Using $\mathcal{B}$ for the standard Brownian Motion. I`m supposed to prove the following:

$$\limsup_{\delta\to0^+}\frac{1}{\delta}\mathbb{P}\Bigg(\sup_{v\in[0,1]}|\mathcal{B}(v)|>\frac{\eta}{\sqrt{\delta}}\Bigg)=0,$$ for every $\eta>0$. I`ve tried to use the distribution of the $\sup$ of a Brownian Motion together with Chebyshev's Inequality, but I couldn't just get to this result. Can anyone give a hint?

  • $\begingroup$ What happens if you evaluate the left hand side with $v=1$ instead of a sup? This would hopefully be a worst-case scenario that is drastically simplified. If it doesn't satisfy your desired equation it might give some hints for the general case. $\endgroup$ – mathematician Feb 16 '17 at 17:29
  • $\begingroup$ Tried to bound this $\sup$ by the case when $v=1$. I might have used some bad bounds, because $\delta$ just disappeared and I can't use the limit now xD $\endgroup$ – Gabriel Sanfins Feb 16 '17 at 17:40

Can use Doob's submartingale inequality and a trick to get $$ \frac{1}{\delta}P\left(\sup_{0<t<1}|B_t|>\frac{\eta}{\sqrt{\delta}}\right) = \frac{1}{\delta}P\left(\sup_{0<t<1}|B_t|^3>\frac{\eta^3}{\delta^{3/2}}\right) \le \frac{1}{\delta}\frac{E(|B_1|^3)\delta^{3/2}}{\eta^3}$$

  • $\begingroup$ Yeah, this solves the problem. Thank you very much $\endgroup$ – Gabriel Sanfins Feb 16 '17 at 17:47

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