Think of an airline running many itineraries. Each starts at a random airport (law \(\mu\)), hops along random routes (transition probabilities \(P\)), and ends by a rule that cannot peek ahead (a stopping time \(\tau\)). The final airport has law \(\nu\).
Fix an airport \(y\) and count every stop there. Each stop is either the start of an itinerary or an arrival from an earlier flight, and it is either followed by a departure or is the itinerary’s end. So, on average per itinerary,
departures from \(y\) + itineraries ending at \(y\) = arrivals into \(y\) + itineraries starting at \(y,\)
flowchart LR
S(["Itineraries
starting at y"]):::src --> Y
A["Airport x1"] -->|"Flights x1 to y"| Y
B["Airport x2"] -->|"Flights x2 to y"| Y
Y(("Airport y"))
Y -->|"Departures from y"| C["Other airports"]
Y --> E(["Itineraries
ending at y"]):::snk
classDef src fill:#d8f0e0,stroke:#2e7d4f,color:#12351f
classDef snk fill:#fbe0dc,stroke:#b3402f,color:#4a1610
Formally, let \(\{X_t\}_t\) be a Markov chain (the walk over desatinations); define occupancy as \(g(y) = \mathbb{E}\sum_{t=0}^{\tau-1} \mathbf{1}\{X_t=y\}\) (the expected departure counts from \(y\)). Then, the occuapancy solves the following dual Poisson-type equation in the space of measures
Theorem (Poisson-type Occupancy Equation). With the notation as above:
the second identity over the effective domain of \(g\).
Proof. On every itinerary, the visits to \(y\) before the end, plus the final stop if there is one, equal the start plus the arrivals:
\(\displaystyle \sum_{t<\tau}\mathbf{1}\{X_t=y\}+\mathbf{1}\{\tau<\infty,\,X_\tau=y\}=\mathbf{1}\{X_0=y\}+\sum_{t<\tau}\mathbf{1}\{X_{t+1}=y\}.\)
Take expectations. On the right, \(\{t<\tau\}\) is known at time \(t\), before the next flight, so the Markov property gives \(\mathbb{E}\big[\mathbf{1}\{t<\tau\}\mathbf{1}\{X_{t+1}=y\}\big]=\mathbb{E}\big[\mathbf{1}\{t<\tau\}P(X_t,y)\big]\). Summing over \(t\) (Tonelli, all terms are \(\ge0\)) gives \((gP)(y)\). \(\square\).
Applications. A number of classical results follow. For the first two, assume the chain is irreducible and positive recurrent with stationary law \(\pi\).
Corollary (stationary occupation). If \(\mu=\nu\) (which forces \(\tau<\infty\) a.s.) and \(\mathbb{E}\tau<\infty\), then \(gP=g\), so \(g=\pi\,\mathbb{E}\tau\). The time spent in any set \(S\) is \(\pi(S)\,\mathbb{E}\tau\) in expectation, even from a non-stationary start.
Corollary (Kac). Start at \(z\) and stop at the first return \(T_z^+\). The itinerary visits \(z\) once, so \(g(z)=1\) and \(\displaystyle \mathbb{E}_zT_z^+=\frac{1}{\pi(z)}\). Every other airport \(y\) is visited \(\pi(y)/\pi(z)\) times per loop.
Corollary (Poisson equation). For a finite irreducible chain with \(\mathbb{E}_\mu\tau<\infty\), let \((I-P)h=f-\pi(f)\). Then
\(\displaystyle \mathbb{E}_\mu\sum_{t<\tau}f(X_t)=\pi(f)\,\mathbb{E}_\mu\tau+(\mu-\nu)h.\)
When the endpoint laws differ, the mismatch becomes an explicit correction. Pair \(g(I-P)=\mu-\nu\) with \(h\).
Credit. The balance identity is due to Pitman (Prop. 4.1). Its equal-endpoint case, Cor. 3 appears as Prop. 2.4 in Aldous and Fill . Kac’s return-time formula is the case of a single-state excursion. Reading the identity as a dual Poisson equation and pairing it with \(h\) gives the Poisson corollary above, which for finite chains needs no aperiodicity, unlike Pitman’s Prop. 4.6(iii).