Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 33 / 3

If ${\sigma }_1>n$, all the integer-time positions through $n$ lie in $(-1,1)$, so every increment over an intervening unit interval has absolute value at most two. The unit-time increments of Brownian motion are independent standard normal random variables. PutThe inclusion of events and independence giveIn particular ${\sigma }_1$ is finite almost surely. For real $t\ge 0$, its tail is at most ${\rho }^{\lfloor t\rfloor }$. The tail integral formula for moments therefore gives, for every $p>0$,An exponential geometric factor dominates the polynomial factors in this sum. This proves the requested moment finiteness for $p>1$, and in fact for every positive $p$.

The process $W_t=a^{-1}{B}_{a^{2}t}$ is standard Brownian motion, by Brownian scaling. Its first exit from $(-1,1)$ occurs at ${\sigma }_a/a^2$, soThe first part ensures integrability. Set $C=\mathbb{E}{\sigma }_1$; it is finite and strictly positive because path continuity at zero gives ${\sigma }_1>0$ almost surely. ThereforeWe keep $C$ unspecified until the comparison of limits requested below.

At each finite stopping time ${\sigma }_a^k$, the Strong Markov property says that $B_{{\sigma }_a^k+t}-B_{{\sigma }_a^k}$ is a standard Brownian motion independent of the past sigma-algebra. Its exit time from $(-a,a)$ is exactly ${\sigma }_a^{k+1}-{\sigma }_a^k$. By induction the stopping times are finite almost surely, and these durations are independent with common law ${\sigma }_a$, not merely identically distributed. This is the Brownian exit-time skeleton.

Put $v=\mathrm{Var}({\sigma }_1)<\infty$, which follows from the moment estimate. Brownian scaling gives $\mathrm{Var}({\sigma }_a)=a^{4}v$. Forthere are $2^{2n}$ independent durations of mean $C{2}^{-2n}$ and variance $v{2}^{-4n}$. ConsequentlyFor each $\epsilon >0$, Chebyshev's inequality givesApply the Borel-Cantelli first lemma for a countable sequence of positive errors tending to zero. We obtain the Brownian exit-skeleton clock convergence:No independence between the clocks for different values of $n$ is required.

Let ${\xi }_j=B_{{\sigma }_1^j}-B_{{\sigma }_1^{j-1}}$. Continuity makes each displacement equal to $+1$ or $-1$. Reflection symmetry of a fresh Brownian motion gives probability one half to each sign, and the Strong Markov property makes these signs independent across successive exits. Thus $B_{{\sigma }_1^n}={\sum }_{j=1}^n{\xi }_j$.

For the normalized position, the characteristic function issince $\log \cos z=-z^2/2+O(z^4)$ near zero. By the Lévy continuity theorem, this provesThis is the Gaussian limit of a Brownian exit skeleton.

Compare along the subsequence $n=2^{2m}$ with ${\tau }_m={\sigma }_{2^{-m}}^{2^{2m}}$. Joint Brownian scaling of the successive exits givesThe right-hand side converges weakly to $N(0,1)$. The preceding part and path continuity give $B_{{\tau }_m}\to B_C$ almost surely on the original Brownian probability space. Hence its weak limit is also the law $N(0,C)$. Uniqueness of the weak limit forces $C=1$, so finally $\mathbb{E}{\sigma }_a=a^2$. Independence of the position and its random clock was not assumed in this comparison.