Serving the Quantitative Finance Community

 
User avatar
contradanza
Topic Author
Posts: 0
Joined: March 25th, 2008, 7:19 pm

Strong Markov property

August 15th, 2009, 8:21 am

Hello,I have a question on stochastic calculus, which I became interested about. I am now studying the subject with 'Stochastic calculus and Brownian motion' book of Karatzas and Shreve. When I started with weak solutions of SDE's I was quite intriguied by the fact that a weak solution is a strong Markov process only when the SDE has a unique (in law) weak solution. To my mind one could expect all weak solutions to be strong Markov. Can somebody explain me this fact intuitively?Thanks in advance.
 
User avatar
RDK
Posts: 0
Joined: September 14th, 2008, 10:45 pm

Strong Markov property

August 15th, 2009, 2:28 pm

This is just what I thought, I'm not sure if it's right. When you have a weak solution to a diffusion, you have a measure on the space of continuous functions, which is equivalent to specifying all finite dimensional distributions of the process. But if a stopping time takes on uncountably many values, then the finite dimensional distributions don't give you enough information. In other words, if you have a strong markov process and you take a modification of it, the new process need not be strong markov, so a priori weak solutions don't tell you anything about the strong markov property.When you do have uniqueness, you want to show that under (say) has the same law as X under , where T is some stopping time. But KS show in one of the preceding lemmas that the measure induced by the first process solves the martingale problem for the second process. By uniqueness, they conclude that those two measures must be the same. I don't really have an intuitive feel for this part though.
 
User avatar
Alan
Posts: 3050
Joined: December 19th, 2001, 4:01 am
Location: California
Contact:

Strong Markov property

August 15th, 2009, 5:30 pm

contradanza,What's the specific theorem (and page) number in K&S for your statement?
 
User avatar
contradanza
Topic Author
Posts: 0
Joined: March 25th, 2008, 7:19 pm

Strong Markov property

August 19th, 2009, 8:38 am

Thank you for comments, RDK. QuoteWhat's the specific theorem (and page) number in K&S for your statement?Section 5.5.4.C Theorem 4.20In this theorem the authors assume that the martingale problem is well posed, which is equivalent to the well-posedness of the SDE in the context of locally bounded coefficients.Looked it up in some other books as well. Uniqueness seems to be an indispensable requirement.
 
User avatar
list
Posts: 0
Joined: October 26th, 2005, 2:08 pm

Strong Markov property

August 19th, 2009, 1:03 pm

Is that a known fact that solution of the SDE is Markovian in case when the uniqueness ( either in strong or weak sense ) does not exist?