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

Question

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?

Answer 1

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.

Answer 2

I think, I just made it! basing me on the proof of @Thomas and something of my knowledge. Because @Thomas is not using the same hypothesis, but very similar ones (Thanks Thomas!).

Define $Y_i=\sum_{j=1}^i C_j$. We claim that

\begin{equation*} E[e^{\lambda Y_i}] \leq e^{\frac{1}{8} \lambda^2 \sum_{j=1}^i(b_j-a_j)^2} \end{equation*}

For all $i$ and $\lambda$ we have

\begin{equation*} E[e^{\lambda Y_i}] = E[e^{\lambda Y_{i-1}}\cdot e^{\lambda C_i}] = E[e^{\lambda Y_{i-1}}\cdot E[e^{\lambda C_i}|C_1,...,C_{i-1}]] \end{equation*}

By assumption $\mu _{C_i}=E[C_i|C_1=c_1,...,C_{i-1}=c_{i-1}]\leq \beta$ and $Pr[C_i \in [-\alpha,\alpha]]=1$($Pr[C_i \in [-\alpha,\alpha]]=1 \Longleftrightarrow -\alpha \leq C_i \leq \alpha$) (Absolute value definition). By Hoeffding Lemma , where $a_i=-\alpha$ y $b_i=\alpha$. We can get:

\begin{equation*} E \left[e^{\lambda C_i}\middle|C_1=c_1,...,C_{i-1}=c_{i-1}\right] \leq e^{\lambda\mu _{C_i}+\frac18 \lambda^2(2\alpha)^2} \end{equation*}

because $\mu _{C_i} \leq \beta$,

\begin{equation*} \begin{aligned} E[e^{\lambda Y_i}] &= E[e^{\lambda Y_{i-1}}\cdot E[e^{\lambda C_i}|C_1,...,C_{i-1}]] \\ &\leq E[e^{\lambda Y_{i-1}}\cdot e^{\lambda\beta +\frac18 \lambda^2(2\alpha)^2}]\\ &.\\ &. \ \text{(Iterating)} \\ &. \\ &\leq e^{\sum_{i=1}^k \lambda \beta + \frac18 \lambda^2(2\alpha)^2 }\\ &= e^{\lambda k \beta + k \frac18 \lambda^2(2\alpha)^2}\\ &= e^{\lambda k \beta + k \frac12 \lambda^2(\alpha)^2}\ \end{aligned} \end{equation*}

In that way :

\begin{equation*} \begin{aligned} Pr \Big[ \sum_{i=1}^k C_i > k\beta +z \sqrt{k}\cdot \alpha \Big] &= Pr \Big[ \lambda Y_n > \lambda (k\beta +z \sqrt{k}\cdot \alpha) \Big] \\ &= Pr \Big[ e^{ \lambda Y_n} > e^{ \lambda (k\beta +z \sqrt{k}\cdot \alpha)} \Big] \ \ \ \text{(By Markov inequality)}\\ &\leq \frac{E[e^{Y_n}]}{e^{ \lambda(k\beta +z \sqrt{k}\cdot \alpha)}}\\ &= e^{- \lambda(k\beta +z \sqrt{k}\cdot \alpha)}E[e^{ \lambda Y_n}]\\ & \leq e^{- \lambda(k\beta +z \sqrt{k}\cdot \alpha)} e^{\lambda k \beta + k \frac12 \lambda^2(\alpha)^2} \end{aligned} \end{equation*}

When we minimize $\lambda $, we get that $\lambda = \frac{z}{\sqrt{k}\alpha}$, replacing it on the equation we have :

\begin{equation*} Pr \Big[ \sum_{i=1}^k C_i > k\beta +z \sqrt{k}\cdot \alpha \Big] \leq e^{\frac{-z^2}{2}} \end{equation*}