Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 33 / 2 / c

Work in the usual joint natural filtration of $B$ and $W$, in which both are Brownian motions. The integrand is $\mathrm{sgn}(B_s)$, as printed in the PDF. It is predictable and bounded, so $X$ is a zero-starting continuous local martingale. The quadratic variation of a stochastic integral givesIf the sign is defined to be zero at zero, the final equality still holds: Tonelli's theorem gives $\mathbb{E}{\int }_0^t{1}_{\{B_s=0\}}ds={\int }_0^t\mathbb{P}(B_s=0)ds=0$. Thus the Brownian zero set has zero Lebesgue time on every compact interval. The Lévy characterization of Brownian motion proves that $X$ is standard Brownian motion. This is the Brownian sign transform.

Apply the two-variable Itô formula to $e^w\cos b$. The independent Brownian motions satisfy $[B,W]=0$, while the two diagonal second derivatives sum to zero. ConsequentlyThus $Y$ is a continuous local martingale, but it is not Brownian motion. Brownian motion from any fixed starting point has quadratic variation $t$. If the displayed bracket were identically $t$, differentiating its continuous density would force $e^{2W_t}=1$ for every $t$, hence $W_t=0$ identically, an event of probability zero. This is the cosine-exponential Brownian local martingale. Its initial value $Y_0=1$ is retained in the next part.

Put $C_u={\int }_0^u{e}^{2W_s}ds$. Its paths are continuous and strictly increasing. We also need $C_{\infty }=\infty$ to ensure that the printed inverse is finite for every $t$.

Here is a duration argument for Divergence of a driftless Brownian exponential clock. Start at zero and choose successive return times $T_j$ to zero, each at least one time unit after the preceding return. Recurrence of one-dimensional Brownian motion makes all these times finite. By the Strong Markov property, each unit interval after $T_j$ has the same fixed positive probability that the Brownian path stays above $-1$. The intervals are disjoint, and repeated strong Markov conditioning makes these trials independent. Infinitely many succeed almost surely, by the Borel-Cantelli lemma. Each successful interval contributes at least $e^{-2}$ to $C_{\infty }$, proving divergence.

Therefore ${\tau }_t$ is the unique finite inverse of $C$, with $C_{{\tau }_t}=t$. It is an increasing family of stopping times, since $\{{\tau }_t\le u\}=\{C_u\ge t\}$. The continuous-local-martingale time-change theorem states that inverse-bracket time change preserves the local-martingale property in the filtration ${\mathcal{G}}_t={\mathcal{F}}_{{\tau }_t}$ and transforms the bracket by composition. Applied to $Y$, this givesThe Lévy characterization of Brownian motion now proves that $Z$ is Brownian motion starting at one; $Z-1$ is standard Brownian motion, in the time-changed filtration. If the term Brownian motion is reserved for a process starting at zero, $Z$ itself fails only that normalization. The substantive Brownian property, including independent Gaussian increments in the new time, holds.