# Implications of Girsanov Theorem

I am confused by the role Girsanov Theorem plays in deducing absolute continuity of laws of certain processes. Say $$B$$ is a Brownian motion in $$(\Omega, \mathcal{F}_{\infty},\mathcal{F}_t,P)$$ and let $$K$$ be a well-behaved enough process that $$H_t=\int_0^t K_s ds$$ and $$N_t=\int_0^t K_s dB_s$$ are both defined and well-behaved and $$X_t=\exp(N_t -\frac{\langle N \rangle}{2})$$ is UI. Then by applying Girsanov we find $$Q$$ such that $$B'_t=(B_t - H_t)_{t \geq 0}$$ is a local martingale under this new probability measure $$Q$$ - actually a Brownian motion. But then the following statements confuse me: apparently, the law of $$(B_t-H_t)_{t\geq 0}$$ under $$P$$ is absolutely continuous wrt the law of a Q-Brownian motion and, conversely, the law of $$(B'_t+H_t)_{t \geq 0}$$ under $$Q$$ is absolutely continuous wrt to the law of a $$P$$-Brownian motion.

From Girsanov theorem, the new probability measure $$Q$$ is absolutely continuous wrt $$P$$ by construction as far as I understood. Why then does this absolutely continuity of laws hold? Just because the underlying probability measures are absolutely continuous and the laws are just the push-forward of these probabilities?

• I think I remember this on this matter, as long as $X_t$ isn't null with positive probability everything is good and for all time marginals are mutually absolutely continuous but Girsanov might not entail absolute continuity at the end of time because 0 proba events for P in the tail might have positive probability on Q. For example take a positively drifted BM on P and make it a martingale with Girsanov on Q, then P(B<0) = 0 and Q(B<0) = 1/2 unless mistaken so even in this simple case things can go wrong. Commented Aug 30, 2021 at 14:27
• If $X_t$ is UI, $Q << P$ is guaranteed at all times including the end as far as I know. My doubts revolve more around the implications for the laws in the cases such as the one I mentioned. But thank you for your example.
– Karl
Commented Aug 30, 2021 at 14:31