Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Background

I work in a subfield of computability theory called algorithmic randomness. We have been using martingales as long as probability theory (going back to work of von Mises). However, since algorithmic randomness is often about sequences of $0$s and $1$s, the martingales are usually dyadic martingales. Also they are usually nonnegative. We've developed our our terminology and tricks separately from probability theory.

The savings property

A common transformation in algorithmic randomness is called the "savings trick" or "savings property". The main idea is that the gambler saves all but one dollar of her money and bets with the rest using the strategy of the original martingale. More formally, if $(M_n)$ is a nonnegative martingale, consider the two processes defined recursively as follows (here $N_n$ is the new martingale and $L_n$ is the "savings account").

  • $N_0=M_0+1, \quad L_0=0$,
  • $\displaystyle{N_{n+1}=N_n+(M_{n+1}-M_n)\cdot \frac{N_n-L_n}{M_n+1}}, \quad L_{n+1}=\max \{N_{n+1}-1,L_n\}$

It can be shown that

  1. $N_n$ is a nonnegative martingale.
  2. $\limsup_n M_n = \infty$ iff $\lim_n N_n = \infty$.
  3. $N_n$ is uniformly integrable, and hence $N_n = \mathbb{E}[X\mid \mathcal {G}_n]$ for some integrable $X$.
  4. (I think it is true that...) if $d\mathbb{Q} = X\, d\mathbb{P}$ then there is a modulus of absolute continuity for $\mathbb {Q} \ll \mathbb{P}$ that depends only on the filtered probability space $(\Omega, \mathcal{B}, \mathbb{P}, (\mathcal{G}_n))$.

My question

Is there anything like this in probability theory? Namely a martingale transformation that takes in a nonnegative martingale and returns a new martingale satisfying (1) through (3) (or (1) through (4)).

I am hoping to find something similar to the one above, but that is cleaner and easier to work with, or just more studied.

share|improve this question

Your Answer

 
discard

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

Browse other questions tagged or ask your own question.