Characteristic exponent after Girsanov transformation

Question

Let $B$ be a standard Brownian motion. Its characteristic exponent (or Fourier transform) is easily calculated to be $$ \mathbb E [e^{ixB_t}] = e^{-\frac 12 x^2 t}. $$ Now I want to apply a Girsanov transformation with density $\exp(\int_0^t A_s d B_s - \int_0^t A_s^2 ds)$ for some predictable process $A_s$ satisfying Novikov's condition. It is well-known that $B_t-\int_0^t A_sds$ is Brownian motion under the new measure. However, I am more interested in the question what the characteristic exponent of the old Brownian motion under the new measure is. Is there an explicit way to calculate this? Same question, but differently phrased: Is there an elegant way to calculate $$ \mathbb E [ e^{\int_0^t A_s dB_s} e^{ixB_t}]? $$

Answer 1

By Doob-Dynkin we have that $A_t=f(t,\{B_s\}_{0\leq s\leq t})$.

If $\mu_0$ is the law of Brownian motion then under the measure $\mu=\exp\left(\int_0^T A_sdB_s-1/2\int_0^T A_s^2 ds\right)\mu_0$ we have that the law of $B$ is the law of the solution to the (possibly path dependent) SDE $dX_t=f(t, \{X_s\}_{0\leq s\leq t}))dt+dB_t$. So computing the characteristic function is the same as finding the law of $X$.