= Killed-walk occupation voltage
{title2=$v_{az}(u)=g_{az}(u)/\deg(u)$}
For <simple random walk> on a finite connected loopless unweighted <graph>, start at $a\ne z$ and stop on first hitting $z$. If
$$
g_{az}(u)=\mathbb E_a\sum_{t<T_z}\mathbf1_{\{X_t=u\}},\qquad v_{az}(u)=\frac{g_{az}(u)}{\deg(u)},
$$
then $v_{az}(z)=0$ and $Lv_{az}=\delta_a-\delta_z$ for the <Graph Laplacian> $L$. The incoming-visit balance equation is $g_{az}(u)=\mathbf1_{\{u=a\}}+\sum_{w\sim u}g_{az}(w)/\deg(w)$ for $u\ne z$; at $z$ the Laplacian value is $-1$ because all its coordinates sum to zero. Thus $v_{az}$ is the <voltage> for unit current from $a$ to $z$, and $v_{az}(a)=R_{\mathrm{eff}}(a,z)$. In particular, the expected number of directed transitions $u\to w$ before $T_z$ is $v_{az}(u)$.
Back to article page
