Consider simple diffusion $dX_t = \sigma dw_t$ and a parameter $a>0$ and $X_0=x$. Let us denote $Y_t = X_{at}$  thus we made a change of time. Let us denote an original measure as $P$. How to find measure $Q$ such that the process $Y$ is obtained through the process $X$ not with the change of time but with a change of measure, i.e. $$ \mathcal{Law}(Y\text{ under }P) = \mathcal{Law}(X\text{ under } Q). $$ Maybe it is possible to find the density process $$ Z_t = \frac{dQ_t}{dP_t} $$ like for the case when we're looking for the martingale measure? Unfortunately, now it seems the Girsanov theorem is useless for my case.

for example, if $\sigma$ is a constant, then $Y$ satisfies $dY_t = \sqrt{a} \sigma dW_t$ so that the law $\mathbb{Q}_Y$ and $\mathbb{Q}_X$ of the processes $Y$ and $X$ on the Wiener space $C([0;T],\mathbb{R})$ are generaly singular: in other words $\frac{d \mathbb{Q}_Y}{d \mathbb{Q}_X} = 0$ if $a \neq 1$. 


Hi, I think what you are looking for is not attainable (at least I can't see how) because the time change changes the filtration of the processes you are studying. Nevertheless you can get by the transform below an equivalent process that the one you get by time change. (from X_t=\sigma W_t getting $Y_t=dW_t$) Under some conditions on the volatility of the diffusion you can (by socalled Lamperti's Transformation) get a process with unit volatility + drift term, then use change of measure technique (i.e. Girsanov) to get back to pure diffusion term and you are done. NB: You can adapt Lamperti's transform (+ Girsanov) to get a process equal in law to your time change in more general case. with this operation you still don't get the same law under different measure (because you have to transform the process first) but at least you stay within the same filtation. Regards, Reference for Lamperti's Transform : Theorem 2 in http://www.imm.dtu.dk/English/Research/Scientific_Computing/People.aspx?lg=showcommon&id=271164 

