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Alternative routing lets a call try another resource path when its preferred route is blocked. Overflow onto longer paths can create feedback and multiple solutions of the Erlang fixed point approximation, even though an exact finite irreducible occupancy Markov chain has a unique stationary distribution.

An unpinned uniform height driven by Gaussian white noise is Brownian motion. Its variance grows as ${\sigma }^{2}t$ and it has no stationary distribution on the real height axis for $\sigma >0$. Indeed, a stationary characteristic function would obey $\hat{\mu }(k)=\hat{\mu }(k)e^{-{\sigma }^2{k}^{2}t/2}$, incompatible with continuity at zero. Quotienting out the uniform height can still leave stationary shape fluctuations.

For an irreducible Markov chain on a finite or countable state space, a positive recurrent state implies a unique stationary distribution $\pi$. If the chain is also an aperiodic Markov chain, its transition probabilities satisfy the displayed limit for every pair of states. The same limit holds from any initial probability distribution by bounded convergence. Positive recurrence cannot be omitted in the countably infinite case, and aperiodicity is required for ordinary, rather than averaged, convergence. This is the standard discrete-time convergence theorem; the finite-state statement alone does not cover unbounded renewal countdowns.

The Ehrenfest urn model moves one uniformly chosen ball between two boxes at each step. For $k\ge 1$ balls, the count in one box is a Markov chain on $\{0,\dots ,k\}$. Its stationary distribution is $2^{-k}\left(\genfrac{}{}{0ex}{}{k}{i}\right)$, and detailed balance makes it a reversible Markov chain. The chain has period two, but its mean recurrence time to the all-in-one-box state is still $2^k$.

If $r={\max }_{j\ge 2}|{\lambda }_j|$ for a finite irreducible Markov chain with at least two states, then its worst-case total variation distance from its stationary distribution satisfies $d(t)\ge r^t/2$. Choose a possibly complex eigenfunction $f$ for an eigenvalue of modulus $r$, normalize ${\max }_x|f(x)|=1$, and start where the maximum is attained. Since $\pi (f)=0$, $|P^{t}f(x)-\pi (f)|=r^t$; the defining sum for total variation distance bounds this difference by $2d(t)$. For $0<r<1$, this yields $t_{\mathrm{mix}}(\epsilon )\ge (r/(1-r))\log (1/(2\epsilon ))$ for $0<\epsilon <1/2$, because $-r\log r\le 1-r$.

For a finite irreducible Markov chain in continuous time, time averages converge almost surely to averages under its unique stationary distribution, from any initial state. Applying this to an activation indicator yields the corresponding throughput.

Place one distinct label at each vertex of a finite connected weighted graph. At rate $c_{uv}=c_{vu}\ge 0$, interchange the labels at $u$ and $v$. This continuous-time Markov chain has a uniform distribution on a finite set as its stationary distribution. The position of one label is the random walk with the same rates.

For an irreducible positive recurrent Markov chain with stationary distribution $\pi$, the expected return time to state $i$ is $1/{\pi }_i$. More generally, expected occupation rewards in a return cycle equal their stationary rate times the expected cycle length.

The mean recurrence time of $i$ is ${\mathbb{E}}_i{T}_i^{+}$. In a finite irreducible Markov chain with stationary distribution $\pi$, it equals $1/{\pi }_i$.

For a finite Markov chain with transition matrix $P$, define $\bar{d}(t)={\max }_{x,y}\Vert P^t(x,\cdot )-P^t(y,\cdot ){\Vert }_{\mathrm{TV}}$. If $\pi$ is a stationary distribution and $d(t)={\max }_x\Vert P^t(x,\cdot )-\pi {\Vert }_{\mathrm{TV}}$, then $d(t)\le \bar{d}(t)\le 2d(t)$. The first bound follows by writing $\pi ={\sum }_y\pi (y)P^t(y,\cdot )$ and using the triangle inequality; the second uses the triangle inequality through $\pi$.