# What is the proof of this nonstandard version of Azuma's inequality?

In Appendix B of Boosting and Differential Privacy by Dwork et al., the authors state the following result without proof and refer to it as Azuma's inequality:

Let $C_1, \dots, C_k$ be real-valued random variables such that for every $i \in [k]$,

1. $\Pr[|C_i| \leq \alpha] = 1$, and
2. for every $(c_1, \dots, c_{i - 1}) \in \text{Supp}(C_1, \dots, C_{i - 1})$, we have $\text{E}[C_i \mid C_1 = c_1, \dots, C_{i - 1} = c_{i - 1}] \leq \beta$.

Then for every $z > 0$, we have $\Pr[\sum_{i = 1}^k C_i > k\beta + z \sqrt{k} \cdot \alpha] \leq e^{-z^2/2}$.

I'm having trouble proving this. The standard version of Azuma's inequality says:

Suppose $\{X_0, X_1, \dots, X_k\}$ is a martingale, and $|X_i - X_{i - 1}| \leq \gamma_i$ almost surely. Then for all $t > 0$, we have $\Pr[X_k \geq t] \leq \exp(-t^2/(2 \sum_{i = 1}^k \gamma_i^2))$.

To prove the version of Azuma's inequality stated by Dwork et al., I figured we should take $X_0 = 0$ and $X_i = X_{i - 1} + C_i - \text{E}[C_i \mid C_1, C_2, \dots, C_{i - 1}]$. That way, I think $\{X_0, \dots, X_k\}$ is a martingale. But all we can say is that $|X_i - X_{i - 1}| \leq 2\alpha$ almost surely, right? That factor of two causes trouble, because it means that after substituting, we merely find that $\Pr[\sum_{i = 1}^k C_i > k\beta + z\sqrt{k} \cdot 2\alpha] \leq e^{-z^2/2}$, which is weaker than the conclusion stated by Dwork et al.

Is there a simple trick I'm missing? Is the statement by Dwork et al. missing a factor of two?

• The statement in the paper is true, but does not follow from the "usual" version of Azuma's inequality. The issue is that the usual statement assumes $X_i-X_{i-1}\in[-a,a]$ but any interval of the same length will do; there is no reason to assume a symmetric interval. – Thomas supports Monica Sep 25 '17 at 23:44

I can't find a reference, so I'll just sketch the proof here.

Theorem. Let $X_1, \cdots, X_n$ be real random variables. Let $a_1, \cdots, a_n, b_1, \cdots, b_n$ be constants. Suppose that, for all $i \in \{1,\cdots,n\}$ and all $(x_1,\cdots,x_{i-1})$ in the support of $(X_1, \cdots, X_{i-1})$, we have

1. $\mathbb{E}[X_i | X_1=x_1, \cdots, X_{i-1}=x_{i-1}] \leq 0$ and
2. $\mathbb{P}[X_i \in [a_i,b_i]]=1$.

Then, for all $t\geq0$, $$\mathbb{P}\left[\sum_{i=1}^n X_i \geq t\right] \leq \exp\left(\frac{-2t^2}{\sum_{i=1}^n (b_i-a_i)^2}\right).$$

Proof. Define $Y_i = \sum_{j=1}^i X_j$. We claim that $$\forall i \in \{1,\cdots,n\}~\forall \lambda \geq 0 ~~~~~ \mathbb{E}\left[e^{\lambda Y_i}\right] \leq e^{\frac18 \lambda^2 \sum_{j=1}^i (b_j-a_j)^2}.\tag*{(*)}$$ For all $i$ and $\lambda$, we have $$\mathbb{E}\left[e^{\lambda Y_i}\right] = \mathbb{E}\left[e^{\lambda Y_{i-1}}\cdot e^{\lambda X_i}\right] = \mathbb{E}\left[e^{\lambda Y_{i-1}}\cdot \mathbb{E}\left[e^{\lambda X_i}\middle|Y_{i-1}\right]\right].$$ By assumption, $\mu(y_{i-1}):=\mathbb{E}[X_i|Y_{i-1}=y_{i-1}]\leq 0$ and $\mathbb{P}[X_i \in [a_i,b_i]] =1$ for all $y_{i-1}$ in the support of $Y_{i-1}$. (Note that $Y_{i-1}=X_1+\cdots+X_{i-1}$.) Thus, by Hoeffding's lemma, $$\mathbb{E}\left[e^{\lambda X_i}\middle|Y_{i-1}=y_{i-1}\right] \leq e^{\lambda\mu(y_{i-1})+\frac18 \lambda^2(b_i-a_i)^2}$$ for all $y_{i-1}$ in the support of $Y_{i-1}$ and all $\lambda \in \mathbb{R}$. Since $\mu(y_{i-1})\leq 0$, we have, for all $\lambda \geq 0$, $$\mathbb{E}\left[e^{\lambda Y_i}\right] \leq \mathbb{E}\left[e^{\lambda Y_{i-1}}\cdot e^{0+\frac18 \lambda^2(b_i-a_i)^2}\right].$$ Now induction yields the claim (*) above.

Now we apply Markov's inequality to $e^{\lambda Y_n}$ and use our claim (*). For all $t, \lambda > 0$, $$\mathbb{P}\left[\sum_{i=1}^n X_i \geq t\right]=\mathbb{P}[Y_n \geq t] = \mathbb{P}\left[e^{\lambda Y_n} \geq e^{\lambda t}\right] \leq \frac{\mathbb{E}\left[e^{\lambda Y_n}\right]}{e^{\lambda t}} \leq \frac{e^{\frac18 \lambda^2 \sum_{i=1}^n (b_i-a_i)^2}}{e^{\lambda t}}.$$ Finally, set $\lambda=\frac{4t}{\sum_{i=1}^n (b_i-a_i)^2}$ to minimize the right hand expression and obtain the result. $\tag*{$\blacksquare$}$

As I mentioned in my comment, the key difference between this and the "usual" statement of Azuma's inequality is requiring $X_i \in [a_i,b_i]$, rather than $X_i \in [-a,a]$. The former allows more flexibility and this saves a factor of 2 in some cases.

Note that the $Y_i$ random variables in the proof are a supermartingale. You can obtain the usual version of Azuma's inequality by taking a Martingale $Y_1, \cdots, Y_n$, setting $X_i=Y_i-Y_{i-1}$ and $[a_i,b_i]=[-c_i,c_i]$ (where $\mathbb{P}[|Y_i-Y_{i-1}|\leq c_i]=1$), and then applying the above result.

• In the first line of the proof, it should presumably be $Y_i = \sum_{j=1}^i X_j$ (upper bound of the sum as $i$ rather than $n$).... – Dougal Sep 26 '17 at 2:12
• The proof is also given in the monograph by Dubhashi and Panconesi. – Kristoffer Arnsfelt Hansen Sep 27 '17 at 8:01
• @KristofferArnsfeltHansen: Great. Do you have a link? – Thomas supports Monica Sep 27 '17 at 18:17