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One can in the case of the Fokker-Planck / forward Kolmogorov equation, set the time derivative term to zero, and solve the remaining ODE to obtain the "forward-time" stationary distribution.

Does it make any sense to solve for the stationary distribution of the backward Kolmogorov equation? If so, what would be the interpretation?

I was just thinking about this, couldn't see to wrap my head around it.

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The equation $Lu=0$ where $L$ is the generator of your process (the formal adjoint of the Fokker-Planck operator) does indeed have significance, but its significance depends on the boundary conditions that you impose. In the case of the stationary FPE, the natural "boundary condition" is just the normalization and positivity conditions. About the only other thing that can make sense is dropping the normalization condition (so that you are finding a stationary measure which is not a distribution).

One case is when the boundary condition is just a Dirichlet condition on the exterior of a domain; then the problem can be interpreted as the average of the boundary function against the hitting distribution of the boundary. This can in particular be used to find the probability of hitting the boundary in a particular region.

A special case of this is when you have two open sets $A,B$ with disjoint closures which are in the domain of the process but are excluded from the domain of your BKE. You impose the Dirichlet constraint of zero on the boundary of $A$ and of one on the boundary of $B$. Then the solution is called the committor, and it measures the probability of hitting $B$ before hitting $A$ starting from a given point $x$. It can be used to compute the rate of the process of going from $A$ to $B$ without returning to $A$, which is a kind of "reaction rate". This is used in "transition path theory".

One point to keep in mind here is that the Fokker-Planck operator evolves measures: given a measure, it tells you how that measure is changing under the dynamics of your process. Its formal adjoint evolves functions: given a function and a point, it tells you how the mean of that function changes as the process is run starting from that point.

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  • $\begingroup$ Thank you. That is very helpful. $\endgroup$ – Thomas Moore Jul 28 '16 at 19:03

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