Serving the Quantitative Finance Community

 
User avatar
islandboy
Topic Author
Posts: 0
Joined: December 21st, 2005, 4:35 pm

stopping time, Fn measurability, and the stochastic process

August 1st, 2006, 8:53 pm

I'm having difficulty answering this question posed to us by our teacher. It deals generally with proving something is Fn measurable which I really have a hard time grasping the concept of.Here's the question:Let b a filtered probability space. A random variable on is a stopping time for F if and only if for any , the event . Assume is finite and let be a stochastic process. Let A be a subset of R, the real numbers, and for any .Define to be the random variable defined by:Show thata) (i.e, it is Ft measurable)b) intuitively, both makes sense. Its just I have a hard time writing down the formal proofs needed.For a) Is Ft measurable because St is a stochastic process? I really have a hard time proving Ft measurability. I know the formal definition of Ft measurability is that I must show for any is true for any a, element of real number.So I was thinking, here's how I would approach the problemI make different cases:case (i) if a is not in A but a is less than any elements of A, then w| St(w) >= A is the empty setcase(ii) if a is not in A but a is greater than any elements of A, then w| St(w) >= A is case(iii) if a is in A, then w|St(w) >=A is in St (which is a stochastic process, thus is Ft measurablefor all 3 cases, w|St(w) >= A is Ft measurable, thus I prove that is Ft measurable.Is this correct?
Last edited by islandboy on July 31st, 2006, 10:00 pm, edited 1 time in total.
 
User avatar
islandboy
Topic Author
Posts: 0
Joined: December 21st, 2005, 4:35 pm

stopping time, Fn measurability, and the stochastic process

August 1st, 2006, 9:35 pm

Continuing...for b, I must show that each side of the equation is contained in the other. Here's how I approached the problem: is simply (since T+1 is always greater than t0). This is always a by definition of . And since is just an element of , I have shown that is contained in For the other way, first, I get a general element of . Let's call this element x. By definition, x has the property that St(x) is in A. Thus x is in any w satisfying . If this is so, then x is always contained in any w satisfying since must be true if we want .Is this reasoning/proof correct? Or did I miss something. how can I write out this proff more rigorously?Thanks
Last edited by islandboy on July 31st, 2006, 10:00 pm, edited 1 time in total.
 
User avatar
islandboy
Topic Author
Posts: 0
Joined: December 21st, 2005, 4:35 pm

stopping time, Fn measurability, and the stochastic process

August 6th, 2006, 5:46 am

bump...anyone?