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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.

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

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: 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?

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    $\begingroup$ 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 $d\mathbb{Q}/d\mathbb{P} =\mathcal{E}(-\int_0^t \cos(s) dW_s + a)$, for any $a \in \mathbb{R}$. I think the exercice should have been written the other way around as you mention: show that with the given Radon-Nikodym derivative, the measure $\mathbb{Q}$ is equivalent to the original measure and such that $X_t$ is a $\mathbb{Q}$-martingale. $\endgroup$ – Quantuple Jul 10 '16 at 15:05
  • $\begingroup$ @Quantuple thanks ^-^ post as answer? $\endgroup$ – BCLC Jul 10 '16 at 16:20
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    $\begingroup$ done, but you also raise interesting questions that I do not answer to :) $\endgroup$ – Quantuple Jul 10 '16 at 16:33
  • $\begingroup$ @Quantuple I might set a bounty but at least you got your +10 $\endgroup$ – BCLC Jul 11 '16 at 13:39
  • $\begingroup$ Sure. Don't worry for me. But maybe your question will find an answer on math or mathoverflow SE? Also it would help if you narrow down the questions you want an answer to, or at least better highlight them. $\endgroup$ – Quantuple Jul 11 '16 at 13:46
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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.

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  • $\begingroup$ Is this correct? If $M_t = E[ \frac{dQ}{dP} \rvert \mathcal{F}_t] $ then I believe we need $M_0=1$ to ensure $Q(\Omega) = 1$. $\endgroup$ – Henrik Jul 10 at 10:22

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