For the transition matrix

$P = \begin{pmatrix} 1-\alpha & \alpha \\ \beta & 1-\beta \end{pmatrix},$

we solve $\pi P = \pi$:

$\pi_0 (1-\alpha) + \pi_1 \beta = \pi_0, \quad \pi_0 \alpha + \pi_1 (1-\beta) = \pi_1.$

These reduce to $\pi_1 \beta = \pi_0 \alpha$. With the normalization $\pi_0 + \pi_1 = 1$, we obtain

$\pi_0 = \frac{\beta}{\alpha + \beta}, \quad \pi_1 = \frac{\alpha}{\alpha + \beta}.$

This is the unique stationary distribution (assuming $\alpha, \beta > 0$).

For the simple random walk on $\mathbb{Z}$ with $p > 1/2$ (drift to the right), the chain is transient: it escapes to $+\infty$ and visits each state only finitely many times. There is no probability distribution $\pi$ satisfying $\pi P = \pi$ — the chain has no stationary distribution.

• Dimension 1: The symmetric random walk on $\mathbb{Z}$ (with $p = 1/2$) is recurrent but null recurrent: $\mathbb{P}(\tau_0 < \infty) = 1$ but $\mathbb{E}[\tau_0] = \infty$. Hence, no stationary distribution.
• Dimension 2: The symmetric random walk on $\mathbb{Z}^2$ is also null recurrent. No stationary distribution.
• Dimension $\geq 3$: The symmetric random walk on $\mathbb{Z}^d$ for $d \geq 3$ is transient, so no stationary distribution.

A birth-death chain on $S = \{0, 1, 2, \ldots\}$ has transitions only to neighboring states:

$p_{i,i+1} = p_i, \quad p_{i,i-1} = q_i, \quad p_{ii} = r_i = 1 - p_i - q_i.$

The stationary distribution satisfies the detailed balance equations:

$\pi_i p_{i,i+1} = \pi_{i+1} p_{i+1,i}, \quad \text{for all } i \geq 0.$

Iterating this relation gives

$\pi_n = \pi_0 \prod_{k=0}^{n-1} \frac{p_k}{q_{k+1}}.$

Normalization $\sum_{n=0}^\infty \pi_n = 1$ determines $\pi_0$ (if the series converges).

In an $M/M/1$ queueing system, arrivals occur at rate $\lambda$ and service at rate $\mu$. Modeling the queue length as a birth-death chain with $p_i = \lambda/(\lambda + \mu)$ and $q_i = \mu/(\lambda+\mu)$ gives

$\pi_n = \pi_0 \rho^n, \quad \rho = \frac{\lambda}{\mu}.$

If $\rho < 1$ (arrival rate less than service rate), then $\sum_{n=0}^\infty \rho^n = 1/(1-\rho)$, so

$\pi_0 = 1 - \rho, \quad \pi_n = (1-\rho) \rho^n.$

This is the geometric distribution with parameter $\rho$.

For the two-state chain with stationary distribution $\pi = (\beta/(\alpha+\beta), \alpha/(\alpha+\beta))$, the eigenvalues of $P$ are $1$ and $1 - \alpha - \beta$. Hence,

$P^n = \frac{1}{\alpha+\beta} \begin{pmatrix} \beta & \alpha \\ \beta & \alpha \end{pmatrix} + (1-\alpha-\beta)^n \frac{1}{\alpha+\beta} \begin{pmatrix} \alpha & -\alpha \\ -\beta & \beta \end{pmatrix}.$

As $n \to \infty$, the second term vanishes (assuming $\alpha + \beta \in (0, 2)$), and $P^n \to \mathbf{1} \pi^T$, where each row is $\pi$.

An urn contains $N$ balls distributed between two urns. At each step, a ball is chosen uniformly at random and moved to the other urn. Let $X_n$ be the number of balls in urn 1. Then

$p_{k, k+1} = \frac{N-k}{N}, \quad p_{k, k-1} = \frac{k}{N}.$

The stationary distribution is binomial:

$\pi_k = \binom{N}{k} \frac{1}{2^N}.$

One can verify detailed balance: $\pi_k p_{k, k+1} = \pi_{k+1} p_{k+1, k}$.