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

What are the most fundamental/useful/interesting ways in which the concepts of Brownian motion, martingales and markov chains are related?

I'm a graduate student doing a crash course in probability and stochastic analysis. At the moment, the world of probability is a confusing blur, but I'm starting with a grounding in the basic theory of markov chains, martingales and Brownian motion. While I've done a fair amount of analysis, I have almost no experience in these other matters and while understanding the definitions on their own isn't too difficult, the big picture is a long way away.

I would like to gather together results and heuristics, each of which links together two or more of Brownian motion, martingales and Markov chains in some way. Answers which relate probability to real or complex analysis would also be welcome, such as "Result X about martingales is much like the basic fact Y about sequences".

The thread may go on to contain a Big List in which each answer is the posters' favourite as yet unspecified result of the form "This expression related to a markov chain is always a martingale because blah. It represents the intuitive idea that blah".

Because I know little, I can't gauge the worthiness of this question very well so apologies in advance if it is deemed untenable by the MO police.

share|improve this question
    
How about this: Brownian motion is a Markov process and a martingale. While you're at it, consider the representation of continuous martingales as time changes of Brownian motion: en.wikipedia.org/wiki/Wiener_process#Time_change –  Steve Huntsman Sep 29 '10 at 17:40
2  
If you're a grad student at Cambridge, have you tried looking in some combination of James Norris' Markov Chains book and David Williams' Probability with Martingales? The former grew out of the Cambridge lecture course I think. –  Yemon Choi Sep 29 '10 at 18:21
    
Changed to community - I had intended to but forgot when I posted it. Yes I am familiar with both of those books and am looking at them at the momnet; I don't really intend for this question to be a substitute to learning things 'properly' (from e.g. books), but thought it would be fun and the answers would be nice tid bits with which to test my understanding and flesh out the big picture a bit (not that's really taken off (yet)). –  Spencer Sep 29 '10 at 21:17
4  
There's not really a question here, so this post isn't well-suited to MO. This sort of big-picture exploration is better done in person than on this forum. Since you're at Cambridge, I recommend going to somebody like Geoffrey Grimmett and asking him to regale you with tales of probability. –  Tom LaGatta Oct 1 '10 at 16:22

5 Answers 5

Hi,

Regarding Martingales you can see them as fair games This means that if the (martingale) process represents your (random) wealth, you should not be able to design a strategy to increase your current wealth, no matter what the outcome of the sample space is.

Brownian Motion can be seen as a limit of rather simple random walks but I'm sure that you know about this.

Markov processes "disconnect" Future and Past of the process conditionnally on the present value of the process. Where "disconnect" means that functions of past and of future values of the process are independent conditionnally on the present value of the process.

Does it make things more clear ?

share|improve this answer

Levy's characterisation of Brownian motion:

If $X$ is a continuous martingale and $X$ has quadratic variation process $[ X ]_t = t$ then $X$ is a standard Brownian motion.

share|improve this answer

Let $(B_t)_{t \geq 0}$ be a Brownian motion and $\mathcal B_t:=\sigma(\{B_s : s \leq t\})$ its natural filtration. Then

  • $B_t$ is a martingale,
  • $B_t^2-t$ is a martingale,
  • $\exp(\theta B_t - \frac{1}{2}\theta^2 t)$ is a martingale, and
  • $H_n(t,B_t)$ is a martingale for every $n$, where $H_n(t,x)$ is the Hermite polynomial defined by $$ \exp(\theta x - \frac{1}{2}\theta^2 t) = \sum_{n=0}^{\infty} \frac{\theta^n}{n!} H_n(t,x).$$ Note that $H_1(t,x)=x$ and $H_2(t,x)=x^2-t$, so this latter statement implies the first two.

When writing "is a martingale", it is obviously "with respect to the filtration $(\mathcal B_t)_{t\geq 0}$".

A remarkable consequence of the Levy's characterization of Brownian motion is that every continuous martingale is a time-change of Brownian motion.

Source: L.C.G. Rogers and D. Williams, Diffusion, Markov Processes and Martingales, Vol.1 (2000)

share|improve this answer

If X is a continuous martingale of finite variation such that $X_0 = 0$, then $P(X_t = 0 \ \ \forall t) = 1$.

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.