Let$$M_t = \max\{B_s : 0 \le s \le t\},\text{ }m_t = \min\{B_s : 0 \le s \le t\},$$where $B_t$ is a standard Brownian motion. My question is, does there exist $r$ such that with probability one,$$\limsup_{t \to \infty} {{M_t - m_t}\over{\sqrt{t \log \log t}}} = r?$$If so, what is $r$?
2 Answers
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Does there exist $r$ such that with probability one,$$\limsup_{t \to \infty} {{M_t - m_t}\over{\sqrt{t \log \log t}}} = r?$$
Yes, such an $r$ does exist. First note that by the original LIL, with probability one, $$\sqrt{2} = \limsup_{t \to \infty} {{M_t}\over{\sqrt{t \log \log t}}} \le \alpha := \limsup_{t \to \infty} {{M_t - m_t}\over{\sqrt{t \log \log t}}}$$
$$\le \limsup_{t \to \infty} {{M_t}\over{\sqrt{t \log \log t}}} + \limsup_{t \to \infty} {{- m_t}\over{\sqrt{t \log \log t}}} = 2\sqrt{2}.$$ So with probability 1, $\sqrt{2}\le \alpha\le 2\sqrt{2}$. On the other hand, for each fixed $r$, the event $\alpha\le r$ is a tail event, hence it must have probability either 0 or 1. Done.
answered Nov 1, 2015 at 1:46
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You can use the fact that $B_t-m_t$ is a reflected Brownian motion (see e.g. Revuz-Yor, Chapter VI, Theorem 2.3). I think it shouldn't be difficult to show that $$ \limsup_{t\to\infty} \frac{M_t-m_t}{\sqrt{t\log\log t}} = \limsup_{t\to\infty} \frac{B_t-m_t}{\sqrt{t\log\log t}}. $$
