Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 202 / 4

The printed assumptions omit the orthogonality needed for the claimed local martingales. Merely placing two Brownian motions on the same probability space does not make them independent. Interpret both as Brownian motions relative to a common filtration, and put $C_t=[B,\vartheta {]}_t$. For $X=e^B\cos \vartheta$ and $Y=e^B\sin \vartheta$, the two-variable Itô formula givesThe pure second derivatives cancel because $[B{]}_t=[\vartheta {]}_t=t$, while the mixed partial derivatives are $X_{b\vartheta }=-Y$ and $Y_{b\vartheta }=X$. Thus the mixed quadratic covariation term cannot be omitted.

For a counterexample allowed by the PDF, take $\vartheta =B$. Then $C_t=t$ and $X$ has drift $-Ydt$, while $Y$ has drift $Xdt$. In particular $X_0=1$ makes the latter drift nonzero near zero, so $Y$ is not a local martingale. Nor is $X$: $Y$ cannot be identically zero, so its accumulated continuous drift is not identically zero. Uniqueness of the continuous semimartingale decomposition excludes cancellation by another local martingale term.

The corrected hypothesis is $[B,\vartheta ]\equiv 0$, for example independent Brownian motions in their joint filtration. Conversely, if both were local martingales, uniqueness of the continuous semimartingale decomposition would force $YdC=XdC=0$; since $X^2+Y^2=e^{2B}>0$, this forces $dC=0$. Thus the correction is exactly the condition needed for both local martingales. Under this hypothesis the drift terms vanish and the intended stochastic differential equations are $dX=XdB-Yd\vartheta$ and $dY=YdB+Xd\vartheta$. The continuous integrands are locally bounded, so localization makes these Itô integrals continuous local martingales. Subsequent intended conclusions are proved under this explicitly stated correction.

With the printed assumptions alone, the general quadratic covariation calculation from the two noise coefficients givesA continuous finite-variation process contributes no quadratic covariation, so the drift terms in the previous part do not alter these formulas. Taking $\vartheta =B$ gives $d[X,Y]=e^{2B}\cos (2B)dt$, which is nonzero near zero since $B_0=0$. Also $d([X]-[Y])=-4XYdt$ is not identically zero. Thus the requested assertions are false as printed.

Under the corrected assumption $C=0$, $X^2+Y^2=e^{2B}$ yieldsThe pair is therefore a pair of orthogonal continuous local martingales. This equality also identifies the clock for the intended time change of a continuous process.

Under the corrected orthogonality hypothesis, the preceding quadratic variation calculation gives $[X{]}_t=A_t={\int }_0^t{e}^{2B_s}ds$. We justify the needed infinite occupation time of one-dimensional Brownian motion, rather than inferring an infinite integral from recurrent visits alone.

Let $T_0=0$, and recursively let $S_n$ be the first exit of $B$ from $(-1,1)$ after $T_n$, and $T_{n+1}$ its first return to zero after $S_n$. Recurrence of one-dimensional Brownian motion makes these stopping times finite almost surely. The Strong Markov property shows that $D_n=S_n-T_n$ are independent and identically distributed random variables, each strictly positive. Choose $\epsilon >0$ with $\mathbb{P}(D_n\ge \epsilon )>0$. The Borel-Cantelli lemma gives infinitely many such durations and hence ${\sum }_n{D}_n=\infty$ almost surely. These intervals are disjoint, and $|B_s|\le 1$ on each, soThis proves the intended conclusion. Without the orthogonality correction, the printed setup gives instead the more general quadratic variation in the previous part; it does not justify identifying that quantity with $A$. The proof here is explicitly for the corrected setup.

Under the corrected hypothesis, $N_t=X_t-1$ is a continuous local martingale with $N_0=0$, $[N{]}_t=A_t={\int }_0^t{e}^{2B_s}ds$, and $A_{\infty }=\infty$ almost surely. The Dambis-Dubins-Schwarz theorem states that, for such a process,defines a standard Brownian motion relative to the filtration ${\mathcal{F}}_{{\tau }_u}$, and $N_t=W_{[N{]}_t}$. The divergence of the quadratic variation ensures every ${\tau }_u$ is finite, so no extension of the probability space is needed. Here $A$ is continuous and strictly increasing because its derivative is $e^{2B_t}>0$, so the inverse is particularly straightforward. Consequently,The constructed Brownian motion need not be independent of its random clock. Under the literal PDF assumptions $X$ need not be a local martingale, as the counterexample in part (a) shows, so this Dambis-Dubins-Schwarz theorem conclusion cannot be asserted without the correction.