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  "body": "\n## Example\n\nLet $\\lambda \u003e 0$ and $\\mu \u003e 0$.\nIn the M/M/1 queue of [[queueing theory|P64V0HTE78]], customers arrive as a Poisson process of rate $\\lambda$ and there is one server.\nService times are independent [[exponential|0E99F51GVB]] random variables of rate $\\mu$.\nThe number $Q(t)$ of customers in the system is a [[birth–death process|E8DAD7052E]] on $\\set{0, 1, 2, \\ldots}$.\nIts birth rates are $\\lambda_n = \\lambda$ for $n \\ge 0$ and its death rates are $\\mu_n = \\mu$ for $n \\ge 1$.\nThis holds because the exponential law has no memory.\nAt each moment, the time to the next arrival and the remaining service time are again exponential, of rates $\\lambda$ and $\\mu$.\n\nLet $\\rho = \\lambda / \\mu$.\nThe balance equations $\\pi_n \\lambda = \\pi_{n+1} \\mu$ give $\\pi_n = \\rho^n \\pi_0$.\nSo a [[stationary distribution|02KYGSCTCN]] exists if and only if $\\rho \u003c 1$, and then\n$$\\pi_n = (1 - \\rho) \\rho^n, \\qquad n \\ge 0,$$\na geometric law.\nThe mean number in the system is $L = \\sum_n n \\pi_n = \\rho / (1 - \\rho)$.\nBy [[Little's law|JN2YME6F4A]], the mean time in the system is $W = L / \\lambda = 1/(\\mu - \\lambda)$.\nIf $\\rho \\ge 1$ the queue has no stationary distribution: it is null recurrent for $\\rho = 1$ and transient for $\\rho \u003e 1$.\n\n## Explanations\n\n### L1 human\n\nOne clerk serves customers who arrive at random, at $\\lambda$ per hour on average, and serves $\\mu$ per hour on average when busy.\nIf $\\lambda = 3$ and $\\mu = 4$, the clerk is busy three quarters of the time, there are $3$ customers in the system on average, and each spends $1$ hour there on average.\nAs $\\lambda$ comes close to $\\mu$, the average line grows without bound.\n",
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