currently I am learning the stochastic differential equation and their numerical approximations and something is bothering me. Some processes has stationary distributions if stationary Fokker-Planck equation is solved. However, we need the Fokker-Planck equation for this. We can also simulate different realizations by solving numerically the SDE:
$dX =a(X)*dt+b(X)*dB$
My question: is it possible somehow to sample the stationary distribution if I solve the following equation:
$0 =a(X_{eq})*dt+b(X_{eq})*dB$
The reason why I am asking is that in deterministic case one solves always the
$ A*x_{eq} = f $
linear equation (or system) to calculate the stationary solution $x_{eq}$. This is obviously impossible in stochastic case because the $f$ source term is random. But if we do this step enough then we should see the $X_{eq}$ when the $dX=0$ (it is stationary), should we not?
Thank you for your answers!
If there is a solution $X_{\mathrm{eq}}$ that satisfies your equation $$ 0\, =\, a(X_{\mathrm{eq}}) dt + b(X_{\mathrm{eq}}) dB_t $$ and which also satisfies $dX_{\mathrm{eq}} = a(X_{\mathrm{eq}}) dt + b(X_{\mathrm{eq}}) dB_t$ then that would mean that $X_{\mathrm{eq}}$ is not only stationary: it is non-random and constant in $t$. That is fine. There are such solutions to some SDE's. For example $dX = X\, dt + X\, dB_t$. Then one solution is $X=0$. But a more general family of solutions is $X=A \exp(B_t+\frac{1}{2}t)$ with $A=0$ giving the only solution which is stationary. (But it is also constant, and $0$.)