= Mean spanning-tree path current
{title2=$J(u,v)=\mathbb E[\mathbf1_{\{u\to v\in\gamma_T\}}-\mathbf1_{\{v\to u\in\gamma_T\}}]$}
In a finite unweighted <connected graph>, take distinct <graph vertices> $a\ne z$ and orient the unique <graph path> from $a$ to $z$ in a <uniform spanning tree> and assign signed unit current to its <edges>. Its mean over <trees> is the electrical <unit flow> from $a$ to $z$. The <Kirchhoff node law> follows by averaging <graph path> divergences. For the <graph cycle> law, let $\mathcal F_{az}$ be the <two-component spanning forests> separating the terminals and $A_F$ the component containing $a$. Deleting a tree-path <edge> and conversely adjoining an <edge> across a forest cut are inverse operations. Hence the mean signed current on an <edge> $u\to v$ is
$$
J(u,v)=\frac1{|\mathcal T(G)|}\sum_{F\in\mathcal F_{az}}\bigl(\mathbf1_{A_F}(u)-\mathbf1_{A_F}(v)\bigr).
$$
This is the gradient of $h(u)=|\mathcal T(G)|^{-1}\sum_F\mathbf1_{A_F}(u)$, proving the <graph cycle> law. In particular $R_{\mathrm{eff}}(a,z)=|\mathcal F_{az}|/|\mathcal T(G)|$. For an oriented existing <edge> $e=(a,z)$, that <edge> is used by the <tree> <graph path> exactly when it belongs to the <tree>, and then has positive orientation. Thus its mean current is $\mathbb P(e\in T)$, giving the <edge-inclusion formula for a uniform spanning tree> by <Ohm's law>.
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