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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.

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.

Choose a spanning tree $T$ of $G$, rooted at $a$. A depth-first traversal of a tree gives a deterministic closed walkthat traverses each edge of $T$ once in each direction and visits every vertex. Define stopping times $S_0=0$ andThe simple random walk is at $v_j$ at time $S_j$. All these hitting times have finite mean on the finite connected graph. By the Strong Markov property,The cover time is no larger than $S_{2(n-1)}$. Grouping the summands by the two traversals of each tree edge and applying the commute time identity givesThe effective resistances are measured in the original graph, where each tree-edge pair is adjacent. Part (a) therefore bounds every summand by one, soFor the one-vertex graph, the cover time is zero and the same bound is immediate.

Let $d=d_G(a,b)>0$, and for $i=1,\dots ,d$ putEach edge cutset ${\Pi }_i$ separates $a$ from $b$. Adjacent vertices have graph distances from $a$ differing by at most one. Hence an edge can cross at most one of these level boundaries, so the edge cutsets are pairwise edge-disjoint. Write $k_i=|{\Pi }_i|\ge 1$; then ${\sum }_i{k}_i\le |E|$.

The Nash-Williams inequality for unit conductances giveswhere the middle inequality is the Cauchy-Schwarz inequality. One can also derive the first inequality directly: every unit flow has signed net flux one through each ${\Pi }_i$, so its energy of a flow on that cut is at least $1/k_i$; sum over the disjoint cuts and use the Thomson principle.

Combining with the commute time identity yieldsThe case $a=b$ is trivial. A path graph, with $a,b$ its endpoints, attains equality.

Thomson principle states that the effective resistance is the minimum energy of a unit flow from $a$ to $z$:An edge cutset separating $a$ and $z$ is a set of edges whose removal disconnects them. Every unit flow has net flux one across each such cutset ${\Pi }_k$. By Cauchy-Schwarz inequality,For disjoint cutsets, summing these energy lower bounds and applying Thomson's principle gives the Nash-Williams inequalityThe commute time identity is

There are $\asymp n^3$ edges. The $n^2$ horizontal cutsets between successive rows perpendicular to the long direction are disjoint and each contains $\asymp n$ unit-conductance edges. Nash-Williams inequality givesConversely, spread a unit flow nearly uniformly across the width $n$ while moving it through the $n^2$ long-direction layers, with bounded extra energy to fan out from and collect at the two corner vertices. Its energy is $O(n^2/n)+O(\log n)=O(n)$, so Thomson's principle gives the matching upper bound $R_{\mathrm{eff}}\asymp n$.

The graph is invariant under a half-turn exchanging the two corners, so the two directional hitting-time expectations are equal. The commute time identity therefore yields