Regarding calculation the moments of a random variable whose probability distribution obeys the Fokker Planck equation

I was going through Van Kampen's 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 $$\dfrac{<\Delta y>^j}{\Delta t}$$. 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!

Try the integration by parts bearing in mind that you know the distribution $$P(y,t_0|y_0,t_0)=\delta(y-y_0)$$.
If the boundary conditions are not given, then it is usually assumed that the probability and the probability flux are zero at $$\pm \infty$$. This is a way to obtain normalizable probabilities.
• I think I got it! Also, do you mean that $P(y,t|y_0, t_0)=\delta(y-y_0)$ only at $t=t_0$ (because $P$ can definitely have a non-trivial dependency on $t$). Because this condition allows me to in some sense 'collect' whichever moment I want from the FPE. Commented May 26, 2022 at 9:52