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Sep 24, 2019 at 8:49 comment added Davide Giraudo Indeed, there was a typo. The proofs works in the same way as for Doob's inequality. I also made the reference more precise.
Sep 24, 2019 at 8:48 comment added Thomas Dybdahl Ahle Concerning only the second term, it seems that if all $\mathbb E[S_i^2]=i$, we need $c_i^2\approx i \log i$ to get a probability below 1. This seems similar to the answer of @Mark, but worse than the $\sqrt{\log\log n}$ answer by Tanguy. (Just for my own understanding and comparison)
Sep 24, 2019 at 8:48 history edited Davide Giraudo CC BY-SA 4.0
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Sep 24, 2019 at 8:41 comment added Thomas Dybdahl Ahle Should the $c_n^p$ on the right-hand side be in the denominator? It seems that if we scale the $S_k$ down by a larger factor, the probability that they exceed some value should decrease rather than increase. I would check the paper, but it doesn't seem to have a Theorem 1, so I'm not really sure.
Nov 3, 2013 at 11:47 vote accept Adrien
Nov 3, 2013 at 11:47 vote accept Adrien
Nov 3, 2013 at 11:47
Oct 31, 2013 at 10:51 history edited Davide Giraudo CC BY-SA 3.0
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Oct 31, 2013 at 10:17 comment added Adrien Thanks for the answer. I didn't know about this generalization of the Hajek-Renyi inequality. I need to think about it to see if it can be generalized further with exponential functions.
Oct 30, 2013 at 19:18 history edited Davide Giraudo CC BY-SA 3.0
I missed the $\max$.
Oct 30, 2013 at 12:39 history answered Davide Giraudo CC BY-SA 3.0