expectation via martingale

motszi · February 23rd, 2009, 12:56 pm

I have a dumb question, but can't seem to find the answer. I come from a pde background and I'm new to martingales. I know how they are defined and how to check whether a process is a martingale, but how do you use it as a tool to calculate the expectation of a function? For concreteness, let's say:I know the answer is from doing the integral, but using martingales?

brucebo · February 23rd, 2009, 1:38 pm

I think the right answer is . In fact, if we define , then M_t admits SDE , hence it is a martingale. Thus . Using Ito rule, .

motszi · February 23rd, 2009, 2:15 pm

Yeah sorry, mine was a typo in the exponent.Right, thanks for the explanation. So the key in general is to find a known martingale form similar to your function, rather than a general method which works every time?

brucebo · February 23rd, 2009, 11:05 pm

Oh, yes. It is not a general approach which works in any forms.

motszi · February 24th, 2009, 9:45 am

Strange though that people say how much quicker it is via martingales. In this case doing the integral is slightly quicker, even though you can arrange a product of two martingales.

brucebo · February 24th, 2009, 10:36 am

It's known that B.M. is a distributed r.v.. This makes the interal method available. However, the martingale argument seems to work in a gereral case , where M is a martingale with non-random quadratic variance.

motszi · February 24th, 2009, 11:22 am

yeah that's a good point. I also see how even a function like starts to get messy with the integral approach, whereas using and gets us there painlessly.Well thanks for your help brucebo.