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Each emitted symbol identifies the state of a stationary finite Markov chain. Here the alphabet is the state space itself. Its transition matrix $P$ and stationary distribution $\pi$ determine the entropy rate of a fully observed Markov source.

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.

A harmonic function assigns a value to each state of a Markov chain such that one transition leaves its expected value unchanged. This turns a question about many possible future sample paths, such as which endpoint is reached first, into a local averaging rule. For a transition matrix $P$, a real-valued function $h$ is harmonic at state $i$ whenHere ${\mathbb{E}}_i$ means expected value conditional on starting at $X_0=i$; on a countable state space the sum must be absolutely convergent. If this holds at every state, write $Ph=h$. At an absorbing state the equation is automatic, so prescribed boundary conditions supply information that the averaging rule cannot.

For a finite Markov chain, choose a boundary set $B$ that can be reached from every state, and assign boundary values $b$. If $T_B$ is its first hitting time, the harmonic extension isChoosing $b$ to be the indicator function of one target within $B$ gives a hitting probability. This is a discrete Dirichlet problem: enforce the averaging rule outside $B$ and fix the values on $B$. See Hao Wu's MIT lecture 8, harmonic extension, slide 5. The construction extends the mean-value idea of a harmonic function; for a random walk on a graph it is a discrete harmonic function.

For a finite, time-homogeneous Discrete-time Markov chain with row-stochastic transition matrix $P$, irreducibility meansEquivalently, all states form one communicating class. The step count may depend on $i,j$; irreducibility does not require every one-step transition probability to be positive. Drawing an arrow $i\to j$ whenever $P_{ij}>0$ makes the condition exactly that of a strongly connected directed graph. It is a property of the entire chosen state space, independent of the initial probability distribution.

Irreducibility describes where the Markov chain can travel; aperiodicity describes the possible lengths of return paths. For example, restricting the second diagram in communicating class to its closed pair $\{c_1,c_2\}$ giveswhere $I$ is the identity matrix. This Markov chain is irreducible, but each state's period of a state in a Markov chain is two. Every destination is reachable, yet a given destination cannot be reached at every sufficiently large step count. Starting at $c_1$ produces alternating one-time probability distributions. See Hao Wu's MIT lecture 2, irreducibility and periods for the definitions.

Adding a holding probability of a Markov chain $\epsilon$ at each state, with $0<\epsilon <1$, givesThe arrows between the two states remain, so the Markov chain remains irreducible. A positive one-step return makes it an aperiodic Markov chain. At $\epsilon =1/2$, both rows are $(1/2,1/2)$ and $C_{1/2}^n=C_{1/2}$ for every $n\ge 1$: the initial state is forgotten after one step. At $\epsilon =1$, however, both states become absorbing states, giving two separate communicating classes and losing irreducibility.

For a finite Markov chain, irreducibility guarantees a unique stationary distribution with positive weight at every state. Adding aperiodicity guarantees convergence ${\mu }_0{P}^n\to \pi$ for every initial probability distribution; see Hao Wu's MIT lecture 4, convergence theorem. The Markov chain ergodic theorem gives convergence of empirical visit frequencies even for an irreducible periodic Markov chain. These results explain why checking connectivity is a useful first step before studying equilibrium. Conversely, any absorbing state prevents irreducibility if there is another state: it cannot reach that state. The one-state Markov chain is the exception.

A Markov information source uses a stationary finite Markov chain to emit symbols. The emitted symbol can be a function of the state. If it does not identify that state, the observed sequence need not itself be a Markov chain. The examples here use a fully observed Markov source, which makes the next-symbol uncertainty directly computable from the transition matrix.

The restriction $Q_C$ is a finite Markov chain that is an irreducible Markov chain. It is aperiodic, because its upper states have self-transition probability $a/(a+b)>0$, and the period is constant on a communicating class. We now find its stationary distribution.

Every column of $Q_C$ sums to one. Here is the full incoming-mass check, with each total measured in units of $1/(a+b)$. For $j\in L$, the incoming contributions are $a$ from $j+b\in I$ and $b$ from the self-transition. For $j\in U$, they are $b$ from $j-a\in I$ and $a$ from the self-transition.

For $j\in I$, the upward predecessor $j-a$ lies in $I$ exactly when $j\ge -c+a$, contributing $b$ units. When $j\le -c+a-1$, that predecessor is replaced by resets from all $b$ states of $L$, again contributing $b$ units. Independently, the downward predecessor $j+b$ contributes $a$ units when $j\le c-b$; when $j\ge c-b+1$, resets from the $a$ states of $U$ contribute those $a$ units instead. Thus every column total is $a+b$ units.

Therefore $Q_C$ is a doubly stochastic matrix, and the uniform distribution on $C$ is stationary:The countable-state Markov chain convergence theorem, applied to the finite closed class $C$, gives $Q_C^n(i,j)\to {\pi }_j$ for every $i\in C$. Its hypotheses hold: $C$ is finite and irreducible, hence positive recurrent, and it is aperiodic.

If $i_0>c+a$, the hitting time $H$ of $C$ has $\mathbb{P}(H>n)=(a/(a+b){)}^n$. If $i_0<-c-b$, the corresponding tail is $(b/(a+b){)}^n$. Thus $H<\infty$ almost surely from every exterior initial state. Conditioning on $H$ and $R_H$, use convergence within $C$ after entrance and the bounded convergence theorem; equivalently, first restrict to $H\le K$ and then let $K\to \infty$. This proves, for every $i_0\in \mathbb{Z}$,The limiting distribution is independent of $i_0$.

A stationary distribution is a probability distribution $\pi$ preserved by a Markov chain's evolution. For a Discrete-time Markov chain in the row convention,Thus $\pi$ is a normalized nonnegative left eigenvector of the transition matrix for eigenvalue one. Starting with $X_0$ distributed as $\pi$ makes every $X_n$ have that same probability distribution. For a time-homogeneous Markov chain, the path probabilities are also unchanged by shifting all observation times together: the process has strict stationarity. This does not mean its individual paths are constant. See Hao Wu's MIT lecture 2, stationary distributions.

For the sunny/rainy transition matrix, write $\pi =(a,1-a)$. The sunny component of $\pi P=\pi$ gives $a=0.8a+0.4(1-a)$, so $0.6a=0.4$ andMoreover,These equal stationary flows verify detailed balance and show that the Markov chain is reversible. The diagonal detailed balance equations hold automatically. Only $1/3$ of the stationary days are rainy, but $0.6$ of the days following a rainy day are rainy: stationarity permits dependence.

A finite Markov chain that is an irreducible Markov chain has a unique stationary distribution. If it is also an aperiodic Markov chain, ${\mu }_0{P}^n\to \pi$ for every initial probability distribution. The sunny/rainy example satisfies both conditions: both directions are possible and each state has a positive holding probability of a Markov chain. See Hao Wu's MIT lecture 4, convergence theorem. For an infinite state space, the additional positive recurrence hypothesis in the countable-state Markov chain convergence theorem matters.

Existence of a stationary distribution alone does not give this convergence. Consider the deterministic two-cycleHere $\pi C=\pi$, but starting in the first state gives alternating probability distributions $(1,0),(0,1),(1,0),\dots$. Its period of a state in a Markov chain is two. The running fraction of visits to either state nevertheless tends to $1/2$, illustrating the Markov chain ergodic theorem. Starting with $\pi$ instead keeps the one-time probability distributions stationary even though every path alternates.

Every finite Markov chain has a stationary distribution, without requiring Markov chain irreducibility or aperiodicity. For a row-stochastic matrix $P$, put $B=P^T$ and choose a column probability vector $q$. Every $B^{j}q$ lies in the probability simplex. Their arithmetic averages $x_N=N^{-1}{\sum }_{j=0}^{N-1}{B}^{j}q$ therefore lie in this compact set, while telescoping givesA convergent subsequence has a limit $x$ in the probability simplex. Continuity of matrix multiplication gives $Bx=x$, so $\pi =x^T$ is a stationary distribution. The individual iterates need not converge: a deterministic two-state swap alternates its point masses forever.

An irreducible Markov chain on a finite or countably infinite state space has at most one stationary distribution. On a countably infinite state space it exists exactly when the chain is positive recurrent. In that case, for every state $i$,where $T_i^{+}$ is the first return time. The expression determines every component uniquely. The formula follows by dividing the one visit to $i$ in each return cycle by the mean recurrence time; the strong law of large numbers for successive cycles makes this occupation fraction precise.

Every irreducible finite Markov chain is positive recurrent and has exactly one stationary distribution. On an infinite state space, an irreducible null recurrent Markov chain or transient Markov chain has none. For example, the simple symmetric random walk on the integers is null recurrent. Aperiodicity affects convergence of the one-time laws, but is unnecessary for existence or uniqueness of a stationary distribution.