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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.

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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.

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

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