0
$\begingroup$

Prove Bernstein's inequality for any $t>0$. $P(X>y) \leq e^{-ty} E(e^{tX})$

This is for homework, but we did not go over Bernstein's inequality in class. We were going over Markov's and Chebyshev's inequalities. From what I have seen from looking it up on the internet there are several different ones, but I am not sure what applies to this particular problem.

1 Answers 1

3

By Markov inequality, if $Z \geq 0$ $ P(Z > u) \leq \frac{EZ}{u}$

Put $Z = e^{tX}$ and $u=e^{ty}$ (they are non -ve) to get $ P(e^{tX} > e^{ty}) \leq \frac{Ee^{tX}}{e^{ty}}$

But $P(e^{tX} > e^{ty}) = P(X > y)$. Hence we are done.

  • 0
    That is what I was thinking, but I was unsure because it brought up Bernstein's inequality.2012-11-19