Bounding the variance of a sum of independent random variables

Question

Suppose $\{X_i\}_{i=1}^n$ is a sequence of independently distributed random variables that take values in $[0,1]$. Let $\bar{X}_n = \frac{1}{n}\sum_{i=1}^n X_i$ denote the average of the sequence. I'd like to find an upper bound for $\text{Var}(\bar{X})$.

My strategy was to use Hoeffding's inequality, which states that $$ \Pr(|\bar X_n - E\bar X_n| \geq t) \leq e^{-2nt^2} $$

We therefore have \begin{align} E\left(|\bar X_n - E\bar X_n|^2\right) &= \int_{x \in [0,1]:\, \left(x - E\bar X_n\right)^2 \geq t}|\bar X_n - E\bar X_n|^2dP + \int_{x \in [0,1]:\, \left(x - E\bar X_n\right)^2 < t}|\bar X_n - E\bar X_n|^2dP \\ &\leq e^{-2nt^2} + t(1-e^{-2nt^2}) \end{align} for all $t$. Minimizing the right-hand side with respect to $t$ gives a bound for any $n$.

Is it possible to provide a tighter bound than this?

Thanks!

Answer 1

Try Bernstein's inequality? http://www.cs.cornell.edu/~sridharan/concentration.pdf

That or the Efron Stein inequality (which is the tightest bound I know), although it can be difficult to understand how to implement correctly (imo).