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For a recurrent Markov chain, start at $k$ and stop at its first positive return $T_k$. The expected visits to each state before that return form the excursion occupation measure. Its $k$th coordinate is one. Counting the return endpoint instead of the start gives the same measure. It is invariant by shifting the occupation sum one time step and using the Markov property. Any Markov chain invariant measure normalized by $x_k=1$ dominates it: iterate the stationarity equations on the state space with $k$ removed, then retain the nonnegative excursion-path terms.

In a classical risk model, before the first claim at time $t$, available capital is $u+ct$. A claim of size $x\le u+ct$ leaves future survival probability $\phi (u+ct-x)$ by the Markov property. Integrating over the independent first-arrival exponential distribution and claim density gives $\phi (u)={\int }_0^{\infty }\lambda e^{-\lambda t}{\int }_0^{u+ct}\phi (u+ct-x)f(x)dxdt$.

For a stationary autoregressive process of order one with mean $\mu$, autoregressive coefficient $\varphi$ and innovation variance ${\sigma }^2$, an interior observation conditional on its neighbors hasThe Markov property makes the same law valid when all other observations are also conditioned on.

The hitting probability of a set is one on that set and transition-harmonic elsewhere by the Markov property. Finite-horizon hitting probabilities begin at the boundary indicator and increase through the same positive transition recursion. Any nonnegative solution dominates every iterate; monotone passage to infinite time shows that it dominates the hitting probability. This proves minimality without requiring almost-sure hitting.