (Also edited because I forgot the absolute value around $S_{a,b}$.)

Ok, I figured this out.  The Bernstein inequality states that if $|X_i| \leq M$, 
$$\mathbf{P} \left (\sum_{i=1}^n X_i \geq t \right ) \leq \exp \left ( -\frac{\frac{1}{2} t^2}{\sum \mathbf{E} \left[X_j^2 \right ]+\frac{1}{3} Mt} \right) $$. 
This gives exponential control of any window $|S_{a,b}|$ being greater than $\delta (b-a)^{1/2+\epsilon}$, and so now a union bound over all pairs $a<b \leq n$ can be made as small as you wish.  
