It is known that there are multiplicative version concentration inequalities for sums of independent random variables. For example, the following multiplicative version Chernoff bound.

Chernoff bound:

Let $X_1,\ldots,X_n$ be independent random variables and $X_i \in$ $[0,1]$. Let $Y=\sum_{i=1}^n X_i$. Then for any $\delta>0$,

$\Pr\left(Y \ge (1+\delta)EY \right) \le e^{-c\cdot(EY)\delta ^2},$

where $c$ is some absolute constant, e.g., c=1/3.

Now consider dependent random variables. A slight variant of Azuma's inequality states the following.

Azuma's Inequality:

Let $X_1,\ldots,X_n$ be (dependent) random variables and $X_i \in [0,1]$. Assume that there exists $\mu$, such that $ \Pr \left( \sum_{i=1}^n \mathbb{E}[X_i|X_{1},\ldots,X_{i-1}] \le \mu\right) = 1$. Let $Y=\sum_{i=1}^n X_i$. Then for any $\lambda > 0$,

$\Pr\left(Y \ge n\mu+\lambda \right) \le e^{-2 \lambda^2/n}.$

Azuma's inequality is additive. My question is that does a multiplicative version of Azuma's inequality such as the following hold?

My question: does the following hold?

Let $X_1,\ldots,X_n$ be (dependent) random variables and $X_i \in [0,1]$. Assume that there exists $\mu$, such that $\Pr\left( \sum_{i=1}^n \mathbb{E}[X_i|X_1,\ldots,X_{i-1}] \le \mu\right) = 1.$ Let $Y=\sum_{i=1}^n X_i$. Then for any $\delta>0$

$\Pr\left(Y \ge (1+\delta)n\mu \right) \le e^{-c\cdot n\mu \delta^2},$

where $c$ is some absolute constant.

Note: the standard Azuma's inequality does not imply the multiplicative version when $n\mu \ll \sqrt{n}$.

  • $\begingroup$ Something is broken. Maybe you want $Y$ to be the mean of $X_1,\ldots,X_n$ instead of the sum? $\endgroup$ – Brendan McKay Jan 22 '13 at 23:56
  • $\begingroup$ Are you only interested in the case of small $\delta$? You statement of Chernoff does not seem right to me for large $\delta$. $\endgroup$ – Ori Gurel-Gurevich Jan 23 '13 at 1:12
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    $\begingroup$ Also, take a look at arxiv.org/abs/0901.4056, section 2, particularly proposition 2.3, and see if that helps. $\endgroup$ – Ori Gurel-Gurevich Jan 23 '13 at 1:15
  • $\begingroup$ Ori, Thanks a lot! Your result is very relavant to my question. On the one hand, the proposition in your paper is stronger than my question; it's a uniform bound over n. But on the other hand, your result does not directly give an affirmative answer to my question. Using the notion in your paper, if m is an upper bound of $\sum_{i=1}^t Y_i$, can I say $\sum_{i=1}^t X_i < 3m/2$ w.h.p.? $\endgroup$ – Liwei Wang Jan 23 '13 at 16:22
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    $\begingroup$ If by w.h.p. you mean a bound going to 0 with $m$, then I think the answer is yes. Just add more $X_i$ at the end that will make sure that $\sum Y_i$ is equal to $m$ and then apply proposition 2.3. $\endgroup$ – Ori Gurel-Gurevich Jan 23 '13 at 22:45

Yes, such bounds are possible. You can adapt the proof of Azuma's inequality to the multiplicative-error case, if you set it up correctly. For example:

Lemma 10 [this paper]. Let $Y=\sum_{t=1}^T x_t$ and $Z=\sum_{t=1}^T z_t$ be sums of non-negative random variables, where $T$ is a random stopping time with finite expectation, and, for all $t$, $|x_t-z_t|\le 1$ and $$\textstyle E\big[\,x_{t} - z_{t} ~|\, \sum_{s< t} x_s, \sum_{s< t} z_s\,\big] ~\le~ 0.$$ Let $\epsilon\in[0,1]$ and $A\in\mathbb{R}$. Then $$\Pr\big[\,(1-\epsilon) Y \,\ge\, Z + A\, \big] ~\le~ \exp({-\epsilon}A).$$

To apply this to your question, take $T=n$, $z_t=\mu$, and $A=\epsilon\, n\, \mu$. Then you get $$\Pr\big[\,Y \,\ge\, \frac{1+\epsilon}{1-\epsilon}n\mu\, \big] ~\le~ \exp({-\epsilon^2}n\mu).$$

  • $\begingroup$ I should add that for your question you should be able to adapt the proof of Chernoff (with bound $\exp(-\epsilon^2 n \mu/3)$) directly to show that same bound. For example, define $$\phi_t = \frac{\prod_{i=1}^t (1+\epsilon x_i) \times \exp(-\epsilon \mu (n-t))}{(1+\epsilon)^{(1+\epsilon)\mu n}}.$$ Then $\phi_0, \phi_1, \ldots, \phi_n$ is a super-martingale, so $E[\phi_n] \le \phi_0$. From this inequality the desired bound follows as in the proof of standard Chernoff. $\endgroup$ – Neal Sep 14 '17 at 15:16

$\newcommand{\de}{\delta}$ The "dependent" version of the multiplicative Chernoff bound can be proved quite similarly to the "independent" case. Indeed, let $E_{i-1}$ denote the conditional expectation given $X_1,\dots,X_{i-1}$, so that $E_{i-1}X_i\le\mu$ almost surely (a.s.) for $i=1,\dots,n$. Take any real $t\ge0$. By the convexity of $e^{tx}$ in $x$ and the conditions that $0\le X_i\le1$ and $E_{i-1}X_i\le\mu$, we have $e^{tX_i}\le1+(e^t-1)X_i$ and hence \begin{equation*} E_{i-1}e^{tX_i}\le1+(e^t-1)E_{i-1}X_i\le1+(e^t-1)\mu\le\exp\{(e^t-1)\mu\} \end{equation*} for $i=1,\dots,n$. So, by induction, for $j=1,\dots,n$ and $Y_j:=\sum_1^j X_i$ we have \begin{equation*} Ee^{tY_j}=EE_{j-1}e^{tY_j}=Ee^{tY_{j-1}}E_{j-1}e^{tX_j}\le Ee^{tY_{j-1}}\exp\{(e^t-1)\mu\}, \end{equation*} whence, by induction, \begin{equation*} Ee^{tY}=Ee^{tY_n}\le\exp\{n(e^t-1)\mu\}. \end{equation*} So, using Markov's inequality and then choosing $t=\ln(1+\de)$, we get \begin{align*} P(Y\ge(1+\de)n\mu)&\le e^{-(1+\de)n\mu t}Ee^{tY} \le\exp\{-(1+\de)n\mu t+n(e^t-1)\mu\} \\ &=\exp\{-n\mu\psi(\de)\}, \end{align*} where $\psi(u):=u-(1+u)\ln(1+u)$. Up to notation, this bound is the same as the known multiplicative Chernoff bound in the "independent" case.

Since $\psi(u)\le-u^2/3$ for $u\in[0,3/2]$, we have \begin{equation*} P(Y\ge(1+\de)n\mu)\le\exp\{-n\mu\de^2/3\} \tag{1} \end{equation*} if $\de\in[0,3/2]$.

Note that (1) cannot hold for all $\de\ge0$, even in the "independent" case. Indeed, suppose that the $X_i$'s are iid with $P(X_1=1)=\mu=1-P(X_i=0)$, where $\mu:=1/n$ and $n\to\infty$. Then $Y$ will converge in distribution to a random variable $\Pi$ with the Poisson distribution with parameter $1$, and (1) will yield \begin{equation*} P(\Pi\ge1+\de)\le\exp\{-\de^2/3\}. \end{equation*} Since Poisson distributions are not subgaussian, the latter inequality cannot hold for all $\de\ge0$. So, (1) cannot hold for all $\de\ge0$.


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