Foundations of Data Science

Random walks on lattices
We now apply the analogy between random walks and current to lattices. Consider
a random walk on a finite segment −n, . . . , −1, 0, 1, 2, . . . , n of a one dimensional lattice
starting from the origin. Is the walk certain to return to the origin or is there some probability
that it will escape, i.e., reach the boundary before returning? The probability of
reaching the boundary before returning to the origin is called the escape probability. We
shall be interested in this quantity as n goes to infinity.
Convert the lattice to an electrical network by replacing each edge with a one ohm
resister. Then the probability of a walk starting at the origin reaching n or –n before
returning to the origin is the escape probability given by
p escape = c eff
c a
where c eff is the effective conductance between the origin and the boundary points and c a
is the sum of the conductance’s at the origin. In a d-dimensional lattice, c a = 2d assuming
that the resistors have value one. For the d-dimensional lattice
p escape =
1
2d r eff
In one dimension, the electrical network is just two series connections of n one ohm resistors
connected in parallel. So as n goes to infinity, r eff goes to infinity and the escape
probability goes to zero as n goes to infinity. Thus, the walk in the unbounded one dimensional
lattice will return to the origin with probability one. This is equivalent to flipping
a balanced coin and keeping tract of the number of heads minus the number of tails. The
count will return to zero infinitely often.
Two dimensions
For the 2-dimensional lattice, consider a larger and larger square about the origin for
the boundary as shown in Figure 5.9a and consider the limit of r eff as the squares get
larger. Shorting the resistors on each square can only reduce r eff . Shorting the resistors
results in the linear network shown in Figure 5.9b. As the paths get longer, the number
of resistors in parallel also increases. The resistor between vertex i and i + 1 is really
4(2i + 1) unit resistors in parallel. The effective resistance of 4(2i + 1) resistors in parallel
is 1/4(2i + 1). Thus,
r eff ≥ 1 + 1 + 1 + · · · = 1(1 + 1 + 1 + · · · ) = Θ(ln n).
4 12 20 4 3 5
Since the lower bound on the effective resistance and hence the effective resistance goes
to infinity, the escape probability goes to zero for the 2-dimensional lattice.
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