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From Wikipedia:

the conductance of a graph G=(V,E) measures how "well-knit" the graph is: it controls how fast a random walk on G converges to a uniform distribution.

I was wondering why a random walk on a graph G converges to a uniform distribution? Any references on this?

Thanks!

Tim
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2 Answers2

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It doesn't, in general. That article needs correction.

A random walk starting at any vertex will (assuming G is connected and [as Nate pointed out] gives an aperiodic walk) converge to the stationary distribution, which is given by the values of the left eigenvector associated with the first eigenvalue of the transition matrix. This depends on both the connections in the graph and the probabilities, at any vertex, with which you will move to each of its neighbors.

The stationary distribution will be uniform if, for example, the graph is regular (all vertices have the same degree) and at each step you choose one of the neighboring nodes with equal probability, since then the left eigenvector is proportional to $\mathbf{1} = (1,1,\dotsc,1)$.

DonAntonio
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Jonathan Christensen
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A random walk on a graph $G$ with vertices $\{1,2,\ldots,n\}$ has transition matrix $P(i,j)=\frac{n_{ij}}{d_i}$ where $n_{ij}$ is 1 for $i,j$ being neighbors and 0 otherwise and $d_i$ is the degree of $i$. This transition matrix is reversible with respect to the measure $\pi(i)=\frac{d_i}{\sum_i d_i}=\frac{d_i}{2m}$, where $m$ is the number of edges:

$\pi(i)P(i,j)=\pi(j)P(j,i)$.

This easily implies $\pi$ is a stationary measure of the Markov chain: $\pi P= \pi$, in that $\sum_i \pi(i)P(i,j)=\sum_i\pi(j)P(j,i)=\pi(j)$. Assuming the Markov chain is both irreducible and aperiodic, this implies we have convergence to the stationary distribution. Note that it is not quite "uniform" in that not all points are equally likely, rather each point is as likely as it's degree.

Alex R.
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    Thanks! (1) I wonder why $\sum_i d_i = 2n$? It seems require there are $n$ edges in $G$, where $n$ is the number of vertices in $G$. (2) Why is a random walk on a graph $G$ is irreducible and aperiodic? Being irreducible seems equivalent to require $G$ to be connected, and being aperiodic seems also require some extra conditions? – Tim Dec 01 '12 at 13:24
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    @Tim Your concern (1) is probably a notation problem--Alex probably indented $\sum d_i = 2m$ where $m$ is the number of edges, or some such. It is not correct as written.

    In addition, the random walk is, as you noted, only irreducible if the graph is connected (and with some conditions on the transition probabilities, too), and not all random walks on graphs are aperiodic (an easy counterexample is bipartite graphs, on which random walks always have period two).

     – Jonathan Christensen Dec 01 '12 at 16:12
  • @JonathanChristensen: whoops thanks for the correction – Alex R. Dec 01 '12 at 19:15
  • @Alex: Thanks for the correction. Why is a random walk on a graph G irreducible and aperiodic? Being irreducible seems equivalent to require G to be connected, and being aperiodic seems also require some extra conditions? – Tim Dec 01 '12 at 19:39
  • @Tim: I've changed the wording slightly. Irreducibility does indeed require connectivity. Aperiodicity is definitely a property that needs to be verified of the graph. A typical example is looking a circle graph of points, which will be aperiodic/periodic depending on the odd/even number of vertices. – Alex R. Dec 01 '12 at 20:26
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    @Tim: I believe the walk is aperiodic if and only if every component contains an odd cycle (or a loop, which you can consider as a 1-cycle). – Nate Eldredge Dec 01 '12 at 20:38
  • @NateEldredge: Thanks! That makes sense. – Tim Dec 01 '12 at 20:48
  • @NateEldredge: "the walk is aperiodic if and only if every component contains an odd cycle". Suppose the graph is just a cycle of length 3 with 3 vertices. The period of each vertex will be 3, and a random walk on the graph isn't aperiodic (which requires all the periods to be 1)? – Tim Dec 11 '12 at 22:40
  • @Tim: A cycle of length 3 is aperiodic. Odd length cyclic graphs are aperiodic. You can step away and immediately back to the starting vertex in 2 steps, or walk around in 3 steps and already $gcd(2,3)=1$. Even length cyclic graphs are periodic with period 2. – Alex R. Dec 11 '12 at 23:36
  • @Alex: Right. Thanks! – Tim Dec 12 '12 at 00:02