Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 3 / 29K (Stochastic Financial Models)

For every finite collection of positive times, $(B_{t_1},\dots ,B_{t_m})$ is a linear transformation of the corresponding values of $W$, so $B$ is a centered Gaussian process. If $0<s\le t$, thenThus $B$ has the covariance of Brownian motion.

Its paths are continuous for $t>0$. As $t\downarrow 0$, put $u=1/t$. The given almost-sure limit givesThe Gaussian-process characterization of Brownian motion now proves the time inversion of Brownian motion:

Use the time-inverted Brownian motion $B_u=u{W}_{1/u}$. For $s>0$, set $u=1/s$. ThenAs $0<s\le t$ corresponds exactly to $u\ge 1/t$, and the inequality at $s=0$ is automatic because $b>0$, this is the pathwise identitySince $B$ and $W$ have the same law by time inversion of Brownian motion,

Let $X_s=W_s-as$ and $M_t={\sup }_{0\le s\le t}{X}_s$. Split the crossing event according to its endpoint:Under the Cameron-Martin theorem for a linear drift, the law of $X$ relative to standard Wiener measure has endpoint densityOn paths that cross $b$ and end at $x\le b$, the Brownian reflection principle reflects the path after its first hit of $b$ and changes the endpoint to $2b-x\ge b$. Under this reflection the density becomesThe remaining exponential is the Cameron--Martin density for drift $+a$. Thereforewhere the last equality uses the symmetry of the normal distribution. Subtracting the crossing probability from one givesEquivalently, in terms of the standard normal distribution function,This is the finite-horizon maximum of Brownian motion with negative drift formula.

By continuity of Brownian motion, the event that the last strict crossing of $b$ occurs by time $t$ isApply part (b) with $a$ and $b$ interchanged and its time parameter replaced by $1/t$. This turns the right-hand side intoPart (c), now with drift $b$, barrier $a$, and horizon $1/t$, givesThis is the distribution function of the last passage time above a level for Brownian motion with negative drift. Taking $t\downarrow 0$ givesthe probability that the negatively drifted Brownian path never exceeds $b$, consistent with the infinite-horizon crossing probability for Brownian motion with negative drift.