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Recall McShane's Identity.

Take a Brownian Motion on this punctured torus constrained to return to the original point. Now take the "curve" you get this way and try and contract it as much as possible. Maybe the probability you get of ending with a given simple closed geodesic of length $l$ is $2/(1+e^l)$ or something. This would give a probabilistic proof of McShane's identity or something like this.

Can one construct this?

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  • $\begingroup$ On a circle the probability that a Brownian loop has length ell (is homotopic to a geodesic with length ell) is more like exp(-ell^2). $\endgroup$
    – David Treumann
    Commented Aug 17, 2017 at 12:51
  • $\begingroup$ Another big problem with this argument is that he probability it contracts to a point should be positive. $\endgroup$
    – Will Sawin
    Commented Aug 17, 2017 at 15:21

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Not that I know of, but there is a Brownian motion argument for the (philosophically related) Basmajian identity:

Calegari, Danny, Chimneys, Leopard spots and the identities of Basmajian and Bridgeman, Algebr. Geom. Topol. 10, No. 3, 1857-1863 (2010). ZBL1196.57010.

The Brownian motion thing seems to be not really essential, as shown in my preprint. But make of all this what you will.

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