For a nontrivial example of the use of Kolmogorov’s forward equation (the Fokker-Planck equation) where it can be contrasted with Kolmogorov’s backward equation, we look beyond Brownian motion, since the generator of the Brownian motion process is its own adjoint and applied to the transition density is the same in both the $x$ and $y$ variables. So we consider the next simplest example, the geometric Brownian motion process, which is given by
$$
dX_t = \mu X_t dt + \sigma X_t dW_t
$$
where we will assume $\sigma=1$ and $\mu=0$.

Generators and their adjoints

The generator for the GBM process in the $x$ variable is
$$
A = \frac{1}{2}x^2\frac{\partial^2}{\partial x^2}
$$
Following Øksendal’s book we need to find the adjoint $A^*$, satisfying $\langle Af,g\rangle = \langle f,A^* g\rangle$. The adjoint of $\partial/\partial x$ is $-\partial/\partial x$; this follows by integration by parts and the theory of distributions (where test functions are $C^\infty$ with compact support and hence the boundary terms vanish). Thus the adjoint of $A$, in the $y$ variable, is
$$
f\mapsto \frac{1}{2}\frac{\partial^2}{\partial y^2} (y^2 f)
$$
Using the product rule several times, we have
$$
A^*_y p = \frac{1}{2}\frac{\partial^2}{\partial y^2} (y^2 p) = p + 2y\frac{\partial p}{\partial y} + \frac{1}{2}y^2\frac{\partial^2 p}{\partial y^2}
$$

The pdf

By Shreve volume II page 119 the transition density for GBM is
$$
p(t,x,y)=\frac{1}{y\sqrt{2\pi T}}\exp\left(-\frac{(\log (y/x)-\nu T)^2}{2T}\right)
$$
where $\nu=\mu-\sigma^2/2=-1/2$, and where it is important to keep $x$ vs. $y$ straight.

Backward equation

If we wanted the backward equation we would replace $T$ by $T-t$ where $T>t$ and we are looking at the transition from $(t,x)$ to $(T,y)$ (instead of from $(0,x)$ to $(T,y)$). The resulting $p=p(t,x;T,y)$ gives the pdf of $X(T)$, as a function of $y$, given that $X(t)=x$. As $t\rightarrow T$ and with $x:=X(t)$ this pdf in $y$ will converge toward a Dirac delta function at the point $X(T)$. The backward equation is an expression of the fact that $p(t,X(t);T,y)$ is a martingale for fixed $T$ and $y$. It is the kind of martingale that keeps betting too much and hence will almost surely lose all its money. For a while the pdf at $y$ may be going up because $X$ approaches $y$, but this will only last for so long.

The backward equation can be verified and states that
$$
\frac{\partial p}{\partial t} = -A_x p
$$

Not-famous versions

On the other hand, since $p$ is a function of $t$ and $T$ only through $T-t$, it also follows that
$$
\frac{\partial p}{\partial T} = A_x p
$$
although this latter equation is not as famous as the Forward and Backward ones. It connects moving the future distribution forward in time, with space derivatives at the past time. Nor is it famous that
$$
\frac{\partial p}{\partial t} = – A^*_y p
$$

Verification of the Kolmogorov forward equation

In order to verify Kolmogorov’s forward equation we will have to take $\partial/\partial T$ of the transition density expression, which turns out to be less tedious if we use the functions $Q=\frac{1}{T}\log(y/x)$ and $R=Q+\frac{1}{2}$. If we ignore the trivial $\sqrt{2\pi}$ factor, after a while we obtain:
\begin{eqnarray}
p&=&\frac{1}{y\sqrt{T}}\exp(-R^2T/2) \\
\frac{\partial p}{\partial t} &=& p\cdot \left(\frac{-1}{2T} +RQ-R^2/2\right) \\
2y\frac{\partial p}{\partial y} &=& 2p(-1-R)\\
\frac{1}{2}y^2\frac{\partial^2 p}{\partial y^2} &=& \frac{p}{2}\left((-1-R)^2 +1+R-\frac{1}{T}\right)
\end{eqnarray}
By adding lines we get
$$
A^*_y p = \frac{\partial p}{\partial T}
$$
which is Kolmogorov’s forward equation.