For Brownian motion $B_t,$ we have $\lim_{t\to \infty}\frac{B_t}{t^{\alpha}}=0$ almost surely.

Question

This might be a silly problem. I am trying to show that $\lim_{t\to \infty}\frac{B_t}{t^{\alpha}}=0$ almost surely if $\alpha>\frac{1}{2}.$ I can apply Chebyshev's inequality to show that $P(|B_{t}/t^{\alpha}|>\epsilon)\to 0$ as $t\to \infty.$ This tells me that $\frac{B_t}{t^{\alpha}}$ converges to $0$ in probability. But I am not sure how to argue the almost sure limit. I found at the very beginning of a book I am reading, so I am looking forward to something which does not use $0-1$ laws if that's possible. I am wondering if this can be done cheaply using the Borel-Cantelli lemma.

Answer 1

I do not think this is easy at all from the ground up, but it can be derived from some other well known facts, which are perhaps proved in your text.

Let $W_t=tB_{1/t}$. It is well known that $W_t$ is also a Brownian motion. Then $$\lim_{t\to \infty}B_{t}/t^{\alpha}=\lim_{t\to\infty}W_{1/t}/t^{\alpha-1}=\lim_{s\to0}W_s/s^{1-\alpha}$$ The fact that the last limit is zero is the definition of $W_s$ being Hölder continuous at $0$ with exponent $1-\alpha$. The fact that $W_s$ is Hölder continuous for all exponents less than $\frac12$ is often proved early on in textbooks.

Answer 2

Let $\beta=(\alpha-1/2)/2$. By the law of iterated logarithm, $$ \limsup_{t\to\infty}\frac{|B_t|}{t^\alpha}\le \limsup_{t\to\infty}\frac{|B_t|}{t^{ \beta}\sqrt{t\ln\ln t}}=\lim_{t\to\infty}\frac{1}{t^{\beta}}\times\limsup_{t\to\infty}\frac{|B_t|}{\sqrt{t\ln\ln t}}=0\quad\text{a.s.} $$