Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 214 / 3

Fix an edge $e=\{u,v\}$ and orient it from $u$ to $v$. For each spanning tree $T$, let ${\gamma }_T$ be its unique graph path from $u$ to $v$. Assign a signed unit graph path flow $j_T$ to oriented edges: value $+1$ in the direction used by ${\gamma }_T$, value $-1$ on the reversed orientation, and zero otherwise. Define its meanEach $j_T$ has net divergence one at $u$, minus one at $v$ and zero elsewhere, so $J$ is a unit flow. The mean spanning-tree path current also satisfies the Kirchhoff cycle law. Although its verification is permitted to be omitted, a short counting argument makes it explicit. Let ${\mathcal{F}}_{uv}$ be the two-component spanning forests separating $u$ and $v$, write $A_F$ for the component containing $u$, and let $N_T$ be the number of spanning trees. Deleting an edge used by the oriented tree graph path gives one such forest. Conversely adjoining an edge across its two components gives a unique tree with that edge on its $u$-to-$v$ graph path. ConsequentlyThe gradient form proves the Kirchhoff cycle law directly and is Ohm's law for unit edge resistance. The Kirchhoff node law then gives $Lh={\delta }_u-{\delta }_v$, and the electrical unit flow is unique: the difference between two such voltages is a harmonic function on a graph and hence constant on the connected finite graph.

If $e\in T$, its single-edge route is the unique tree graph path from $u$ to $v$, so $j_T(u,v)=1$. If $e\notin T$, that tree graph path does not use $e$, so $j_T(u,v)=0$. ThereforeBy Ohm's law, $J(u,v)=h(u)-h(v)$; for a unit terminal current, this voltage difference is exactly the effective resistance between the terminals. We have proved the requested edge-inclusion formula for a uniform spanning tree:Now use linearity of expectation and the fact that every spanning tree has exactly $n-1$ edges. This gives Foster's theorem:The sums here use unoriented edges exactly once. For unit conductance this is the displayed identity; with conductances $c_e$ the inclusion probability and summand become $c_e{R}_{\mathrm{eff}}(e)$.

Use simple random walk on the loopless undirected unweighted graph, with transition probability $1/\mathrm{deg}(x)$ to each graph neighbour, and take $a\ne z$. Set $T_z=\inf \{t\ge 0:X_t=z\}$. Returns to $a$ before visiting $z$ do not end the commute. After its first visit to $z$, stop at the subsequent first visit to $a$, so $\tau =T_z+T_a\circ {\vartheta }_{T_z}$. Count transitions with departure times $0\le t<\tau$, including the final step arriving at $a$. The expected hitting times are finite: in a finite connected graph there is a uniformly positive probability to reach any fixed target within a fixed number of steps, giving a geometric tail bound.

We prove the needed killed-walk occupation voltage identity from its visit balance equation. DefineFor $x\ne z$, counting arrivals before absorption givesHence, for the graph Laplacian $Lv(x)=\mathrm{deg}(x)v(x)-{\sum }_{y\sim x}v(y)$, we have $Lv(x)=1_{\{x=a\}}$ away from $z$. Since the sum of all Laplacian coordinates is zero, $Lv(z)=-1$. Thus $v$ is the electrical voltage of a unit flow from $a$ to $z$, grounded at $z$, and $v(a)=R_{\mathrm{eff}}(a,z)$.

Construct $w(x)=g_{za}(x)/\mathrm{deg}(x)$ for the opposite killed leg. It satisfies $Lw={\delta }_z-{\delta }_a$ and $w(a)=0$. Therefore $L(v+w)=0$. The harmonic maximum principle on a finite graph makes $v+w$ constant: at a maximum its value equals the average over its graph neighbours, forcing all of them to share that value, and connectedness propagates it. At $a$ the constant equals $R_{\mathrm{eff}}(a,z)$.

For any oriented edge $(x,y)$, each visit to $x$ before the target hit produces a transition to $y$ with conditional probability $1/\mathrm{deg}(x)$. The expected traversals during the first killed leg are therefore $v(x)$, and those during the second leg are $w(x)$ by the Strong Markov property at $T_z$. This proves directed edge occupation in a random-walk commute:In particular, both directions of every unoriented edge have this same expected value, although their counts on an individual commute need not coincide. Summing over the $2|E|$ directed edges also recovers the commute time identity ${\mathbb{E}}_a\tau =2|E|R_{\mathrm{eff}}(a,z)$.

The distinct-terminal convention is necessary for the printed “returns” interpretation. If $a=z$ and $\tau$ is the first positive return, a graph with two graph vertices and one edge gives $S(a,y)=1$ while $R_{\mathrm{eff}}(a,a)=0$. Thus either take $a\ne z$, as intended, or explicitly define the degenerate commute as $\tau =0$. A general nonuniform random walk is not covered by the unweighted simple-walk formula.

Group the random resistance cost by unoriented edges. For $e=\{x,y\}$, its resistance contributes once whenever the walk traverses it in either direction. Pathwise,The graph is finite and the commute has finite expected value. Linearity of expectation, or Tonelli theorem for these nonnegative terms, therefore permits taking expectations of the sum. By the directed edge occupation in a random-walk commute, both directed traversal counts have expected value $R_{\mathrm{eff}}(a,z)$. ThusApply Foster's theorem from part (a) to obtainThe factor two counts the two orientations of each edge. It would be lost by treating the expected directed count in part (b) as an expected count for both directions together. The same distinct-terminal and simple-random-walk conventions from part (b) remain in force.