= Commute time identity
For a finite connected loopless undirected <electrical network> without killing, give each <edge> a positive <electrical conductance> $c_e$, put $W=\sum_e c_e$, and use <transition probabilities> $P(x,y)=c_{xy}/c_x$, where $c_x=\sum_yc_{xy}$. Then the <commute time identity> is
$$
\mathbb E_aT_z+\mathbb E_zT_a=2W R_{\mathrm{eff}}(a,z).
$$
The factor is total <electrical conductance>, not total <electrical resistance>. A killed <random walk> can fail to reach its target and is outside this statement.
Here is a direct proof. Let $h_z(x)=\mathbb E_xT_z$ and define the network operator $Lh(x)=\sum_yc_{xy}(h(x)-h(y))$. <First-step analysis> gives $Lh_z(x)=c_x$ off $z$. Summing $Lh_z$ over all <graph vertices> gives zero, so $Lh_z(z)=c_z-2W$. Therefore $u=(h_z-h_a)/(2W)$ has divergence $+1$ at $a$, $-1$ at $z$, and zero elsewhere. Its <voltage> drop is $(h_z(a)+h_a(z))/(2W)$, which is the <effective resistance> for unit <electric current>.
For an unweighted path with $n$ <edges>, $W=n$ and endpoint <effective resistance> is $n$, giving commute time $2n^2$. Uniformly multiplying all <electrical conductances> changes $W$ and <effective resistance> inversely and leaves the <Markov chain> unchanged. <The electrical resistance of a graph captures its commute and cover times by Ashok K. Chandra and others>[Section 2, Theorems 2.1–2.2: unweighted and weighted commute-time identities]{ref}
Back to article page
