Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 202 / 4

On $[\tau ,\sigma )$ the Brownian path does not meet zero, so the power function $x\mapsto |x{|}^{\delta }$ is twice continuously differentiable along the path. The Itô formula givesSince $Y_t=|B_t{|}^{\delta }$, this becomeswhere ${\hat{B}}_t={\int }_0^t\mathrm{sign}(B_s)d{B}_s$ is a standard Brownian motion by the Lévy characterization of Brownian motion. Thus the displayed equation in the paper is valid after the customary renaming of $\hat{B}$ as $B$.

LetThe clock $A$ is an absolutely continuous function and is strictly increasing: its derivative is positive away from the Brownian zero set, which has zero Lebesgue measure. It also tends to infinity. This is immediate for $\delta =1$; for $\delta >1$, recurrence and the Strong Markov property imply that the Brownian occupation time of, for example, $\{x:1<|x|<2\}$ is unbounded, while the integrand is bounded below there by ${\delta }^2$.

Thus $A$ is continuous, strictly increasing, and maps $[0,\infty )$ onto itself. Its inverse function ${\tau }_s=A^{-1}(s)$ is finite, continuous, and strictly increasing.

Part (b) shows that $s\mapsto {\tau }_s$ is finite and continuous. Since both $t\mapsto B_t$ and the power function $x\mapsto |x{|}^{\delta }$ are continuous, their compositionis continuous. This is an instance of a time change of a continuous process.

As printed, the requested conclusion is false for $\delta \ne 1$. On an interval on which $X$ stays positive, the time change of a continuous process satisfies $d{\tau }_s=ds/({\delta }^2{X}_s^{2(\delta -1)/\delta })$. The time-changed martingale term has quadratic variation $s$, so the Lévy characterization of Brownian motion identifies it with a standard Brownian motion $\tilde{B}$. Dividing the drift in part (a) by the derivative of the clock givesConsequently the construction actually satisfieswhich is the Bessel process equation of dimension $2-1/\delta$. It equals the paper's claimed drift $(\delta -1)/(2X_s)$ only when $\delta =1$. The mismatch between the specified power, clock, and conclusion is therefore a typographical error in the question.

For the equation actually produced by the preceding construction, namely the Bessel process equation of dimension $2-1/\delta$, the drift has Lipschitz continuity on every compact subset of $(0,\infty )$. Starting at any positive time and position, pathwise uniqueness therefore makes the time-changed process agree until its first hit of zero with the maximal local solution of a stochastic differential equation. For $\delta >1$, its dimension lies in $(1,2)$, so it can hit zero; the time-change construction then supplies further excursions, whereas the maximal local solution on $(0,\infty )$ stops at that first hit.

For $\delta =1$, $X_s=|B_s|$ is Reflected Brownian motion; away from zero it agrees with the maximal local solution of $dX=d\tilde{B}$. For the dimension-$\delta$ equation printed in the paper, the preceding construction does not agree with the maximal local solution unless $\delta =1$, for the coefficient mismatch established in part (d).

Let $Z=\{t:B_t=0\}$. Since $A$ is the clock from part (b) and $\tau =A^{-1}$,For $\delta >1$, the absolutely continuous function $A$ has derivative $A^{\prime }(t)={\delta }^2|B_t{|}^{2\delta -2}=0$ on $Z$. The one-dimensional area bound for an absolutely continuous function therefore givesFor $\delta =1$, $A_t=t$ and the conclusion follows directly because the Brownian zero set has zero Lebesgue measure. Hence the zero set of $X$ has zero Lebesgue measure almost surely.