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.

**Answer**

I don’t know if this answers the question but here are my two cents :

If we start from the “SDE” definition of Doléans-Dade exponential for a general semi-martingale $X_t$, then the Doléans-Dade exponential is the process $Z_t$ the solution of the following equation :

$$

\begin{cases}

dZ_t&=Z_{t-}dX_t,

\\

Z_0 &=1.

\end{cases}

$$

In discrete time this gives an anology which allows us to define the Doléans-Dade exponential as the only dicrete process s.t. :

$$

\begin{cases}

\Delta Z_n&=Z_{n-1}\Delta X_n,

\\

Z_0 &=1.

\end{cases}

$$

where $\Delta Y_n$ means $Y_n-Y_{n-1}$ for any discrete process $(Y_n)_{n\ge 0}$. That can be solved by recurence in the form :

$$

Z_n=\prod_{i=0}^{n}(1 +\Delta X_i)

$$

with the convention $\Delta X_0=0$, so that $Z_0=1$.

Notice that when expressing the solution for time continuous pure jumps semi-martingale you get almost the same answer ( check Jacod, Shiryaev “Limit Theorem for Stochastic Processes” at the end of Chapter 1).

(note that this exponential can take négative values !!!)

Don’t know if this helps

Best regards

**Attribution***Source : Link , Question Author : Ben Derrett , Answer Author : Ilya*