Distribution of local martingale is absolutly continuous to that of the Brownian motion?

Question

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 $\phi$?

Answer 1

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.

Answer 2

Since Brownian motion is absolutely continuous with Lebesgue, this question is the same as asking when do Ito integrals/diffusions have a density.

In terms of a counterexample, see here A non-degenerate martingale

and also as mentioned in "Probability density function of SDEs with unbounded and path–dependent drift coefficient"

It is worth noting that Fabes and Kenig [13] provided an example of diffusion coefficient $\sigma$ such that the law of $X_t = x + \int_{0}^{t}\sigma(s, X_s)dW_s$ is purely singular with respect to Lebesgue measure.

from "Examples of singular parabolic measures and singular transition probability densities".

As mentioned here,https://math.stackexchange.com/questions/1072086/is-the-distribution-of-an-ito-diffusion-at-time-t-absolutely-continuous-wrt-lebe here is a general density result in "Absolute continuity for some one-dimensional processes"