Let $W$ be a standard one dimensional Brownian motion, $\mathcal F\_t$ its natural filtration, and $\mathbb P$ be the induced Wiener measure on $\Omega := C[0, 1]$.


Given a $C[0, 1] $ valued random variable $F$, define the translation map $T\_F: \Omega \to \Omega$ by $T\_F (\omega) = \omega + F(\omega)$, and denote the pushforward of $\mathbb P$ under this map by $T\_F^{\ast} \, \mathbb P$.


We view $F$ as a stochastic process and say that $F$ is adapted to $W$ if as a stochastic process, $F\_t$ is $\mathcal F\_t$ adapted. Denote by $W^{1, 2}$ the space of absolutely continuous functions on $[0, 1]$ with derivative in $L^2$.



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> **Question:** Is it true that $\mathbb Q$ is equivalent to $\mathbb P$ if and only if $\mathbb Q = T\_F ^\ast \, \mathbb P$ for some adapted $F$ such that $F(0) = 0$ and $F \in W^{1, 2}$ almost surely?
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Converse Cameron-Martin theorem for shifts by adapted processes
If $F\_t(\omega)=\int\_0^t h\_s(\omega)\,ds$, with $h$ progressive and such that $\int\_0^1 h\_s^2(\omega)\,ds\le C$ for some finite constant $C$, then indeed $T^\*\_F\Bbb P$ is equivalent to $\Bbb P$.


Let $X\_t$ denote the coordinate process on $\Omega$, a Brownian motion under $\Bbb P$. Now define the martingale $M\_t:=\int\_0^t h\_s\,dX\_s$ and the associated exponential martingale $L\_t:=\exp(-M\_t-{1\over 2}\langle M\rangle\_t)$. (This is a martingale by Novikov's criterion.)
Finally define $\Bbb Q$ to be the probability measure on $\Omega$ with density $L\_1$ with respect to $\Bbb P$. By Girsanov's theorem, the process $Y\_t:=X\_t+F\_t$ is a $\Bbb Q$-Brownian motion. Then for any bounded measurable $f:\Omega\to\Bbb R$,
$$
\int\_\Omega f\,d\Bbb P = \int\_\Omega f(Y)\,d \Bbb Q=
\int\_\Omega L\_1 f(Y)\,d \Bbb P.
$$
Thus, $\int\_\Omega f\,d\Bbb P=0$ iff $\int\_\Omega L\_1 f(Y)\,d \Bbb P=0$, and in turn iff $\int\_\Omega f\,dT^\*\_F\Bbb P=\int\_\Omega f(Y)\,d \Bbb P=0$ because the density $L\_1$ is strictly positive and finite a.s. $\Bbb P$.


The boundedness condition on $h$ can be partially relaxed by localization. Suppose $\int\_0^1 h^2\_s\,ds<\infty$, $\Bbb P$-a.s. Define, for each positve integer $n$, $T\_n:=\inf\{t: \int\_0^t h\_s^2\,ds>n\}$, and then $F^{(n)}\_t:=\int\_0^{t\wedge T\_n} h\_s\,ds$. By the preceding paragraph, $\Bbb P$ is equivalent to $T^\*\_{F^{(n)}}\Bbb P$ for each $n$. But $F^{(n)}=F$ on the event $G\_n:=\{\int\_0^1 h^2\,ds\le n\}$. Because $\Bbb P(\cup\_nG\_n)=1$, it follows that $T^\*\_F\Bbb P \ll\Bbb P$.


