Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 33 / 4

For $x>0$, reflect a Brownian path after its first hitting of $x$. The Strong Markov property and symmetry preserve its law and exchange the events $\{S_1\ge x,B_1<x\}$ and $\{B_1>x\}$. The terminal equality event has probability zero, so the Brownian reflection principle givesContinuity of this distribution also gives no atom at zero. ThereforeA useful global bound follows from the maximum of the standard normal distribution density:

Monotonicity of $f$ gives, for each $n\ge 1$,Hence ${\sum }_{n\ge 1}f(2^{-n})<\infty$; the additional $n=0$ term is finite by continuity at one. Also $f(t)\to 0$ as $t\downarrow 0$, since a positive limiting lower bound would make the integral diverge.

By Brownian scaling, $S_{2^{-n-1}}$ has law $2^{-(n+1)/2}{S}_1$. Apply the small-value bound for the Brownian running maximum to obtainSumming yields the requested finite probability sum. The argument needs no independence of these nested-time events.

The Borel-Cantelli first lemma applied to that finite sum says that almost surely, for all sufficiently large $n$,For $2^{-n-1}\le t\le 2^{-n}$, both the Brownian running maximum and the positive function $\sqrt{t}f(t)$ are nondecreasing. Thus eventuallyThis proves the lower limit bound one, with interpolation across all small real times rather than only the dyadic sequence.

For every positive integer $m$, the function $mf$ satisfies the same assumptions and integrability condition. Repeating the argument gives an eventual lower bound $S_t\ge m\sqrt{t}f(t)$ on a probability-one event. Take the countable intersection of these events. On it, the ratio is eventually at least any prescribed positive integer, proving the integral lower envelope for the Brownian maximum: