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Bennett's Inequality is stated with a rather unintuitive function,

$$ h(u) = (1+u) \log(1+u) - u $$

See here. I have seen in multiple places that Bernstein's Inequality, while slightly weaker, can be obtained by bounding $h(u)$ from below,

$$ h(u) \ge \frac{ u^2 }{ 2 + \frac{2}{3} u} $$

and plugging it back into Bennett's Inequality. However, I can't see where this expression comes from. Could someone point me in the right direction?

  • 0
    Have you tried using a Taylor expansion for $\log(1 + u)$?2012-02-29
  • 0
    Which expression are you asking about? That of $h(u)$ or that of the lower bound for $h(u)$?2012-02-29
  • 0
    I am concerned with how to get the lower bound.2012-03-01

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