Foundations of Data Science

v ij = 1 + ∑ k adj
to i
1
d i
v kj = ∑ k adj
to i
1
d i
(1 + v kj ). (5.11)
Now the hitting time from i to j is the average time over all paths from i to k adjacent
to i and then on from k to j. This is given by
h ij = ∑ k adj
to i
Subtracting (5.12) from (5.11), gives v ij − h ij
1
d i
(1 + h kj ). (5.12)
= ∑ k adj
to i
1
d i
(v kj − h kj ). Thus, the function
v ij − h ij is harmonic. Designate vertex j as the only boundary vertex. The value of
v ij − h ij at i = j, namely v jj − h jj , is zero, since both v jj and h jj are zero. So the function
v ij −h ij must be zero everywhere. Thus, the voltage v ij equals the expected time h ij from
i to j.
To complete the proof, note that h ij = v ij is the voltage from i to j when currents are
inserted at all vertices in the graph and extracted at vertex j. If the current is extracted
from i instead of j, then the voltages change and v ji = h ji in the new setup. Finally,
reverse all currents in this latter step. The voltages change again and for the new voltages
−v ji = h ji . Since −v ji = v ij , we get h ji = v ij .
Thus, when a current is inserted at each vertex equal to the degree of the vertex
and the current is extracted from j, the voltage v ij in this set up equals h ij . When we
extract the current from i instead of j and then reverse all currents, the voltage v ij in
this new set up equals h ji . Now, superpose both situations, i.e., add all the currents and
voltages. By linearity, for the resulting v ij , which is the sum of the other two v ij ’s, is
v ij = h ij + h ji . All currents cancel except the 2m amps injected at i and withdrawn at j.
Thus, 2mr ij = v ij = h ij + h ji = commute(i, j) or commute(i, j) = 2mr ij where r ij is the
effective resistance from i to j.
The following corollary follows from Theorem 5.9 since the effective resistance r uv is
less than or equal to one when u and v are connected by an edge.
Corollary 5.10 If vertices x and y are connected by an edge, then h xy + h yx ≤ 2m where
m is the number of edges in the graph.
Proof: If x and y are connected by an edge, then the effective resistance r xy is less than
or equal to one.
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