I am getting confused about some important point of Girsanov theorem used for diffusion process. Starting with the diffusion $$dX_t=a(X_t)dt+b(X_t)dW_t$$ where $W_t$ is a P-Brownian motion. One can define a measure Q under which $$ d\tilde{W}_t=dW_t+\frac{a(X_t)}{b(X_t)}dt$$ is a Q-Brownian Motion. Rewriting $dX_T$ $$dX_t=a(X_t)dt+b(X_t)(d\tilde{W}_t-\frac{a(X_t)}{b(X_t)}dt)=b(X_t)d\tilde{W}_t$$ The RN derivative is given by $$ Z_t:=\frac{d Q}{d P} |_{\mathcal{F}_t} = \exp \left (-\int_0^t \frac{a(X_s)}{b(X_s)}dWs - \frac{1}{2} \int_0^t(\frac{a(X_s)}{b(X_s)})^2ds \right ) $$ and that is where I have many uncertainties:
- The fact that the original BM is used here means that deriving the distribution of $Z_t$ is easy enough under P but that if I were to derive its distribution under Q then would I have to do the substitution $dW_s\rightarrow d\tilde{W}_s - \frac{a(X_s)}{b(X_s)}$ ?
- In Oksendal "SDE an intro with applications", exercise 8.15 (the content of the exercise is not critically relevant), since b=1 the RN derivative $Z_t$ reads $\exp \left (-\int_0^t a(X_s)dWs - \frac{1}{2} \int_0^ta(X_s)^2ds \right )$ and the solution to work out entails the following steps: $$ E_P[f(X_t)]=E_Q[f(\tilde{W}_t)]=E_P[Z_tf(\tilde{W}_t)]=E_P[\exp \left (-\int_0^t a(X_s)dWs - \frac{1}{2} \int_0^ta(X_s)^2ds \right )f(\tilde{W}_t)]$$ and at this point, the solution that I could access (*) diverge from what I expect. The solution substitutes $a(X_s)\rightarrow a(W_s)$, therefore they write the following equality $$ E_P[\exp \left (-\int_0^t a(X_s)dWs - \frac{1}{2} \int_0^ta(X_s)^2ds \right )f(\tilde{W}_t)]=E_P[\exp \left (-\int_0^t a(W_s)dWs - \frac{1}{2} \int_0^ta(W_s)^2ds \right )f(\tilde{W}_t)] $$but the expectation is taken under P and under this measure $X_t$ is not a Brownian motion, isnt it ?
Sorry for the long post, I'd appreciate any pointers. Thanks