Brownian particles on a circle

Today, James norris gave a talk related to Diffusion-limited aggregation processes and mentioned, in passing, the following amusing fact: put {N} equidistant Brownian particles {W_1, \ldots, W_N} on the circle with unit circumference and let them evolved. When two of them collide they get stuck to each other and continue together afterwards: after a certain amount of time {T_N}, only one particle remains. Perhaps surprisingly, this is extremely easy to obtain the first few properties of {T_N}. For example, {\lim_N \mathop{\mathbb E}\left[ T_N \right] = \frac{1}{6}}.

To see that, define {D_k} the distance between {W_k} and {W_{k+1}} (modulo {N}) so that {D_k = \frac{1}{N}} for {k=1,2 \ldots, N}. Notice then (It\^o’s formula) that

\displaystyle  M_t = e^{\lambda t} \sum_{k=1^N} \sin(\sqrt{\lambda} D_k)

is a (local) martingale that starts from {M_0 = N \sin(\frac{\sqrt{\lambda}}{N})}. Also, at time {T_N}, exactly {N-1} of the distances {D_1, D_2, \ldots, D_N} are equal to {0} while one of them is equal to {1}: this is why {M_{T_N} = e^{\lambda T_N} \sin(\sqrt{\lambda})}. The end is clear: apply the optional sampling theorem (to be rigorous, take {\lambda} not too big, or do some kind of truncations to be sure that the optional sampling theoem applies) to conclude that

\displaystyle  \mathop{\mathbb E}\left[ e^{\lambda T_N} \right] = \frac{N \sin(\frac{\sqrt{\lambda}}{N})}{\sin(\sqrt{\lambda})}.

This gives for example {\mathop{\mathbb E}\left[ T_N \right] = \frac{1}{6}(1-\frac{1}{N^2})}. I just find it cute!

So what if we do that on a segment ?

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