Local Diffusion Versus Random Relocation in Random Walks

Abstract

We study a class of random walks on graphs with two mechanisms: local diffusion and random relocation. Such mechanisms are common in search algorithms, an example being the PageRank. The rate with which two mechanisms are mixed is called damping factor. It determines the first-passage time, defined as the time required for a walker starting from a source node to find a given target node. We provide bounds on the stationary distribution of random walks. Global mean first-passage time as a function of the damping factor is computed in closed form for a simple example. The results provide new insights on the search engines.

Appendix

We now assume that 1 is absorbing state/node; then the transition matrix P transforms to

$$\begin{aligned} P = \left[ \begin{array}{ccc} 1 &{} 0 &{} 0 \\ 1-b &{} 0 &{} b \\ c &{} 1-c &{} 0 \end{array} \right] , \end{aligned}$$

and $P_1$, the matrix of one-step transition probabilities of the (sub)Markov chain on the set $\{2, 3\}$, is given by

$$\begin{aligned} P_1 = \left[ \begin{array}{cc} 0 &{} b \\ 1-c &{} 0 \end{array} \right] . \end{aligned}$$

The Perron – Frobenius eigenvalue of $P_1$ is $\lambda = \sqrt{b(1-c)}<1$ and its normalized eigenvector $\mathbf {u}$ is

$$\begin{aligned} \mathbf {u}^T = \left( \frac{\lambda - b}{1-c-b}, \frac{1 - c - \lambda }{1-c-b} \right) . \end{aligned}$$

Since the set of transient states $S = \{2, 3\}$ is irreducible, $\mathbf {u}$ the (unique) quasi-stationary distribution for the discrete-time Markov chains with one absorbing state and a finite set S of transient states.

The fundamental matrix for the absorbing chain is

$$\begin{aligned} I + P_1 + P_1^2 + \ldots= & {} I \left[ 1+ b(1-c) + b^2(1-c)^2 + \ldots \right] + \\&+ P_1 \left[ 1+ b(1-c) + b^2(1-c)^2 + \ldots \right] \\= & {} \frac{1}{1-b(1-c)} \left[ \begin{array}{cc} 1 &{} b \\ 1-c &{} 1 \end{array} \right] \equiv \left[ \begin{array}{cc} n_{22} &{} n_{23} \\ n_{32} &{} n_{33} \end{array} \right] . \end{aligned}$$

recovering $m_{21}$ and $m_{31}$ from the matrix M, that is

$$\begin{aligned} m_{21} = \sum _{k\in S} n_{2k} = \frac{1+b}{1-b(1-c)}, \quad \; m_{31} = \sum _{k \in S} n_{3k} = \frac{2-c}{1-b(1-c)}. \end{aligned}$$

The other elements of the matrix M can be computed in a similar way.