Change of variables in stochastic PDE I have the following stochastic partial differential equation (SPDE):
$d v = -\mu \frac{\partial v}{\partial x} dt + \frac{1}{2} \frac{\partial^2 v}{\partial x^2} dt - \sqrt{\rho} \frac{\partial v}{\partial x} d M_t$,
with $M_t$ standard Brownian motion and $\mu, 0 \leq \rho \leq 1$ real-valued parameters. I read that the solution (without boundary conditions) can be written as the solution of the PDE
$ \frac{\partial u}{\partial t} = \frac{1}{2} (1-\rho) \frac{\partial^2 u}{\partial x^2} - \mu \frac{\partial u}{\partial x}$
shifted by the current value of the Brownian driver
$v(t,x) = u(t, x-\sqrt{\rho} M_t)$
How can I show this explicitly?
NOTE: It might be useful to keep in mind that the SPDE describes the density of the particles
$ d X_t^i = \mu dt + \sqrt{1-\rho} d W_t^i + \sqrt{\rho} d M_t$,
with $W_t^i$ independent Brownian motions. $M_t$ somehow represents a "common noise".
NOTE: is there maybe a mistake in the SPDE? Should we have $\frac{1}{2} (1-\rho) \frac{\partial^2 v}{\partial x^2}$?

Proving these kind of phase shifts can indeed be difficult as the Ito calculus does not provide a nice chain rule as we have in deterministic calculus. There is however a trick to solve this problem. Choose a smooth test function $\zeta$ and define the functional 
\begin{align}
\phi(v_t,M_t)=\langle v_t,\zeta\rangle_{H}=\langle u(t,\cdot),\zeta(\cdot+\sqrt{\rho}M_t)\rangle_{H},
\end{align}
where I assume your solution lives in some suitable Hilbert space $H$. Now using standard Ito calculus you can derive an SDE for $\phi$ (because you know the (S)PDE for $u$ and the SDE for $M_t$). Then, you can shift the $\sqrt\rho M_t$ back to $u$ (which results in $v$) and then drop the inner product with $\zeta$. The resulting SPDE should be equal to your original problem. 
I hope this answer is still useful for you.
