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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 $r$?

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  • $\begingroup$ This is a homework exercise, please do not respond. $\endgroup$ Nov 2 '15 at 20:39
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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.

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  • $\begingroup$ so what is $r$? $\endgroup$
    – Fan Zheng
    Nov 1 '15 at 2:33
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    $\begingroup$ Assume CH if you like... $\endgroup$
    – Fan Zheng
    Nov 1 '15 at 2:56
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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 $$ \limsup_{t\to\infty} \frac{M_t-m_t}{\sqrt{t\log\log t}} = \limsup_{t\to\infty} \frac{B_t-m_t}{\sqrt{t\log\log t}}. $$

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  • $\begingroup$ so what is $r $? $\endgroup$ Nov 1 '15 at 16:51
  • $\begingroup$ $\sqrt{2}$, I guess $\endgroup$ Nov 1 '15 at 17:12
  • $\begingroup$ I agree, using the strassen LIL $\endgroup$
    – Michael
    Nov 2 '15 at 12:08

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