Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 3 down vote accepted

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.

share|cite|improve this answer
That is what I was thinking, but I was unsure because it brought up Bernstein's inequality. – Sprock Nov 19 '12 at 20:33

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.