Let $\{B_t:t\in [0,1]\}$ be the standard one-dimensional Brownian motion on the closed unit interval. Fix $\gamma\in (0,1/2)$. It is well known that there is a positive random variable $K\equiv K(\gamma)$ such that for any pair $s,t\in [0,1]$ we have $$ |B_t-B_s|\leq K|t-s|^{\gamma} \qquad \text{a.s.} $$ I would like to know if $K$ can be chosen so that $\mathbb{E}[K]<+\infty$.
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$\begingroup$ Where you wrote "$K$ can be chosen", might you have meant "$\gamma$ can be so chosen"? $\qquad$ $\endgroup$– Michael HardyMar 4, 2016 at 18:34
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$\begingroup$ @MichaelHardy I would like to fix $\gamma$ and ask if the "best constant" $K$ we can chose for this fixed choice of $\gamma$ has finite first moment. $\endgroup$– LeandroMar 4, 2016 at 19:01
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$\begingroup$ You write of an expected value of $K$. That means $K$ is a random variable. How does it make sense to speak of choosing the value of $K$ if it's a random variable? $\qquad$ $\endgroup$– Michael HardyMar 5, 2016 at 0:45
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$\begingroup$ @MichaelHardy you can think of $K$ and $B_t$ as measurable functions on some measurable space $(\Omega,\mathcal{F})$, the Wiener space, for example. I think the point is if a r.v. $K$ fits the inequality $|B_t-B_s|\leq K|t-s|^{\gamma} \ \text{a.s.}$ for all $t,s\in [0,1]$, then the r.v. $K\cdot P$, where $P\geq 1$ is any positive r.v. also fits the above inequality. But a more interesting question is whether is possible to find a positive r.v. $0<P<1$ such that $K\cdot P$ fits the inequality and $\mathbb{E}[K\cdot P]<\infty$. Since $K>K\cdot P$ then $KP$ is a better random constant than $K$. $\endgroup$– LeandroMar 8, 2016 at 4:07
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In the text Brownian Motion by R. Schilling L. Partzsch, one can see a demonstration of the fact that, for fixed $\gamma\in(0,1/2)$, the random variable $$ K:=\sup\{|B_t-B_s|/|t-s|^\gamma: 0\le s,t\le 1\} $$ has finite moments of all orders; see Theorem 10.1 on page 150.
answered Mar 4, 2016 at 19:50