Is the truncated Brownian motion of the class DL? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T03:15:04Z http://mathoverflow.net/feeds/question/27042 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/27042/is-the-truncated-brownian-motion-of-the-class-dl Is the truncated Brownian motion of the class DL? kenneth 2010-06-04T12:26:24Z 2010-09-29T05:43:35Z <p>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, $W^a(t) = W(t \wedge T^a)$.<br> We want to consider the following question: Is the process $W^1$ of <a href="http://almostsure.wordpress.com/2009/12/22/class-d-processes/" rel="nofollow">the class DL</a>?</p> <p>(Solution1): Yes. Indeed, for any fixed $t>0$, we can prove the collection of random variables $( W(s), 0&lt; s&lt; t)$ is uniformly integrable by definition, since $E [|W^1(t)|] &lt; \infty$.</p> <p>We provide completely different answer using the following proposition from the Problem 1.5.19 (i) of Book [Karazas and Shereve 98].</p> <p>[Proposition] A local martingale of class DL is martingale.</p> <p>(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.</p> <p>In the above, we obtained completely two different solutions. Where is wrong?</p> http://mathoverflow.net/questions/27042/is-the-truncated-brownian-motion-of-the-class-dl/27046#27046 Answer by Steve Huntsman for Is the truncated Brownian motion of the class DL? Steve Huntsman 2010-06-04T13:40:07Z 2010-06-04T13:40:07Z <p>$W^a$ is in fact a martingale. To see this, write $W^a(t) = W(t \land T_a)$. See also Theorem 3.39 <a href="http://books.google.com/books?id=JYzW0uqQxB0C&amp;pg=PA85" rel="nofollow">here</a>.</p> <p>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).</p> http://mathoverflow.net/questions/27042/is-the-truncated-brownian-motion-of-the-class-dl/28238#28238 Answer by The Bridge for Is the truncated Brownian motion of the class DL? The Bridge 2010-06-15T11:02:47Z 2010-06-16T15:14:19Z <p>Hi kenneth</p> <p>Have a look at the following document <a href="http://www.ma.utexas.edu/users/gordanz/teaching/10_Spring_M385D/lecture16.pdf" rel="nofollow">http://www.ma.utexas.edu/users/gordanz/teaching/10_Spring_M385D/lecture16.pdf</a> (In particular to Propositions 16.24,16.25, 16.26, and 16.30)</p> <p>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)</p> <p>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).</p> <p>We have $X_t$ is a martingale if $E[M_\tau]=E[M_0]$ for every bounded stopping time $\tau$.</p> <p>This is the criteria you are trying to apply to get your contradiction.</p> <p>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.</p> <p>I hope I didn't make any mistake </p> <p>Regards</p>