Is the truncated Brownian motion of the class DL? - MathOverflow

Is the truncated Brownian motion of the class DL?

2010-06-04T12:26:24Z

Let $W$ be a standard Brownian motion under given probability space. For a given constant $a$, $W^a$ is a truncated Brownian motion by stopping time $T^a = \inf(t>0:W(t) = a)$. That is, $W^a(t) = W(t \wedge T^a)$.
We want to consider the following question: Is the process $W^1$ of the class DL?

(Solution1): Yes. Indeed, for any fixed $t>0$, we can prove the collection of random variables $( W(s), 0< s< t)$ is uniformly integrable by definition, since $E [|W^1(t)|] < \infty$.

We provide completely different answer using the following proposition from the Problem 1.5.19 (i) of Book [Karazas and Shereve 98].

[Proposition] A local martingale of class DL is martingale.

(Solution2): No. $W^1$ is strict local martingale, since $E [W^1(T^1)] = 1> E [W(0)]$. By [Proposition], $W^1$ is not of class DL.

In the above, we obtained completely two different solutions. Where is wrong?

Answer by Steve Huntsman for Is the truncated Brownian motion of the class DL?

2010-06-04T13:40:07Z

$W^a$ is in fact a martingale. To see this, write $W^a(t) = W(t \land T_a)$. See also Theorem 3.39 here.

When you write an expression like $\mathbb{E}(W^a(T^a))$ you are implicitly assuming that $W^a(T^a)$ is measurable. This requires $t \ge T_a$ (and trivializes the expectation).

Answer by The Bridge for Is the truncated Brownian motion of the class DL?

2010-06-15T11:02:47Z

Hi kenneth

Have a look at the following document http://www.ma.utexas.edu/users/gordanz/teaching/10_Spring_M385D/lecture16.pdf (In particular to Propositions 16.24,16.25, 16.26, and 16.30)

First $W^1$ will be of class DL as soon as it is a martingale by proposition 16.25. So showing that $W^1$ is a martingale is sufficient to prove your claim (which follows from Proposition 16.30 for example if you know that a Brownian Motion is a martingale)

Second here is why Solution 2 doesn't work By proposition 16.26 if $M_\tau$ is in L^1 for every bounded stopping times (which is the case here).

We have $X_t$ is a martingale if $E[M_\tau]=E[M_0]$ for every bounded stopping time $\tau$.

This is the criteria you are trying to apply to get your contradiction.

The problem with this, is that $T_1$ is not bounded almost surely so you cannot apply the preceding criteria to show that $W^1$ is not of class DL.

I hope I didn't make any mistake

Regards