Memorylessness

Summary

A random waiting time $X \ge 0$ is memoryless if, for all $s, t \ge 0$ with $P(X > s) > 0$, $$P(X > s + t \mid X > s) = P(X > t).$$ So the time already waited does not change the law of the remaining waiting time. Among the distributions on $[0, \infty)$ with $P(X > t) > 0$ for each $t$, only the exponential distributions are memoryless. Among the distributions on $\{0, 1, 2, \dots\}$, with $s$ and $t$ integers, only the geometric distributions are memoryless.

Remarks

For $X \sim \operatorname{Exp}(\lambda)$, $P(X > s + t) = e^{-\lambda(s + t)} = P(X > s) \, P(X > t)$. This is the memoryless property. The proof of the characterization uses that $G(t) = P(X > t)$ 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 $\lambda \ge 0$.

Explanations

L1 human

A memoryless waiting time does not "age". If a bus comes at exponential times and you have already waited ten minutes, your expected further wait is the same as when you arrived.