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 Answer 1


As you say (how did $\sin t$ in the initial problem become $\cos t$?), $\mathbb{Q}$ is a measure under which $W_t$ becomes $\tilde{W}_t - \sin t$, where $\tilde{W}_t$ is a $\mathbb{Q}$-Brownian motion. So Girsanov's theorem would say

$$ L_t = E[\frac{ d \mathbb{Q} }{ d \mathbb{P} }|\mathcal{F}_t] = e^{ - \int_0^t \sin s dW_s - \frac{1}{2} \int_0^t \sin^2 s ds} $$

is one such measure.

As for uniqueness, if we take $\Omega$ to be the canonical path space $C[0, T]$ with $\mathcal{F}$ generated by the cylinder sets. Then the Wiener measure would be unique, simply by the fact that the Brownian law says any two such measures must agree on the cylinder sets and therefore all of $\mathcal{F}$.

In general, the Girsanov $\mathbb{Q}$ is unique up to the Brownian filtration $\mathcal{F}_{W} \subset \mathcal{F}$, since it is constructed $W$-pathwise.

  • $\begingroup$ thanks. will read more later. integral of cosine is sine? $\endgroup$
    – BCLC
    Apr 16, 2017 at 19:38
  • $\begingroup$ 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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