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I'm getting completely bogged down by sign errors when using Girsanov's theorem. Keeping it simple, suppose $W_t$ is a standard Brownian motion under a probability measure $\mathbb{P}$, and we have a Brownian motion with constant drift

$\tilde W_t = W_t+\theta t$

Then Girsanov's theorem says that $\tilde W_t$ is a Brownian motion with no drift under a probability measure $\mathbb{Q}$, which has Radon-Nikodym derivative

$Z_t=\frac{d\mathbb{Q}}{d\mathbb{P}}=\exp(-\theta W_t - \theta^2 t/2)$

My confusion arises when you invert this, so that writing $W_t=\tilde W_t-\theta t$, the Girsanov theorem says that $W_t$ is a Brownian motion with no drift under $\mathbb{P}$, where

$\frac{d\mathbb{P}}{d\mathbb{Q}}=\exp(\theta W_t - \theta^2 t/2)\neq \frac{1}{Z_t}$

Can someone explain what's going on here?

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Actually, writing $W_t=\tilde W_t-\theta t$, (the) Girsanov theorem says that $W_t$ is a Brownian motion with no drift under $\mathbb P$, where... $$\left.\frac{d\mathbb P}{d\mathbb Q}\right|_{\mathcal F_t}=\exp\left(\theta \color{red}{\tilde W_t} - \tfrac12\theta^2 t\right)=\exp\left(\theta W_t +\tfrac12\theta^2 t\right)=\frac{1}{Z_t}$$

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  • $\begingroup$ Thanks so much for clearing that up $\endgroup$ – ndrue Nov 8 '14 at 8:41

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