2
$\begingroup$

I consider the version of Girsanov's theorem presented in this question.

Let us take the particular case that $\mathbb{F}$ is the filtration generated by standard Brownian motion $(W)_{t\in[0;T]}$ of $(\Omega, \mathcal{F}, \mathbb{P})$. We want to show that $(\widetilde{W}_t)_{t\in[0;T]}$ is a standard Brownian motion in $(\Omega, \mathcal{F}, \widetilde{\mathbb{P}})$, where $$ \frac{d\widetilde{\mathbb{P}}}{d\mathbb{P}}=Z_T=e^{-\int_0^T\Theta_udW_u-\frac{1}{2}\int_0^T\Theta^2_udu} $$ and $(\Theta_t)_{t\in[0;T]}$ is some $\mathbb{F}$-adapted process

The proof proceeds to show this using the Lévy characterization of Brownian motion:

  1. $(\widetilde{W}_t)_{t\in[0;T]}$ is a continuous process, starting at 0, with quadratic variation $t$ (this is all evident)
  2. All that remains to be shown is that $(\widetilde{W}_t)_{t\in[0;T]}$ is a $(\widetilde{\mathbb{P}}, \mathbb{F})$-martingale.

First we show that $(Z_t)_{t\in[0;T]}$ is a $(\mathbb{P}, \mathbb{F})$-martingale. But then, in order to prove point (2.) above, the proof tries to show that $(\widetilde{W}_tZ_t)_{t\in[0;T]}$ is a $(\mathbb{P}, \mathbb{F})$-martingale by explaining that: $$ d(\widetilde{W}_tZ_t)=\cdots=(-\widetilde{W}\Theta_t+1)Z_tdW_t. $$ has no drift.

However, for $(\widetilde{W}_tZ_t)_{t\in[0;T]}$ to be a $(\mathbb{P}, \mathbb{F})$-martingale we also must know that this process is $\mathbb{F}$-adapted. Can we prove this fact? This seems contrary to the comment made after Corollary 5.3.2 that states:

the filtration generated by $\widetilde{W}$ may be different from the filtration generated by $W$. Is this not a contradiction?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.