3
$\begingroup$

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.

  • 0
    See [this answer](http://math.stackexchange.com/a/133971/822) and let $t \to \infty$.2012-08-09
  • 0
    @Nate Eldredge I don`t know how to do with $\frac{b+X(t)}{1+t}$. It is not a Brownian motion.2012-08-09

1 Answers 1

4

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.