Let $B(t, \omega)$ be a Brownian motion defined on a probability space $(\Omega, \mathcal{F}, \mathbb{P})$, adapted to a filtration $\{\mathcal{F}\_t\}$. Let $\phi(t, \omega)$ be a $\{\mathcal{F}\_t\}$-adapted quadratic variation process such that the Ito integral


$$
Z(t, \omega) := \int\_0^t \phi(s, \omega) d B(s, \omega)
$$


is a local martingale.


My question: is the (path) distribution/pushforward measure $\mathbb{P}\circ Z^{-1}$ absolutely continuous with respect to the distribution $\mathbb{P} \circ B^{-1}$ of the Brownian motion? If so, how to prove? If not, do we have a concrete counter example of $\phi$?



Distribution of local martingale is absolutly continuous to that of the Brownian motion?
Let $ B $ be a Brownian Motion, $\mu\_{B}=\mathsf{P}\circ B^{-1} $ and $ Z $ be a continuous local martingale, $ \mu\_{Z}=\mathsf{P}\circ Z^{-1}$.
Then $ \mu\_{Z}\ll \mu\_{B} $ if and only if $ \mu\_{Z}=\mu\_{B} $.
In this case, if $ Z $ may be expressed as Ito stochastic integral of a predicatble $ \phi $ with respect to $ B $:
\begin{equation\*}
 Z(t,\omega)=\int\_0^t \phi(s,\omega)\,\mathrm{d}B(s,\omega)
\end{equation\*}
Then processes $ |\phi|=(|\phi(s,\omega)|,s\ge 0) $ and 1 are indistinguishable, i.e., for almost all $ \omega $ trajectories $ \phi(\cdot,\omega)\equiv 1 $.


Now we prove above facts. Since $ Z $ and $ B $ are continuous local martingale, for its predictable variation process $ \langle Z,Z\rangle$ and $\langle M,M\rangle $ the following facts hold, (cf. D. Revuz, M. Yor, *Continuous Martingales and Brownian Motion*, Corrected 3rd Ed. Springer, 2005, p124. Th.4.1.8.)
\begin{gather\*}
 \textrm{pr}-\lim\_{n\to\infty}\sum\_{k=1}^{2^n}\Big[Z\Big(\frac{kt}{2^n}\Big)- Z\Big(\frac{(k-1)t}{2^n}\Big)\Big]^2=\langle Z,Z\rangle\_t,\qquad \forall t>0.\tag{1}\\
 \textrm{pr}-\lim\_{n\to\infty}\sum\_{k=1}^{2^n}\Big[B\Big(\frac{kt}{2^n}\Big)- B\Big(\frac{(k-1)t}{2^n}\Big)\Big]^2=t,\qquad \forall t>0. \tag{2}
\end{gather\*}
(2) is equivalent to the following,
\begin{equation\*} \lim\_{n\to\infty}\mu\_{B}\Big(\Big|\sum\_{k=1}^{2^n}\Big[x\Big(\frac{kt}{2^n}\Big)- x\Big(\frac{(k-1)t}{2^n}\Big)\Big]^2 - t \Big|>\epsilon\Big)=0,\quad \forall \epsilon >0,\quad \forall t>0. \tag{3} 
\end{equation\*}
Due to $ \mu\_{Z}\ll \mu\_{B} $ and (3), as $ n\to\infty $,
\begin{align\*}
&\mathsf{P}\Big(\Big|\sum\_{k=1}^{2^n}\Big[Z\Big(\frac{kt}{2^n}\Big)- Z\Big(\frac{(k-1)t}{2^n}\Big)\Big]^2-t \Big|>\epsilon\Big)\\ 
&\quad =\mu\_{Z}\Big(\Big|\sum\_{k=1}^{2^n}\Big[x\Big(\frac{kt}{2^n}\Big)- x\Big(\frac{(k-1)t}{2^n}\Big)\Big]^2 - t \Big|>\epsilon\Big)\\
& \quad \longrightarrow 0,\quad \forall \epsilon >0,\quad \forall t>0. \tag{4} 
\end{align\*}
comparing (1) and (4), get
\begin{gather\*}
 \mathsf{P}(\langle Z,Z\rangle\_t=t)=1, \quad\forall t>0,\\
 \mathsf{P}(\langle Z,Z\rangle\_t=t, \forall t>0 )=1 \tag{5}
\end{gather\*}
Now from the Lévy's characterization of Brownian Motion, $Z$ is a Brownian motion and
$\mu\_Z=\mu\_{B} $.


Remark: The conclusion in following book is also useful: C. Dellacherie & P. Meyer, *Probabilities and Potential B*, volume 72 of *North-Holland Mathematics Studies*.
North-Holland, Amsterdam, 1982.


