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

Subtract $M_0$, which does not affect either the integral or its quadratic variation. Let $N=M-M_0$. Choose a localizing sequence ${\tau }_n$ for $N$. By local boundedness of the previsible process $H$, there are increasing stopping times ${\rho }_n\uparrow \infty$ and finite deterministic constants $K_n$ with $|H_s|\le K_n$ on $0<s\le {\rho }_n$. PutContinuity and finiteness on compact intervals imply ${\sigma }_n\uparrow \infty$ almost surely. The stopped process $N^{{\sigma }_n}$ is a bounded square-integrable martingale, its bracket is bounded by $n$, and its stopped integrand is bounded by $K_n$. The indicator $1_{\{s\le {\sigma }_n\}}$ is predictable, so the stopped stochastic integral identity is valid.

Fix $n$ and abbreviateThe Itô isometry gives $\mathbb{E}{J}_t^2=\mathbb{E}{Q}_t\le K_n^{2}n$, so $J$ is a square-integrable martingale. For $F\in {\mathcal{F}}_s$, apply the same isometry to the predictable integrand $1_F{1}_{(s,t]}H$, stopped at ${\sigma }_n$. This yieldsAlso $\mathbb{E}[J_t-J_s\mid {\mathcal{F}}_s]=0$. Expanding $J_t^2-J_s^2$ therefore proves that $J^2-Q$ is a true martingale. The process $Q$ is continuous adapted increasing and starts at zero. By the characterization of continuous martingale quadratic variation as its increasing square compensator, and uniqueness of an increasing square compensator, we have $[J]=Q$.

Stochastic integration and quadratic variation commute with stopping; for the latter this follows either from squared-increment sums or from the same square-compensator characterization. Hence the desired identity holds up to every ${\sigma }_n$. Since these times increase to infinity, they exhaust every finite time interval almost surely, and the continuous identities patch to giveThis is the localized isometry proof of stochastic-integral quadratic variation. No unstopped second-moment or bracket-integrability assumption has been made.

The one-dimensional Lévy characterization of Brownian motion states: a continuous adapted process $L$, with $L_0=0$, is standard Brownian motion relative to its filtration if and only if it is a local martingale and $[L{]}_t=t$ for all $t$ almost surely.

For necessity, the centered independent Gaussian increments of Brownian motion give both the martingale property and the martingale $L_t^2-t$. The uniqueness characterization of quadratic variation then gives $[L{]}_t=t$.

For sufficiency fix $\theta \in \mathbb{R}$. The Itô formula givesThe second-order term cancels the time derivative because $d[L{]}_t=dt$. Thus $E$ is a complex local martingale. Its modulus is the deterministic quantity $e^{{\theta }^{2}t/2}$, bounded on every finite horizon, so its real and imaginary parts are true martingales by the bounded local martingale criterion. For $s<t$,Uniqueness of characteristic functions says that the increment has normal distribution $N(0,t-s)$ independently of ${\mathcal{F}}_s$. Indeed multiplying the identity by any bounded ${\mathcal{F}}_s$-measurable variable gives the product characteristic-function identity expressing independence. Iterating over deterministic times gives independent increments. Together with continuity and $L_0=0$, these are precisely the defining properties of standard Brownian motion. If the starting value is a fixed $x$, the same argument identifies $L-x$ as standard Brownian motion.

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.