Can Kolmogorov backward and forward equations hold at the same time?

Question

I'm bit confused at Kolmogorov forward and backward equations: can they hold at the same time?

I understand that Kolmogorov forward equation (KFE) describes the evolution of the pdf of $X(t)$ given a starting pdf (the initial condition) $$ \frac{\partial}{\partial t} p(t,x)+ \frac{\partial}{\partial x} (\mu\cdot p(t,x))-\frac12 \frac{\partial^2}{\partial x^2} (\sigma^2 \cdot p(t,x)) =0$$ ; while Kolmogorov backward equation (KBE), for $X(t)$ given a terminal condition $p(T,x) = P_T(x)$ (here $T$ is the terminal time), describes the pdf of $X(t)$ as $$ \frac{\partial}{\partial t} p(t,x)+ \mu \frac{\partial}{\partial x} (p(t,x))+\frac12 \sigma^2 \frac{\partial^2}{\partial x^2} (p(t,x)) =0$$

They are different things and are related by the fact that the partial differential operators are adjoint of each other.

But consider a diffusion process $\mathrm dX = \mu \mathrm dt+ \sigma \mathrm dW$, let's give the initial condition first, for example everything starts from $X(0)=0$, then define the final time $T$, so $X(t)$ will arrive at the final condition.

Now both initial and final condition are known, including with each moment's probability density function. Then if I look at a time $t$, $0<t<T$, trying to figure out the probability density function $p(t,x)$ at this point, it seems that I can either look forward from time $0$, or look back from time $T$. So here arrive 2 results -- which one is right?

Or both stand at the same time?

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Background:

Given diffusion process (1) $$\mathrm dX(t)=\mu(t,X(t))\mathrm dt+\sigma(t,X(t))\mathrm dW$$ Denote the probability density function at time $t$ as $p(t,x)$.

Denote $p_{t,x|s,w}(t,x)$ the transition probability density function for the process to go from $(s,w)$ to $(t,x)$, here $s<t$.

On one hand, I'm happy with Kolmogorov forward equation ontransition probability density function (2): $$ \frac{\partial}{\partial t} p_{t,x|s,w}(t,x)+ \frac{\partial}{\partial x} (\mu\cdot p_{t,x|s,w}(t,x))-\frac12 \frac{\partial^2}{\partial x^2} (\sigma^2 \cdot p_{t,x|s,w}(t,x)) =0$$

Since we can choose any $s <t $ to compute as (3)

$$p(t,x) = \int p(s,w) p_{t,x|s,w}(t,x) \mathrm dw $$

Multiply (2) by $p(s,w)$ then integrate by $w$ we have the Kolmogorov forward equation (4)

$$ \frac{\partial}{\partial t} p(t,x)+ \frac{\partial}{\partial x} (\mu\cdot p(t,x))-\frac12 \frac{\partial^2}{\partial x^2} (\sigma^2 \cdot p(t,x)) =0$$

On the other hand, I'm also happy with Kolmogorov backward equation on transition probability density function (5): $$ \frac{\partial}{\partial t} p_{u,y|t,x}(t,x)+ \mu\frac{\partial}{\partial x} p_{u,y|t,x}(t,x)+\frac12 \sigma^2 \frac{\partial^2}{\partial x^2} p_{u,y|t,x}(t,x) =0$$ , where $p_{u,y|t,x}(t,x)$ is the transition probability density function for the process to go from $(t,x)$ to $(u,y)$, here $t<u$.

Since we can choose any $u>t $ to "look back" to compute $p(t,x)$ as (6)

$$p(t,x) = \int p(u,y) p_{u,y|t,x}(t,x) \mathrm dy $$

Multiply (5) by $p(u,y)$ then integrate by $u$ we have the Kolmogorov backward equation (7)

$$ \frac{\partial}{\partial t} p(t,x)+ \mu \frac{\partial}{\partial x} (p(t,x))+\frac12 \sigma^2 \frac{\partial^2}{\partial x^2} (p(t,x)) =0$$

Answer 1

Part (if not all) of the confusion seems to come from the fact that you got a sign wrong. The Kolmogorov forward equation is $$\tag{KFE} \frac{\partial}{\partial t} p(t,x)+ \frac{\partial}{\partial x} (\mu\,p(t,x))\color{red}{-}\frac12 \frac{\partial^2}{\partial x^2} (\sigma^2 \, p(t,x)) =0 $$ (see [1]). The Kolmogorov backward equation is $$\tag{KBE} \frac{\partial}{\partial t} p(t,x)+ \mu\frac{\partial}{\partial x} p(t,x)\color{red}{+}\sigma^2\frac12 \frac{\partial^2}{\partial x^2} p(t,x) =0\, $$ (this $\color{red}{+}$ sign you had right).

Even if $\mu$ and $\sigma$ are constant the two equations are in general not satisfied by the same function. In [2] we can read that (KFE) (also known as Fokker Planck equation) is satisfied by the probability density of the stochastic process $X$ that satisfies the SDE $$ dX_t=\mu(t,X_t)\,dt+\sigma(t,X_t)\,dW_t,\quad\quad X_0=x\,. $$ In [1] we can read that (KBE) with final condition $u(x)$ at $t=T$ is satisfied by the conditional expectation $$ p(t,x)=\mathbb E[u(X_T)|X_t=x]\,. $$ This $p(t,x)$ is clearly different from the $p(t,x)$ solving (KFE), i.e., from the density of $X_t\,$.

Answer 2

I think I found my mistake. Actually equation (6) does not hold: with $t<u$

$$p(t,x) = \int p(u,y) p_{u,y|t,x}(t,x) \mathrm dy $$

is wrong!

A simpler example would be, when $p_t(x)$ and $p_u(y)$ are discrete. Let $$p_t(0)= p_t(1)=\frac12$$

or $$P_t = \begin{pmatrix}\frac12 \\ \frac12 \end{pmatrix}$$

and

$$T= \begin{pmatrix}\frac12 & \frac14\\ \frac12 & \frac34\end{pmatrix}$$

Then $$P_u = T\cdot P_t = \begin{pmatrix}\frac38 \\ \frac58 \end{pmatrix}$$

Obviously $$P_t \ne T\cdot P_u$$, but $$P_t = T^{-1}\cdot P_u$$