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I’m looking for a version of the Cameron Martin theorem for the Brownian motion under random shifts. Here is the precise statement:

Let $\mathbb P$ be 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 $.

Denote by $W^{1, 2}$ the space of absolutely continuous functions on $[0, 1]$ with derivative in $L^2$.

Theorem: $\mathbb Q$ is equivalent to $\mathbb P$ if and only if $\mathbb Q = T_F ^\ast \, \mathbb P$ for some $F$ such that $F(0) = 0$ and $F \in W^{1, 2}$ almost surely.

My questions are two-fold:

  1. Is this statement true?

  2. If so, where can I find a proof?

So far the statements I could find were only for deterministic shifts $F$, these are what is usually called the Cameron Martin theorem for Brownian motion. On the other hand, statements of Girsanov’s theorem deal with random shifts, but typically are concerned with the Radon Nikodym derivative, and are silent on the interpretation as shifts by $W^{1, 2}$ processes.

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    $\begingroup$ Did you check the monograph of Lipster and Shiryaev? They proved various results of this type characterizing measures absolutely cts wrt bm. They give a structure theorem for such measures of similar flavor to what you posted above. See this chapter: link.springer.com/chapter/10.1007/978-3-662-13043-8_8 $\endgroup$
    – shalop
    Commented Dec 21, 2022 at 19:59
  • $\begingroup$ Oh, I have not seen that - seems pretty comprehensive indeed. Thanks! $\endgroup$
    – Nate River
    Commented Dec 22, 2022 at 5:13

1 Answer 1

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Let $\varphi$ be a smooth function such that $\varphi(x)=0$ for $x\leq 0$ and $\varphi(x)=1$ for $x\geq 1$, and $0\leq\varphi(x)\leq 1$ else. If $\tau_1$ and $\tau_2$ are is the first times $B_t$ hits $1,2$ respectively, and $M$ is a maximum of the $B_t$ on $[0,1]$, put $F_t=-M\varphi(\frac{t-\tau_1}{\tau_2-\tau_1})$. Then, $\sup (B_t+F_t)\leq 2$ almost surely, hence $\mathbb{P}$ is not absolutely continuous with respect to $\mathbb{Q}$.

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  • $\begingroup$ Ah, so it is not true I see. Thank you for your answer! $\endgroup$
    – Nate River
    Commented Dec 10, 2022 at 8:18
  • $\begingroup$ @NateRiver, I don't know though whether $\mathbb{Q}$ is always a.c. w.r.t $\mathbb{P}$, or whether the converse implication holds, but I would be very surprised if it did. $\endgroup$
    – Kostya_I
    Commented Dec 10, 2022 at 11:08
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    $\begingroup$ @Kostya_I see fabricebaudoin.wordpress.com/2012/10/02/… for a "converse" Girsanov $\endgroup$
    – user479223
    Commented Dec 10, 2022 at 13:13

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