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Foundations of Data Science

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and then reaching a from y before reaching b. That is<br />

p x = ∑ y<br />

p xy p y .<br />

Hence, p x is the same harmonic function as the voltage function v x and v and p satisfy the<br />

same boundary conditions at a and b.. Thus, they are identical functions. The probability<br />

<strong>of</strong> a walk starting at x reaching a before reaching b is the voltage v x .<br />

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

In a moment, we will set the current into the network at a to have a value which we will<br />

equate with one random walk. We will then show that the current i xy is the net frequency<br />

with which a random walk from a to b goes through the edge xy before reaching b. Let<br />

u x be the expected number <strong>of</strong> visits to vertex x on a walk from a to b before reaching b.<br />

Clearly u b = 0. Every time the walk visits x, x not equal to a, it must come to x from<br />

some vertex y. Thus, the number <strong>of</strong> visits to x before reaching b is the sum over all y <strong>of</strong><br />

the number <strong>of</strong> visits u y to y before reaching b times the probability p yx <strong>of</strong> going from y<br />

to x. For x not equal to b or a<br />

u x = ∑ y≠b<br />

u y p yx .<br />

Since u b = 0 and c x p xy = c y p yx<br />

u x = ∑ all y<br />

u y<br />

c x p xy<br />

c y<br />

and hence ux<br />

c x<br />

= ∑ u y<br />

c y<br />

p xy . It follows that ux<br />

c x<br />

is harmonic with a and b as the boundary<br />

y<br />

where the boundary conditions are u b = 0 and u a equals some fixed value. Now, u b<br />

c b<br />

= 0.<br />

Setting the current into a to one, fixed the value <strong>of</strong> v a . Adjust the current into a so that<br />

v a equals ua<br />

c a<br />

. Now ux<br />

c x<br />

and v x satisfy the same boundary conditions and thus are the same<br />

harmonic function. Let the current into a correspond to one walk. Note that if the walk<br />

starts at a and ends at b, the expected value <strong>of</strong> the difference between the number <strong>of</strong> times<br />

the walk leaves a and enters a must be one. This implies that the amount <strong>of</strong> current into<br />

a corresponds to one walk.<br />

Next we need to show that the current i xy is the net frequency with which a random<br />

walk traverses edge xy.<br />

(<br />

ux<br />

i xy = (v x − v y )c xy = − u )<br />

y c xy c xy<br />

c xy = u x − u y = u x p xy − u y p yx<br />

c x c y c x c y<br />

The quantity u x p xy is the expected number <strong>of</strong> times the edge xy is traversed from x to y<br />

and the quantity u y p yx is the expected number <strong>of</strong> times the edge xy is traversed from y to<br />

x. Thus, the current i xy is the expected net number <strong>of</strong> traversals <strong>of</strong> the edge xy from x to y.<br />

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