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It is well known that under mild assumptions a scalar diffusion $dX_t = a(X_t) dt + \sigma(X_t) dW_t$ with invariant probability distribution $\pi$ is reversible. This is indeed not true for multidimensional diffusions. The usual proofs consists in writing down generators, speed functions etc...

I am trying to intuitively understand this result, and the only (not very satisfying) argument that I have found is the following. It is straightforward to check that any Markov chain on $\mathbb{Z}$ that has an invariant probability $\pi$ and that can only make jumps of size $+1$ or $-1$ is reversible: notice for example that $$F(k) = \pi(k)p(k,k+1)-\pi(k+1)p(k+1,k)$$ is independent of $k$ and is thus equal to $0$. If $a(\cdot)$ and $\sigma(\cdot)$ are regular enough, a diffusion can be seen as a limit of such Markov chains on $\epsilon \mathbb{Z}$ so that this makes the result plausible.

question: what are arguments/proofs/examples that could shed light on why a one dimensional ergodic diffusion is automatically reversible.

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The identity you are referring to can be interpreted as saying that if you run your Markov chain on $\mathbb Z$ from the initial distribution $\pi$, then the probability of crossing the edge $[k,k+1]$ in the positive direction is the same as the probability of crossing it in the negative direction (as otherwise the measure $\pi$ won't be stationary).

The same idea works in the general case as well. For any nice test function $f$ $$ \lim \frac1t \int [f(\xi_t) - f(\xi_0)] d{\mathbf P}(\xi) = \int Df(x) d\pi(x) , $$ where ${\mathbf P}$ is the measure in the space of sample paths $\xi$ of the diffusion process with the initial stationary distribution $\pi$, and $D$ is the diffusion generator. The left-hand side of the above equation is 0 by stationarity of the measure $\pi$, so that $\int Df(x)d\pi(x)=0$. In the same way it is true for the reverse generator $D^{\ast}$ as well. Now, the operators $D$ and $D^{\ast}$ may differ by their first-order parts only (denote these vector fields by $v$ and $v^{\ast}$, respectively). Taking a step-like function $f$ whose derivative is non-zero only on a small set where $v\neq v^*$ (it is at this place that dim $=1$ is used) then gives a contradiction.

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thanks: this is exactly the kind of simple approaches that I am looking for. – Alekk Nov 8 '10 at 17:51

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