Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 30 / 2 / d / Solution

On $I=g(\mathbb{R})$ define $h(y)=g^{\prime }(g^{-1}(y))$. Differentiation givesThe mean value theorem proves Lipschitz continuity of $h$ on $I$, with constant $2B$.

If an endpoint of $I$ is finite, the Lipschitz bound gives a finite limiting value of $h$ there. That value must be zero. Otherwise $h$ would be bounded below by a positive number near the endpoint, andwould make $g^{-1}$ approach a finite limit, contradicting its tending to $+\infty$ or $-\infty$. The zero extension of a scale diffusion coefficient at finite endpoints therefore gives a globally Lipschitz function $\bar{h}$ on $\mathbb{R}$: set it to zero beyond each finite endpoint and retain $h$ on $I$.

Use the following standard global existence theorem for stochastic differential equations with Lipschitz coefficients: globally Lipschitz drift and diffusion coefficients give a nonexplosive, pathwise unique strong stochastic solution for every prescribed initial state and driving Brownian motion. Global Lipschitz continuity also gives the required linear-growth bound. Apply it toIt remains to check that $Y$ stays in $I$, so the inverse transform is defined.

Before the first boundary time, put $X=g^{-1}(Y)$. Since $(g^{-1}{)}^{\prime }=1/h$ and $(g^{-1}{)}^{\prime \prime }=-h^{\prime }/h^2$, the Itô formula, stopped inside compact subintervals of $I$, yieldsFor each finite $T$, up to that boundary time,A finite endpoint of $I$ would require $X$ to diverge, which this bound excludes. Therefore $Y$ stays in $I$ at every finite time, and $X=g^{-1}(Y)$ is a global strong stochastic solution.

Finally, any two solutions driven by the same $W$ transform into solutions of the globally Lipschitz $Y$ equation. Its pathwise uniqueness makes the transforms, and hence their inverses, indistinguishable. Thus