Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 26 / 3

First prove the strong law for Brownian motion, $B_t/t\to 0$ with probability one. For each $\epsilon >0$, the Gaussian tail bound givesBy stationary increments and the Brownian reflection principle, followed by the same tail bound,Both bounds are summable in $n$. The First Borel-Cantelli lemma therefore implies that, eventually, both quantities inside these probability events are at most $\epsilon n$. For $n\le t\le n+1$ this gives $|B_t|/t\le 2\epsilon$. Intersect over a sequence of positive rational $\epsilon$ tending to zero to obtain the asserted continuous-time limit.

On this one almost sure event,so $B_t+at\to +\infty$ for $a>0$ and to $-\infty$ for $a<0$. For each real $x$, the path is eventually strictly on the corresponding side of $x$. HenceThe same event works simultaneously for all levels $x$, giving the requested transience of Brownian motion with drift.

For $t>0$, write $\Phi$ for the standard normal distribution function. Since the Brownian running maximum is nonnegative, $a\le 0$ gives simply $\mathbb{P}(B_t\le b)=\Phi (b/\sqrt{t})$.

For $a>0$ and $b<a$, reflect the Brownian motion after its first hit of $a$. This is a stopping time, and the Strong Markov property together with symmetry of Brownian increments shows that reflection preserves its law. On paths that hit $a$, it sends the endpoint $B_t$ to $2a-B_t$. Thus the event with endpoint at most $b$ maps to endpoints at least $2a-b$, all of which necessarily hit $a$. This provesThe same reflection gives $\mathbb{P}(M_t\ge a)=2(1-\Phi (a/\sqrt{t}))$. If $b\ge a$, every path with $B_t>b$ has hit $a$, so subtracting that endpoint tail gives the complete answer:The expressions agree at $b=a$. This is the joint distribution of Brownian motion and its running maximum. In particular $M_t$ has the distribution of $|B_t|$ and has no atoms for $t>0$. If $t=0$, the pair is $(0,0)$ deterministically, so the requested probability is $1_{\{a\le 0,b\ge 0\}}$.

Continuity ensures that the maximum on $[0,1]$ is attained, so its first attainment time $\tau$ lies in $[0,1]$. To exclude the endpoint, use Brownian time reversal on a finite interval:This is a standard Brownian motion on this interval. Indeed, its increments are increments of $B$ on disjoint intervals taken in reverse order, hence are independent centered normal variables with the required variances, and its paths are continuous.

If $B_1$ is the maximum, then $R_s\ge 0$ for every $s$. But by the Brownian reflection principle and symmetry,the running maximum of $-R$ has the distribution of $|R_1|$, whose probability of being zero is zero. In particular the event $\tau =1$ has probability zero. Thus $\tau <1$ with probability one. Uniqueness of the maximizing time is not needed for this proof.

Put $Y_t=B_{\tau +t}-B_{\tau }$. Because $B_{\tau }$ is the maximum on $[0,1]$,The preceding part makes this a strictly positive random interval with probability one. A standard Brownian motion cannot have this path property. For each deterministic $h>0$, the Brownian reflection principle gives $\mathbb{P}({\sup }_{0\le t\le h}{B}_t\le 0)=0$. Taking the countable union over $h=1/k$ shows that the probability of staying nonpositive on any initial interval of positive length is zero.

Therefore $Y$ is not a Brownian motion. This is the obstruction for Brownian motion shifted at its finite-horizon maximum. It concerns its path law, irrespective of any proposed filtration for the shifted process.

Fix $0<t<1$. The event $\{\tau \le t\}$ means that the future path through time one never exceeds the maximum already attained by time $t$. Given ${\mathcal{F}}_t$, future increments form an independent Brownian motion $W$, soHere the Brownian running maximum distribution is the one derived above. On $\{B_t<0\}$, an event of probability $1/2$, the finite quantity $M_t-B_t$ is strictly positive, so this conditional probability lies strictly between zero and one.

If $\tau$ were a stopping time, $\{\tau \le t\}$ would be ${\mathcal{F}}_t$-measurable, and the displayed conditional expectation of its indicator would equal that indicator, taking only zero and one. The contradiction proves $\tau$ is not an $({\mathcal{F}}_t)$-stopping time. Thus the Strong Markov property cannot be invoked at $\tau$ to assert a Brownian shifted process; the direct argument here does not assume that property at this random time.