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 $\theta_t = \cos t$:

  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

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

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.

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


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

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?

  • 1
    $\begingroup$ Given a Radon-Nikodym derivative and a base measure, the new measure is clearly unique since you can define its value on any set via an integral with respect to the base measure. And yes, implicit in the application of Girsanov theorem here is the uniqueness of RN derivative. $\endgroup$
    – John Jiang
    Jul 12, 2016 at 21:14
  • $\begingroup$ @JohnJiang So who's right? my prof (solution given)? me (tried something slightly different)? both? $\endgroup$
    – BCLC
    Jul 12, 2016 at 21:25
  • 1
    $\begingroup$ Your solution is more explicit, and less confusing than your professor's. But the question is implicitly assuming uniquness of dQ/dP. I was confused about this at one point as well, but it just shows lack of understanding of the theorem. I would also recommend you try math.stackexchange.com first next time. $\endgroup$
    – John Jiang
    Jul 12, 2016 at 21:40
  • $\begingroup$ @JohnJiang ah thanks. So what about the part at the end re the uniqueness or lack thereof of Q? $\endgroup$
    – BCLC
    Jul 13, 2016 at 16:10
  • 1
    $\begingroup$ see my first comment. $\endgroup$
    – John Jiang
    Jul 13, 2016 at 17:36

1 Answer 1


Uniqueness of the RN derivative, in this context, can be seen as a consequence of the Martingale Representation Theorem.

Taking the RN derivative, we can write $Z_t = E[d\mathbb{Q}/d\mathbb{P}|\mathcal{F}_t]$ for $t\leq T$. This is clearly a martingale (at least after taking a continuous modification) and nonnegative and therefore must be able to be written

$Z_t =1 +\int_0^t H_s Z_s dW_s$

for some predictable process $H$.

Applying Girsanov's theorem, you obtain a unique definition of $H$ which makes the stated process $W^Q$ a martingale. Hence $Z$ is uniquely defined, so $\mathbb{Q}$ is uniquely defined.

  • $\begingroup$ Do you disagree with Quantuple who says the problem is not stated correctly? $\endgroup$
    – BCLC
    Nov 25, 2022 at 19:54

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