Given $\mathbb Q$ and $X_t$ is $\mathbb Q$-Brownian, find $\frac{d\mathbb Q}{d\mathbb P}$ / Uniqueness of Brownian or Radon-Nikodym derivative

Question

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.

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The solution given:

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

Let $\theta_t = \cos t$:

1. It is $\mathscr F_t$-adapted

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

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

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

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

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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.

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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?

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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?

Answer 1

IMHO the problem isn't stated correctly indeed, in the sense that the Radon-Nikodym derivative provided as the "solution" is not the unique way to define a measure $\mathbb{Q}$ equivalent to $\mathbb{P}$ and under which $X_t$ is a martingale. Just take $$\frac {d\mathbb{Q}}{d\mathbb{P}} =\mathcal{E}\left(-\int_0^t \cos(s) dW_s + a\right)$$ for any $a \in \mathbb{R}$ (or any finite variation process for that matter, the point being that $a $ should not contribute in terms of quadratic (co)variation) and $\mathcal {E}(.) $ figures the stochastic (or Doleans-Dade) exponential. Note that the solution provided corresponds to picking $a=0$.

As such, I think that the exercice should have been written the other way around as you mention in your question. Something like: show that, given the following Radon-Nikodym derivative specification, the resulting measure $\mathbb{Q}$ is equivalent to the original measure $\mathbb{P}$ and such that $(X_t)_{t\in [0,T]}$ is a $(\mathbb{Q},\mathcal{F}_t^W)$-martingale.