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Foundations of Data Science

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There are important connections between electrical networks and random walks on<br />

undirected graphs. Choose two vertices a and b. For reference purposes let the voltage<br />

v b equal zero. Attach a current source between a and b so that the voltage v a equals<br />

one. Fixing the voltages at v a and v b induces voltages at all other vertices along with a<br />

current flow through the edges <strong>of</strong> the network. The analogy between electrical networks<br />

and random walks is the following. Having fixed the voltages at the vertices a and b, the<br />

voltage at an arbitrary vertex x equals the probability <strong>of</strong> a random walk starting at x<br />

reaching a before reaching b. If the voltage v a is adjusted so that the current flowing into<br />

vertex a corresponds to one walk, then the current flowing through an edge is the net<br />

frequency with which a random walk from a to b traverses the edge.<br />

Probabilistic interpretation <strong>of</strong> voltages<br />

Before showing that the voltage at an arbitrary vertex x equals the probability <strong>of</strong> a<br />

random walk starting at x reaching a before reaching b, we first show that the voltages<br />

form a harmonic function. Let x and y be adjacent vertices and let i xy be the current<br />

flowing through the edge from x to y. By Ohm’s law,<br />

i xy = v x − v y<br />

r xy<br />

= (v x − v y )c xy .<br />

By Kirchh<strong>of</strong>f’s law the currents flowing out <strong>of</strong> each vertex sum to zero.<br />

∑<br />

i xy = 0<br />

y<br />

Replacing currents in the above sum by the voltage difference times the conductance<br />

yields<br />

∑<br />

(v x − v y )c xy = 0<br />

or<br />

y<br />

∑<br />

v x c xy = ∑<br />

y<br />

y<br />

v y c xy .<br />

Observing that ∑ c xy = c x and that p xy = cxy<br />

c x<br />

, yields v x c x = ∑ v y p xy c x . Hence,<br />

y<br />

y<br />

v x = ∑ v y p xy . Thus, the voltage at each vertex x is a weighted average <strong>of</strong> the voltages<br />

at the adjacent vertices. Hence the voltages form a harmonic function with {a, b} as<br />

y<br />

the boundary.<br />

Let p x be the probability that a random walk starting at vertex x reaches a before b.<br />

Clearly p a = 1 and p b = 0. Since v a = 1 and v b = 0, it follows that p a = v a and p b = v b .<br />

Furthermore, the probability <strong>of</strong> the walk reaching a from x before reaching b is the sum<br />

over all y adjacent to x <strong>of</strong> the probability <strong>of</strong> the walk going from x to y in the first step<br />

162

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