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 $c_x={\sum }_y{c}_{xy}$. Then the commute time identity isThe 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 $h_z(x)={\mathbb{E}}_x{T}_z$ and define the network operator $Lh(x)={\sum }_y{c}_{xy}(h(x)-h(y))$. First-step analysis gives $L{h}_z(x)=c_x$ off $z$. Summing $L{h}_z$ over all graph vertices gives zero, so $L{h}_z(z)=c_z-2W$. 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. ^[ref]

Let simple random walk travel from $a\ne z$ to its first visit to $z$, then stop on its subsequent first hit of $a$ in a finite connected unweighted loopless graph. Every directed edge $u\to v$ has expected traversal countThe Strong Markov property splits the count into the two killed legs. By killed-walk occupation voltage, their contributions are $v_{az}(u)$ and $v_{za}(u)$. Their Laplacians cancel, so the harmonic maximum principle on a finite graph makes the sum constant, with value $v_{az}(a)+v_{za}(a)=R_{\mathrm{eff}}(a,z)$. Summing over all directed edges recovers the commute time identity.