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For a nearest-neighbour Markov chain on $\{L,\dots ,U\}$ with positive right-step probabilities in its interior, let $m=U-L$ and $p_{\ast }>0$ be their minimum. From every interior state, at most $m$ consecutive right steps reach $U$, with probability at least $p_{\ast }^m$. The Markov property then gives $\mathbb{P}(T>km)\le (1-p_{\ast }^m{)}^k$ for the first boundary hitting time. Thus $T$ is almost surely finite and has finite expectation, regardless of the holding probability of a Markov chain at each interior state.

// Target: probability-theory.bigb

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.

Write $s_i=p_i+q_i>0$ and $d_i=p_i-q_i$ for $i=0,1$. For the parity correction $h_{\theta }(k)=k-\theta 1_{\{k\text{ odd}\}}$, the conditional expectations of the increments of $h_{\theta }(X_n)$ areThe holding move contributes zero. Thus the period-two martingale corrector for a birth-death chain exists precisely whenThis is the condition that $h_{\theta }$ be a harmonic function for a Markov chain. The process has integrable values from a fixed initial state because $|X_n|\le |X_0|+n$; the Markov property therefore turns the zero conditional increments into the required martingale property. Conversely, both parities are reached with positive probability because all right and left probabilities are positive, so the two conditional equations are necessary. Expanding their numerator also gives the equivalent balance condition $p_0{p}_1=q_0{q}_1$. Positivity makes $|\theta |<1$.

Put $L=-aN$, $U=bN$. The exit time $T_N$ is almost surely finite. Indeed, set $m=U-L$ and $p_{\ast }=\min (p_0,p_1)>0$. From any interior point, consecutive right steps reach $U$ in at most $m$ moves, with probability at least $p_{\ast }^m$. The Markov property gives $\mathbb{P}(T_N>km)\le (1-p_{\ast }^m{)}^k$. This is the finite-interval exit times have geometric tails argument. The walk cannot skip a boundary because its jumps have size at most one.

Under the balance condition, the stopped martingale $h_{\theta }(X_{n\wedge T_N})$ is bounded, because the stopped position stays in $[L,U]$. Apply the optional stopping theorem at $n\wedge T_N$, then dominated convergence as $n\to \infty$. Starting from zero givesWith ${\epsilon }_a=1_{\{aN\text{ odd}\}}$ and ${\epsilon }_b=1_{\{bN\text{ odd}\}}$,The bounded parity corrections disappear after division by $N$; the holding probability of a Markov chain may be positive or zero and may differ between parities.

For failure of balance, use scale increments for a birth-death chain rather than the nonexistent bounded correction. Let $u(k)$ be the upper exit probability from $k$ for boundaries $L,U$. The Markov property yieldswhere $p_k,q_k$ depend only on parity. Choose positive numbers $w_j$ with $w_{j+1}=(q_j/p_j)w_j$. Then $u(k)={\sum }_{j=L+1}^k{w}_j/{\sum }_{j=L+1}^U{w}_j$ satisfies these equations and boundary values. Stopping this bounded harmonic function proves that it is the exit probability. In particular,If $\rho <1$, the positive-index sum stays bounded as $N\to \infty$, while the negative-index contribution diverges geometrically, so ${\pi }_N\to 0$. If $\rho >1$, the negative-index sum stays bounded while the positive-index sum diverges, so ${\pi }_N\to 1$. Therefore the complete classification isThus a fixed nonzero imbalance drives the macroscopic exit to the corresponding side, for every positive $a,b$.

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.