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Suppose we have a circumference divided in N arcs of the same length. A particle can move on the circumference jumping from an arc to the adjacent, with probability $P_{k \to k-1}=P_{k\to k+1}=\frac{1}{2}$. I have some difficulty to find the probability that the particle fulfilled at least one complete turn on the circumference after $M$ jumps with $M\gt N$. Thanks

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Are you asking about the first time when the particle visited at least once every arc? – Did Jul 12 '12 at 13:41
@did: yes the interpretetion is correct – Riccardo.Alestra Jul 12 '12 at 15:54
up vote 5 down vote accepted

The time needed to visit at least once each vertex of a graph is called the cover time of the graph. Let $C_n$ denote the cover time of the discrete circle of size $n$. Then $\mathrm E(C_n)=\frac12n(n-1)$. The expression of the distribution of $C_n$ for any fixed $n$ is cumbersome but a result due to Jean-Pierre Imhof in 1985 describes the asymptotics as follows: when $n\to\infty$, $n^{-2}C_n\to C$ in distribution, where the distribution of $C$ has density $$ \sqrt{\frac{8}{\pi t^3}}\cdot\sum_{k=1}^{+\infty}(-1)^{k-1}k^2\mathrm e^{-k^2/(2t)}\cdot[t\gt0]. $$ A different, equivalent, expression of the limit distribution is in Amine Shihi's PhD thesis.

A related (and somewhat counterintuitive) result might be worth mentioning: for every $n\geqslant2$, the last vertex visited, that is, the position of the particle at time $C_n$ is uniformly distributed on the $n-1$ vertices which are not the starting one.

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