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We work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\geq 0},\mathbb{P})$. Let $W_t$ be a Brownian Motion adapted and $\theta(t)$ an appropriate $\mathcal{F}_t$-adapted process. According to Girsanov theorem, the following process:

$$\tilde{W}_t=W_t+\int_0^t\theta(t)\text{d}t$$

is a Brownian Motion under the measure $\tilde{\mathbb{P}}$ induced by the Doléans-Dade exponential of $\theta(t)$.

My question is, if we assume that the adapted process is a function of the Brownian Motion, $\theta(t,W_t)$, does Girsanov theorem still hold?

I don't see anything in Girsanov theorem's statement that would imply the opposite but I want to be sure.

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    $\begingroup$ We have to be careful with Girsanov in general settings like this because the theorem requires the Doleans-Dade exponential of $\theta(t, W_t)$ to be a martingale in the first place. If you introduce randomness to $\theta$, it's not so clear. If $\theta$ is deterministic, the Doleans-Dade exponential is in fact a martingale. One way to ensure it's a martingale is to impose Novikov condition. $\endgroup$ – James Yang Apr 11 '18 at 1:27
  • $\begingroup$ Thank you @JamesYang. I was implicitly assuming the usual technical conditions for Girsanov hold. $\endgroup$ – Morris Fletcher Apr 11 '18 at 1:47

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