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We know (see here ) that the random walk generated in $R^1$ can converge in distribution to the standard Brownian motion $B_t$ in $R^1$. Could anybody provide a rigorous mathematical proof, how a random walk generated in $R^3$ can converge in distribution to the standard Brownian motion on a sphere $S^2$ using an appropriate mapping?

Denis
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  • How do you get from a random walk in $\Bbb R^3$ to a random walk on $S^2$ or at least close to it? – Lutz Lehmann Aug 15 '20 at 17:42
  • For example here , they argue that they can make the random walk on a tangent plane and then projecting on sphere. There are also other examples of such approach, where they map the walk onto the surface. – Denis Aug 15 '20 at 18:00
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    Yes, that is the usual definition, project the step via the exponential map from the tangent space to the manifold. You would need to tell how you define the "standard Brownian motion on a sphere", with that I think the proof is along the same lines as in flat space, only that charts enter the play, the variable metric and its curvature give additional terms in the equations of the motion. – Lutz Lehmann Aug 16 '20 at 08:29
  • I think one example can be defining as $d\theta_t=dB_t^{\theta}+\frac{1}{2}\cot(\theta_t)dt, \qquad d\phi_t=\frac{1}{\sin(\theta_t)}dB_t^{\phi}.$, the next steps are not clear for me... – Denis Aug 18 '20 at 15:39
  • you may want to look into A Brief Introduction to Brownian Motion on a Riemannian Manifold by Elton P. Hsu , in particular Example 1.4.1 – am70 Jun 14 '22 at 13:55

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