4
$\begingroup$

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?

Improve this question
$\endgroup$
4
  • 1
    $\begingroup$ BTW, you should define DL. I know it's in Karatzas and Shreve (and viewable on Amazon) but it's not a common (or perhaps even standard?) property. $\endgroup$
    – Steve Huntsman
    Commented Jun 4, 2010 at 14:07
  • 1
    $\begingroup$ Note that $E[|W^1(t)|] < \infty$ is not sufficient for uniform integrability. However, $\sup_{0 < s < t} E[|W^1(t)|^2] < \infty$ is. $\endgroup$
    – Nate Eldredge
    Commented Jun 4, 2010 at 14:49
  • $\begingroup$ You are right, $E[|W^1(t)|]<\infty$ is not sufficient. But, it would be sufficient together with following proposition from Durrett book: Suppose $X\in L^1(\Omega, \mathbb F, \mathbb P)$ and $\mathbb F_{\tau} \subset \mathbb F$ for all $\tau$, then $Y_\tau = E[X|\mathbb F_{\tau}]$ is uniformly integrable. In the above, we may treat $W^1(\tau)=E[W^1(t)|\mathbb F_{\tau}]$ for any $\tau< t$. $\endgroup$
    – kenneth
    Commented Jun 4, 2010 at 15:35
  • $\begingroup$ Yes, there are lots of reasons why this process is UI on any interval $[0,t]$. But one should keep the definitions straight, in particular so that it is clear that it is not UI on $[0,\infty)$ (although $E[W^1(t)] < \infty$ for each $t$). $\endgroup$
    – Nate Eldredge
    Commented Jun 6, 2010 at 0:23

2 Answers 2

Reset to default
3
$\begingroup$

$W^a$ 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).

Improve this answer
$\endgroup$
2
  • 2
    $\begingroup$ Perhaps the OP is thinking that if $W^a$ were a martingale, we should have $E[W^a(T^a)]=E[W^a(0)]=0$ by the optional sampling theorem (e.g. Karatzas and Shreve 1.3.22). But the optional sampling theorem is not applicable here because $W^a$ does not have a "last element". $\endgroup$
    – Nate Eldredge
    Commented Jun 4, 2010 at 14:46
  • $\begingroup$ Thank you, Nate. I agree that $W^1$ is martingale, and optional sampling does not apply here. Hence, although $\mathbb{E} [W^1(T^1)] = 1 > W^1(0)$ is correct, it shall not become a reason for the strict local martingale. $\endgroup$
    – kenneth
    Commented Jun 4, 2010 at 15:57
1
$\begingroup$

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

Improve this answer
$\endgroup$
3
  • $\begingroup$ Hi, Bridge Thank you for your careful answer, and I totally agree with you. But I still have a question on the Optional Sampling Theorem in book [Karazas and Shreve 1998]: For a martingale $X_t$ with last element, $E X_T = X_0$ for stopping time $T$. My understanding is that, optional sampling works on "bounded" $T$. Otherwise, we reach following contradiction: $W^1$ is a martingale with last element $W^1_\infty = 1$. $T^1$ is a stopping time. But $E W^1(T^1) > W^1(0)$. Did I misread something from the book [Karazas and Shreve 1998]? $\endgroup$
    – kenneth
    Commented Jun 16, 2010 at 7:15
  • $\begingroup$ Kenneth, You are right, Optional Sampling Theorem works for bounded stopping for general martingales. But if you had Unifor Integrability condition on your martingales, then you don't need the boundedness condition on the stopping times and the result is applicable in full generality for any stopping time Regards $\endgroup$
    – The Bridge
    Commented Jun 16, 2010 at 7:43
  • $\begingroup$ Yeah, a martingale of class D satisfies optional sampling as well. So $W^1$ is of class DL but not of class D. But, I did not see either of two requirements in this theorem from the book: bounded stopping time or uniform integrability. Thank you, Bridge. $\endgroup$
    – kenneth
    Commented Jun 16, 2010 at 13:17

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .