Waiting paradox

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The waiting time paradox is a paradox from queuing theory , a branch of probability theory . In English literature, it is also called hitchhiker's paradox , after a frequently used example (from English hitchhiker - hitchhiker ).

Clear formulation

If buses run every minute on average , one intuitively expects a waiting time of minutes if the arrival time at the bus stop is random . However, this is only correct if the buses come every minute. The more the distances vary, the more the expected waiting time varies and then it is more likely to have a longer distance between two buses than an average or shorter one. The interval that we observe has become longer through observation , a short one we would have caught with less probability. ${\displaystyle t}$ ${\displaystyle t/2}$ ${\displaystyle t}$

Mathematical formulation

The random variables of the distances between two buses are independent and equally distributed with expected value and standard deviation . The first buses then need . To get between the -th and -th bus, the waiting time decreases linearly from to . The expected value of the waiting time is thus ${\displaystyle X_{1},X_{2},\dots }$ ${\displaystyle \mu }$ ${\displaystyle \sigma }$ ${\displaystyle n}$ ${\displaystyle X_{1}+X_{2}+\dotsb +X_{n}}$ ${\displaystyle (k-1)}$ ${\displaystyle k}$ ${\displaystyle X_{k}}$ ${\displaystyle 0}$

${\displaystyle {\frac {X_{1}^{2}/2+X_{2}^{2}/2+\dotsb +X_{n}^{2}/2}{X_{1}+X_{2}+\dotsb +X_{n}}}={\frac {1}{2}}\cdot {\frac {\frac {X_{1}^{2}+X_{2}^{2}+\dotsb +X_{n}^{2}}{n}}{\frac {X_{1}+X_{2}+\dotsb +X_{n}}{n}}}}$

If one now forms the limit value , the numerator converges against and the denominator against . The expected value is therefore: ${\displaystyle n\to \infty }$ ${\displaystyle E(X^{2})=E(X)^{2}+V(X)=\mu ^{2}+\sigma ^{2}}$ ${\displaystyle E(X)=\mu }$

${\displaystyle {\frac {1}{2}}\cdot {\frac {\mu ^{2}+\sigma ^{2}}{\mu }}={\frac {\mu }{2}}+{\frac {\sigma ^{2}}{2\mu }}}$

The expectation is therefore always greater than , except for . In particular, the expectation value can become infinite if . ${\displaystyle \mu /2}$ ${\displaystyle \sigma =0}$ ${\displaystyle \sigma =\infty }$

Examples

• If the buses arrive exactly at a distance , then the expected value of the waiting time is and therefore is . ${\displaystyle \mu }$ ${\displaystyle \sigma =0}$ ${\displaystyle \mu /2}$
• If the buses arrive with probability at a distance and with probability at a distance , then and , thus is the expected value of the waiting time . ${\displaystyle 1/2}$ ${\displaystyle a}$ ${\displaystyle 1/2}$ ${\displaystyle b}$ ${\displaystyle \mu =(a+b)/2}$ ${\displaystyle \sigma =|a-b|/2}$ ${\displaystyle {\frac {a+b}{4}}+{\frac {(a-b)^{2}}{4(a+b)}}}$
• If the distances are evenly distributed in , then is . So the expected value of the waiting time is . ${\displaystyle [\mu -\varepsilon \mu ,\mu +\varepsilon \mu ]}$ ${\displaystyle \sigma =\varepsilon \mu /{\sqrt {3}}}$ ${\displaystyle {\frac {\mu }{2}}+{\frac {\varepsilon ^{2}\mu }{6}}=\left({\frac {1}{2}}+{\frac {\varepsilon ^{2}}{6}}\right)\mu }$
• If the distances are exponentially distributed with parameter , then is . Thus, the expected value of the waiting time , i.e. H. although the buses come every minute on average , you still have to wait minutes on average ! (See section Poisson process.) ${\displaystyle \lambda }$ ${\displaystyle \mu =\sigma ={\frac {1}{\lambda }}}$ ${\displaystyle {\frac {\mu }{2}}+{\frac {\mu ^{2}}{2\mu }}=\mu }$ ${\displaystyle \mu }$ ${\displaystyle \mu }$
• If the distances are with probability for , then is , but . So: Although a bus leaves on average every 4 minutes, the expected waiting time is infinitely large. ${\displaystyle 2^{n}}$ ${\displaystyle 2\cdot 3^{-n}}$ ${\displaystyle n=1,2,3,\dots }$ ${\displaystyle \mu =4}$ ${\displaystyle \sigma =\infty }$

Poisson process

Often the waiting time paradox is only described for Poisson processes , where there is a more natural explanation for the paradox.

In the Poisson process, the distances between two increases are exponentially distributed (see example above), so the expected value of the waiting time agrees with the expected value of the distances. This is due to the homogeneity of the Poisson process; in other words: the expected value of the waiting time is independent of when the last bus left. In particular, you get this expected value when the bus has just left, and at this moment the waiting time is the distance to the next bus. The expected values ​​therefore agree.

This paradox also exists in the discrete model, namely when rolling the dice. On average, you have to roll the dice 6 times to get a six. However, it doesn't matter how often you've tried - you still have to roll the dice 6 times on average.