The proof shares similar idea of the proof for the general form of Azuma's inequality listed below. Actually, this can be viewed as a direct corollary of the general form of Azuma's inequality.
A general form of Azuma's inequality
Limitation of the vanilla Azuma's inequality
Note that the vanilla Azuma's inequality requires symmetric bounds on martingale increments, i.e.. So, if known bound is asymmetric, e.g., to use Azuma's inequality, one need to choose which might be a waste of information on the boundedness of. However, this issue can be resolved and one can obtain a tighter probability bound with the following general form of Azuma's inequality.
Statement
Let be a martingale with respect to filtration. Assume there are predictable processes and with respect to, i.e. for all, are -measurable, and constants such that almost surely. Then for all, Since a submartingale is a supermartingale with signs reversed, we have if instead is a martingale, If is a martingale, since it is both a supermartingale and submartingale, by applying union bound to the two inequalities above, we could obtain the two-sided bound:
Proof
We will prove the supermartingale case only as the rest are self-evident. By Doob decomposition, we could decompose supermartingale as where is a martingale and is a nonincreasing predictable sequence. From, we have Applying Chernoff bound to, we have for, For the inner expectation term, since as is a martingale; ; and are both -measurable as is a predictable process; and , by applying Hoeffding's lemma, we have Repeating this step, one could get Note that the minimum is achieved at, so we have Finally, since and as is nonincreasing, so event implies, and therefore
Remark
Note that by setting, we could obtain the vanilla Azuma's inequality. Note that for either submartingale or supermartingale, only one side of Azuma's inequality holds. We can't say much about how fast a submartingale with bounded increments rises. This general form of Azuma's inequality applied to the Doob martingale gives McDiarmid's inequality which is common in the analysis of randomized algorithms.
Simple example of Azuma's inequality for coin flips
Let Fi be a sequence of independent and identically distributed random coin flips. Defining yields a martingale with |Xk − Xk−1| ≤ 1, allowing us to apply Azuma's inequality. Specifically, we get For example, if we set t proportional to n, then this tells us that although the maximum possible value of Xn scales linearly with n, the probability that the sum scales linearly with ndecreases exponentially fast with n. If we set we get: which means that the probability of deviating more than approaches 0 as n goes to infinity.
Remark
A similar inequality was proved under weaker assumptions by Sergei Bernstein in 1937. Hoeffding proved this result for independent variables rather than martingale differences, and also observed that slight modifications of his argument establish the result for martingale differences.
