**Proof plan.** The stationary [distribution](/page/Distribution) satisfies $Q\pi = 0$. For a tridiagonal generator (birth-death chain), this system can be solved recursively. The normalisation constraint $\sum_n \pi_n = 1$ then determines $\pi_0$. **Step 1 (The balance equations).** Writing out $Q\pi = 0$ row by row: \begin{align*} n = 0: \quad & -\lambda_0 \pi_0 + \mu_1 \pi_1 = 0, \\ n \geq 1: \quad & \lambda_{n-1}\pi_{n-1} - (\lambda_n + \mu_n)\pi_n + \mu_{n+1}\pi_{n+1} = 0. \end{align*} **Step 2 (Detailed balance).** The $n = 0$ equation gives $\pi_1 = (\lambda_0/\mu_1)\pi_0$. For $n = 1$, substituting the expression for $\pi_1$: \begin{align*} \lambda_0 \pi_0 - (\lambda_1 + \mu_1)\frac{\lambda_0}{\mu_1}\pi_0 + \mu_2 \pi_2 = 0. \end{align*} Solving for $\pi_2$: \begin{align*} \pi_2 = \frac{1}{\mu_2}\left[(\lambda_1 + \mu_1)\frac{\lambda_0}{\mu_1} - \lambda_0\right]\pi_0 = \frac{\lambda_0\lambda_1}{\mu_1\mu_2}\pi_0. \end{align*} [claim:Recursive Formula] For all $n \geq 1$, $\pi_n = \frac{\lambda_0 \lambda_1 \cdots \lambda_{n-1}}{\mu_1 \mu_2 \cdots \mu_n}\,\pi_0$. [/claim] [proof] By induction. The base case $n = 1$ is $\pi_1 = (\lambda_0/\mu_1)\pi_0$, established above. Suppose the formula holds for all indices up to $n$. From the balance equation at level $n$: \begin{align*} \mu_{n+1}\pi_{n+1} = (\lambda_n + \mu_n)\pi_n - \lambda_{n-1}\pi_{n-1}. \end{align*} Substituting the inductive hypothesis for $\pi_n$ and $\pi_{n-1}$: \begin{align*} \mu_{n+1}\pi_{n+1} &= (\lambda_n + \mu_n)\frac{\lambda_0 \cdots \lambda_{n-1}}{\mu_1 \cdots \mu_n}\pi_0 - \lambda_{n-1}\frac{\lambda_0 \cdots \lambda_{n-2}}{\mu_1 \cdots \mu_{n-1}}\pi_0 \\ &= \frac{\lambda_0 \cdots \lambda_{n-2}}{\mu_1 \cdots \mu_{n-1}}\left[\frac{(\lambda_n + \mu_n)\lambda_{n-1}}{\mu_n} - \lambda_{n-1}\right]\pi_0 \\ &= \frac{\lambda_0 \cdots \lambda_{n-2}}{\mu_1 \cdots \mu_{n-1}} \cdot \frac{\lambda_{n-1}\lambda_n}{\mu_n}\,\pi_0 \\ &= \frac{\lambda_0 \cdots \lambda_n}{\mu_1 \cdots \mu_n}\,\pi_0. \end{align*} Dividing by $\mu_{n+1}$ gives $\pi_{n+1} = \frac{\lambda_0 \cdots \lambda_n}{\mu_1 \cdots \mu_{n+1}}\pi_0$, completing the induction. [/proof] **Step 3 (Normalisation).** Summing over all states: \begin{align*} 1 = \sum_{n=0}^\infty \pi_n = \pi_0\left(1 + \sum_{n=1}^\infty \frac{\lambda_0 \lambda_1 \cdots \lambda_{n-1}}{\mu_1 \mu_2 \cdots \mu_n}\right). \end{align*} This determines $\pi_0$ provided the [series](/page/Series) converges. If the series diverges, no stationary distribution exists and $P_n(t) \to 0$ for all $n$ as $t \to \infty$ (transience).