Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 35 / 1 / d

Differentiate once less than might seem necessary: writing $v=h^{\prime }$ turns the equation intoUniqueness for this first-order linear equation also covers the solution $v\equiv 0$. Thus every solution has $h^{\prime }(z)=C[z(z-1){]}^{-4/\kappa }$. Since $4/\kappa <1$, the derivative is integrable at $1$. The specified boundary value fixes the additive constant, givingThe strict positivity forces $C>0$; $C=0$ gives the zero function. These are all the positive solutions. The function $s_{\kappa }$ is the increasing scale function of a one-dimensional diffusion for the ratio process.

As $v\to \infty$, $[v(v-1){]}^{-4/\kappa }\sim v^{-8/\kappa }$. ThereforeThus every positive solution is unbounded when κ ≥ 8, including the logarithmically divergent critical case. For $4<\kappa <8$, by contrast,For example the substitution $r=1-1/v$ identifies this normalizing constant as $B(1-4/\kappa ,8/\kappa -1)$, with both beta function parameters positive.

For any $C^2$ function $h$, the Itô formula and the ratio equation giveThe bracket is zero for the functions just found. On compact localized intervals the stochastic integrand is bounded, so the stochastic integral is a martingale. Removing localization proves that $h(Z_t)$, on $0\le t<T(x)$, is a local martingale. It is nonnegative because $Z_t>1$ and $C>0$.

In particular it is a nonnegative local martingale, hence a supermartingale after the usual stopping and extension by its endpoint limit. This is the justification for using its maximal inequality; it is not automatically an unstopped uniformly integrable martingale.

The time-changed diffusion has generator $\mathcal{A}=(\kappa /2){\partial }_{zz}+2(1/z+1/(z-1)){\partial }_z$. For $1<z<R$, its boundary hitting probability from a diffusion scale function isTo see this directly, stop the scale local martingale on $[1+ϵ,R]$ and pass to $ϵ\downarrow 0$. Exit from $(1,R)$ occurs in finite clock time: the scale density behaves as $(z-1{)}^{-4/\kappa }$ and the speed density as $(z-1{)}^{4/\kappa }$, so the usual scale-times-speed exit integral is finite at $1$. More explicitly, with $m(v)=2/(\kappa s_{\kappa }^{\prime }(v))$ the finite-interval diffusion exit Green kernel givesThe equation $\mathcal{A}u=-1$ follows by differentiation and the derivative jump at $v=z$, so stopping the corresponding Itô identity gives this exit bound. The integrand near $1$ is $O(v-1)$; elsewhere it is bounded. At large $z$ the drift is bounded, excluding explosion at infinity in finite clock time.

Here is why these exits describe the original swallowing event. In clock time,If $\hat{Z}$ hits $1$ at finite $u$, its integrated stochastic differential equation shows ${\int }_0^{{\tau }_1}({\hat{Z}}_v-1{)}^{-1}dv<\infty$: the Brownian term has a finite limit and both positive drift integrals must then have finite limits. Consequently $D({\tau }_1)>0$, $X$ hits zero and $Y$ stays positive. This gives $T(x)<T(y)$.

Conversely $T(x)$ is finite by part (b). If its clock ends at a finite value, the diffusion must hit $1$; an interior endpoint with positive $D$ would leave $X$ positive. If the clock runs forever without hitting $1$, the standard one-dimensional scale classification gives escape to infinity whenever $S_{\kappa }<\infty$. This cannot represent a strict swallowing event, which has $D\to Y_{T(x)}>0$ and $Z\to 1$. Hence the surviving event represents $T(x)=T(y)$. This reasoning also explains why a simultaneous collision must be treated separately rather than assigned the boundary value at $1$.

For $\kappa \ge 8$, the scale is unbounded. The maximal inequality for a nonnegative supermartingale gives $\mathbb{P}(\sup s_{\kappa }(\hat{Z})\ge s_{\kappa }(R))\le s_{\kappa }(z)/s_{\kappa }(R)$; letting $R\to \infty$ rules out escape. Finite-interval exit then forces a hit at $1$. For $4<\kappa <8$, taking $R\to \infty$ givesThe strict-event probability is also positive, since $0<s_{\kappa }(z)<S_{\kappa }$ for finite $z$.

For $\kappa \ge 8$, first intersect the probability-one strict-order events over rational $0<a<b$. The real boundary flow gives nondecreasing swallowing times on $(0,\infty )$. For any $0<x<y$, choose rational $x<a<b<y$; then $T(x)\le T(a)<T(b)\le T(y)$. This proves the simultaneous assertionThe range is positive points, as in the definition of $F$; reflection reverses the ordering on the negative axis. At the printed endpoint $\kappa =4$, each fixed positive point has $T=\infty$. Thus $\mathbb{P}(T(x)=T(y))=1$ if equality of extended times is allowed, but there is no finite simultaneous-swallowing assertion there. The preceding formulas assume $\kappa >4$.