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Sep 25, 2020 at 21:33 comment added Iosif Pinelis @Siam : Oops! I had given a wrong link. This is now corrected. Also, the last displayed inequality in the answer differs from Azuma's in that the inequality in the answer assumes a bound on the sum of the squares of the $X_i$'s, whereas Azuma's assumes bounds on the each of the squares of the $X_i$i's. When the $X_i$'s are independent, these two assumptions are equivalent, but not so in general. (On the other hand, Azuma's inequality does not assume the conditional symmetry). Also, Azuma's inequality is due entirely to Hoeffding '63 (see the last paragraph of Section 2 of Hoeffding's paper).
Sep 25, 2020 at 19:44 history edited Iosif Pinelis CC BY-SA 4.0
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Sep 25, 2020 at 15:37 comment added Siam thank you very much for such a comprehensive answer. I went through the paper that you have cited, but I couldn't find Theorem 3.6. Perhaps you meant Theorem 3? I have a couple of more questions :) 1. Is $B^2$ here equal to $A_1^2$ (and $A_2=0$) according to the notation of paper? 2. And isn’t the inequality you wrote at the end the same as Azuma’s inequality or am I missing something?
Sep 25, 2020 at 11:23 vote accept Siam
Sep 25, 2020 at 5:51 history edited Iosif Pinelis CC BY-SA 4.0
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Sep 25, 2020 at 3:40 history edited Iosif Pinelis CC BY-SA 4.0
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Sep 25, 2020 at 1:51 history answered Iosif Pinelis CC BY-SA 4.0