For a continuous martingale $X$, we have the Doléans-Dade exponential: $$\epsilon(X)_t=\exp\left(X_t-\frac{1}{2}[X]_t\right)$$

What is the "correct" analogue, if one exists, for some discrete-time martingale $M$? (Unfortunately in discrete time this is no longer a (local) martingale.)

The motivation is as follows. On page 68 of these notes, the author proves the Exponential Martingale Inequality (EMI) (quoted in my words):

Let $X$ be a continuous local martingale with $X_0=0$. Then for all $x>0$,$u>0$,

$$\mathbb{P}\left[\sup_{t\geq 0}~ X_s\geq x, [X]_t\leq u\right]\leq \exp\left(-\frac{x^2}{2u}\right)$$

This reminded me of the Azuma-Hoeffding Inequality, which states:

Let $M_n$ be a martingale with $M_0=0$ and $|M_i-M_{i-1}|\leq c_i$ for all $i$. Then, for $x>0$ $$\mathbb{P}\left[\sup_{k\leq n}~M_k\geq x\right]\leq \exp\left(-\frac{x^2}{2\sum_{k=1}^nc_k^2}\right)$$

Well, we have $[M]_n\leq \sum_{k=1}^nc_k^2$, so the former inequality would imply the second if only it were true for discrete-time (discontinuous) martingales (setting $u=\sum_{k=1}^nc_k^2$).

To prove EMI, we apply the optional stopping theorem to $\epsilon(\theta X)$ at the time when the $X$ first hits $x$. This gives a set of bounds parameterized by $\theta$ for the probability we want to bound, and optimizing over $\theta$ gives the result (see notes).

For an appropriate definition of $\epsilon(M)$, would the same argument would work in discrete time (and give a proof of Azuma-Hoeffding)?

Thanks.