Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 38 / 4

Put $A_t=[M{]}_t$. Its continuous, strictly increasing paths start at zero and tend to infinity. Thus the inverse $T_s$ is finite and continuous, andso each $T_s$ is a stopping time. Set ${\mathcal{G}}_s={\mathcal{F}}_{T_s}$ and ${\beta }_s=M_{T_s}$. This is the inverse-clock construction in the Dambis-Dubins-Schwarz theorem.

To verify its martingale property, stop $M$ at times ${\tau }_n$ bounding its path, its bracket and time itself. The stopped $M$ and $M^2-A$ are bounded martingales. Optional sampling at $T_s\wedge {\tau }_n$ and $T_t\wedge {\tau }_n$ shows that the time-changed process, stopped at $A_{{\tau }_n}$, is a martingale and that its square minus the stopped clock $s\wedge A_{{\tau }_n}$ is a martingale. These clock stopping times tend to infinity because $A_{\infty }=\infty$. Therefore $\beta$ is a continuous local martingale with ${\beta }_0=0$ and quadratic variation $[\beta {]}_s=s$. The Lévy characterization of Brownian motion identifies it as a ${\mathcal{G}}_s$-Brownian motion.

Strict monotonicity and continuity give $A_{T_s}=s$ and $T_{A_t}=t$, soThe inverse-clock argument proves the asserted representation; the random clock need not be independent of the Brownian motion it produces.

The terminal bracket value printed as zero is incompatible with a strictly increasing bracket starting at zero. If the intended value is infinity, solution applies directly. In fact the substantive conclusion holds for every deterministic continuous bracket $a(t)=[M{]}_t$, whether its terminal value is finite or infinite, and without strict increase.

For $u\in \mathbb{R}$, the Itô formula makesa complex local martingale. Its modulus is the deterministic value $e^{u^{2}a(t)/2}$, bounded on every finite horizon, so its real and imaginary parts are true martingales there. Conditioning between $s$ and $t$ givesThis deterministic conditional characteristic function shows that the increment is independent of ${\mathcal{F}}_s$ and is normal with mean zero and variance $a(t)-a(s)$. Induction over ordered times gives independent Gaussian increments and hence jointly Gaussian coordinate vectors. Since $a(t)$ is finite at every finite time, the integrable terminal bracket criterion applied on each fixed horizon also upgrades $M$ to a true square-integrable martingale. This proves the deterministic quadratic variation characterizes a Gaussian continuous local martingale result in a form covering the corrected question.

A simple example isfor a standard Brownian motion $B$. The Itô formula gives $M_t=2{\int }_0^t{B}_{s}d{B}_s$, a continuous square-integrable martingale on every finite horizon. For $h>0$, write $\Delta =B_{s+h}-B_s$, independent of the past and normal with variance $h$. ThenSince $B_s^2=M_s+s$, this conditional second moment is already a nonconstant function of the martingale's own past for $s>0$. The tower property of conditional expectation gives the same expression on conditioning on that own past. Were the increments independent, that conditional second moment would be constant. Thus the increments are not independent.

For $t>1$, $B_{t^2}$ is not measurable with respect to the original Brownian filtration ${\mathcal{F}}_t$: its conditional variance is $t^2-t>0$. Thus it is not even adapted on the full half-line. It also fails the martingale condition before time one. With $s=1/4$, $t=3/4$,The appropriate filtration isBrownian independent increments make $M_t=B_{t^2}$ a continuous ${\mathcal{G}}_t$-martingale, with quadratic variation $[M{]}_t=t^2$. Define the stochastic integralThe value of the integrand at zero is immaterial; it is square-integrable against the bracket because ${\int }_0^t(2s{)}^{-1}d(s^2)=t$. Consequently $[W{]}_t=t$, and the Lévy characterization of Brownian motion makes $W$ a ${\mathcal{G}}_t$-Brownian motion. Associativity of stochastic integrals gives the Brownian motion under a quadratic time change representationThe function $f$ is continuous at zero. One can first verify the inverse-integrand identity on $[\epsilon ,t]$ and then let $\epsilon \downarrow 0$ using the Itô isometry, since $\mathbb{E}{M}_{\epsilon }^2={\epsilon }^2$.