Assume the time inhomogeneous SDE 
$dX(t)=\mu(t,X(t))dt+\sigma(t,X(t))dW(t)$
has a solution $X(t)$. If we replace $\sigma$ with its absolute value, does the new SDE
$dY(t)=\mu(t,Y(t))dt+|\sigma(t,Y(t))|dW(t)$
have a solution $Y(t)$ which has the same distribution as $X(t)$? Here, the initial condition for both SDEs are assumed to have the same distribution.


My guess is yes. And I have a non rigorous proof coming from the numerical solution of this SDE
$X(t\_{k+1})=X(t\_k)+\mu(t\_k,X(t\_k))(t\_{k+1}-t\_k)+\sigma(t\_k,X(t\_k))(W(t\_{k+1})-W(t\_{k}))$.
For any path of $X(t\_k)$ such that $X(t\_k)=x$, on the corresponding path of $Y(t\_k)$ such that $Y(t\_k)=x$, the drift parts are the same, and the volality parts should have the same distribution. But these are all purely heuristic. Can someone offer some ideas?


Thank you! 



change the sign of volatility
There is no change into the probability distribution due to the absolute value. The reason can be traced back on the Ito's lemma applied to the derivation of the (Fokker-Planck) equation for the probability distribution: This will depend on $\sigma^2$.


