most recent 30 from http://mathoverflow.net 2013-06-19T07:24:22Z http://mathoverflow.net/feeds/question/110585 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/110585/modification-of-doob-inequality Higgs88 2012-10-24T20:36:05Z 2012-10-24T22:52:06Z

<p>Hi everyone, please consider the following problem:</p> <p>Let $(M_t)_{t\geq 0}$ be a continuous and positive submartingale and $S_t=\sup_{0\leq s\leq t}M_s$. Please prove that for any $\lambda>0$ we have</p> <p>$$\lambda P(S_t>2\lambda)\leq E[M_t1_{{M_t>\lambda}}]$$</p> <p>This inequality makes me remember the Doob inequality</p> <p>$$\lambda P(S_t>\lambda)\leq E[M_t1_{{S_t>\lambda}}]$$</p> <p>So it is enough to show that </p> <p>$$E[M_t1_{{S_t>\lambda}}]\leq 2E[M_t1_{{M_t>\lambda}}]$$</p> <p>But I have no idea to deal about the term $1_{{M_t>\lambda}}$, even by introducing a stopping time $T_{\lambda}$. Could someone help me prove this inequality or give some idea? Many thanks!</p>

http://mathoverflow.net/questions/110585/modification-of-doob-inequality/110593#110593 Pascal Maillard 2012-10-24T22:52:06Z 2012-10-24T22:52:06Z

<p>Nice little exercise, it should go to math.stackexchange.com, though. I'll give you a hint: Let $T$ be the hitting time of $2\lambda$. Then, $E[M_t 1_{M_t\ge \lambda}] \ge E[M_t 1_{M_t\ge \lambda,\ T\le t}] \ge E[M_t1_{T\le t}]- E[M_t1_{M_t\le \lambda}1_{T\le t}].$</p>