# Converse Cameron-Martin theorem for shifts by adapted processes

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?

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

• Thanks, really interesting if this checks out - I haven’t found any books that actually address this. I will check again tomorrow but it seems to be correct as far as I can tell. Mar 18 at 21:26
• This definitely checks out actually, nice. Just one thing, where precisely did we use that $h$ was progressive? The result fails without that but I haven’t been able to see where it’s used. Mar 18 at 21:37
• $h$ needs to be progressive so that the stochastic integral $\int_0^t h_s\,dX_s$ is defined. Mar 18 at 22:07

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.

• Could you explain in more detail how Girsanov's theorem yields $F$ such that $\Bbb Q=T^*_F\Bbb P$? Mar 18 at 17:35