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For $\mathcal{L}=\frac{1}{2}\Delta -\lambda x\cdot \nabla$ and potential $V={\lambda }^2|x{|}^2/2$, the Feynman-Kac formula with initial value one gives $u={\mathbb{E}}_x^{OU}{e}^{{\int }_0^{t}V(X_s)ds}$. The Ornstein-Uhlenbeck likelihood relative to Wiener measure cancels the quadratic time integral, leaving a Gaussian endpoint integral. Completing the square gives the displayed answer for $1+\lambda t>0$. It is finite for all nonnegative times when $\lambda \ge 0$; when $\lambda <0$ the path expectation diverges at and beyond $t=-1/\lambda$.

If $X$ is a P-Brownian motion, its starting value is deterministic, so Q5(a) applies. The correction $[X,M]$ has finite variation and does not alter quadratic variation; Q1(b) gives $[\tilde{X}{]}_t=t$ under Q. The Lévy characterization of Brownian motion therefore identifies $\tilde{X}$ as a Q-Brownian motion with the same starting value.

For the likelihood, use Wiener measure ${\mathbb{Q}}^x$ as the reference measure and write its coordinate as ${\omega }_t=x+W_t$. Setand let $L=\mathcal{E}(M)$ be its stochastic exponential. We must prove it is a true martingale; unverified large-horizon Novikov estimates would not suffice.

Fix $T$ and stop at ${\tau }_n=\inf \{s:|{\omega }_s|\ge n\}$, with $n>|x|$. The bounded integrand satisfies the Novikov condition, so $L_{T\wedge {\tau }_n}$ defines a probability $P_n$ on ${\mathcal{F}}_T$. By the result just established,is Brownian under $P_n$ on this horizon. Therefore $d{\omega }_s=d{W}_s^{(n)}-\mu 1_{\{s\le {\tau }_n\}}{\omega }_{s}ds$. The Gronwall inequality givesThe Doob L2 maximal inequality then bounds its squared expectation by $C_T=2e^{2|\mu |T}(x^2+4T)$, uniformly in $n$. In particular $P_n({\tau }_n\le T)\le C_T/n^2$. On $\{{\tau }_n>T\}$ the stopped and unstopped densities agree, soContinuous Brownian paths are bounded on $[0,T]$, and the indicators increase to one. The monotone convergence theorem proves $\mathbb{E}{L}_T=1$. Since a nonnegative local martingale is a supermartingale, constant expectation at every time makes $L$ a true martingale.

Under its tilted measure, ${\hat{W}}_t=W_t+\mu {\int }_0^t{\omega }_{s}ds$ is Brownian and $d{\omega }_t=d{\hat{W}}_t-\mu {\omega }_{t}dt$. The explicit linear solution gives uniqueness of this law, identifying it with ${\mathbb{P}}^x$. Consequently the Ornstein-Uhlenbeck likelihood relative to Wiener measure isThis is a finite-horizon density statement; it does not assert absolute continuity of the full infinite-horizon path laws.