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
$$ \mathbb E [ e^{\int\_0^t A\_s dB\_s} e^{ixB\_t}]? $$



Characteristic exponent after Girsanov transformation
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 $X$.


