**The problem:**

Let $T >0$, and let $(\Omega, \mathscr F, \{ \mathscr F_t \}_{t \in [0,T]}, \mathbb P)$ be a filtered probability space where $\mathscr F_t = \mathscr F_t^W$ where $W = \{W_t\}_{t \in [0,T]}$ is standard $\mathbb P$-Brownian motion.

Let $X = \{X_t\}_{t \in [0,T]}$ be a stochastic process where $X_t = W_t + \sin t$, and let $\mathbb Q$ be an equivalent probability measure s.t. $X$ is standard $\mathbb Q$-Brownian motion.

Give $\frac{d \mathbb Q}{d \mathbb P}$.

**Girsanov Theorem:**

Let $T >0$, and let $(\Omega, \mathscr F, \{ \mathscr F_t \}_{t \in [0,T]}, \mathbb P)$ be a filtered probability space where $\mathscr F_t = \mathscr F_t^W$ where $W = \{W_t\}_{t \in [0,T]}$ is the standard $\mathbb P$-Brownian motion.

Let the Girsanov kernel $\{\theta_t\}_{t \in [0,T]}$ be a $\mathscr F_t$-adapted stochastic process s.t. $\int_0^T \theta_s^2 ds < \infty$ a.s. and $\{L_t\}_{t \in [0,T]}$ is a $( \mathscr F_t , \mathbb P)$ martingale where

$$L_t := \exp(-\int_0^t \theta_s dW_s - \frac 1 2 \int_0^t \theta_s^2 ds)$$

Let $\mathbb Q$ be the probability measure defined by

$$Q(A) = \int_A L_T dP \ \forall A \in \ \mathscr F$$

or $$L_T = \frac{d \mathbb Q}{d \mathbb P}$$

Then $\{W_t^Q\}_{t \in [0,T]}$ defined by

$$W_t^Q := W_t + \int_0^t \theta_s ds$$

is standard $\mathbb Q$-Brownian motion.

**The solution given:**

$$X_t = W_t + \int_0^t \cos s ds$$

Let $\theta_t = \cos t$:

It is $\mathscr F_t$-adapted

$\int_0^T \theta_s^2 ds < \infty$ a.s.

$E[\exp(\frac 1 2 \int_0^T \theta_t^2 dt)] < \infty$

Then $\{L_t\}_{t \in [0,T]}$ is a $( \mathscr F_t , \mathbb P)$ martingale, by Novikov's condition, where

$$L_t := \exp(-\int_0^t \cos s dW_s - \frac 1 2 \int_0^t \cos^2 s ds)$$

Thus, by Girsanov's Theorem, we have

$$\frac{d\mathbb Q}{d\mathbb P} = L_T...?$$

How exactly does that last line follow?

What I find strange is that the Girsanov Theorem defines $\mathbb Q$ and then concludes $X_t$ is standard $\mathbb Q$-Brownian motion while the problem says there is some $\mathbb Q$ s.t. $X_t$ is standard $\mathbb Q$-Brownian motion and then asks about $\frac{d \mathbb Q}{d \mathbb P}$. Is the problem maybe stated wrong?

To say that $L_T$ is indeed the required density $\frac{d \mathbb Q}{d \mathbb P}$, I think we need to use the converse of the Girsanov Theorem), or maybe the problem should instead give us $\frac{d \mathbb Q}{d \mathbb P}$ and then ask us to show that $L_T = \frac{d \mathbb Q}{d \mathbb P}$ possibly showing that $E[\frac{d \mathbb Q}{d \mathbb P} | \mathscr F_t] = L_t$ or some other route.

**I tried something slightly different:**

I define $\hat{\mathbb P}$ s.t.

$$L_T = \frac{d\hat{\mathbb P}}{d\mathbb P}$$

or

$$\hat{\mathbb P} = \int_A L_T d\mathbb P$$

It follows by Girsanov Theorem that $X_t$ is standard $\hat{\mathbb P}$-Brownian motion. Since we are given that there is some $\mathbb Q$ equivalent to $\mathbb P$ s.t. $X_t$ is also standard $\mathbb Q$-Brownian motion, it follows by the uniqueness of the Radon-Nikodym derivative that

$$\frac{d\hat{\mathbb P}}{d\mathbb P} = \frac{d\mathbb Q}{d\mathbb P}$$

$\therefore, \frac{d\mathbb Q}{d\mathbb P}$ is given by $L_T$.

Is that right? I think I'm missing a step somewhere.

So, is that indeed what the solution given is meant to be but just omitted pointing out uniqueness of the Radon-Nikodym derivative, if such justification is right?

Edit based on this: Even if Radon-Nikodym derivative is unique, $\mathbb Q$ may not be unique? If so, is it then that $\hat{\mathbb P}$ is merely a candidate for one of many possible $\mathbb Q$'s?

I think we conclude $\hat{\mathbb P} = \mathbb Q$ based on $X_t$ being standard Brownian motion under both measures. Is there a proposition for that? Uniqueness of Brownian motion measure or something?