# Proof of Khintchine's inequality

I'm trying to understand the proof of Khintchine's inequality in these lecture notes: http://www.math.ubc.ca/~ilaba/wolff/notes_march2002.pdf

On page 27, second display-style equation after (51), the author claims

$$\mathrm{Prob}\left(\sum_n a_n\omega_n\ge \lambda\right)\le e^{-t\lambda+\frac{t^2}{2}\sum_n a_n^2}$$ I don't understand this conclusion. Where does the exponential function come from? I would be glad if someone could shed light on this. Please don't assume any knowledge in probability, as I don't have any.

• Never mind, I figured it out. Trivially for any $t>0$: $$\mathrm{Prob}\left(\sum_n a_n\omega_n\ge \lambda\right)=\mathrm{Prob}\left(e^{t\sum_n a_n\omega_n}\ge e^{t\lambda}\right)$$ then use Markov's inequality and (51). Please somebody close this question. – Anonymous999 Apr 22 '13 at 17:18
• You can answer your question. – Davide Giraudo Apr 22 '13 at 17:22
• But only after 8 hours -.-.. – Anonymous999 Apr 22 '13 at 18:08

As noted in the comment we have for $t>0$: $$\mathrm{Prob}\left(\sum_n a_n\omega_n\ge \lambda\right)=\mathrm{Prob}\left(e^{t\sum_n a_n\omega_n}\ge e^{t\lambda}\right)$$ So by Markov's inequality $$\mathrm{Prob}\left(\sum_n a_n\omega_n\ge \lambda\right)\le e^{-t\lambda}\mathbb{E}(e^{t\sum_n a_n\omega_n})\le e^{-t\lambda+\frac{t^2}{2}\sum_n a_n^2}$$