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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)?


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2 Answers 2

I found a reference to a discrete time result similar the EMI in Exercise 3.11 of the excellent lecture notes Ramon van Handel, Probability in High Dimension (Princeton University, 2014)

The statement of the result is: Let $(M_n)$ be an $(\mathcal F_n)$-martingale, such that for all $n \geq 1$ there exist $\mathcal F_{n-1}$-measurable random variables $A_n, B_n$ such that $A_n \leq M_n - M_{n-1} \leq B_n$, almost surely. Then \begin{equation} \mathbb P\left[ M_n \geq t \ \mbox{and} \ \sum_{k=1}^n (B_k - A_k)^2 \leq c^2 \right] \leq e^{-2t^2/c^2}. \end{equation}

The exercise gives the hint to consider $\lambda M_n - \frac{\lambda^2}{8} \sum_{k=1}^n (B_k - A_k)^2$.

Indeed, we have \begin{align*} \mathbb P \left[ M_n \geq t \ \mbox{and} \ \sum_{k=1}^n (B_k - A_k)^2 \leq c^2 \right] & \leq \mathbb P \left[ \lambda M_n - \frac{\lambda^2}{8} \sum_{k=1}^n (B_k-A_k)^2 \geq \lambda t - \frac{\lambda^2 c^2}{8}\right] \\ & \leq \exp \left( -\lambda t+ \frac{\lambda^2 c^2}{8}\right) \mathbb E\left[ \exp\left( \lambda M_n - \frac{\lambda^2}{8} \sum_{k=1}^n (B_k-A_k)^2 \right) \right]. \end{align*} The Hoeffding lemma states that for $a \leq X \leq b$ a.s., $\mathbb E \left[ \exp( \lambda(X - \mathbb E X ) ) \right] \leq e^{\lambda^2(b-a)^2/8}$. Therefore \begin{align*} \mathbb E \left[ \exp\left( \lambda M_n - \frac{\lambda^2}{8} \sum_{k=1}^n (B_k-A_k)^2 \right)\mid \mathcal F_{n-1} \right] \leq \exp \left( \lambda M_{n-1} - \frac{\lambda^2}{8} \sum_{k=1}^{n-1} (B_k-A_k)^2 \right) \end{align*} and iterating gives that the expectation above is $\leq 1$. Optimizing over $\lambda$ gives the result.

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See also Dzhaparidze, van Zanten - On Bernstein-type inequalities for martingales, 2001. –  Joris Bierkens Nov 14 '14 at 17:39

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

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