Let $W_t$ be a Brownian motion. How do I show the following?

$$ \alpha > \frac{1}{2} \Rightarrow \lim_{t\rightarrow\infty} \frac{W_t}{t^{\alpha}} = 0 \text{ a.s.} $$

Showing convergence of this in probability is easy enough, but I am not sure of how I could get the stronger a.s. result. By (super)martingale convergence I know that the limit is almost surely finite, but how do I show that it is equal to $0$?

  • 2
    $\begingroup$ If $X \to Y$ almost surely and $E(X^2) \to 0$, we have $Y=0$ by Fatou's lemma $\endgroup$ – Petite Etincelle Mar 20 '15 at 8:30
  • $\begingroup$ Thanks, you are right, that does it. If you wish to post that as an answer I will mark it as the solution. $\endgroup$ – em70 Mar 20 '15 at 9:48
  • $\begingroup$ What is the super martingale under consideration? $\endgroup$ – minion May 11 '16 at 1:26

Exploiting the martingale property of the Brownian motion, we can show that for each positive $\varepsilon$, the inequality $$\mathbb P\left\{\sup_{2^n\leqslant t\leqslant 2^{n+1} }\left|W_t\right| \gt \varepsilon 2^{n\alpha} \right\}\leqslant 4 \varepsilon^{-2} 2^{n(1-2\alpha)},$$ holds. We then conclude by the Borel-Cantelli lemma.

  • $\begingroup$ Are you using Kolmogorov Maximal Inequality? $\endgroup$ – minion May 11 '16 at 1:36
  • $\begingroup$ @minion I don't know the name, maybe it refers to Doob's maximal inequality. $\endgroup$ – Davide Giraudo May 11 '16 at 8:01

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