I am doing an exercise question: define $$ M_{t}=\max \left\{B_{s}: 0 \leq s \leq t\right\}, \quad m_{t}=\min \left\{B_{s}: 0 \leq s \leq t\right\} $$ we are asked to find a number $r$ such that with probability one $$ \limsup _{t \rightarrow \infty} \frac{M_{t}-m_{t}}{\sqrt{t \log \log t}}=r $$ My attempt is that: due to the symmetry of $M_t$ and $m_t$, we can reduce the problem to find a number $r$ such that with probability one $$ \limsup _{t \rightarrow \infty} \frac{2M_{t}}{\sqrt{t \log \log t}}=r $$ I guess the constant $r$ is somehow related with $\sqrt{2}$ by "Law of the Iterated Logarithm", and perhaps we need to prove from two direction: "LHS $\leq r$" and "LHS $\geq r$", first direction is relatively easy because we can pick a large number but the other direction is no idea. Can anybody help? Thanks a lot!
I run the simulation from which I guess $r=2\sqrt{2}$, and I am able to show "LHS$\leq 2\sqrt{2}$" but I cannot show the other side.