Different Fokker-Plank Equations after change of variables

Question

Let consider a real-valued stochastic process described by the stochastic differential equation (SDE) in the Ito formulation: \begin{align} dx= x \mu dt + x \sigma dW \tag{1} \label{1} \end{align} where $\mu$ and $\sigma$ are real-valued constants and $W$ is a Wiener process. The corresponding Fokker Plank equation (FPE) reads (https://en.wikipedia.org/wiki/Fokker%E2%80%93Planck_equation): \begin{align} \frac{\partial P}{\partial t}=-\frac{\partial }{\partial x}(x \mu P)+ \frac{1}{2} \frac{\partial^2 }{\partial x^2} (x^2 \sigma^2 P) \tag{2} \label{2} \end{align} Eq. \ref{2} can be rewritten as: \begin{align} \frac{\partial P}{\partial t}=(\sigma^2-\mu)P+(2 \sigma^2-\mu)x \frac{\partial P }{\partial x} + \frac{x^2 \sigma^2}{2} \frac{\partial^2 P}{\partial x^2} \tag{3} \label{3} \end{align} If we perform a change of variables in eq. \ref{1}, $y=\log x$, applying Ito change of variables formula (https://en.wikipedia.org/wiki/It%C3%B4%27s_lemma) we get: \begin{align} dy= \biggl(\mu- \frac{\sigma^2}{2}\biggr) dt + \sigma dW \tag{4} \label{4} \end{align} which corresponds to the FPE: \begin{align} \frac{\partial P}{\partial t}=\biggl(\frac{\sigma^2}{2} -\mu \biggr)\frac{\partial P }{\partial y}+ \frac{\sigma^2}{2}\frac{\partial^2 P}{\partial y^2} \tag{5} \label{5} \end{align} If we now take eq. \ref{3} and perform the same change of variables of eq. \ref{4}, we get: \begin{align} \frac{\partial P}{\partial t}=(\sigma^2-\mu)P+\biggl(\frac{3}{2} \sigma^2-\mu \biggr) \frac{\partial P }{\partial y} + \frac{ \sigma^2}{2} \frac{\partial^2 P}{\partial y^2} \tag{6} \label{6} \end{align} I expect eq. \ref{5} and \ref{6} to coincide, but they do not. Can someone please explain me the reason? Where am I wrong? Thanks in advance

Answer 1

Let $ P (x) $ and $ Q (y) $ be the densities with respect to the two different parametrizations.

They are related by $ P (x)=Q (y)\frac{dy}{dx} $.

This means that your second approach is the wrong one. Instead of deriving a PDE for $ Q(y)=P(x(y))\frac{dx}{dy}(y)$ you derived a PDE for $ \tilde{Q}(y):=P(x(y)) $, which is not a probability density.

PS: that your second approach is the wrong one can of course also be seen by the fact that it doesn't preserve the mass of $ P $ over time