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. ${\displaystyle -c_t\le X_t-X_{t-1}\le c_t}$. So, if known bound is asymmetric, e.g. ${\displaystyle a_t\le X_t-X_{t-1}\le b_t}$, to use Azuma's inequality, one need to choose ${\displaystyle c_t=max(|a_t|,|b_t|)}$ which might be a waste of information on the boundedness of ${\displaystyle X_t-X_{t-1}}$. 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 ${\displaystyle \left\{X_0,X_1,\cdots \right\}}$ be a martingale (or supermartingale) with respect to filtration ${\displaystyle \left\{{\mathcal{F}}_0,{\mathcal{F}}_1,\cdots \right\}}$. Assume there are predictable processes ${\displaystyle \left\{A_0,A_1,\cdots \right\}}$ and ${\displaystyle \left\{B_0,B_1,\dots \right\}}$ with respect to ${\displaystyle \left\{{\mathcal{F}}_0,{\mathcal{F}}_1,\cdots \right\}}$, i.e. for all ${\displaystyle t}$, ${\displaystyle A_t,B_t}$ are ${\displaystyle {\mathcal{F}}_{t-1}}$-measurable, and constants ${\displaystyle 0<c_1,c_2,\cdots <\mathrm{\infty }}$ such that

${\displaystyle A_t\le X_t-X_{t-1}\le B_t\text{and}{B}_t-A_t\le c_t}$

almost surely. Then for all ${\displaystyle ϵ>0}$,

${\displaystyle \text{P}(X_n-X_0\ge ϵ)\le \exp \left(-\frac{2{ϵ}^2}{\sum _{t=1}^n{c}_t^2}\right).}$

Since a submartingale is a supermartingale with signs reversed, we have if instead ${\displaystyle \left\{X_0,X_1,\dots \right\}}$ is a martingale (or submartingale),

${\displaystyle \text{P}(X_n-X_0\le -ϵ)\le \exp \left(-\frac{2{ϵ}^2}{\sum _{t=1}^n{c}_t^2}\right).}$

If ${\displaystyle \left\{X_0,X_1,\dots \right\}}$ 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:

${\displaystyle \text{P}(|X_n-X_0|\ge ϵ)\le 2\exp \left(-\frac{2{ϵ}^2}{\sum _{t=1}^n{c}_t^2}\right).}$

Proof

We will prove the supermartingale case only as the rest are self-evident. By Doob decomposition, we could decompose supermartingale ${\displaystyle \left\{X_t\right\}}$ as ${\displaystyle X_t=Y_t+Z_t}$ where ${\displaystyle \left\{Y_t,{\mathcal{F}}_t\right\}}$ is a martingale and ${\displaystyle \left\{Z_t,{\mathcal{F}}_t\right\}}$ is a nonincreasing predictable sequence (Note that if ${\displaystyle \left\{X_t\right\}}$ itself is a martingale, then ${\displaystyle Z_t=0}$). From ${\displaystyle A_t\le X_t-X_{t-1}\le B_t}$, we have

${\displaystyle -(Z_t-Z_{t-1})+A_t\le Y_t-Y_{t-1}\le -(Z_t-Z_{t-1})+B_t}$

Applying Chernoff bound to ${\displaystyle Y_n-Y_0}$, we have for ${\displaystyle ϵ>0}$,

${\displaystyle \begin{array}{cc}\text{P}(Y_n-Y_0\ge ϵ)& \le \underset{s>0}{min}{e}^{-sϵ}\mathbb{E}[e^{s(Y_n-Y_0)}]\\ & =\underset{s>0}{min}{e}^{-sϵ}\mathbb{E}\left[\exp \left(s\sum _{t=1}^n(Y_t-Y_{t-1})\right)\right]\\ & =\underset{s>0}{min}{e}^{-sϵ}\mathbb{E}\left[\exp \left(s\sum _{t=1}^{n-1}(Y_t-Y_{t-1})\right)\mathbb{E}\left[\exp \left(s(Y_n-Y_{n-1})\right)\mid {\mathcal{F}}_{n-1}\right]\right]\end{array}}$

For the inner expectation term, since

(i) ${\displaystyle \mathbb{E}[Y_t-Y_{t-1}\mid {\mathcal{F}}_{t-1}]=0}$ as ${\displaystyle \left\{Y_t\right\}}$ is a martingale;

(ii) ${\displaystyle -(Z_t-Z_{t-1})+A_t\le Y_t-Y_{t-1}\le -(Z_t-Z_{t-1})+B_t}$;

(iii) ${\displaystyle -(Z_t-Z_{t-1})+A_t}$ and ${\displaystyle -(Z_t-Z_{t-1})+B_t}$ are both ${\displaystyle {\mathcal{F}}_{t-1}}$-measurable as ${\displaystyle \left\{Z_t\right\}}$ is a predictable process;

(iv) ${\displaystyle B_t-A_t\le c_t}$;

by applying Hoeffding's lemma, we have

${\displaystyle \mathbb{E}\left[\exp \left(s(Y_t-Y_{t-1})\right)\mid {\mathcal{F}}_{t-1}\right]\le \exp \left(\frac{s^2(B_t-A_t{)}^2}{8}\right)\le \exp \left(\frac{s^2{c}_t^2}{8}\right).}$

Repeating this step, one could get

${\displaystyle \text{P}(Y_n-Y_0\ge ϵ)\le \underset{s>0}{min}{e}^{-sϵ}\exp \left(\frac{s^2\sum _{t=1}^n{c}_t^2}{8}\right).}$

Note that the minimum is achieved at ${\displaystyle s=\frac{4ϵ}{\sum _{t=1}^n{c}_t^2}}$, so we have

${\displaystyle \text{P}(Y_n-Y_0\ge ϵ)\le \exp \left(-\frac{2{ϵ}^2}{\sum _{t=1}^n{c}_t^2}\right).}$

Finally, since ${\displaystyle X_n-X_0=(Y_n-Y_0)+(Z_n-Z_0)}$ and ${\displaystyle Z_n-Z_0\le 0}$ as ${\displaystyle \left\{Z_n\right\}}$ is nonincreasing, so event ${\displaystyle \left\{X_n-X_0\ge ϵ\right\}}$ implies ${\displaystyle \left\{Y_n-Y_0\ge ϵ\right\}}$, and therefore

${\displaystyle \text{P}(X_n-X_0\ge ϵ)\le \text{P}(Y_n-Y_0\ge ϵ)\le \exp \left(-\frac{2{ϵ}^2}{\sum _{t=1}^n{c}_t^2}\right).◻}$

Remark

Note that by setting ${\displaystyle A_t=-c_t,B_t=c_t}$, 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 (or a supermartingale falls).

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 F_i be a sequence of independent and identically distributed random coin flips (i.e., let F_i be equally likely to be −1 or 1 independent of the other values of F_i). Defining ${\displaystyle X_i=\sum _{j=1}^i{F}_j}$ yields a martingale with |X_k − X_k−1| ≤ 1, allowing us to apply Azuma's inequality. Specifically, we get

${\displaystyle \mathrm{P}(X_n>t)\le \exp \left(\frac{-t^2}{2n}\right).}$

For example, if we set t proportional to n, then this tells us that although the maximum possible value of X_n scales linearly with n, the probability that the sum scales linearly with n decreases exponentially fast with n.

If we set ${\displaystyle t=\sqrt{2n\ln n}}$ we get:

${\displaystyle \mathrm{P}\left(X_n>\sqrt{2n\ln n}\right)\le \frac{1}{n},}$

which means that the probability of deviating more than ${\displaystyle \sqrt{2n\ln n}}$ 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 (see page 9 of his 1963 paper).