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For a finite Markov chain in which an absorbing state is accessible from every state, order the nonabsorbing states first and writeHere $I$ is the identity matrix and $N$ is the matrix inverse of $I-Q$. With $T$ the first hitting time of the absorbing set,Thus $N$ gives the expected occupation count before absorption, including the initial visit; $B_{ia}$ is the absorption probability of state $a$; and $t_i$ is the expected hitting time of the absorbing set. The vector $1$ has every entry one. See Grinstead and Snell, section 11.2, theorems 11.4-11.6.

For the toy routine in absorbing state, direct matrix multiplication givesThe first row of $N$ says that starting at $u$ gives $2.4$ visits to $u$ and $0.8$ to $v$ before termination. Their sum is the mean duration $3.2$. The identities $(I-Q)N=I$, $B=QB+R$, and $t=1+Qt$ independently verify the inverse, outcome, and duration calculations. The absorbing set must be reachable from every state for these finite formulas: applying them to all nonabsorbing states in the closed-cycle example would leave a persistent cycle in $Q$ and make $I-Q$ singular.

Write $w_{xy}=w_{yx}>0$ for an edge conductance and $w_x={\sum }_{y\sim x}{w}_{xy}$. Take $s\ne t$, as implicit in a unit source-to-sink electrical network. An oriented current flow is antisymmetric, $i_{yx}=-i_{xy}$. Kirchhoff's first law is conservation of current: ${\sum }_y{i}_{xy}=0$ at every vertex other than the source and sink, with the divergences there equal to the prescribed total flow and its negative. Kirchhoff's second law says that the sum of oriented voltage drops around any closed cycle is zero. Ohm's law isThus the resistance of an edge is $1/w_{xy}$, and the cycle law reads ${\sum }_{j=0}^{m-1}{i}_{x_j{x}_{j+1}}/w_{x_j{x}_{j+1}}=0$ for $x_m=x_0$.

Let $\tau =\inf \{n\ge 0:X_n=t\}$. Its expectation is finite. Indeed, from each state there is a path to $t$ of length at most $|V|-1$ with positive probability; finiteness gives a common lower bound $a>0$ on reaching $t$ within that many steps. The Markov property then gives $\mathbb{P}(\tau >k(|V|-1))\le (1-a{)}^k$, proving the assertion. Define the expected occupation count before absorptionConditional expectation at every pre-stopping visit to $x$ gives ${\mathbb{E}}_s{N}_{xy}=G(s,x)p_{xy}$ for the number $N_{xy}$ of directed traversals. Hence the expected signed count isThis proves Ohm's law with voltage $v_x=G(s,x)/w_x$, grounded at $v_t=0$. Summing its voltage drops along a cycle telescopes, proving Kirchhoff's second law.

For Kirchhoff's first law, count entrances and exits on each stopped path. Their net difference at $x$ is the initial indicator minus the final indicator. Taking expectations,Thus the source emits exactly one unit, the sink receives one unit, and all other vertices conserve current.

We also establish the uniqueness required for the conclusion. The cycle law implies that edge voltage drops are differences of vertex potentials: define a potential by summing drops along a path to $t$, and cycle sums make this independent of the path. If two unit flows obey both laws, their difference has the form $j_{xy}=w_{xy}(h_x-h_y)$ with zero divergence at every vertex. ThereforeStrict positivity of the conductances and connectedness force $h$ to be constant, so $j=0$. We conclude the electrical current from stopped random-walk traversals identity