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Take the usual Brownian motion on $R^d$ and project it to $T^d$, for almost every individual trajectory, will it be equidistributed on the torus? Does this depend on $d$?

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up vote 1 down vote accepted

Yes - it is equidistributed (which is the case for the Brownian motion on any compact Riemannian manifold).

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Could you please give some references? Thanks! – Jerry Jan 30 '13 at 23:12
The Riemannian volume is stationary for the Brownian motion on any Riemannian manifold. For a compact manifold (more generally, for a finite volume one) this is the only stationary measure (which follows, for instance, from the fact that the transition probabilities are absolutely continuous with respect to the Riemannian volume), in particular, it is ergodic. Equidistribution of a.e. sample path is then essentially just the usual ergodic theorem for the Brownian motion. – R W Jan 31 '13 at 6:38

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