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  "body": "\n## Definition\n\nLet $S = \\set{0, 1, 2, \\ldots}$, and let $(\\lambda_n)_{n \\ge 0}$ and $(\\mu_n)_{n \\ge 1}$ be nonnegative real numbers, with $\\mu_0 = 0$.\nLet $(X_t)_{t \\ge 0}$ be a [[continuous-time Markov chain|TWQFE0FZZS]] on $S$ with generator matrix $Q = (q_{mn})$.\nIt is a *birth–death process* with birth rates $\\lambda_n$ and death rates $\\mu_n$ if\n$$q_{n, n+1} = \\lambda_n, \\qquad q_{n, n-1} = \\mu_n \\ (n \\ge 1), \\qquad q_{nn} = -(\\lambda_n + \\mu_n),$$\nand $q_{mn} = 0$ when $\\abs{m - n} \\ge 2$.\nA finite version uses $S = \\set{0, 1, \\ldots, N}$ with $\\lambda_N = 0$.\n\n## Remarks\n\nIn state $n$ the process waits an exponential time of rate $\\lambda_n + \\mu_n$, then jumps to $n + 1$ with probability $\\lambda_n / (\\lambda_n + \\mu_n)$ and to $n - 1$ otherwise.\nExample: the Poisson process of rate $\\lambda$ is the pure birth process with $\\lambda_n = \\lambda$ and $\\mu_n = 0$ for all $n$; each row of $Q$ sums to $0$, as required.\nNon-example: a chain on $S$ with $q_{n, n+2} \u003e 0$ for some $n$ is not a birth–death process, because it jumps by two.\nSuppose all $\\lambda_n \u003e 0$ and all $\\mu_n \u003e 0$ for $n \\ge 1$, and the chain does not explode.\nThen a [[stationary distribution|02KYGSCTCN]] exists if and only if\n$$\\sum_{n \\ge 0} \\frac{\\lambda_0 \\lambda_1 \\cdots \\lambda_{n-1}}{\\mu_1 \\mu_2 \\cdots \\mu_n} \u003c \\infty,$$\nand then $\\pi_n$ is proportional to the $n$-th term of this sum.\nIt follows from the balance equations $\\pi_n \\lambda_n = \\pi_{n+1} \\mu_{n+1}$.\n",
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