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Sharp maximal inequalities for conditionally symmetric martingales and Brownian motion


Author: Gang Wang
Journal: Proc. Amer. Math. Soc. 112 (1991), 579-586
MSC: Primary 60G42; Secondary 60J65
DOI: https://doi.org/10.1090/S0002-9939-1991-1059638-8
MathSciNet review: 1059638
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Abstract: Let $ B = {({B_t})_{t \geq 0}}$ be a standard Brownian motion. For $ c > 0$, $ k > 0$, let

\begin{displaymath}\begin{gathered}T(c,k) = \inf \{ t \geq 0:{\max _{s \leq t}}{... ...}\vert - c\vert{B_t}\vert \geq k\} . \hfill \\ \end{gathered} .\end{displaymath}

We show that for $ c > 0$ and $ k > 0$, both $ T(c,k)$ and $ {T^*}(c,k)$ are finite almost everywhere. Moreover, $ T(c,k)$ and $ {T^*}(c,k) \in {L^{p/2}}$ if and only if $ c < p/(p - 1)$ for $ p > 1$, and for all $ c > 0$ when $ p \leq 1$. These results have analogues for simple random walks. As a consequence, if $ T$ is any stopping time of $ {B_t}$ such that $ {({B_{T \wedge t}})_{t \geq 0}}$ is uniformly integrable, then both of the inequalities

\begin{displaymath}\begin{array}{*{20}{c}} {\vert\vert{{\sup }_{s \leq T}}{B_s}\... ...{{p - 1}}{{\left\Vert {{B_T}} \right\Vert}_p},} \\ \end{array} \end{displaymath}

are sharp. This implies that $ q = p/(p - 1)$ is not only the best constant for Doob's maximal inequality for general martingales but also for conditionally symmetric martingales (in particular, for dyadic martingales), and for Brownian motion.

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Additional Information

DOI: https://doi.org/10.1090/S0002-9939-1991-1059638-8
Keywords: Martingale, conditionally symmetric martingale, dyadic martingale, Brownian motion, martingale maximal function
Article copyright: © Copyright 1991 American Mathematical Society

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