# 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

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 here), 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 comment below and 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?

• You are right nothing in Girsanov's Theorem states that the $\theta$-transformation gives the unique probability under which $X$ is a BM. The fact that you build one solution using Girsanov Theorem doesn't guarantee that it is indeed equal to $\mathbb{Q}$ you are lookign for. I am not sure about htis but looking at the correction I would say that it is more a blunder in the redaction than anything else. I am not sure if we can conclude on the uniqueness part, it might be possible but it might be quite technical. Best regards. Commented Dec 17, 2015 at 15:34
• @TheBridge Thanks. Is it at least true that $\hat{\mathbb P}$ is one of several (infinite?) candidates for $\mathbb Q$?
– BCLC
Commented Dec 17, 2015 at 15:36
• @ BCLC : Yes that's totally true. Regards. Commented Dec 17, 2015 at 17:36
• @TheBridge Clarification: Infinite?
– BCLC
Commented Dec 18, 2015 at 9:24
• @ BCLC : I don't know, what is sure is that Girsanov Theorem authorizes you to find one. By the way unicity should be viewed modulo equivalence class of negligible sets. In the ned my intuition is that it might be true under general hypotheses that unicity holds but it is only a guess and as I said this kind of questions can be sometimes quite tricky and counter intuitive. Best regards Commented Dec 18, 2015 at 9:55

The R-N derivative process $Z_t:=d\Bbb Q|_{\mathcal F_t}/d\Bbb P|_{\mathcal F_t}$ is a strictly positive $\Bbb P$-martingale. As such it can be written as a stochatic exponential $\exp(M_t-{1\over 2}\langle M\rangle_t)$, with $M$ a $(\mathcal F_t^W,\Bbb P)$-local martingale. Thus $M$ admits a stochastic integral representation as $M_t=\int_0^t H_s\,dW_s$, with $H$ predicatable and $\int_0^T H_s^2\,ds<\infty$ $\Bbb P$-a.s. By Girsanov's theorem, the process $W_t-\int_0^t H_s\,ds$ is a $\Bbb Q$-local martingale. By hypothesis, $W_t+\sin t$ is also a $\Bbb Q$-local martingale. Subtracting we find that the process $\int_0^t H_sds+\sin t$ is a continuous $\Bbb Q$-local martingale that is also of finite variation. Consequently, $\int_0^t H_sds+\sin t=0$ for all $t\ge0$, $\Bbb Q$-a.s. (hence also $\Bbb P$-a.s.). It follows that $H_t(\omega)=-\cos t$ for $\Bbb P\otimes \lambda$-a.e $(\omega,t)\in \Omega\times[0,T]$. (Here $\lambda$ is Lebesgue measure.) In particular, $M_t=-\int_0^t\cos s\,dW_s$, and $d\Bbb Q/d\Bbb P=\exp(-\int_0^T\cos s\,dW_s-{1\over 2}\int_0^T\cos^2s\,ds)$.

• Will read later but going to award bounty in mean time hahaha. Thanks John Dawkins
– BCLC
Commented Dec 26, 2015 at 4:40
• So the ae/as-uniqueness of the representation of $\frac{dQ}{dP}$ is based on $$\int_0^t H_s ds + \sin t = 0 \ \forall t \in [0,T] \ or \ t \ge 0$$. How do you know that?
– BCLC
Commented Dec 27, 2015 at 9:10
• Btw, is there some integrability condition that allows to remove the word 'local' in your answer? I've heard of local martingales, but they weren't discussed in classes
– BCLC
Commented Dec 27, 2015 at 9:11
• Do you disagree with Quantuple who says the problem is not stated correctly?
– BCLC
Commented Nov 25, 2022 at 19:54