How can I prove it?
For $b>a>0$, show that $ \operatorname{Pr}\left({\sup_{t\geqslant 0}\left(\frac{b+X(t)}{1+t}\right)\geqslant a}\right)=e^{-2a(a-b)} $ where $X(t)$ is a Brownian motion.
How can I prove it?
For $b>a>0$, show that $ \operatorname{Pr}\left({\sup_{t\geqslant 0}\left(\frac{b+X(t)}{1+t}\right)\geqslant a}\right)=e^{-2a(a-b)} $ where $X(t)$ is a Brownian motion.
Inequality $\sup_{t\geqslant 0}\left(\frac{b+X(t)}{1+t}\right)\geqslant a$ implies that there exists a time, for which $X(t) > a t + (a-b)$, i.e. the hitting time $T_{a,a-b}$ (when the Brownian motion crosses line $a t + (a-b)$) is finite: $ \operatorname{Pr}\left({\sup_{t\geqslant 0}\left(\frac{b+X(t)}{1+t}\right)\geqslant a}\right) = \operatorname{Pr}\left( T_{a,a-b} < \infty \right) = \lim_{t \to \infty} \operatorname{Pr}\left( T_{a,a-b} < t \right) $ Now use the result from the answer of mine as suggested by Nate.