Bennett inequality

Theorem:(Bennett inequality, 1962):

Let ${\{X_i\}}_{i=1}^n$ be a collection of independent random variables satisfying the conditions:

a) $\forall i$, so that one can write ${\sum }_{i=1}^{n}E[X_i^2]=v^2$
b) $\mathrm{Pr}\left\{\left|X_i\right|\le M\right\}=1$  $\forall i$.

Then, for any $\epsilon \ge 0$,

$$\mathrm{Pr}\left\{\sum _{i=1}^n\left(X_i-E[X_i]\right)>\epsilon \right\}\le \exp \left[-\frac{v^2}{M^2}\theta \left(\frac{\epsilon M}{v^2}\right)\right]\le \exp \left[-\frac{\epsilon }{2M}\ln \left(1+\frac{\epsilon M}{v^2}\right)\right]$$

where

$$\theta \left(x\right)=\left(1+x\right)\ln \left(1+x\right)-x$$

Remark: Observing that $\left(1+x\right)\ln \left(1+x\right)-x\ge 9\left(1+\frac{x}{3}-\sqrt{1+\frac{2}{3}x}\right)\ge \frac{3x^2}{2\left(x+3\right)}$ $\forall x\ge 0$, and plugging these expressions into the bound, one obtains immediately the Bernstein inequality under the hypotheses of boundness of random variables, as one might expect. However, Bernstein inequalities, although weaker, hold under far more general hypotheses than Bennett one.

Title	Bennett inequality
Canonical name	BennettInequality
Date of creation	2013-03-22 16:12:25
Last modified on	2013-03-22 16:12:25
Owner	Andrea Ambrosio (7332)
Last modified by	Andrea Ambrosio (7332)
Numerical id	10
Author	Andrea Ambrosio (7332)
Entry type	Theorem
Classification	msc 60E15