<p>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</p>

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

<p>is one such measure.</p>

<p>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 $\mathcal{F}$.</p>

<p>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.    </p>

CC BY-SA 3.0
2017-04-16T19:38:04.347Z
thanks. will read more later. integral of cosine is sine?
BCLC
CC BY-SA 4.0
2022-11-25T19:54:39.073Z
Do you disagree with [Quantuple](https://quant.stackexchange.com/questions/28003/given-mathbb-q-and-x-t-is-mathbb-q-brownian-find-fracd-mathbb-qd-m) who says the problem is not stated correctly?
BCLC
CC BY-SA 3.0
2017-04-14T10:05:36.373Z
2017-04-14T10:05:36.373Z
Michael
<p><strong>The problem:</strong></p>
<blockquote>
<p>Let <span class="math-container">$T &gt;0$</span>, and let <span class="math-container">$(\Omega, \mathscr F, \{ \mathscr F_t \}_{t \in [0,T]}, \mathbb P)$</span> be a filtered probability space where <span class="math-container">$\mathscr F_t = \mathscr F_t^W$</span> where <span class="math-container">$W = \{W_t\}_{t \in [0,T]}$</span> is standard <span class="math-container">$\mathbb P$</span>-Brownian motion.</p>
<p>Let <span class="math-container">$X = \{X_t\}_{t \in [0,T]}$</span> be a stochastic process where <span class="math-container">$X_t = W_t + \sin t$</span>, and let <span class="math-container">$\mathbb Q$</span> be an equivalent probability measure s.t. <span class="math-container">$X$</span> is standard <span class="math-container">$\mathbb Q$</span>-Brownian motion.</p>
<p>Give <span class="math-container">$\frac{d \mathbb Q}{d \mathbb P}$</span>.</p>
</blockquote>
<p><strong>Girsanov Theorem:</strong></p>
<blockquote>
<p>Let <span class="math-container">$T &gt;0$</span>, and let <span class="math-container">$(\Omega, \mathscr F, \{ \mathscr F_t \}_{t \in [0,T]}, \mathbb P)$</span> be a filtered probability space where <span class="math-container">$\mathscr F_t = \mathscr F_t^W$</span> where <span class="math-container">$W = \{W_t\}_{t \in [0,T]}$</span> is the standard <span class="math-container">$\mathbb P$</span>-Brownian motion.</p>
<p>Let the Girsanov kernel <span class="math-container">$\{\theta_t\}_{t \in [0,T]}$</span> be a <span class="math-container">$\mathscr F_t$</span>-adapted stochastic process s.t. <span class="math-container">$\int_0^T \theta_s^2 ds &lt; \infty$</span> a.s. and <span class="math-container">$\{L_t\}_{t \in [0,T]}$</span> is a <span class="math-container">$( \mathscr F_t , \mathbb P)$</span> martingale where</p>
<p><span class="math-container">$$L_t := \exp(-\int_0^t \theta_s dW_s - \frac 1 2 \int_0^t \theta_s^2 ds)$$</span></p>
<p>Let <span class="math-container">$\mathbb Q$</span> be the probability measure defined by</p>
<p><span class="math-container">$$Q(A) = \int_A L_T dP \ \forall A \in \ \mathscr F$$</span></p>
<p>or <span class="math-container">$$L_T = \frac{d \mathbb Q}{d \mathbb P}$$</span></p>
<p>Then <span class="math-container">$\{W_t^Q\}_{t \in [0,T]}$</span> defined by</p>
<p><span class="math-container">$$W_t^Q := W_t + \int_0^t \theta_s ds$$</span></p>
<p>is standard <span class="math-container">$\mathbb Q$</span>-Brownian motion.</p>
</blockquote>
<hr />
<p><strong>The solution given:</strong></p>
<p><span class="math-container">$$X_t = W_t + \int_0^t \cos s ds$$</span></p>
<p>Let <span class="math-container">$\theta_t = \cos t$</span>:</p>
<ol>
<li><p>It is <span class="math-container">$\mathscr F_t$</span>-adapted</p>
</li>
<li><p><span class="math-container">$\int_0^T \theta_s^2 ds &lt; \infty$</span> a.s.</p>
</li>
<li><p><span class="math-container">$E[\exp(\frac 1 2 \int_0^T \theta_t^2 dt)] &lt; \infty$</span></p>
</li>
</ol>
<p>Then <span class="math-container">$\{L_t\}_{t \in [0,T]}$</span> is a <span class="math-container">$( \mathscr F_t , \mathbb P)$</span> martingale, by Novikov's condition, where</p>
<p><span class="math-container">$$L_t := \exp(-\int_0^t \cos s dW_s - \frac 1 2 \int_0^t \cos^2 s ds)$$</span></p>
<p>Thus, by Girsanov's Theorem, we have</p>
<p><span class="math-container">$$\frac{d\mathbb Q}{d\mathbb P} = L_T...?$$</span></p>
<hr />
<p>How exactly does that last line follow?</p>
<p>What I find strange is that the Girsanov Theorem defines <span class="math-container">$\mathbb Q$</span> and then concludes <span class="math-container">$X_t$</span> is standard <span class="math-container">$\mathbb Q$</span>-Brownian motion while the problem says there is some <span class="math-container">$\mathbb Q$</span> s.t. <span class="math-container">$X_t$</span> is standard <span class="math-container">$\mathbb Q$</span>-Brownian motion and then asks about <span class="math-container">$\frac{d \mathbb Q}{d \mathbb P}$</span>. Is the problem maybe stated wrong?</p>
<p>To say that <span class="math-container">$L_T$</span> is indeed the required density <span class="math-container">$\frac{d \mathbb Q}{d \mathbb P}$</span>, I think we need to use the <a href="https://books.google.com.hk/books?id=YoMr5Mbo6coC&amp;pg=PT324&amp;lpg=PT324&amp;dq=converse%20of%20the%20girsanov&amp;source=bl&amp;ots=iniu5xWy6e&amp;sig=w7uyixE0bbaGdT9shskEMhle3rM&amp;hl=en&amp;sa=X&amp;ved=0ahUKEwif1sS3nePJAhXC2aYKHcInBccQ6AEIOTAE#v=onepage&amp;q=converse%20of%20the%20girsanov&amp;f=false" rel="nofollow noreferrer">converse of the Girsanov Theorem</a>), or maybe the problem should instead give us <span class="math-container">$\frac{d \mathbb Q}{d \mathbb P}$</span> and then ask us to show that <span class="math-container">$L_T = \frac{d \mathbb Q}{d \mathbb P}$</span> possibly showing that <span class="math-container">$E[\frac{d \mathbb Q}{d \mathbb P} | \mathscr F_t] = L_t$</span> or some other route.</p>
<hr />
<p><strong>I tried something slightly different:</strong></p>
<p>I define <span class="math-container">$\hat{\mathbb P}$</span> s.t.</p>
<p><span class="math-container">$$L_T = \frac{d\hat{\mathbb P}}{d\mathbb P}$$</span></p>
<p>or</p>
<p><span class="math-container">$$\hat{\mathbb P} = \int_A L_T d\mathbb P$$</span></p>
<p>It follows by Girsanov Theorem that <span class="math-container">$X_t$</span> is standard <span class="math-container">$\hat{\mathbb P}$</span>-Brownian motion. Since we are given that there is some <span class="math-container">$\mathbb Q$</span> equivalent to <span class="math-container">$\mathbb P$</span> s.t. <span class="math-container">$X_t$</span> is also standard <span class="math-container">$\mathbb Q$</span>-Brownian motion, it follows by the uniqueness of the Radon-Nikodym derivative that</p>
<p><span class="math-container">$$\frac{d\hat{\mathbb P}}{d\mathbb P} = \frac{d\mathbb Q}{d\mathbb P}$$</span></p>
<p><span class="math-container">$\therefore, \frac{d\mathbb Q}{d\mathbb P}$</span> is given by <span class="math-container">$L_T$</span>.</p>
<p>Is that right? I think I'm missing a step somewhere.</p>
<p>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?</p>
<hr />
<p>Edit based on <a href="https://quant.stackexchange.com/questions/15159/uniqueness-of-equivalent-martingale-measure-in-black-scholes-model#comment31938_15167">this</a>: Even if Radon-Nikodym derivative is unique, <span class="math-container">$\mathbb Q$</span> may not be unique? If so, is it then that <span class="math-container">$\hat{\mathbb P}$</span> is merely a candidate for one of many possible <span class="math-container">$\mathbb Q$</span>'s?</p>
<p>I think we conclude <span class="math-container">$\hat{\mathbb P} = \mathbb Q$</span> based on <span class="math-container">$X_t$</span> being standard Brownian motion under both measures. Is there a proposition for that? Uniqueness of Brownian motion measure or something?</p>

<mathematical-economics><finance><probability><stochastic-processes><stochastic-calculus>
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
BCLC
# 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

(Asked by: username_0 on 2016-07-09)

**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](https://books.google.com.hk/books?id=YoMr5Mbo6coC&pg=PT324&lpg=PT324&dq=converse%20of%20the%20girsanov&source=bl&ots=iniu5xWy6e&sig=w7uyixE0bbaGdT9shskEMhle3rM&hl=en&sa=X&ved=0ahUKEwif1sS3nePJAhXC2aYKHcInBccQ6AEIOTAE#v=onepage&q=converse%20of%20the%20girsanov&f=false)), 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](https://quant.stackexchange.com/questions/15159/uniqueness-of-equivalent-martingale-measure-in-black-scholes-model#comment31938_15167): 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?

**Tags:** mathematical-economics, finance, probability, stochastic-processes, stochastic-calculus


---

## 1 answers

### Answer by username_1 on 2017-04-14. Score: 2

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.

> **Comments:**
> **username_0**: thanks. will read more later. integral of cosine is sine?  
> **username_0**: Do you disagree with [Quantuple](https://quant.stackexchange.com/questions/28003/given-mathbb-q-and-x-t-is-mathbb-q-brownian-find-fracd-mathbb-qd-m) who says the problem is not stated correctly?  

---

