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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, $W^a(t) = W(t ) 1_{(t\le T^a)} + W(T^a) 1_{(t> T^a)}$\wedge T^a)$. (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? |
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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
$T^a = \inf{t>0: W(t) inf(t>0:W(t) = a}$a)$.
That is, $W^a(t) = W(t) 1_{{t\le T^a}1_{(t\le T^a)} + W(T^a) 1_{{t> T^a}}$1_{(t> T^a)}$. [Solution1]: (Solution1): Yes. Indeed, for any fixed $t>0$, we can prove the collection of random variables ${W(s):0 ( W(s), 0< s< t)$ 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]: (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. Which one Where is not correctwrong? |
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