Probability on Trees and Networks by Russell Lyons, Yuval Peres

By Russell Lyons, Yuval Peres

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Impose a voltage of v(a) at a and 0 on Z. Since v(•) is linear in v(a) by the superposition principle, we have that Px [τa < τZ ] = v(x)/v(a), whence ( ) ∑ c(a, x) [ v(x) ] p(a, x) 1 − Px [τa < τZ ] = 1− π(a) v(a) x x ∑ ∑ 1 1 = c(a, x)[v(a) − v(x)] = i(a, x) . v(a)π(a) x v(a)π(a) x P[a → Z] = ∑ In other words, ∑ v(a) = i(a, x) . 4) c ⃝1997–2014 by Russell Lyons and Yuval Peres. Commercial reproduction prohibited. Version of 27 February 2014. DRAFT §2. More Probabilistic Interpretations 29 where the last notation indicates the dependence on a and Z.

Series Law. Two resistors* r1 and r2 in series are equivalent to a single resistor r1 + r2 . In other words, if w ∈ V(G) \ (A ∪ Z) is a node of degree 2 with neighbors u1 , u2 and we replace the edges (ui , w) by a single edge (u1 , u2 ) having resistance r(u1 , w) + r(w, u2 ), then all potentials and currents in G \ {w} are unchanged and the current that flows from u1 to u2 equals i(u1 , w). u1 w u2 Proof. It suffices to check that Ohm’s and Kirchhoff’s laws are satisfied on the new network for the voltages and currents given.

Following the transformations indicated in the figure, we obtain C (a ↔ z) = 7/12, so that P[a → z] = † DRAFT 7/12 C (a ↔ z) 7 = = . π(a) 3 36 A conductor c is an edge with conductance c. c ⃝1997–2014 by Russell Lyons and Yuval Peres. Commercial reproduction prohibited. Version of 27 February 2014. DRAFT 34 Chap. 2: Random Walks and Electric Networks 1 1/2 1 1 1 1 1 1 1 1/2 a z 1 1/2 a 1/2 1/2 1 1 z 1 1 1 1 1/2 1/2 1 1 1 1/4 1 1/2 a 1 a z z 1 1 1/3 a 1 z 7/12 Note that in any network G with voltage applied from a to z, if it happens that v(x) = v(y), then we may identify x and y to a single vertex, obtaining a new network G/{x, y} in which the voltages at all vertices are the same as in G.

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