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I have problem to show the following. Maybe someone has an idea :/

Let $X_1,...,X_n$ be i.i.d. copies of a random variable X with $|X|<1$. Let $S_n=X_1+..+X_n$. Then for any $A>0$: $\mathbb{P}(|S_n-n\mathbb{E}[X]|\geq An)\leq C_A e^{-c_An}$ for some constants $C_A,c_A>0$ depending on $A$.

I have tried this before:

$\mathbb{P}(|S_n-n\mathbb{E}[X]|\geq An)= \mathbb{P}(|\sum_{i=1}^nX_i-\mathbb{E}[X]|\geq An)= \mathbb{P}(|\frac{1}{n}\sum_{i=1}^n X_i-\mathbb{E}[X]|\geq A)\leq \mathbb{P}(\frac{1}{n}\sum_{i=1}^n| X_i-\mathbb{E}[X]|\geq A)$

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    $\begingroup$ What have you tried so far? $\endgroup$ – Mark May 4 at 9:41
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    $\begingroup$ I have tried this before:) (see above) $\endgroup$ – toni_iva May 4 at 9:48
  • $\begingroup$ I have adjusted my answer, based on your try. Good luck! $\endgroup$ – Mark May 5 at 9:24
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Hint: Markov's inequality

(Extended version for monotonically increasing functions)

Edit (based on your try):

Let us bring you closer to the right direction. Observe that your problem is a special case of the famous

Hoeffding's inequality, see https://en.wikipedia.org/wiki/Hoeffding%27s_inequality

Essentially, the statement is slightly more general than your scenario, but you will get definitely inspired by its proof! Observe, it makes use of Markov's inequality (as I hinted).

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    $\begingroup$ Thank you very much! I have it! :-) $\endgroup$ – toni_iva May 6 at 10:09
  • $\begingroup$ You're welcome. It is customary to (upvote and) accept this answer, since the answer has helped you. It would be very kind. $\endgroup$ – Mark May 6 at 12:32

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