Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 29 / 6

The diffusion generator is $\mathcal{L}f=\frac{1}{2}{\sigma }^2{f}^{\prime \prime }+b{f}^{\prime }$. On a compact subinterval of $I$, the coefficient $s^{\prime }\sigma$ is bounded, and the Itô formula up to its exit time $\rho$ givesbecause $\mathcal{L}s=0$. Increasing compact subintervals proves the local-martingale property up to $T$. The notation $s(X_{t\wedge T})$ at an exit endpoint is understood with its continuous endpoint extension when needed; stopping on $[a,b]\subset I$ below is always well-defined. This is the martingale property of a scale function of a one-dimensional diffusion.

For the hitting ratio there are two necessary qualifications: $s(b)\ne s(a)$ and $\tau =T_a\wedge T_b<\infty$ almost surely. A usual scale function is strictly monotone, which ensures the first condition, but the printed hypotheses alone ensure neither. Under these conditions, $s(X_{t\wedge \tau })$ is a bounded local martingale and hence a uniformly integrable martingale. The optional stopping theorem givesContinuity prevents simultaneous hitting of distinct endpoints, and almost sure exit makes these two probabilities sum to one. Solving yields the boundary hitting probability from a diffusion scale function

The hypotheses for a diffusion scale hitting formula cannot be omitted here. Take $\sigma \equiv 0$, drift $b\equiv 0$, $s(y)=y$ and $X_t\equiv x\in (a,b)$. All printed coefficient and differential conditions hold, but neither endpoint is ever reached, so the left side is zero whereas the displayed ratio is strictly positive. A constant $s$ is another allowed solution of the printed differential equation and makes its denominator zero. In general the bounded stopped martingale has a limit $Y_{\infty }$, and optional stopping instead givesThe following parts have nondegenerate diffusion and strictly monotone powers, and their finite-exit properties are justified explicitly.

Write $\gamma =1-2a\ne 0$ and interpret the positive solution up to its first hit of zero, where the displayed drift is singular. For $s(x)=x^{\gamma }$,Thus $s$ is a scale function of a one-dimensional diffusion on every $[\epsilon ,\infty )$ with $\epsilon >0$. It is strictly decreasing if $a>1/2$; multiplying by $-1$ gives the conventionally increasing scale, without changing any hitting ratio.

Fix $0<\epsilon <x<R$. Exit from $(\epsilon ,R)$ is almost surely finite. Indeed the coefficients are locally Lipschitz there, so there is no lifetime obstruction in this compact interval. If the process stayed inside forever, the equation and $a/X>0$ would give $B_t\le R-x$ for every $t$, contradicting the almost sure unboundedness above of Brownian motion. Therefore part (a)'s bounded optional-stopping argument applies and givesIf $a>1/2$, then $\gamma <0$ and ${\epsilon }^{\gamma }\to \infty$, so ${\mathbb{P}}_x(T_{\epsilon }<T_R)\to 0$. Hitting zero before $R$ would entail hitting every $\epsilon$ before $R$; it therefore has probability zero. Every continuous path reaching zero in finite time is bounded up to that time, so it belongs to this event for some integer $R>x$. A countable union proves

If $0<a<1/2$, then $\gamma \in (0,1)$. Taking $\epsilon \downarrow 0$ in the probabilities alone does not yet exclude approaching zero only at infinite time. To establish finite-time access to zero for Bessel dimensions below two, defineIt is continuous, nonnegative, zero at $0,R$, and satisfies $\mathcal{L}{v}_R=-1$ on $(0,R)$, since $\mathcal{L}(y^{\gamma })=0$ and $\mathcal{L}(y^2)=1+2a$. Itô's formula stopped at ${\tau }_{\epsilon ,R}=T_{\epsilon }\wedge T_R$, followed first by expectation at $t\wedge {\tau }_{\epsilon ,R}$, givesLet $t\to \infty$ and then $\epsilon \downarrow 0$. The stopping times increase and have uniformly bounded expectation, so their limit ${\tau }_{0,R}$ is finite almost surely. On this finite interval $X\le R$, and the increasing drift integral is bounded by $R-x-{\inf }_{u\le {\tau }_{0,R}}{B}_u$. Hence it has a finite limit, and the stochastic equation makes $X$ extend continuously to ${\tau }_{0,R}$. If the upper endpoint was not reached, the lower stopped values tend to zero, so the terminal value is zero. Thus ${\tau }_{0,R}=T_0\wedge T_R$ genuinely denotes a finite hitting time.

The lower-exit events decrease to $\{T_0<T_R\}$ as $\epsilon \downarrow 0$, giving ${\mathbb{P}}_x(T_0<T_R)=1-(x/R{)}^{\gamma }$. Let $R\to \infty$ to obtainThere is no finite explosion to $+\infty$: on a finite horizon, any excursion above level $1$ has drift at most $a$, so its height is bounded by $\max \{x,1\}+2{\sup }_{u\le t}|B_u|+at$. The two alternatives therefore describe the zero-boundary lifetime, giving the Hitting-zero classification for a Bessel process of dimension $2a+1$.

Fix $x=X_0>0$ and $a>1/2$. For the construction take $0<\epsilon <x$, the range relevant to localization at zero, and defineThis drift is bounded by $a/\epsilon$ and globally Lipschitz with constant at most $a/{\epsilon }^2$. Together with the constant diffusion coefficient $1$, the global existence theorem for stochastic differential equations with Lipschitz coefficients gives a unique strong stochastic solution $U^{\epsilon }$ offor the prescribed driving Brownian motion. Let $T_{\epsilon }^{\epsilon }=\inf \{t:U_t^{\epsilon }=\epsilon \}$ and choose $X_t^{\epsilon }=U_{t\wedge T_{\epsilon }^{\epsilon }}^{\epsilon }$ as the stopped continuation. Before this time the drift is exactly $a/X_t^{\epsilon }$. Uniqueness here means uniqueness up to the stopping time; the equation imposes no restriction on arbitrary continuations after it.

For $0<\delta <\epsilon <x$, the two equations have the same coefficients while both solutions exceed $\epsilon$. By local pathwise uniqueness they coincide until that level is reached. Thus the first hit of $\epsilon$ by $X^{\delta }$ is $T_{\epsilon }^{\epsilon }$, and its later first hit of $\delta$ satisfiesPatch along a deterministic sequence ${\epsilon }_n\downarrow 0$: for $t<T:={\lim }_n{T}_{{\epsilon }_n}^{{\epsilon }_n}$, choose $n$ with $t<T_{{\epsilon }_n}^{{\epsilon }_n}$ and set $X_t=X_t^{{\epsilon }_n}$. Compatibility makes the definition independent of $n$. It gives a continuous adapted positive process satisfying the original equation on every compact time interval before $T$. The same compatibility holds for other threshold choices, so the lifetime is the limit as $\epsilon \downarrow 0$, not a property of the chosen sequence.

To prove $T=\infty$, put $\nu =2a-1>0$ and $s(y)=y^{-\nu }$. Its generator vanishes. On each finite horizon, Itô's formula makes $s(X_{t\wedge T_{\epsilon }^{\epsilon }}^{\epsilon })$ a true martingale, because its stochastic integrand is bounded by $\nu {\epsilon }^{-\nu -1}$. Positivity therefore impliesSince $\{T\le t\}$ is contained in each $\{T_{\epsilon }^{\epsilon }\le t\}$, letting $\epsilon \downarrow 0$ gives $\mathbb{P}(T\le t)=0$. Taking a countable union over integer $t$ proves $T=\infty$ almost surely. This is the truncation construction of a positive Bessel strong solution.

For two positive solutions driven by the same Brownian motion, localize at their first hits of $\epsilon$. The cutoff equations and pathwise uniqueness make them equal up to those times. On every finite interval both positive continuous paths have positive minima, so decreasing $\epsilon$ removes the localization and proves indistinguishability. ThusThe resulting global solution also supplies the stopped solution for any threshold $\epsilon \ge x$, completing the literal all-threshold existence request. The monotonicity used to construct the zero-boundary lifetime concerns the thresholds $\epsilon <x$ tending to zero; upper-threshold hitting times have the opposite ordering. Strong adaptation is preserved because every cutoff solution is strong and the patch uses their stopping times in the given filtration.