## 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 ?