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The drift $b(z)=\tanh z$ has derivative $b^{\prime }(z)=1/{\cosh }^{2}z$, so $|b^{\prime }(z)|\le 1$. Hence it is globally Lipschitz. The diffusion coefficient $\sigma (z)=1$ is globally Lipschitz as well, and both coefficients satisfy a linear growth bound.

The global existence theorem for stochastic differential equations with Lipschitz coefficients states that globally Lipschitz coefficients with linear growth give, for each deterministic initial point, an adapted continuous strong solution of a stochastic differential equation on every finite interval, with pathwise uniqueness and no finite-time explosion. Applying this theorem gives a unique strong solution for every $X_0=x\in \mathbb{R}$, satisfying

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