Birth–death process

Remarks

In 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. Example: 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. Non-example: a chain on $S$ with $q_{n, n+2} > 0$ for some $n$ is not a birth–death process, because it jumps by two. Suppose all $\lambda_n > 0$ and all $\mu_n > 0$ for $n \ge 1$, and the chain does not explode. Then a stationary distribution exists if and only if $$\sum_{n \ge 0} \frac{\lambda_0 \lambda_1 \cdots \lambda_{n-1}}{\mu_1 \mu_2 \cdots \mu_n} < \infty,$$ and then $\pi_n$ is proportional to the $n$-th term of this sum. It follows from the balance equations $\pi_n \lambda_n = \pi_{n+1} \mu_{n+1}$.