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For a deterministic path $h$ in the Cameron-Martin space of Wiener measure, the law of $T_h\mathbf{B}$ is equivalent to the law of enhanced Brownian motion. The Cameron-Martin theorem gives equivalence for the underlying Brownian paths; the canonical enhancement of the translated path is exactly the rough-path translation, including its mixed area terms. Continuity of this translation transfers small-ball positivity at the identity to every control signature.

If $h(t)={\int }_0^t{g}_{s}ds$ with deterministic $g\in L^2(0,\infty )$, the finite-time Cameron-Martin theorem densities have second moments $e^{{\int }_0^t{g}_s^{2}ds}$ bounded uniformly in time. Their uniformly integrable martingale limit is the strictly positive displayed density, giving equivalence on the full infinite continuous-path space. Finite-time densities alone do not guarantee such a global density: a nonzero constant drift has infinite energy and its infinite-horizon path law is singular, distinguished by the limiting ratio $B_t/t$.

For the Cameron-Martin space of Wiener measure path $h(s)=\mu s$, the Cameron-Martin theorem says that translation of Brownian motion by this linear function changes its law through time $t$ by the density

Here is a finite-horizon Girsanov theorem statement with the sign fixed explicitly. Let $B$ be a Brownian motion under $P$, and let $\theta$ be predictable with ${\int }_0^T{\theta }_s^{2}ds<\infty$ almost surely. Ifis a true martingale with $\mathbb{E}{L}_T=1$, then the measure $Q$ defined by $dQ=L_{T}dP$ makes $B_t-{\int }_0^t{\theta }_{s}ds$ a Brownian motion up to $T$. The sufficient Novikov condition is $\mathbb{E}\exp (\frac{1}{2}{\int }_0^T{\theta }_s^{2}ds)<\infty$. Merely being a local martingale is not a sufficient density condition.

For a deterministic absolutely continuous function $h$ with $h(0)=0$ and derivative $g\in L^2[0,T]$, take $\theta =g$. Novikov's condition holds because its exponential is deterministic and finite. Under $Q$, the canonical process has the law of a Brownian motion plus $h$. Thus for every bounded measurable path functional $F$,This derives the Cameron-Martin theorem density formula from the Girsanov theorem, rather than assuming the two measures have the same drift convention. Replacing $h$ by $-h$ changes the sign of the stochastic integral.

The question's path space is the whole half-line, not just one finite horizon. When $g\in L^2(0,\infty )$, the density martingale satisfiesIt has a uniformly integrable limit $L_{\infty }=\exp ({\int }_0^{\infty }{g}_{s}d{B}_s-\Vert g{\Vert }_2^2/2)$ of mean one, strictly positive almost surely. Integrals here converge in $L^2$ and almost surely. The finite-time identities then identify $L_{\infty }$ as the density on the full path sigma-algebra, since finite-time cylinder events generate it. This is Cameron-Martin shifts on infinite Wiener path space. Finite-horizon absolute continuity of measures alone does not imply infinite-horizon absolute continuity of measures.

The Cameron-Martin theorem on Wiener space says that the translated measure ${\mathbb{P}}_h(A)=\mathbb{P}(X+h\in A)$ is equivalent to Wiener measure precisely when $h$ is absolutely continuous, $h(0)=0$, and $\dot{h}\in L^2({\mathbb{R}}_{+})$. In that case

For such $h$, the exponential is a uniformly integrable stochastic exponential. Under the measure defined by this density, the Girsanov theorem makes $X_t-{\int }_0^t{\dot{h}}_{s}ds=X_t-h(t)$ a Brownian motion. This identifies the translated law and proves equivalence; replacing $h$ by $-h$ gives the inverse density.

If $h$ fails the Cameron-Martin condition on some finite interval, the finite-horizon theorem gives singularity there. If it belongs locally but ${\int }_0^{\infty }{\dot{h}}_s^{2}ds=\infty$, the log likelihood is a Brownian motion run at that diverging energy clock minus half the clock. It tends to $-\infty$ under one measure and to $+\infty$ under the translate, producing disjoint full-measure events. Thus the measures are singular.

False for a general continuous $f$. For example, take $f(t)=t^{1/3}$. The Brownian Hölder regularity gives $B_t/t^{1/3}\to 0$ almost surely as $t\downarrow 0$, whilealmost surely. The original and translated path laws therefore concentrate on disjoint measurable events and are mutually singular measures.

More generally, the Cameron-Martin theorem gives the exact criterion: translation by $f$ is absolutely continuous precisely when $f$ is an absolutely continuous function, $f(0)=0$, and $f^{\prime }\in L^2[0,1]$.

True. Since $f\in C^1[0,1]$, it lies in the Cameron-Martin space of Wiener measure. The Cameron-Martin theorem says that the translated law is equivalent, and in particular absolutely continuous, with respect to Wiener measure. Its Radon-Nikodym derivative is