Adaptedness in the Martingale Proof of Girsanov’s Theorem
Summary
This note asks whether the product of the density process and a shifted Brownian motion is adapted to the original filtration, as needed to show it is a martingale under the original measure. The resolution is that the shifted process is defined from the original Brownian motion and the adapted drift process, so it is adapted to the original filtration. Multiplying it by the density process preserves adaptedness. The distinction between filtrations generated by the original and shifted Brownian motions does not prevent either process from being adapted to the shared filtration used in the theorem.
The argument uses the density-process martingale and a product calculation with no drift, followed by the change-of-measure martingale criterion. The note highlights that adaptedness and absence of drift are separate requirements; integrability conditions are also needed for a true martingale. It presents the Brownian-filtration setting and does not spell out all regularity assumptions on the drift or density.
Key ideas
- The shifted Brownian process can be adapted to the original filtration even if it generates a different filtration.
- The density process and shifted process are both adapted to the filtration used in the proof.
- A zero-drift product calculation alone does not establish a true martingale without suitable integrability.
- The change-of-measure criterion connects the product martingale to the shifted process under the new measure.
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# Last step step in Girsanov's theorem proof
# Last step step in Girsanov's theorem proof
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:
- $(\widetilde{W}_t)_{t\in[0;T]}$ is a continuous process, starting at 0, with quadratic variation $t$ (this is all evident)
- 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?Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.