Cover time

A simple symmetric random walk on a cycle of $N\ge 2$ vertices, started at vertex zero, has its last newly visited vertex uniformly distributed on the other $N-1$ vertices. To prove this for a specified vertex $j\ne 0$, cut the cycle at $j$, representing it by a path with absorbing endpoints $0,N$ both corresponding to $j$. All other vertices are visited before $j$ exactly when both interior endpoints $1,N-1$ are visited before absorption. After the first hit on either interior endpoint, the probability of reaching the other before the adjacent absorbing endpoint is $1/(N-1)$ by gambler's ruin. The Strong Markov property makes this the desired probability, independent of the original start position. For $N=2$ the sole unvisited vertex has probability one.

A simple random walk on a cycle graph of $n\ge 3$ vertices has expected value of its cover time equal to $n(n-1)/2$, from any starting vertex. Lift the walk to the integers using the same $\pm 1$ increments. Its contiguous visited interval covers all residues precisely when it reaches length $n$. Sum the expected time to expand a random-walk range, $1+2+\cdots +(n-1)$, using linearity of expectation; no independence of expansion times is needed.