Regarding calculation the moments of a random variable whose probability distribution obeys the Fokker Planck equation
I was going through <PERSON> Stochastic Processes in Physics and Chemistry, and I was trying to solve the exercises from Chapter 8 about the Fokker Planck equation (just in case context could help anyone understand my question).
There, we have been asked to assume that some $P(y,t|y_0, t_0)$ does solve the Fokker Planck equation. We further assume that $t=t_0+\Delta t$ and that $\Delta t \rightarrow 0$. Then we have been asked to calculate the moments of $\Delta y$, which in turn is defined as $\Delta y \equiv y-y_0$, and thence calculate $\dfrac{<\Delta y>^j}{\Delta t}$. 563
We have to show that $\dfrac{<\Delta y>}{\Delta t}=A(y_0)$, $\dfrac{<\Delta y>^2}{\Delta t}=B(y_0)$, and $\dfrac{<\Delta y>^j}{\Delta t}=0$ for $j \geq 3$.
Now, we have our Fokker Planck equation expressed as: $$ \frac{\partial P(y,t)}{\partial t}=-\frac{\partial}{\partial y}[A(y) P(y,t)]+\frac{1}{2}\frac{\partial^2}{\partial y^2}[B(y) P(y,t)] $$
This is how far I got: we can write down the definitions of our moments from first principles as (I will just write down the first moment of $\Delta y$).
$$ \frac{<\Delta y>}{\Delta t}=\int \frac{(y-y_0)}{\Delta t}P(y,t+\Delta t|y_0, t_0)dy $$ Now, given that $\Delta t$ is small we can write $$ \begin{align} P(y,t_0+\Delta t|y_0, t_0)&=P(y,t_0|y_0, t_0)+\Delta t \frac{\partial}{\partial t}P(y,t|y_0, t_0)\vert_{y_0, t_0}\ &=\delta(y-y_0)+\Delta t \frac{\partial}{\partial t}P(y,t|y_0, t_0)\vert_{y_0, t_0} \end{align} $$ When we plug this back into the integral, the delta function term won't contribute anything and we can work only with the second term:
$$ \frac{<\Delta y>}{\Delta t}=\int \frac{(y-y_0)}{\Delta t}\Delta t \frac{\partial}{\partial t}P(y,t|y_0, t_0)\vert_{y_0, t_0}dy $$
We can cancel out the $\Delta t$ and then use the fact that our $P$ solves the FPE to get:
$$ \frac{<\Delta y>}{\Delta t}=\int (y-y_0) \Big[-\frac{\partial}{\partial y}[A(y) P(y,t|y_0, t_0)]+\frac{1}{2}\frac{\partial^2}{\partial y^2}[B(y) P(y,t|y_0, t_0)]\Big]\Big\vert_{y_0, t_0}dy $$
I am not sure how to proceed from here. My first instinct is to try integrating it by parts, but I don't know the limits of integration, so I am unaware of how to proceed. Additionally, I do have a feeling there is a more elegant way to arrive at this result because these moments do describe convection and diffusion, and $A$ and $B$ are the convection and diffusion terms (am not sure if I am putting the cart before the horse here, but it does seem quite intuitive).
Any help would be appreciated! 374