 Solution

Because $v\in L^1(0,1)$, its indefinite integral $V$ is bounded, so $e^{-V}$ is bounded above and away from zero. Hence $\psi$ extends continuously and strictly increasingly to $[0,1]$. The quadratic-variation clock of $\psi (X)$ iswhose rate is bounded above and away from zero before exit. The Dambis-Dubins-Schwarz theorem therefore identifies $\psi (X)$, up to an equivalent time change, with Brownian motion in the bounded interval $(0,\psi (1))$; in particular, $\mathcal{T}<\infty$ almost surely.

The bounded stopped local martingale $\psi (X_{t\wedge \mathcal{T}})$ is a martingale. If $q={\mathbb{P}}_x(X_{\mathcal{T}}=0)$, the optional sampling theorem givesThereforeThis is the boundary hitting probability from a diffusion scale function.