Could someone please explain how to find the limit as $t \rightarrow \infty$ of the running maximum of a brownian motion $B_t =\max W_t-ut$? Is there a way to calculate the limit itself and not just find the distribution of the maximum? How would this intuitively make sense to be a finite limit, since wouldn't $B$ approach either positive or negative infinity?
Many thanks
We know that $max W_t = |N(0,t)|$ in distribution thus $max W_t = \sqrt{t}|\xi|$ in distribution, where $\xi \sim N(0,1)$. So $$max W_t - ut = t\Bigg(\frac{|\xi|}{\sqrt{t}} - u \Bigg)$$ where $\frac{|\xi|}{\sqrt{t}} \to 0$ in distribution (and thus in probability) and thus $\frac{|\xi|}{\sqrt{t}} - u \to -u$ in probability. We have $$max W_t - ut = t (-u + o_p(1)) \to - \infty$$in probability.