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Suppose we have an Ito diffusion $$ dX_t = b(X_t)dt + \sigma(X_t) dB_t, \tag{1}$$ where $dB_t$ is Brownian motion. Also assume we know that this diffusion process converges to a stationary distribution $\pi$. I am interested in the quantity $$u_{\infty}(x) = \lim_{t \rightarrow \infty} \mathbb E[\psi(X_t)] = \mathbb E_\pi[\psi(X)],$$ for some function $\psi : \mathbb R \rightarrow \mathbb R$; assuming the expectation exists.

I am wondering if there is a way to connect this quantity with the backward Kolmogorov equation, i.e. define $u(x,t) = \mathbb E^x[\psi(X_t)]$; then the backward Kolmogorov equation says that $u$ solves \begin{align} \frac{\partial u}{\partial t} &= \mathcal A u, \tag{2}\\ u(0,x) &= \psi(x) \end{align} where $\mathcal A$ is the infitesimal generator for (1).

To me it seems like $u_\infty(x)$ should be the stationary solution of (2), so we might be able to set $\frac{\partial u}{\partial t} = 0$ to reduce things to an ODE. But I can't wrap my mind around the boundary conditions that should be in place for this to work, or if I've missed an obvious reason for why this wouldn't work.

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The stationary distribution is a time-invariant solution of the Fokker--Planck (forward) equation, which is probably an easier way of looking at it, as it comes out as an exponential-integral (of $b(x)/\sigma^2(x)$).

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  • $\begingroup$ Thanks, I know but I wondered if this might be a way of getting an expression for the expectation out directly; rather than finding the whole stationary distribution itself. $\endgroup$
    – Jack
    Commented Nov 14, 2018 at 13:31

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