 Solution

A finite-horizon form of Girsanov theorem is as follows. If $\theta$ is predictable, ${\int }_0^T{\theta }_s^{2}ds<\infty$ almost surely, and its stochastic exponential $M$ is a true martingale of mean one, then under the probability $d\mathbb{Q}=M_{T}d\mathbb{P}$ the processis Brownian motion. The same conclusion holds on the completed natural filtration, since a strictly positive density gives equivalent probabilities.

Part (b) establishes precisely the required density martingale and integrand ${\theta }_s={\partial }_x\log p(1-s,B_s,y)$. ThereforeIn particular it is the requested local martingale. Equivalently, under $\mathbb{Q}$, $B$ solves $d{B}_t=(y-B_t)(1-t{)}^{-1}dt+d{X}_t$, the Brownian bridge drift coefficient toward $y$ at time one. Keeping $T<1$ avoids the singular endpoint in this density change.