Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

1 Answer 1

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.