Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 201 / 5

For a stopping time $T$ of a Brownian filtration, define the reflected stochastic processThe Brownian reflection at a stopping time form of the Brownian reflection principle states that $\hat{B}$ is again standard Brownian motion. If $T=\infty$, leave the path unchanged; the second branch is used only when $T<t$.

For an almost surely finite $T$, the Strong Markov property says that $(B_{T+s}-B_T{)}_{s\ge 0}$ is a fresh Brownian motion independent of ${\mathcal{F}}_T$. Its negative has the same probability distribution, so reflecting the future preserves the full path probability distribution. Allowing $T=\infty$ follows by applying this argument to $T\wedge n$ and restricting to each fixed finite time interval.

For $a>0$, the Brownian reflection principle at the first-passage time of $a$ gives the Brownian running maximum identitywhere $\Phi$ is the distribution function of the standard normal distribution. Indeed, reflection pairs paths that have reached $a$ and end below $a$ with paths ending above $a$; $B_t$ has no atom at $a$.

As $t\to \infty$, this probability tends to one. The events $\{S_n\ge a\}$ increase with integer $n$, so with probability one the path reaches $a$ in finite time. Taking a countable intersection over positive integer $a$ givesHere $S_{\infty }={\sup }_{s\ge 0}{B}_s$. This also proves that every positive-level Brownian first-passage time is finite almost surely.

Let $T_y=\inf \{s\ge 0:B_s=y\}$. Reflect at this stopping time, using the Brownian reflection principle from (a). Path continuity gives $\{S_t\ge y\}=\{T_y\le t\}$, and on this event the reflected endpoint is ${\hat{B}}_t=2y-B_t$. ConsequentlySince $x\le y$, an endpoint ${\hat{B}}_t\ge 2y-x\ge y$ forces the reflected path to have hit $y$ by time $t$. Reflection preserves the first hit of $y$, so the condition $T_y\le t$ is redundant on the right. As $\hat{B}$ has the same probability distribution as $B$, the joint distribution of Brownian motion and its running maximum satisfiesThis argument includes $x=y$; at positive $t$ the boundary endpoint has zero probability under the normal distribution. At $t=0$ both sides vanish because $y>0$.

Taking $x=y=a$ in (c) and using that $B_t$ has no atom at $a$ givesThis is the distribution function of the Brownian first-passage time; (b) ensures that it is a proper probability distribution. Comparing these distribution functions gives the Brownian scaling identity $T_{ca}\stackrel{d}{=}{c}^2{T}_a$ for $c>0$.

For an integer $n\ge 1$, set $T_0=0$ and successively hit the levels $a,2a,\dots ,na$. The Strong Markov property at these finite stopping times and spatial translation imply thatare independent random variables, each with the probability distribution of $T_a$. Thus, for independent copies $T_a^{(j)}$,ThereforeThe paper's definition is the strictly stable distribution convention, with no centering term, a special case of a stable distribution.