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 $n\mu \ll
\sqrt{n}$.