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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!

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1 Answer 1

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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 $\sigma^2$.

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Thank you very much, Jon! This exactly answers my question. –  perfectconan Jun 15 '12 at 14:26

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