Fastest Mixing Markov Chain on a Path

We consider the problem of assigning transition probabilities to the edges of a path in such a way that the resulting Markov chain or random walk mixes as rapidly as possible. In this note we prove that fastest mixing is obtained when each edge has a transition probability of 1/2. Although this result is intuitive (it was conjectured in [6]) and can be found numerically using convex optimization methods [2], [3], we give a self-contained proof here. Consider a path with n (≥ 2) nodes, labeled 1, 2, . . . , n, with n − 1 edges connecting pairs of adjacent nodes and with a loop at each node, as shown in Figure 1. We consider a Markov chain (or random walk) on this path, with transition probability from node i to node j denoted Pi j . The requirement that transitions can occur only along an edge or loop of the path is equivalent to Pi j = 0 when |i − j | > 1 (i.e., P is a tridiagonal matrix). Since the Pi j are transition probabilities, we have Pi j ≥ 0 and ∑ j Pi j = 1 (i.e., P is a stochastic matrix). This can be expressed as P1 = 1, where 1 is the vector with all components one. We consider symmetric transition probabilities, meaning those that satisfy Pi j = Pji . Thus, P is a symmetric, (doubly) stochastic, tridiagonal matrix. Since P1 = 1, we have (1/n)T P = 1T /n, which means that the uniform distribution, given by 1T /n, is stationary.