Converse Cameron-Martin theorem for shifts by adapted processes

Question

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 $L^2$.

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?

Answer 1

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 $T^*_F\Bbb P \ll\Bbb P$.

Answer 2

If $\mathbb Q$ is equivalent to $\mathbb P$ then we have that $\mathbb Q=T_F^\ast \mathbb P$ for some $F\in W_0^{1,2}$ a.s. See https://fabricebaudoin.wordpress.com/2012/10/02/lecture-25-girsanov-theore/ for a proof.

The converse however is not true. Let $U$ solve $$dU(t)=22U^2(t) dB(t),$$

with $U(0)=1$. Then $U$ is a local martingale but not a martingale (see https://math.stackexchange.com/questions/2019053/proving-that-a-process-is-a-local-martingale-but-not-a-martingale). Additionally, it is the Doleans-Dade exponential of its stochastic logarithm.

Crucially $E[U_t]\leq Ct^{-1/2}$, so $U$ cannot define a density.