# The law of the iterated logarithm for BM and boundedness of stopping times

My question is regarding the usefulness of the law of the iterated logarithm, and its connection to stopping times. In many answers of this forum, I understand that some people often claim that some stopping times, such as $$\tau_n = \inf\left\{t \geq 0 : B_t = n\right\}$$ are finite almost surely because of the law of the iterated logarithm. I am familiar with the law of course, but I don't really see how we can argue that any stopping time is bounded if, say, in the example above $n$ is very large.

This takes me to ask the following two questions: does the law assure that $\mathscr{F}_{t}$-measurable stopping times are always bounded? Is there an example when the Law of the Iterated Logarithm doesn't help us to conclude that a stopping time is not bounded almost surely?

• For this particular example, using the LIL is overkill. Commented May 19, 2014 at 15:00

The law of the iterated logarithm states that almost surely: $$\limsup_{t \to \infty} \frac{|B_t|}{\sqrt{2t\log(\log(t))}} = 1$$

So asymptotically we have that $|B_t| \approx \sqrt{2t\log(\log(t))}$, and in particular $\limsup_{t \to \infty} |B_t| = \infty$. Since $P(B_t > 0, \forall t > 0) = 0$, the brownian motion must cross infinite times the zero line (use the Markov property), and since $\limsup_{t \to \infty} |B_t| = \infty$ we have that $\limsup B_t = \infty$ and $\liminf B_t = -\infty$. Finally by continuity of the paths you can conclude that any level $c$ must be attained.

• thanks for the answer. Using this same argument, can you give an example of a stopping time that can't be bounded using this law? Commented May 19, 2014 at 10:30
• Hand waving a little bit: An Orstein-Uhlenbeck process driven by a $F_t$-Brownian motion is a a $F_t$-semi martingale and thus its first passage times $\tau_n'$ are $F_t$ stopping times. But the OU processes converges almost surely to its $\mu$ parameter, and there will be first passage times $\tau_n'$ that have positive probability of being infinite. Commented May 19, 2014 at 12:43
• It is not very rigorous to say "and in particular $|B_t|\rightarrow\infty$ as $t\rightarrow\infty$". What do you mean by that? Also, the OU processes converge to a stationary distribution which is centered around $\mu$, they do not "converge almost surely to their parameter". An example of a stopping time of the Brownian motion that "can't be bounded with the law of the iterated logarithm" would be $\inf\{t\ge0\ :\ B_t=t\}$, i.e. the hitting time of the line $y=x$ by the BM. This stopping time is not finite almost surely.
– Ian
Commented May 19, 2014 at 14:15
• Maybe some confusion is also arising because you think the stopping times are bounded. They are not. In general, even the hitting time of a level $n$ is unbounded. It can only be shown to be finite almost surely.
– Ian
Commented May 19, 2014 at 14:18
• @Ian you are right. I'll correct it. Commented May 19, 2014 at 14:40