Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 202 / 6

Let the infinitesimal generator be the second-order operatorwhere $a(x)$ is symmetric positive semidefinite and $b(x)$ is a vector. In the standard bounded-coefficient setting take $a,b$ bounded and measurable. An L-diffusion is a continuous adapted process $X$ for which, for every $g\in C_b^2({\mathbb{R}}^d)$,Here $C_b^2$ includes boundedness of the function and its derivatives through order two. This is the true-martingale version of the diffusion martingale problem. The matrix $a$ is the diffusivity, and $b$ is the drift. An Itô diffusion with coefficient $\sigma$ has $a=\sigma {\sigma }^{\mathsf{T}}$.

A more general local martingale problem may permit unbounded coefficients and require only a local martingale. That convention is weaker and must not silently replace the true martingale property used here and in part (b). The bounded-coefficient convention guarantees all finite-horizon integrability needed below; a sufficient more general replacement is $\mathbb{E}{\int }_0^T({\sum }_i|b_i(X_s)|+{\sum }_i{a}_{ii}(X_s))ds<\infty$ for every $T$.

We prove the extension from spatial test functions to time-dependent test functions for a diffusion martingale problem directly, without assuming a driving Brownian motion. Fix $0\le s<t$ and a deterministic partition $s=t_0<\cdots <t_n=t$. For each $j$, $g_j(x)=f(t_j,x)$ lies in $C_b^2$. The defining martingale problem givesCall the sum $S_n$. Telescoping separates the spatial and temporal changes:The sample path $X$ is uniformly continuous on $[s,t]$ and has compact range. Joint continuity of ${\partial }_{t}f$, the spatial first derivatives, and the spatial second derivatives therefore gives, as the mesh tends to zero,almost surely. For bounded $a,b$ and bounded derivatives of $f$, both integrals are uniformly bounded on this finite horizon, as are the two endpoint terms. The dominated convergence theorem permits passage to the limit in the expectation against $1_A$. Thus $M_t^f$ is integrable and $\mathbb{E}[1_A(M_t^f-M_s^f)]=0$ for every $A\in {\mathcal{F}}_s$, provingAdaptation and continuity follow from the formula, so $M^f$ is a continuous martingale. Under the more general integrability condition in part (a), the same proof uses an integrable bound proportional to $1+{\int }_s^t({\sum }_i|b_i(X_u)|+{\sum }_i{a}_{ii}(X_u))du$; positive semidefiniteness bounds every off-diagonal entry by the diagonal ones. With only local assumptions this reasoning proves a local martingale, and an integrability hypothesis is needed to upgrade that conclusion.

The Itô formula for $f(t,x)=e^{x+at}$ gives the stochastic differential equationWrite $N_t={\int }_0^t{Z}_{u}d{B}_u$ for its continuous local martingale part. Its clock is the quadratic variationBecause $Z_t>0$, $A$ is strictly increasing. For $s<A_{\infty }$, its inverse ${\tau }_s$ satisfies $A_{{\tau }_s}=s$ and $d{\tau }_s/ds=Z_{{\tau }_s}^{-2}$. The Dambis-Dubins-Schwarz theorem, or the Lévy characterization of Brownian motion after the optional time-change theorem, makes ${\beta }_s=N_{{\tau }_s}$ a Brownian motion for the time-changed filtration. With $Y_s=Z_{{\tau }_s}$,This is the Bessel process equation with dimension $\boxed{2+2a}$, on the positive stochastic interval. This establishes the Exponential Brownian-to-Bessel time change.

The lifetime convention is important: the inverse is defined only for $s<A_{\infty }$. If $a<0$, the strong law for Brownian motion gives $B_t/t\to 0$, so $Z_t\to 0$ exponentially and $A_{\infty }<\infty$. Then $Y_s\to 0$ as $s\uparrow A_{\infty }$, and this is its first hit of zero after continuous extension. If $a>0$ the clock diverges by the same strong law. If $a=0$ it diverges by infinite occupation time of one-dimensional Brownian motion on a bounded interval, since $e^{2B_t}$ is bounded below there. Thus the lifetime is infinite when $a\ge 0$. Negative dimensions are interpreted by the same positive-state equation killed at zero; no post-zero Bessel process behavior is asserted.

Work before the first zero of $Z$, with $Z_0=z>0$. For the Bessel process equation, the Itô formula applied to $x^r$ givesIts quadratic variation isThis clock is strictly increasing on the positive-state interval. For $s$ before its terminal clock value, let $Y_s=W_{{\sigma }_s}$ and let ${\beta }_s=({\int }_0^{\cdot }r{Z}_u^{r-1}d{B}_u{)}_{{\sigma }_s}$. The Dambis-Dubins-Schwarz theorem makes $\beta$ a Brownian motion there. Since $d{\sigma }_s=ds/(r^2{Z}_{{\sigma }_s}^{2r-2})$, the drift changes toConsequently the Power time change of a Bessel process givesThe initial value is $z^r$, and this equation identifies a Bessel process of dimension ${\delta }^{\prime }$ until its first zero. Localize both equations on compact subintervals of $(0,\infty )$ to justify the inverse and all stochastic integrals for arbitrary $r>0$.

For completeness, the inverse-clock lifetime really is the positive Bessel process lifetime: the map $z\mapsto z^r$ is a homeomorphism of $(0,\infty )$, and exits from nested compact intervals correspond under the clock. There is no finite explosion at infinity for a positive Bessel process, since the Itô formula for $Y^2$ gives $d(Y^2)=2Yd\beta +{\delta }^{\prime }ds$; stopping at zero or level $R$ yields a probability of hitting $R$ before a fixed $T$ bounded by $(Y_0^2+\max ({\delta }^{\prime },0)T)/R^2$. If the transformed lifetime were finite while $Y$ stayed in a compact positive interval, the inverse clock and the original solution could both be continued. Thus a finite lifetime is approach to zero. For $\delta <2$, also ${\delta }^{\prime }<2$ and the positive-state Bessel process hits zero; for $\delta \ge 2$ the transformed lifetime is infinite. No continuation through zero is needed.

If the initial value is already zero, the interval before its first zero is empty; the positive-state argument concerns starts $z>0$.