# Probability for 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.

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See this answer and let $t \to \infty$. –  Nate Eldredge Aug 9 '12 at 2:37
@Nate Eldredge I don`t know how to do with $\frac{b+X(t)}{1+t}$. It is not a Brownian motion. –  user30795 Aug 9 '12 at 14:31

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.