I’ve recently realised there is a subtlety in Girsanov’s theorem that I don’t really understand.


Consider a standard one dimensional Brownian motion $W$, and consider the SDE


$$dZ\_t = \mu(t, Z\_t) \, dt + \sigma(t, Z\_t) \, dW\_t \, , \,Z\_0 = x\_0 \, \, \, \text{(Equation 1)}$$


for some $x\_0 \in \mathbb R$, where $\mu, \sigma: [0, \infty) \times \mathbb R \to \mathbb R$ are Lipschitz continuous.


Denote by $\mathbb P$ the probability measure under which $W$ is a standard Brownian motion. Suppose we have an equivalent probability measure $\mathbb Q$ under which $W$ is no longer a standard Brownian motion, but a semimartingale.


We may still consider Equation 1 under $\mathbb Q$ as a semimartingale SDE.


Suppose $X$ solves Equation 1 under $\mathbb P$, and $Y$ solves Equation 1 under $\mathbb Q$.


**Question:** Is it true that we still have $X = Y$ up to indistinguishability? That is, do we have $X\_t = Y\_t$ for all $t \in [0, \infty)$, ($\mathbb P$, and hence $\mathbb Q$) almost surely?


It seems that this result is used implicitly in transforming SDE via Girsanov’s theorem, but it is not obvious to me at all.



A question related to Girsanov’s theorem
I managed to find an answer to this problem - indeed the answer is yes. The key ingredient is a theorem in Protter’s *Stochastic Integration and Differential Equations* (Chapter 2, Theorem 14) which states the following:



> 
> For every fixed process $X$, integrable with respect to a semimartingale $S$, the stochastic integrals $\int X\, dS$ under $\mathbb P$ and $\mathbb Q$ are indistinguishable as processes.
> 
> 
> 


The indistinguishability as processes part is key, and allows us to conclude that solutions to SDE too are invariant, as follows:


Since $X$ solves Equation 1 under $\mathbb P$, by definition we have


$$X = X\_0 + \int\_0^\cdot \mu(s, X\_s) \, ds + \int\_0^\cdot \sigma(s, X\_s) \, dW\_s$$


under $\mathbb P$, almost surely.


By indistinguishability of the stochastic integral on the RHS, we have also


$$X = X\_0 + \int\_0^\cdot \mu(s, X\_s) \, ds + \int\_0^\cdot \sigma(s, X\_s) \, dW\_s$$


under $\mathbb Q$, almost surely.


So $X$ solves Equation 1 under $\mathbb Q$. By uniqueness to solutions of SDE, it is *the* solution, i.e. $X = Y$ up to indistinguishability, and we are done.


