{
  "aliases": [
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  "body": "\n## Summary\n\nA random waiting time $X \\ge 0$ is memoryless if, for all $s, t \\ge 0$ with $P(X \u003e s) \u003e 0$,\n$$P(X \u003e s + t \\mid X \u003e s) = P(X \u003e t).$$\nSo the time already waited does not change the law of the remaining waiting time.\nAmong the distributions on $[0, \\infty)$ with $P(X \u003e t) \u003e 0$ for each $t$, only the [[exponential distributions|028Z8WC5Y9]] are memoryless. Among the distributions on $\\{0, 1, 2, \\dots\\}$, with $s$ and $t$ integers, only the [[geometric distributions|0XCVV5Q500]] are memoryless.\n\n## Remarks\n\nFor $X \\sim \\operatorname{Exp}(\\lambda)$, $P(X \u003e s + t) = e^{-\\lambda(s + t)} = P(X \u003e s) \\, P(X \u003e t)$. This is the memoryless property.\nThe proof of the characterization uses that $G(t) = P(X \u003e t)$ satisfies $G(s + t) = G(s) \\, G(t)$ and is right-continuous. The only such functions with values in $(0, 1]$ are $G(t) = e^{-\\lambda t}$ with $\\lambda \\ge 0$.\n\n## Explanations\n\n### L1 human\n\nA memoryless waiting time does not \"age\".\nIf a bus comes at exponential times and you have already waited ten minutes, your expected further wait is the same as when you arrived.\n",
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