I'm sorry but I'm not good at English. If you find any sentence or word doesn't make sense, please comment.
EDIT: I found $(5)$ is wrong. There is no KBE for probability density. There is only KFE for probability density. If this is also wrong, please let me know.
I use [1] as a reference in this question because it summarizes KBE and KFE compactly. I think probability density of a stochastic process below is subject to both Kolmogorov backward equation and Kolmogorov forward equation at the same time. But in [2], there is an denial answer with less logic. I want to know more detailed proof that denies KBE and KFE hold at the same time if I'm wrong.
Question
For a stochastic differential equation: $$\tag{1}dX_t=b(X_t,t)dt+\sigma(X_t,t)dW_t$$
is it possible to use any one of KBE or KFE to get a solution of probability density of a diffusion process for a given boundary condition at time $t$?
Once the probability density is given at $t$, $\rho_t(x)$, we can know $\rho_{t'}(y)$ for $t'<t$ by KBE $(6)$(I derive it in following). Since we can use KFE $(7)$ for the boundary condition $\rho_{t'}(y)$, then, I think we can conclude the $\rho(y,s)$ satisfies KBE $(6)$ and KFE $(7)$ for all $s$. But in another question[2], there is an answer denies my opinion. Which is wrong?
For simplicity, I consider 1-dimentional case. Let $p(x,t|y,s)$ be the transition probability density for $s<t$ with which a particle goes from $y$ at time $s$ to $x$ at time $t$. That is,
$$\tag{2}P(X_t\in A|X_s=y)=\int_{A}p(x,t|y,s)dx.$$
I derive Kolmogorov Backward equation with respect to the probability density $\rho(y,s)$ of the stochastic process. Let $f$ be a function and define
$$u(y,s,t)=\mathbb{E}[f(X_t)|X_s=y].$$
Then $u(y,s,t)$ solves
$$\tag{3}\partial_su(y,s,t)+(b(y,s)\partial_y+\frac{1}{2}\sigma^2(y,s)\partial^2_y)u(y,s,t)=0 $$
for $s<t$ and satisfies $u(y,t,t)=f(y)$ (see (15) of [1]).
If I use a Dirac's delta function for $f$ ( $f(X_t)=\delta(X_t-x)$ ), $u(y,s,t)$ becomes $p(x,t|y,s)$ and $(3)$ becomes
$$\tag{4}\partial_sp(x,t|y,s)+(b(y,s)\partial_y+\frac{1}{2}\sigma^2(y,s)\partial^2_y)p(x,t|y,s)=0 $$
for $s<t$ and satisfies $p(x,t|y,t)=\delta(y-x)$.
Giving the probability density at time $t$, $\rho_t(x)$, then the probability density at time $s$ is given as follows:
$$\color{red}{\tag{5}\rho(y,s)=\int_{-\infty}^{\infty}p(x,t|y,s)\rho(x,t)dx.}$$
EDIT: I noticed (5) is wrong. $\rho(x,t)=\int_{-\infty}^{\infty}p(x,t|y,s)\rho(y,s)dy$ is correct and it just leads to KFE. So, there is no KBE for probability density($(6)$ is wrong).
Multiplying $\rho(x,t)$ to both sides of $(4)$ and integrating with respect to $x$, one gets,
$$\tag{6}\partial_s\rho(y,s)+(b(y,s)\partial_y+\frac{1}{2}\sigma^2(y,s)\partial^2_y)\rho(y,s)=0.$$
for $s<t$ and satisfies $\rho(y,t)=\rho_t(y)$.
On the other hand, we can obtain Kolmogorov Forward equation as on p.7 of [1]. For given probability density at time $t'$, $\rho_{t'}(x)$, the equation is
$$\tag{7}\partial _s\rho(y,s)+\partial_y(b(y,s)\rho(y,s))-\partial_y^2(\dfrac{1}{2}\sigma^2(y,s)\rho(y,s))=0$$
for $t'<s$ and satisfies $\rho(y,t')=\rho_{t'}(y)$.
Once the probability density is given at $t$, $\rho_t(x)$, we can know $\rho_{t'}(y)$ for $t'<t$ by KBE $(6)$. Since we can use $(7)$ for the boundary condition $\rho_{t'}(y)$, then, we can conclude the $\rho(y,s)$ satisfies $(6)$ and $(7)$ for all $s$.
That is, we can use either PDEs, KBE or KFE, to get probability density of $(1)$.
Considering the difference of $(6)$ and $(7)$, we get
$$\tag{8}-\partial _y(b\rho)+\partial^2_y(\frac{1}{2}\sigma \rho)-\frac{1}{2}\sigma\partial^2_y\rho=0$$
and I think it is related to the fact that $\partial_y b(y,s)-\partial_y^2\dfrac{1}{2}\sigma^2(y,s)$ is adjoint operator of $b(y,s)\partial_y+\frac{1}{2}\sigma^2(y,s)\partial^2_y$(see [1] p.6).
Is it correct that we can use either KBE or KFE for the probability density and construct the solution for all $s$ for a boundary condition $\rho(y,\tau)=\rho_\tau(y)$?
Thanks for reading.
I also appreciate it if you share what you come up with about this problem!
