Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 27 / 5 / c

Let $S$ denote this maximum and put $\tau =\tau (-b)$. Brownian motion reaches $-b$ almost surely: the Brownian reflection principle gives crossing probability $2\mathbb{P}(W_t<-b)\to 1$. Continuity gives $W_{\tau }=-b$, with the strict-crossing infimum interpreted as in part (b). The processis a continuous nonnegative local martingale starting at one and tending to zero. Its maximum is $1+S/b$. Apply part (b) at $a=1+x/b$, for $x>0$:Differentiating gives the maximum before a lower Brownian barrier densityThe tail tends to one as $x\downarrow 0$, so there is no atom at zero. The density integrates to one.

Use the positive exponential Brownian martingaleThe strong law for Brownian motion, $W_t/t\to 0$, makes its exponent tend to $-\infty$ and hence $M_t\to 0$. If $S={\sup }_{t\ge 0}(W_t-\lambda t)$, then $\sup M=e^{2\lambda S}$. Part (b) gives, for $x>0$,Thus the maximum is exponentially distributed with rate $2\lambda$:There is no atom at zero, by letting $x\downarrow 0$ in the tail. This is the infinite-horizon crossing probability for Brownian motion with negative drift.